Elementa

Euclid

Elementa

Euclid · Elem 10 · perseus-eng2 · More by Euclid

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Original / primary: Perseus Eng2

10.def.1.1.p3150

Those magnitudes are said to be commensurable which are measured by the same measure, and those incommensurable which cannot have any common measure.

10.def.1.2.p3151

Straight lines are commensurable in square when the squares on them are measured by the same area, and incommensurable in square when the squares on them cannot possibly have any area as a common measure.

10.def.1.3.p3152

With these hypotheses, it is proved that there exist straight lines infinite in multitude which are commensurable and incommensurable respectively, some in length only, and others in square also, with an assigned straight line. Let then the assigned straight line be called rational, and those straight lines which are commensurable with it, whether in length and in square or in square only, rational, but those which are incommensurable with it irrational.

10.def.1.4.p3153

And let the square on the assigned straight line be called rational and those areas which are commensurable with it rational, but those which are incommensurable with it irrational, and the straight lines which produce them irrational, that is, in case the areas are squares, the sides themselves, but in case they are any other rectilineal figures, the straight lines on which are described squares equal to them.

10.prop.1.1.p3154

Two unequal magnitudes being set out, if from the greater there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the lesser magnitude set out.

10.prop.1.1.p3155

Let AB, C be two unequal magnitudes of which AB is the greater: I say that, if from AB there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the magnitude C.

10.prop.1.1.p3156

For C if multiplied will sometime be greater than AB. [cf. v. Def. 4]

10.prop.1.1.p3157

Let it be multiplied, and let DE be a multiple of C, and greater than. AB; let DE be divided into the parts DF, FG, GE equal to C, from AB let there be subtracted BH greater than its half, and, from AH, HK greater than its half, and let this process be repeated continually until the divisions in AB are equal in multitude with the divisions in DE.

10.prop.1.1.p3158

Let, then, AK, KH, HB be divisions which are equal in multitude with DF, FG, GE.

10.prop.1.1.p3159

Now, since DE is greater than AB, and from DE there has been subtracted EG less than its half, and, from AB, BH greater than its half, therefore the remainder GD is greater than the remainder HA.

10.prop.1.1.p3160

And, since GD is greater than HA, and there has been subtracted, from GD, the half GF, and, from HA, HK greater than its half, therefore the remainder DF is greater than the remainder AK.

10.prop.1.1.p3161

But DF is equal to C; therefore C is also greater than AK.

10.prop.1.1.p3162

Therefore AK is less than C.

10.prop.1.1.p3163

Therefore there is left of the magnitude AB the magnitude AK which is less than the lesser magnitude set out, namely C. Q. E. D.

10.prop.1.1.p3164

And the theorem can be similarly proved even if the parts subtracted be halves.

10.prop.1.2.p3165

If, when the less of two unequal magnitudes is continually subtracted in turn from the greater, that which is left never measures the one before it, the magnitudes will be incommensurable.

10.prop.1.2.p3166

For, there being two unequal magnitudes AB, CD, and AB being the less, when the less is continually subtracted in turn from the greater, let that which is left over never measure the one before it; I say that the magnitudes AB, CD are incommensurable.

10.prop.1.2.p3167

For, if they are commensurable, some magnitude will measure them.

10.prop.1.2.p3168

Let a magnitude measure them, if possible, and let it be E; let AB, measuring FD, leave CF less than itself, let CF measuring BG, leave AG less than itself, and let this process be repeated continually, until there is left some magnitude which is less than E.

10.prop.1.2.p3169

Suppose this done, and let there be left AG less than E.

10.prop.1.2.p3170

Then, since E measures AB, while AB measures DF, therefore E will also measure FD.

10.prop.1.2.p3171

But it measures the whole CD also; therefore it will also measure the remainder CF.

10.prop.1.2.p3172

But CF measures BG; therefore E also measures BG.

10.prop.1.2.p3173

But it measures the whole AB also; therefore it will also measure the remainder AG, the greater the less: which is impossible.

10.prop.1.2.p3174

Therefore no magnitude will measure the magnitudes AB, CD; therefore the magnitudes AB, CD are incommensurable.

10.prop.1.2.p3175

Therefore etc. [X. Def. 1]

10.prop.1.3.p3176

Given two commensurable magnitudes, to find their greatest common measure.

10.prop.1.3.p3177

Let the two given commensurable magnitudes be AB, CD of which AB is the less; thus it is required to find the greatest common measure of AB, CD.

10.prop.1.3.p3178

Now the magnitude AB either measures CD or it does not.

10.prop.1.3.p3179

If then it measures it—and it measures itself also—AB is a common measure of AB, CD.

10.prop.1.3.p3180

And it is manifest that it is also the greatest; for a greater magnitude than the magnitude AB will not measure AB.

10.prop.1.3.p3181

Next, let AB not measure CD.

10.prop.1.3.p3182

Then, if the less be continually subtracted in turn from the greater, that which is left over will sometime measure the one before it, because AB, CD are not incommensurable; [cf. X. 2] let AB, measuring ED, leave EC less than itself, let EC, measuring FB, leave AF less than itself, and let AF measure CE.

10.prop.1.3.p3183

Since, then, AF measures CE, while CE measures FB, therefore AF will also measure FB.

10.prop.1.3.p3184

But it measures itself also; therefore AF will also measure the whole AB.

10.prop.1.3.p3185

But AB measures DE; therefore AF will also measure ED.

10.prop.1.3.p3186

But it measures CE also; therefore it also measures the whole CD.

10.prop.1.3.p3187

Therefore AF is a common measure of AB, CD.

10.prop.1.3.p3188

I say next that it is also the greatest.

10.prop.1.3.p3189

For, if not, there will be some magnitude greater than AF which will measure AB, CD.

10.prop.1.3.p3190

Let it be G.

10.prop.1.3.p3191

Since then G measures AB, while AB measures ED, therefore G will also measure ED.

10.prop.1.3.p3192

But it measures the whole CD also; therefore G will also measure the remainder CE.

10.prop.1.3.p3193

But CE measures FB; therefore G will also measure FB.

10.prop.1.3.p3194

But it measures the whole AB also, and it will therefore measure the remainder AF, the greater the less: which is impossible.

10.prop.1.3.p3195

Therefore no magnitude greater than AF will measure AB, CD; therefore AF is the greatest common measure of AB, CD.

10.prop.1.3.p3196

Therefore the greatest common measure of the two given commensurable magnitudes AB, CD has been found. Q. E. D.

10.prop.1.3.p3197

Porism. From this it is manifest that, if a magnitude measure two magnitudes, it will also measure their greatest common measure.

10.prop.1.4.p3198

Given three commensurable magnitudes, to find their greatest common measure.

10.prop.1.4.p3199

Let A, B, C be the three given commensurable magnitudes; thus it is required to find the greatest common measure of A, B, C.

10.prop.1.4.p3200

Let the greatest common measure of the two magnitudes A, B be taken, and let it be D; [X. 3] then D either measures C, or does not measure it.

10.prop.1.4.p3201

First, let it measure it.

10.prop.1.4.p3202

Since then D measures C, while it also measures A, B, therefore D is a common measure of A, B, C.

10.prop.1.4.p3203

And it is manifest that it is also the greatest; for a greater magnitude than the magnitude D does not measure A, B.

10.prop.1.4.p3204

Next, let D not measure C.

10.prop.1.4.p3205

I say first that C, D are commensurable.

10.prop.1.4.p3206

For, since A, B, C are commensurable, some magnitude will measure them, and this will of course measure A, B also; so that it will also measure the greatest common measure of A, B, namely D. [X. 3, Por.]

10.prop.1.4.p3207

But it also measures C; so that the said magnitude will measure C, D; therefore C, D are commensurable.

10.prop.1.4.p3208

Now let their greatest common measure be taken, and let it be E. [X. 3]

10.prop.1.4.p3209

Since then E measures D, while D measures A, B, therefore E will also measure A, B.

10.prop.1.4.p3210

But it measures C also; therefore E measures A, B, C; therefore E is a common measure of A, B, C.

10.prop.1.4.p3211

I say next that it is also the greatest.

10.prop.1.4.p3212

For, if possible, let there be some magnitude F greater than E, and let it measure A, B, C.

10.prop.1.4.p3213

Now, since F measures A, B, C, it will also measure A, B, and will measure the greatest common measure of A, B. [X. 3, Por.]

10.prop.1.4.p3214

But the greatest common measure of A, B is D; therefore F measures D.

10.prop.1.4.p3215

But it measures C also; therefore F measures C, D; therefore F will also measure the greatest common measure of C, D. [X. 3, Por.]

10.prop.1.4.p3216

But that is E; therefore F will measure E, the greater the less: which is impossible.

10.prop.1.4.p3217

Therefore no magnitude greater than the magnitude E will measure A, B, C; therefore E is the greatest common measure of A, B, C if D do not measure C, and, if it measure it, D is itself the greatest common measure.

10.prop.1.4.p3218

Therefore the greatest common measure of the three given commensurable magnitudes has been found.

10.prop.1.4.p3219

Porism. From this it is manifest that, if a magnitude measure three magnitudes, it will also measure their greatest common measure.

10.prop.1.4.p3220

Similarly too, with more magnitudes, the greatest common measure can be found, and the porism can be extended. Q. E. D.

10.prop.1.5.p3221

Commensurable magnitudes have to one another the ratio which a number has to a number.

10.prop.1.5.p3222

Let A, B be commensurable magnitudes; I say that A has to B the ratio which a number has to a number.

10.prop.1.5.p3223

For, since A, B are commensurable, some magnitude will measure them.

10.prop.1.5.p3224

Let it measure them, and let it be C.

10.prop.1.5.p3225

And, as many times as C measures A, so many units let there be in D; and, as many times as C measures B, so many units let there be in E.

10.prop.1.5.p3226

Since then C measures A according to the units in D, while the unit also measures D according to the units in it, therefore the unit measures the number D the same number of times as the magnitude C measures A; therefore, as C is to A, so is the unit to D; [VII. Def. 20] therefore, inversely, as A is to C, so is D to the unit. [cf. V. 7, Por.]

10.prop.1.5.p3227

Again, since C measures B according to the units in E, while the unit also measures E according to the units in it, therefore the unit measures E the same number of times as C measures B; therefore, as C is to B, so is the unit to E.

10.prop.1.5.p3228

But it was also proved that, as A is to C, so is D to the unit; therefore, ex aequali, as A is to B, so is the number D to E. [V. 22]

10.prop.1.5.p3229

Therefore the commensurable magnitudes A, B have to one another the ratio which the number D has to the number E. Q. E. D.

10.prop.1.6.p3230

If two magnitudes have to one another the ratio which a number has to a number, the magnitudes will be commensurable.

10.prop.1.6.p3231

For let the two magnitudes A, B have to one another the ratio which the number D has to the number E; I say that the magnitudes A, B are commensurable.

10.prop.1.6.p3232

For let A be divided into as many equal parts as there are units in D, and let C be equal to one of them; and let F be made up of as many magnitudes equal to C as there are units in E.

10.prop.1.6.p3233

Since then there are in A as many magnitudes equal to C as there are units in D, whatever part the unit is of D, the same part is C of A also; therefore, as C is to A, so is the unit to D. [VII. Def. 20]

10.prop.1.6.p3234

But the unit measures the number D; therefore C also measures A.

10.prop.1.6.p3235

And since, as C is to A, so is the unit to D, therefore, inversely, as A is to C, so is the number D to the unit. [cf. V. 7, Por.]

10.prop.1.6.p3236

Again, since there are in F as many magnitudes equal to C as there are units in E, therefore, as C is to F, so is the unit to E. [VII. Def. 20]

10.prop.1.6.p3237

But it was also proved that, as A is to C, so is D to the unit; therefore, ex aequali, as A is to F, so is D to E. [v. 22]

10.prop.1.6.p3238

But, as D is to E, so is A to B; therefore also, as A is to B, so is it to F also. [V. 11]

10.prop.1.6.p3239

Therefore A has the same ratio to each of the magnitudes B, F; therefore B is equal to F. [V. 9]

10.prop.1.6.p3240

But C measures F; therefore it measures B also.

10.prop.1.6.p3241

Further it measures A also; therefore C measures A, B.

10.prop.1.6.p3242

Therefore A is commensurable with B.

10.prop.1.6.p3243

Therefore etc.

10.prop.1.6.p3244

Porism. From this it is manifest that, if there be two numbers, as D, E, and a straight line, as A, it is possible to make a straight line [F] such that the given straight line is to it as the number D is to the number E.

10.prop.1.6.p3245

And, if a mean proportional be also taken between A, F, as B,

10.prop.1.6.p3246

as A is to F, so will the square on A be to the square on B, that is, as the first is to the third, so is the figure on the first to that which is similar and similarly described on the second. [VI. 19, Por.]

10.prop.1.6.p3247

But, as A is to F, so is the number D to the number E; therefore it has been contrived that, as the number D is to the number E, so also is the figure on the straight line A to the figure on the straight line B. Q. E. D.

10.prop.1.7.p3248

Incommensurable magnitudes have not to one another the ratio which a number has to a number.

10.prop.1.7.p3249

Let A, B be incommensurable magnitudes; I say that A has not to B the ratio which a number has to a number.

10.prop.1.7.p3250

For, if A has to B the ratio which a number has to a number, A will be commensurable with B. [X. 6]

10.prop.1.7.p3251

But it is not; therefore A has not to B the ratio which a number has to a number.

10.prop.1.7.p3252

Therefore etc.

10.prop.1.8.p3253

If two magnitudes have not to one another the ratio which a number has to a number, the magnitudes will be incommensurable.

10.prop.1.8.p3254

For let the two magnitudes A, B not have to one another the ratio which a number has to a number; I say that the magnitudes A, B are incommensurable.

10.prop.1.8.p3255

For, if they are commensurable, A will have to B the ratio which a number has to a number. [X. 5]

10.prop.1.8.p3256

But it has not; therefore the magnitudes A, B are incommensurable.

10.prop.1.8.p3257

Therefore etc.

10.prop.1.9.p3258

The squares on straight lines commensurable in length have to one another the ratio which a square number has to a square number; and squares which have to one another the ratio which a square number has to a square number will also have their sides commensurable in length. But the squares on straight lines incommensurable in length have not to one another the ratio which a square number has to a square number; and squares which have not to one another the ratio which a square number has to a square number will not have their sides commensurable in length either.

10.prop.1.9.p3259

For let A, B be commensurable in length; I say that the square on A has to the square on B the ratio which a square number has to a square number.

10.prop.1.9.p3260

For, since A is commensurable in length with B, therefore A has to B the ratio which a number has to a number. [X. 5]

10.prop.1.9.p3261

Let it have to it the ratio which C has to D.

10.prop.1.9.p3262

Since then, as A is to B, so is C to D, while the ratio of the square on A to the square on B is duplicate of the ratio of A to B, for similar figures are in the duplicate ratio of their corresponding sides; [VI. 20, Por.] and the ratio of the square on C to the square on D is duplicate of the ratio of C to D, for between two square numbers there is one mean proportional number, and the square number has to the square number the ratio duplicate of that which the side has to the side; [VIII. 11] therefore also, as the square on A is to the square on B, so is the square on C to the square on D.

10.prop.1.9.p3263

Next, as the square on A is to the square on B, so let the square on C be to the square on D; I say that A is commensurable in length with B.

10.prop.1.9.p3264

For since, as the square on A is to the square on B, so is the square on C to the square on D, while the ratio of the square on A to the square on B is duplicate of the ratio of A to B, and the ratio of the square on C to the square on D is duplicate of the ratio of C to D, therefore also, as A is to B, so is C to D.

10.prop.1.9.p3265

Therefore A has to B the ratio which the number C has to the number D; therefore A is commensurable in length with B. [X. 6]

10.prop.1.9.p3266

Next, let A be incommensurable in length with B; I say that the square on A has not to the square on B the ratio which a square number has to a square number.

10.prop.1.9.p3267

For, if the square on A has to the square on B the ratio which a square number has to a square number, A will be commensurable with B.

10.prop.1.9.p3268

But it is not; therefore the square on A has not to the square on B the ratio which a square number has to a square number.

10.prop.1.9.p3269

Again, let the square on A not have to the square on B the ratio which a square number has to a square number; I say that A is incommensurable in length with B.

10.prop.1.9.p3270

For, if A is commensurable with B, the square on A will have to the square on B the ratio which a square number has to a square number.

10.prop.1.9.p3271

But it has not; therefore A is not commensurable in length with B.

10.prop.1.9.p3272

Therefore etc.

10.prop.1.9.p3273

Porism. And it is manifest from what has been proved that straight lines commensurable in length are always commensurable in square also, but those commensurable in square are not always commensurable in length also.

10.prop.1.9.p3274

[Lemma. It has been proved in the arithmetical books that similar plane numbers have to one another the ratio which a square number has to a square number, [VIII. 26] and that, if two numbers have to one another the ratio which a square number has to a square number, they are similar plane numbers. [Converse of VIII. 26]

10.prop.1.9.p3275

And it is manifest from these propositions that numbers which are not similar plane numbers, that is, those which have not their sides proportional, have not to one another the ratio which a square number has to a square number.

10.prop.1.9.p3276

For, if they have, they will be similar plane numbers: which is contrary to the hypothesis.

10.prop.1.9.p3277

Therefore numbers which are not similar plane numbers have not to one another the ratio which a square number has to a square number.]

10.prop.1.10.p3278

To find two straight lines incommensurable, the one in length only, and the other in square also, with an assigned straight line.

10.prop.1.10.p3279

Let A be the assigned straight line; thus it is required to find two straight lines incommensurable, the one in length only, and the other in square also, with A.

10.prop.1.10.p3280

Let two numbers B, C be set out which have not to one another the ratio which a square number has to a square number, that is, which are not similar plane numbers; and let it be contrived that, as B is to C, so is the square on A to the square on D —for we have learnt how to do this— [X. 6, Por.] therefore the square on A is commensurable with the square on D. [X. 6]

10.prop.1.10.p3281

And, since B has not to C the ratio which a square number has to a square number, therefore neither has the square on A to the square on D the ratio which a square number has to a square number; therefore A is incommensurable in length with D. [X. 9]

10.prop.1.10.p3282

Let E be taken a mean proportional between A, D; therefore, as A is to D, so is the square on A to the square on E. [V. Def. 9]

10.prop.1.10.p3283

But A is incommensurable in length with D; therefore the square on A is also incommensurable with the square on E; [X. 11] therefore A is incommensurable in square with E.

10.prop.1.10.p3284

Therefore two straight lines D, E have been found incommensurable, D in length only, and E in square and of course in length also, with the assigned straight line A.]

10.prop.1.11.p3285

If four magnitudes be proportional, and the first be commensurable with the second, the third will also be commensurable with the fourth; and, if the first be incommensurable with the second, the third will also be incommensurable with the fourth.

10.prop.1.11.p3286

Let A, B, C, D be four magnitudes in proportion, so that, as A is to B, so is C to D, and let A be commensurable with B; I say that C will also be commensurable with D.

10.prop.1.11.p3287

For, since A is commensurable with B, therefore A has to B the ratio which a number has to a number. [X. 5]

10.prop.1.11.p3288

And, as A is to B, so is C to D; therefore C also has to D the ratio which a number has to a number; therefore C is commensurable with D. [X. 6]

10.prop.1.11.p3289

Next, let A be incommensurable with B; I say that C will also be incommensurable with D.

10.prop.1.11.p3290

For, since A is incommensurable with B, therefore A has not to B the ratio which a number has to a number. [X. 7]

10.prop.1.11.p3291

And, as A is to B, so is C to D; therefore neither has C to D the ratio which a number has to a number; therefore C is incommensurable with D. [X. 8]

10.prop.1.11.p3292

Therefore etc.

10.prop.1.12.p3293

Magnitudes commensurable with the same magnitude are commensurable with one another also.

10.prop.1.12.p3294

For let each of the magnitudes A, B be commensurable with C; I say that A is also commensurable with B.

10.prop.1.12.p3295

For, since A is commensurable with C, therefore A has to C the ratio which a number has to a number. [X. 5]

10.prop.1.12.p3296

Let it have the ratio which D has to E.

10.prop.1.12.p3297

Again, since C is commensurable with B, therefore C has to B the ratio which a number has to a number. [X. 5]

10.prop.1.12.p3298

Let it have the ratio which F has to G.

10.prop.1.12.p3299

And, given any number of ratios we please, namely the ratio which D has to E and that which F has to G, let the numbers H, K, L be taken continuously in the given ratios; [cf. VIII. 4] so that, as D is to E, so is H to K, and, as F is to G, so is K to L.

10.prop.1.12.p3300

Since, then, as A is to C, so is D to E, while, as D is to E, so is H to K, therefore also, as A is to C, so is H to K. [V. 11]

10.prop.1.12.p3301

Again, since, as C is to B, so is F to G, while, as F is to G, so is K to L, therefore also, as C is to B, so is K to L. [V. 11]

10.prop.1.12.p3302

But also, as A is to C, so is H to K; therefore, ex aequali, as A is to B, so is H to L. [V. 22]

10.prop.1.12.p3303

Therefore A has to B the ratio which a number has to a number; therefore A is commensurable with B. [X. 6]

10.prop.1.12.p3304

Therefore etc. Q. E. D.

10.prop.1.13.p3305

If two magnitudes be commensurable, and the one of them be incommensurable with any magnitude, the remaining one will also be incommensurable with the same.

10.prop.1.13.p3306

Let A, B be two commensurable magnitudes, and let one of them, A, be incommensurable with any other magnitude C; I say that the remaining one, B, will also be incommensurable with C.

10.prop.1.13.p3307

For, if B is commensurable with C, while A is also commensurable with B, A is also commensurable with C. [X. 12]

10.prop.1.13.p3308

But it is also incommensurable with it: which is impossible.

10.prop.1.13.p3309

Therefore B is not commensurable with C; therefore it is incommensurable with it.

10.prop.1.13.p3310

Therefore etc.

10.prop.1.13.p3311

Lemma. Given two unequal straight lines, to find by what square the square on the greater is greater than the square on the less.

10.prop.1.13.p3312

Let AB, C be the given two unequal straight lines, and let AB be the greater of them; thus it is required to find by what square the square on AB is greater than the square on C.

10.prop.1.13.p3313

Let the semicircle ADB be described on AB, and let AD be fitted into it equal to C; [IV. 1] let DB be joined.

10.prop.1.13.p3314

It is then manifest that the angle ADB is right, [III. 31] and that the square on AB is greater than the square on AD, that is, C, by the square on DB. [I. 47]

10.prop.1.13.p3315

Similarly also, if two straight lines be given, the straight line the square on which is equal to the sum of the squares on them is found in this manner.

10.prop.1.13.p3316

Let AD, DB be the given two straight lines, and let it be required to find the straight line the square on which is equal to the sum of the squares on them.

10.prop.1.13.p3317

Let them be placed so as to contain a right angle, that formed by AD, DB; and let AB be joined.

10.prop.1.13.p3318

It is again manifest that the straight line the square on which is equal to the sum of the squares on AD, DB is AB. [I. 47] Q. E. D.

10.prop.1.14.p3319

If four straight lines be proportional, and the square on the first be greater than the square on the second by the square on a straight line commensurable with the first, the square on the third will also be greater than the square on the fourth by the square on a straight line commensurable with the third.

10.prop.1.14.p3320

And, if the square on the first be greater than the square on the second by the square on a straight line incommensurable with the first, the square on the third will also be greater than the square on the fourth by the square on a straight line in- commensurable with the third.

10.prop.1.14.p3321

Let A, B, C, D be four straight lines in proportion, so that, as A is to B, so is C to D; and let the square on A be greater than the square on B by the square on E, and let the square on C be greater than the square on D by the square on F; I say that, if A is commensurable with E, C is also commensurable with F, and, if A is incommensurable with E, C is also incommensurable with F.

10.prop.1.14.p3322

For since, as A is to B, so is C to D, therefore also, as the square on A is to the square on B, so is the square on C to the square on D. [VI. 22]

10.prop.1.14.p3323

But the squares on E, B are equal to the square on A, and the squares on D, F are equal to the square on C.

10.prop.1.14.p3324

Therefore, as the squares on E, B are to the square on B, so are the squares on D, F to the square on D; therefore, separando, as the square on E is to the square on B, so is the square on F to the square on D; [V. 17] therefore also, as E is to B, so is F to D; [VI. 22] therefore, inversely, as B is to E, so is D to F.

10.prop.1.14.p3325

But, as A is to B, so also is C to D; therefore, ex aequali, as A is to E, so is C to F. [V. 22]

10.prop.1.14.p3326

Therefore, if A is commensurable with E, C is also commensurable with F, and, if A is incommensurable with E, C is also incommensurable with F. [X. 11]

10.prop.1.14.p3327

Therefore etc.

10.prop.1.14.p3327

1

10.prop.1.15.p3328

If two commensurable magnitudes be added together, the whole will also be commensurable with each of them; and, if the whole be commensurable with one of them, the original magnitudes will also be commensurable.

10.prop.1.15.p3329

For let the two commensurable magnitudes AB, BC be added together; I say that the whole AC is also commensurable with each of the magnitudes AB, BC.

10.prop.1.15.p3330

For, since AB, BC are commensurable, some magnitude will measure them.

10.prop.1.15.p3331

Let it measure them, and let it be D.

10.prop.1.15.p3332

Since then D measures AB, BC, it will also measure the whole AC.

10.prop.1.15.p3333

But it measures AB, BC also; therefore D measures AB, BC, AC; therefore AC is commensurable with each of the magnitudes AB, BC. [X. Def. 1]

10.prop.1.15.p3334

Next, let AC be commensurable with AB; I say that AB, BC are also commensurable.

10.prop.1.15.p3335

For, since AC, AB are commensurable, some magnitude will measure them.

10.prop.1.15.p3336

Let it measure them, and let it be D.

10.prop.1.15.p3337

Since then D measures CA, AB, it will also measure the remainder BC.

10.prop.1.15.p3338

But it measures AB also; therefore D will measure AB, BC; therefore AB, BC are commensurable. [X. Def. 1]

10.prop.1.15.p3339

Therefore etc.

10.prop.1.16.p3340

If two incommensurable magnitudes be added together, the whole will also be incommensurable with each of them; and, if the whole be incommensurable with one of them, the original magnitudes will also be incommensurable.

10.prop.1.16.p3341

For let the two incommensurable magnitudes AB, BC be added together; I say that the whole AC is also incommensurable with each of the magnitudes AB, BC.

10.prop.1.16.p3342

For, if CA, AB are not incommensurable, some magnitude will measure them.

10.prop.1.16.p3343

Let it measure them, if possible, and let it be D.

10.prop.1.16.p3344

Since then D measures CA, AB, therefore it will also measure the remainder BC.

10.prop.1.16.p3345

But it measures AB also; therefore D measures AB, BC.

10.prop.1.16.p3346

Therefore AB, BC are commensurable; but they were also, by hypothesis, incommensurable: which is impossible.

10.prop.1.16.p3347

Therefore no magnitude will measure CA, AB; therefore CA, AB are incommensurable. [X. Def. 1]

10.prop.1.16.p3348

Similarly we can prove that AC, CB are also incommensurable.

10.prop.1.16.p3349

Therefore AC is incommensurable with each of the magnitudes AB, BC.

10.prop.1.16.p3350

Next, let AC be incommensurable with one of the magnitudes AB, BC.

10.prop.1.16.p3351

First, let it be incommensurable with AB; I say that AB, BC are also incommensurable.

10.prop.1.16.p3352

For, if they are commensurable, some magnitude will measure them.

10.prop.1.16.p3353

Let it measure them, and let it be D.

10.prop.1.16.p3354

Since then D measures AB, BC. therefore it will also measure the whole AC.

10.prop.1.16.p3355

But it measures AB also; therefore D measures CA, AB.

10.prop.1.16.p3356

Therefore CA, AB are commensurable; but they were also, by hypothesis, incommensurable: which is impossible.

10.prop.1.16.p3357

Therefore no magnitude will measure AB, BC; therefore AB, BC are incommensurable. [X. Def. 1]

10.prop.1.16.p3358

Therefore etc.

10.prop.1.16.p3359

Lemma. If to any straight line there be applied a parallelogram deficient by a square figure, the applied parallelogram is equal to the rectangle contained by the segments of the straight line resulting from the application.

10.prop.1.16.p3360

For let there be applied to the straight line AB the parallelogram AD deficient by the square figure DB; I say that AD is equal to the rectangle contained by AC, CB.

10.prop.1.16.p3361

This is indeed at once manifest; for, since DB is a square, DC is equal to CB; and AD is the rectangle AC, CD, that is, the rectangle AC, CB.

10.prop.1.16.p3362

Therefore etc.

10.prop.1.17.p3363

If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are commensurable in length, then the square on the greater will be greater than the square on the less by the square on a straight line commensurable with the greater.

10.prop.1.17.p3364

And, if the square on the greater be greater than the square on the less by the square on a straight line commensurable with the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it will divide it into parts which are commensurable in length.

10.prop.1.17.p3365

Let A, BC be two unequal straight lines, of which BC is the greater, and let there be applied to BC a parallelogram equal to the fourth part of the square on the less, A, that is, equal to the square on the half of A, and deficient by a square figure. Let this be the rectangle BD, DC, [cf. Lemma] and let BD be commensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line commensurable with BC.

10.prop.1.17.p3366

For let BC be bisected at the point E, and let EF be made equal to DE.

10.prop.1.17.p3367

Therefore the remainder DC is equal to BF.

10.prop.1.17.p3368

And, since the straight line BC has been cut into equal parts at E, and into unequal parts at D, therefore the rectangle contained by BD, DC, together with the square on ED, is equal to the square on EC; [II. 5]

10.prop.1.17.p3369

And the same is true of their quadruples; therefore four times the rectangle BD, DC, together with four times the square on DE, is equal to four times the square on EC.

10.prop.1.17.p3370

But the square on A is equal to four times the rectangle BD, DC; and the square on DF is equal to four times the square on DE, for DF is double of DE.

10.prop.1.17.p3371

And the square on BC is equal to four times the square on EC, for again BC is double of CE.

10.prop.1.17.p3372

Therefore the squares on A, DF are equal to the square on BC, so that the square on BC is greater than the square on A by the square on DF.

10.prop.1.17.p3373

It is to be proved that BC is also commensurable with DF.

10.prop.1.17.p3374

Since BD is commensurable in length with DC, therefore BC is also commensurable in length with CD. [X. 15]

10.prop.1.17.p3375

But CD is commensurable in length with CD, BF, for CD is equal to BF. [X. 6]

10.prop.1.17.p3376

Therefore BC is also commensurable in length with BF, CD, [X. 12] so that BC is also commensurable in length with the remainder FD; [X. 15] therefore the square on BC is greater than the square on A by the square on a straight line commensurable with BC.

10.prop.1.17.p3377

Next, let the square on BC be greater than the square on A by the square on a straight line commensurable with BC, let a parallelogram be applied to BC equal to the fourth part of the square on A and deficient by a square figure, and let it be the rectangle BD, DC.

10.prop.1.17.p3378

It is to be proved that BD is commensurable in length with DC.

10.prop.1.17.p3379

With the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD.

10.prop.1.17.p3380

But the square on BC is greater than the square on A by the square on a straight line commensurable with BC.

10.prop.1.17.p3381

Therefore BC is commensurable in length with FD, so that BC is also commensurable in length with the remainder, the sum of BF, DC. [X. 15]

10.prop.1.17.p3382

But the sum of BF, DC is commensurable with DC, [X. 6] so that BC is also commensurable in length with CD; [X. 12] and therefore, separando, BD is commensurable in length with DC. [X. 15]

10.prop.1.17.p3383

Therefore etc.

10.prop.1.17.p3383

1

10.prop.1.18.p3384

If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are incommensurable, the square on the greater will be greater than the square on the less by the square on a straight line incommensurable with the greater.

10.prop.1.18.p3385

And, if the square on the greater be greater than the square on the less by the square on a straight line incommensurable with the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it divides it into parts which are incommensurable.

10.prop.1.18.p3386

Let A, BC be two unequal straight lines, of which BC is the greater, and to BC let there be applied a parallelogram equal to the fourth part of the square on the less, A, and deficient by a square figure. Let this be the rectangle BD, DC, [cf. Lemma before X. 17] and let BD be incommensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line incommensurable with BC.

10.prop.1.18.p3387

For, with the same construction as before, we can prove similarly that the square on BC is greater than the square on A by the square on FD.

10.prop.1.18.p3388

It is to be proved that BC is incommensurable in length with DF.

10.prop.1.18.p3389

Since BD is incommensurable in length with DC, therefore BC is also incommensurable in length with CD. [X. 16]

10.prop.1.18.p3390

But DC is commensurable with the sum of BF, DC; [X. 6] therefore BC is also incommensurable with the sum of BF, DC; [X. 13] so that BC is also incommensurable in length with the remainder FD. [X. 16]

10.prop.1.18.p3391

And the square on BC is greater than the square on A by the square on FD; therefore the square on BC is greater than the square on A by the square on a straight line incommensurable with BC.

10.prop.1.18.p3392

Again, let the square on BC be greater than the square on A by the square on a straight line incommensurable with BC, and let there be applied to BC a parallelogram equal to the fourth part of the square on A and deficient by a square figure. Let this be the rectangle BD, DC.

10.prop.1.18.p3393

It is to be proved that BD is incommensurable in length with DC.

10.prop.1.18.p3394

For, with the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD.

10.prop.1.18.p3395

But the square on BC is greater than the square on A by the square on a straight line incommensurable with BC; therefore BC is incommensurable in length with FD. so that BC is also incommensurable with the remainder, the sum of BF, DC. [X. 16]

10.prop.1.18.p3396

But the sum of BF, DC is commensurable in length with DC; [X. 6] therefore BC is also incommensurable in length with DC, [X. 13] so that, separando, BD is also incommensurable in length with DC. [X. 16]

10.prop.1.18.p3397

Therefore etc.

10.prop.1.18.p3398

[Lemma. Since it has been proved that straight lines commensurable in length are always commensurable in square also, while those commensurable in square are not always commensurable in length also, but can of course be either commensurable or incommensurable in length, it is manifest that, if any straight line be commensurable in length with a given rational straight line, it is called rational and commensurable with the other not only in length but in square also, since straight lines commensurable in length are always commensurable in square also.

10.prop.1.18.p3399

But, if any straight line be commensurable in square with a given rational straight line, then, if it is also commensurable in length with it, it is called in this case also rational and commensurable with it both in length and in square; but, if again any straight line, being commensurable in square with a given rational straight line, be incommensurable in length with it, it is called in this case also rational but commensurable in square only.]

10.prop.1.19.p3400

The rectangle contained by rational straight lines commensurable in length is rational.

10.prop.1.19.p3401

For let the rectangle AC be contained by the rational straight lines AB, BC commensurable in length; I say that AC is rational.

10.prop.1.19.p3402

For on AB let the square AD be described; therefore AD is rational. [X. Def. 4]

10.prop.1.19.p3403

And, since AB is commensurable in length with BC, while AB is equal to BD, therefore BD is commensurable in length with BC.

10.prop.1.19.p3404

And, as BD is to BC, so is DA to AC. [VI. 1]

10.prop.1.19.p3405

Therefore DA is commensurable with AC. [X. 11]

10.prop.1.19.p3406

But DA is rational; therefore AC is also rational. [X. Def. 4]

10.prop.1.19.p3407

Therefore etc.

10.prop.1.20.p3408

If a rational area be applied to a rational straight line, it produces as breadth a straight line rational and commensurable in length with the straight line to which it is applied.

10.prop.1.20.p3409

For let the rational area AC be applied to AB, a straight line once more rational in any of the aforesaid ways, producing BC as breadth; I say that BC is rational and commensurable in length with BA. For on AB let the square AD be described; therefore AD is rational. [X. Def. 4]

10.prop.1.20.p3410

But AC is also rational; therefore DA is commensurable with AC.

10.prop.1.20.p3411

And, as DA is to AC, so is DB to BC. [VI. 1]

10.prop.1.20.p3412

Therefore DB is also commensurable with BC; [X. 11] and DB is equal to BA; therefore AB is also commensurable with BC.

10.prop.1.20.p3413

But AB is rational; therefore BC is also rational and commensurable in length with AB.

10.prop.1.20.p3414

Therefore etc.

10.prop.1.21.p3415

The rectangle contained by rational straight lines commensurable in square only is irrational, and the side of the square equal to it is irrational. Let the latter be called medial.

10.prop.1.21.p3416

For let the rectangle AC be contained by the rational straight lines AB, BC commensurable in square only; I say that AC is irrational, and the side of the square equal to it is irrational; and let the latter be called medial.

10.prop.1.21.p3417

For on AB let the square AD be described; therefore AD is rational. [X. Def. 4]

10.prop.1.21.p3418

And, since AB is incommensurable in length with BC, for by hypothesis they are commensurable in square only, while AB is equal to BD, therefore DB is also incommensurable in length with BC.

10.prop.1.21.p3419

And, as DB is to BC, so is AD to AC; [VI. 1] therefore DA is incommensurable with AC. [X. 11]

10.prop.1.21.p3420

But DA is rational; therefore AC is irrational, so that the side of the square equal to AC is also irrational. [X. Def. 4]

10.prop.1.21.p3421

And let the latter be called medial. Q. E. D.

10.prop.1.21.p3422

Lemma. If there be two straight lines, then, as the first is to the second, so is the square on the first to the rectangle contained by the two straight lines.

10.prop.1.21.p3423

Let FE, EG be two straight lines.

10.prop.1.21.p3424

I say that, as FE is to EG, so is the square on FE to the rectangle FE, EG.

10.prop.1.21.p3425

For on FE let the square DF be described, and let GD be completed.

10.prop.1.21.p3426

Since then, as FE is to EG, so is FD to DG, [VI. 1] and FD is the square on FE, and DG the rectangle DE, EG, that is, the rectangle FE, EG, therefore, as FE is to EG, so is the square on FE to the rectangle FE, EG.

10.prop.1.21.p3427

Similarly also, as the rectangle GE, EF is to the square on EF, that is, as GD is to FD, so is GE to EF. Q. E. D.

10.prop.1.22.p3428

The square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied.

10.prop.1.22.p3429

Let A be medial and CB rational, and let a rectangular area BD equal to the square on A be applied to BC, producing CD as breadth; I say that CD is rational and incommensurable in length with CB.

10.prop.1.22.p3430

For, since A is medial, the square on it is equal to a rectangular area contained by rational straight lines commensurable in square only. [X. 21]

10.prop.1.22.p3431

Let the square on it be equal to GF.

10.prop.1.22.p3432

But the square on it is also equal to BD; therefore BD is equal to GF.

10.prop.1.22.p3433

But it is also equiangular with it; and in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; [VI. 14] therefore, proportionally, as BC is to EG, so is EF to CD.

10.prop.1.22.p3434

Therefore also, as the square on BC is to the square on EG, so is the square on EF to the square on CD. [VI. 22]

10.prop.1.22.p3435

But the square on CB is commensurable with the square on EG, for each of these straight lines is rational; therefore the square on EF is also commensurable with the square on CD. [X. 11]

10.prop.1.22.p3436

But the square on EF is rational; therefore the square on CD is also rational; [X. Def. 4] therefore CD is rational.

10.prop.1.22.p3437

And, since EF is incommensurable in length with EG, for they are commensurable in square only, and, as EF is to EG, so is the square on EF to the rectangle FE, EG, [Lemma] therefore the square on EF is incommensurable with the rectangle FE, EG. [X. 11]

10.prop.1.22.p3438

But the square on CD is commensurable with the square on EF, for the straight lines are rational in square; and the rectangle DC, CB is commensurable with the rectangle FE, EG, for they are equal to the square on A; therefore the square on CD is also incommensurable with the rectangle DC, CB. [X. 13]

10.prop.1.22.p3439

But, as the square on CD is to the rectangle DC, CB, so is DC to CB; [Lemma] therefore DC is incommensurable in length with CB. [X. 11]

10.prop.1.22.p3440

Therefore CD is rational and incommensurable in length with CB. Q. E. D.

10.prop.1.23.p3441

A straight line commensurable with a medial straight line is medial.

10.prop.1.23.p3442

Let A be medial, and let B be commensurable with A; I say that B is also medial.

10.prop.1.23.p3443

For let a rational straight line CD be set out, and to CD let the rectangular area CE equal to the square on A be applied, producing ED as breadth; therefore ED is rational and incommensurable in length with CD. [X. 22]

10.prop.1.23.p3444

And let the rectangular area CF equal to the square on B be applied to CD, producing DF as breadth.

10.prop.1.23.p3445

Since then A is commensurable with B, the square on A is also commensurable with the square on B.

10.prop.1.23.p3446

But EC is equal to the square on A, and CF is equal to the square on B; therefore EC is commensurable with CF.

10.prop.1.23.p3447

And, as EC is to CF, so is ED to DF; [VI. 1] therefore ED is commensurable in length with DF. [X. 11]

10.prop.1.23.p3448

But ED is rational and incommensurable in length with DC; therefore DF is also rational [X. Def. 3] and incommensurable in length with DC. [X. 13]

10.prop.1.23.p3449

Therefore CD, DF are rational and commensurable in square only.

10.prop.1.23.p3450

But the straight line the square on which is equal to the rectangle contained by rational straight lines commensurable in square only is medial; [X. 21] therefore the side of the square equal to the rectangle CD, DF is medial.

10.prop.1.23.p3451

And B is the side of the square equal to the rectangle CD, DF; therefore B is medial.

10.prop.1.23.p3452

Porism. From this it is manifest that an area commensurable with a medial area is medial.

10.prop.1.23.p3453

[And in the same way as was explained in the case of rationals [Lemma following X. 18] it follows, as regards medials, that a straight line commensurable in length with a medial straight line is called medial and commensurable with it not only in length but in square also, since, in general, straight lines commensurable in length are always commensurable in square also.

10.prop.1.23.p3454

But, if any straight line be commensurable in square with a medial straight line, then, if it is also commensurable in length with it, the straight lines are called, in this case too, medial and commensurable in length and in square, but, if in square only, they are called medial straight lines commensurable in square only.]

10.prop.1.24.p3455

The rectangle contained by medial straight lines commensurable in length is medial.

10.prop.1.24.p3456

For let the rectangle AC be contained by the medial straight lines AB, BC which are commensurable in length; I say that AC is medial.

10.prop.1.24.p3457

For on AB let the square AD be described; therefore AD is medial.

10.prop.1.24.p3458

And, since AB is commensurable in length with BC, while AB is equal to BD, therefore DB is also commensurable in length with BC; so that DA is also commensurable with AC. [VI. 1, X. 11]

10.prop.1.24.p3459

But DA is medial; therefore AC is also medial. [X. 23, Por.] Q. E. D.

10.prop.1.25.p3460

The rectangle contained by medial straight lines commensurable in square only is either rational or medial.

10.prop.1.25.p3461

For let the rectangle AC be contained by the medial straight lines AB, BC which are commensurable in square only; I say that AC is either rational or medial.

10.prop.1.25.p3462

For on AB, BC let the squares AD, BE be described; therefore each of the squares AD, BE is medial.

10.prop.1.25.p3463

Let a rational straight line FG be set out, to FG let there be applied the rectangular parallelogram GH equal to AD, producing FH as breadth, to HM let there be applied the rectangular parallelogram MK equal to AC, producing HK as breadth, and further to KN let there be similarly applied NL equal to BE, producing KL as breadth; therefore FH, HK, KL are in a straight line.

10.prop.1.25.p3464

Since then each of the squares AD, BE is medial, and AD is equal to GH, and BE to NL, therefore each of the rectangles GH, NL is also medial.

10.prop.1.25.p3465

And they are applied to the rational straight line FG; therefore each of the straight lines FH, KL is rational and incommensurable in length with FG. [X. 22]

10.prop.1.25.p3466

And, since AD is commensurable with BE, therefore GH is also commensurable with NL.

10.prop.1.25.p3467

And, as GH is to NL, so is FH to KL; [VI. 1] therefore FH is commensurable in length with KL. [X. 11]

10.prop.1.25.p3468

Therefore FH, KL are rational straight lines commensurable in length; therefore the rectangle FH, KL is rational. [X. 19]

10.prop.1.25.p3469

And, since DB is equal to BA, and OB to BC, therefore, as DB is to BC, so is AB to BO.

10.prop.1.25.p3470

But, as DB is to BC, so is DA to AC, [VI. 1] and, as AB is to BO, so is AC to CO; [id.] therefore, as DA is to AC, so is AC to CO.

10.prop.1.25.p3471

But AD is equal to GH, AC to MK and CO to NL; therefore, as GH is to MK, so is MK to NL; therefore also, as FH is to HK, so is HK to KL; [VI. 1, V. 11] therefore the rectangle FH, KL is equal to the square on HK. [VI. 17]

10.prop.1.25.p3472

But the rectangle FH, KL is rational; therefore the square on HK is also rational.

10.prop.1.25.p3473

Therefore HK is rational.

10.prop.1.25.p3474

And, if it is commensurable in length with FG, HN is rational; [X. 19] but, if it is incommensurable in length with FG, KH, HM are rational straight lines commensurable in square only, and therefore HN is medial. [X. 21]

10.prop.1.25.p3475

Therefore HN is either rational or medial.

10.prop.1.25.p3476

But HN is equal to AC; therefore AC is either rational or medial.

10.prop.1.25.p3477

Therefore etc.

10.prop.1.26.p3478

4 medial area does not exceed a medial area by a rational area.

10.prop.1.26.p3479

For, if possible, let the medial area AB exceed the medial area AC by the rational area DB, and let a rational straight line EF be set out; to EF let there be applied the rectangular parallelogram FH equal to AB, producing EH as breadth, and let the rectangle FG equal to AC be subtracted; therefore the remainder BD is equal to the remainder KH.

10.prop.1.26.p3480

But DB is rational; therefore KH is also rational.

10.prop.1.26.p3481

Since, then, each of the rectangles AB, AC is medial, and AB is equal to FH, and AC to FG, therefore each of the rectangles FH, FG is also medial.

10.prop.1.26.p3482

And they are applied to the rational straight line EF; therefore each of the straight lines HE, EG is rational and incommensurable in length with EF. [X. 22]

10.prop.1.26.p3483

And, since [DB is rational and is equal to KH, therefore] KH is [also] rational; and it is applied to the rational straight line EF; therefore GH is rational and commensurable in length with EF. [X. 20]

10.prop.1.26.p3484

But EG is also rational, and is incommensurable in length with EF; therefore EG is incommensurable in length with GH. [X. 13]

10.prop.1.26.p3485

And, as EG is to GH, so is the square on EG to the rectangle EG, GH; therefore the square on EG is incommensurable with the rectangle EG, GH. [X. 11]

10.prop.1.26.p3486

But the squares on EG, GH are commensurable with the square on EG, for both are rational; and twice the rectangle EG, GH is commensurable with the rectangle EG, GH, for it is double of it; [X. 6] therefore the squares on EG, GH are incommensurable with twice the rectangle EG, GH; [X. 13] therefore also the sum of the squares on EG, GH and twice the rectangle EG, GH, that is, the square on EH [II. 4], is incommensurable with the squares on EG, GH. [X. 16]

10.prop.1.26.p3487

But the squares on EG, GH are rational; therefore the square on EH is irrational. [X. Def. 4]

10.prop.1.26.p3488

Therefore EH is irrational.

10.prop.1.26.p3489

But it is also rational: which is impossible.

10.prop.1.26.p3490

Therefore etc. Q. E. D.

10.prop.1.27.p3491

To find medial straight lines commensurable in square only which contain a rational rectangle.

10.prop.1.27.p3492

Let two rational straight lines A, B commensurable in square only be set out; let C be taken a mean proportional between A, B, [VI. 13] and let it be contrived that, as A is to B, so is C to D. [VI. 12]

10.prop.1.27.p3493

Then, since A, B are rational and commensurable in square only, the rectangle A, B, that is, the square on C [VI.17], is medial. [X. 21]

10.prop.1.27.p3494

Therefore C is medial. [X. 21]

10.prop.1.27.p3495

And since, as A is to B, so is C to D, and A, B are commensurable in square only, therefore C, D are also commensurable in square only. [X. 11]

10.prop.1.27.p3496

And C is medial; therefore D is also medial. [X. 23, addition]

10.prop.1.27.p3497

Therefore C, D are medial and commensurable in square only.

10.prop.1.27.p3498

I say that they also contain a rational rectangle.

10.prop.1.27.p3499

For since, as A is to B, so is C to D, therefore, alternately, as A is to C, so is B to D. [V. 16]

10.prop.1.27.p3500

But, as A is to C, so is C to B; therefore also, as C is to B, so is B to D; therefore the rectangle C, D is equal to the square on B.

10.prop.1.27.p3501

But the square on B is rational; therefore the rectangle C, D is also rational.

10.prop.1.27.p3502

Therefore medial straight lines commensurable in square only have been found which contain a rational rectangle. Q. E. D.

10.prop.1.28.p3503

To find medial straight lines commensurable in square only which contain a medial rectangle.

10.prop.1.28.p3504

Let the rational straight lines A, B, C commensurable in square only be set out; let D be taken a mean proportional between A, B, [VI. 13] and let it be contrived that, as B is to C, so is D to E. [VI. 12]

10.prop.1.28.p3505

Since A, B are rational straight lines commensurable in square only, therefore the rectangle A, B, that is, the square on D [VI. 17], is medial. [X. 21]

10.prop.1.28.p3506

Therefore D is medial. [X. 21]

10.prop.1.28.p3507

And since B, C are commensurable in square only, and, as B is to C, so is D to E, therefore D, E are also commensurable in square only. [X. 11]

10.prop.1.28.p3508

But D is medial; therefore E is also medial. [X. 23, addition]

10.prop.1.28.p3509

Therefore D, E are medial straight lines commensurable in square only.

10.prop.1.28.p3510

I say next that they also contain a medial rectangle.

10.prop.1.28.p3511

For since, as B is to C, so is D to E, therefore, alternately, as B is to D, so is C to E. [V. 16]

10.prop.1.28.p3512

But, as B is to D, so is D to A; therefore also, as D is to A, so is C to E; therefore the rectangle A, C is equal to the rectangle D, E. [VI. 16]

10.prop.1.28.p3513

But the rectangle A, C is medial; [X. 21] therefore the rectangle D, E is also medial.

10.prop.1.28.p3514

Therefore medial straight lines commensurable in square only have been found which contain a medial rectangle. Q. E. D.

10.prop.1.28.p3515

LEMMA I. To find two square numbers such that their sum is also square.

10.prop.1.28.p3516

Let two numbers AB, BC be set out, and let them be either both even or both odd.

10.prop.1.28.p3517

Then since, whether an even number is subtracted from an even number, or an odd number from an odd number, the remainder is even, [IX. 24, 26] therefore the remainder AC is even.

10.prop.1.28.p3518

Let AC be bisected at D.

10.prop.1.28.p3519

Let AB, BC also be either similar plane numbers, or square numbers, which are themselves also similar plane numbers.

10.prop.1.28.p3520

Now the product of AB, BC together with the square on CD is equal to the square on BD. [II. 6]

10.prop.1.28.p3521

And the product of AB, BC is square, inasmuch as it was proved that, if two similar plane numbers by multiplying one another make some number the product is square. [IX. 1]

10.prop.1.28.p3522

Therefore two square numbers, the product of AB, BC, and the square on CD, have been found which, when added together, make the square on BD.

10.prop.1.28.p3523

And it is manifest that two square numbers, the square on BD and the square on CD, have again been found such that their difference, the product of AB, BC, is a square, whenever AB, BC are similar plane numbers.

10.prop.1.28.p3524

But when they are not similar plane numbers, two square numbers, the square on BD and the square on DC, have been found such that their difference, the product of AB, BC, is not square. Q. E. D.

10.prop.1.28.p3525

LEMMA 2. To find two square numbers such that their sum is not square.

10.prop.1.28.p3526

For let the product of AB, BC, as we said, be square, and CA even, and let CA be bisected by D.

10.prop.1.28.p3527

It is then manifest that the square product of AB, BC together with the square on CD is equal to the square on BD. [See Lemma 1]

10.prop.1.28.p3528

Let the unit DE be subtracted; therefore the product of AB, BC together with the square on CE is less than the square on BD.

10.prop.1.28.p3529

I say then that the square product of AB, BC together with the square on CE will not be square.

10.prop.1.28.p3530

For, if it is square, it is either equal to the square on BE, or less than the square on BE, but cannot any more be greater, lest the unit be divided.

10.prop.1.28.p3531

First, if possible, let the product of AB, BC together with the square on CE be equal to the square on BE, and let GA be double of the unit DE.

10.prop.1.28.p3532

Since then the whole AC is double of the whole CD, and in them AG is double of DE, therefore the remainder GC is also double of the remainder EC; therefore GC is bisected by E.

10.prop.1.28.p3533

Therefore the product of GB, BC together with the square on CE is equal to the square on BE. [II. 6]

10.prop.1.28.p3534

But the product of AB, BC together with the square on CE is also, by hypothesis, equal to the square on BE; therefore the product of GB, BC together with the square on CE is equal to the product of AB, BC together with the square on CE.

10.prop.1.28.p3535

And, if the common square on CE be subtracted, it follows that AB is equal to GB: which is absurd.

10.prop.1.28.p3536

Therefore the product of AB, BC together with the square on CE is not equal to the square on BE.

10.prop.1.28.p3537

I say next that neither is it less than the square on BE.

10.prop.1.28.p3538

For, if possible, let it be equal to the square on BF, and let HA be double of DF.

10.prop.1.28.p3539

Now it will again follow that HC is double of CF; so that CH has also been bisected at F, and for this reason the product of HB, BC together with the square on FC is equal to the square on BF. [II. 6]

10.prop.1.28.p3540

But, by hypothesis, the product of AB, BC together with the square on CE is also equal to the square on BF.

10.prop.1.28.p3541

Thus the product of HB, BC together with the square on CF will also be equal to the product of AB, BC together with the square on CE: which is absurd.

10.prop.1.28.p3542

Therefore the product of AB, BC together with the square on CE is not less than the square on BE.

10.prop.1.28.p3543

And it was proved that neither is it equal to the square on BE.

10.prop.1.28.p3544

Therefore the product of AB, BC together with the square on CE is not square. Q. E. D.

10.prop.1.29.p3545

To find two rational straight lines commensurable in square only and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater.

10.prop.1.29.p3546

For let there be set out any rational straight line AB, and two square numbers CD, DE such that their difference CE is not square; [Lemma 1] let there be described on AB the semicircle AFB, and let it be contrived that, as DC is to CE, so is the square on BA to the square on AF. [X. 6, Por.]

10.prop.1.29.p3547

Let FB be joined.

10.prop.1.29.p3548

Since, as the square on BA is to the square on AF, so is DC to CE, therefore the square on BA has to the square on AF the ratio which the number DC has to the number CE; therefore the square on BA is commensurable with the square on AF. [X. 6]

10.prop.1.29.p3549

But the square on AB is rational; [X. Def. 4] therefore the square on AF is also rational; [id.] therefore AF is also rational.

10.prop.1.29.p3550

And, since DC has not to CE the ratio which a square number has to a square number, neither has the square on BA to the square on AF the ratio which a square number has to a square number; therefore AB is incommensurable in length with AF. [X. 9]

10.prop.1.29.p3551

Therefore BA, AF are rational straight lines commensurable in square only.

10.prop.1.29.p3552

And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [V. 19, Por., III. 31, I. 47]

10.prop.1.29.p3553

But CD has to DE the ratio which a square number has to a square number: therefore also the square on AB has to the square on BF the ratio which a square number has to a square number; therefore AB is commensurable in length with BF. [X. 9]

10.prop.1.29.p3554

And the square on AB is equal to the squares on AF, FB; therefore the square on AB is greater than the square on AF by the square on BF commensurable with AB.

10.prop.1.29.p3555

Therefore there have been found two rational straight lines BA, AF commensurable in square only and such that the square on the greater AB is greater than the square on the less AF by the square on BF commensurable in length with AB.

10.prop.1.30.p3556

To find two rational straight lines commensurable in square only and such that the square on the greater is greater is greater than the square on the less by the square on a straight line incommensurable in length with the greater.

10.prop.1.30.p3557

Let there be set out a rational straight line AB, and two square numbers CE, ED such that their sum CD is not square; [Lemma 2] let there be described on AB the semicircle AFB, let it be contrived that, as DC is to CE, so is the square on BA to the square on AF, [X. 6, Por.] and let FB be joined.

10.prop.1.30.p3558

Then, in a similar manner to the preceding, we can prove that BA, AF are rational straight lines commensurable in square only.

10.prop.1.30.p3559

And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [V. 19, Por., III. 31, I. 47]

10.prop.1.30.p3560

But CD has not to DE the ratio which a square number has to a square number; therefore neither has the square on AB to the square on BF the ratio which a square number has to a square number; therefore AB is incommensurable in length with BF. [X. 9]

10.prop.1.30.p3561

And the square on AB is greater than the square on AF by the square on FB incommensurable with AB.

10.prop.1.30.p3562

Therefore AB, AF are rational straight lines commensurable in square only, and the square on AB is greater than the square on AF by the square on FB incommensurable in length with AB. Q. E. D.

10.prop.1.31.p3563

To find two medial straight lines commensurable in square only, containing a rational rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater.

10.prop.1.31.p3564

Let there be set out two rational straight lines A, B commensurable in square only and such that the square on A, being the greater, is greater than the square on B the less by the square on a straight line commensurable in length with A. [X. 29]

10.prop.1.31.p3565

And let the square on C be equal to the rectangle A, B.

10.prop.1.31.p3566

Now the rectangle A, B is medial; [X. 21] therefore the square on C is also medial; therefore C is also medial. [X. 21]

10.prop.1.31.p3567

Let the rectangle C, D be equal to the square on B.

10.prop.1.31.p3568

Now the square on B is rational; therefore the rectangle C, D is also rational.

10.prop.1.31.p3569

And since, as A is to B, so is the rectangle A, B to the square on B, while the square on C is equal to the rectangle A, B, and the rectangle C, D is equal to the square on B, therefore, as A is to B, so is the square on C to the rectangle C, D.

10.prop.1.31.p3570

But, as the square on C is to the rectangle C, D, so is C to D; therefore also, as A is to B, so is C to D.

10.prop.1.31.p3571

But A is commensurable with B in square only; therefore C is also commensurable with D in square only. [X. 11]

10.prop.1.31.p3572

And C is medial; therefore D is also medial. [X. 23, addition]

10.prop.1.31.p3573

And since, as A is to B, so is C to D, and the square on A is greater than the square on B by the square on a straight line commensurable with A, therefore also the square on C is greater than the square on D by the square on a straight line commensurable with C. [X. 14]

10.prop.1.31.p3574

Therefore two medial straight lines C, D, commensurable in square only and containing a rational rectangle, have been found, and the square on C is greater than the square on D by the square on a straight line commensurable in length with C.

10.prop.1.31.p3575

Similarly also it can be proved that the square on C exceeds the square on D by the square on a straight line incommensurable with C, when the square on A is greater than the square on B by the square on a straight line incommensurable with A. [X. 30]

10.prop.1.32.p3576

To find two medial straight lines commensurable in square only, containing a medial rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater.

10.prop.1.32.p3577

Let there be set out three rational straight lines A, B, C commensurable in square only, and such that the square on A is greater than the square on C by the square on a straight line commensurable with A, [X. 29] and let the square on D be equal to the rectangle A, B.

10.prop.1.32.p3578

Therefore the square on D is medial; therefore D is also medial. [X. 21]

10.prop.1.32.p3579

Let the rectangle D, E be equal to the rectangle B, C.

10.prop.1.32.p3580

Then since, as the rectangle A, B is to the rectangle B, C, so is A to C; while the square on D is equal to the rectangle A, B, and the rectangle D, E is equal to the rectangle B, C, therefore, as A is to C, so is the square on D to the rectangle D, E.

10.prop.1.32.p3581

But, as the square on D is to the rectangle D, E, so is D to E; therefore also, as A is to C, so is D to E.

10.prop.1.32.p3582

But A is commensurable with C in square only; therefore D is also commensurable with E in square only. [X. 11]

10.prop.1.32.p3583

But D is medial; therefore E is also medial. [X. 23, addition]

10.prop.1.32.p3584

And, since, as A is to C, so is D to E, while the square on A is greater than the square on C by the square on a straight line commensurable with A, therefore also the square on D will be greater than the square on E by the square on a straight line commensurable with D.[X. 14]

10.prop.1.32.p3585

I say next that the rectangle D, E is also medial.

10.prop.1.32.p3586

For, since the rectangle B, C is equal to the rectangle D, E, while the rectangle B, C is medial, [X. 21] therefore the rectangle D, E is also medial.

10.prop.1.32.p3587

Therefore two medial straight lines D, E, commensurable in square only, and containing a medial rectangle, have been found such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater.

10.prop.1.32.p3588

Similarly again it can be proved that the square on D is greater than the square on E by the square on a straight line incommensurable with D, when the square on A is greater than the square on C by the square on a straight line incommensurable with A. [X. 30]

10.prop.1.32.p3589

Lemma. Let ABC be a right-angled triangle having the angle A right, and let the perpendicular AD be drawn; I say that the rectangle CB, BD is equal to the square on BA, the rectangle BC, CD equal to the square on CA, the rectangle BD, DC equal to the square on AD, and, further, the rectangle BC, AD equal to the rectangle BA, AC.

10.prop.1.32.p3590

And first that the rectangle CB, BD is equal to the square on BA.

10.prop.1.32.p3591

For, since in a right-angled triangle AD has been drawn from the right angle perpendicular to the base, therefore the triangles ABD, ADC are similar both to the whole ABC and to one another. [VI. 8]

10.prop.1.32.p3592

And since the triangle ABC is similar to the triangle ABD, therefore, as CB is to BA, so is BA to BD; [VI. 4] therefore the rectangle CB, BD is equal to the square on AB. [VI. 17]

10.prop.1.32.p3593

For the same reason the rectangle BC, CD is also equal to the square on AC.

10.prop.1.32.p3594

And since, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the perpendicular so drawn is a mean proportional between the segments of the base, [VI. 8, Por.] therefore, as BD is to DA, so is AD to DC; therefore the rectangle BD, DC is equal to the square on AD. [VI. 17]

10.prop.1.32.p3595

I say that the rectangle BC, AD is also equal to the rectangle BA, AC.

10.prop.1.32.p3596

For since, as we said, ABC is similar to ABD, therefore, as BC is to CA, so is BA to AD. [VI. 4]

10.prop.1.32.p3597

Therefore the rectangle BC, AD is equal to the rectangle BA, AC. [VI. 16] Q. E. D.

10.prop.1.33.p3598

To find two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial.

10.prop.1.33.p3599

Let there be set out two rational straight lines AB, BC commensurable in square only and such that the square on the greater AB is greater than the square on the less BC by the square on a straight line incommensurable with AB, [X. 30] let BC be bisected at D, let there be applied to AB a parallelogram equal to the square on either of the straight lines BD, DC and deficient by a square figure, and let it be the rectangle AE, EB; [VI. 28] let the semicircle AFB be described on AB, let EF be drawn at right angles to AB, and let AF, FB be joined.

10.prop.1.33.p3600

Then, since AB, BC are unequal straight lines, and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB, while there has been applied to AB a parallelogram equal to the fourth part of the square on BC, that is, to the square on half of it, and deficient by a square figure, making the rectangle AE, EB, therefore AE is incommensurable with EB. [X. 18]

10.prop.1.33.p3601

And, as AE is to EB, so is the rectangle BA, AE to the rectangle AB, BE, while the rectangle BA, AE is equal to the square on AF, and the rectangle AB, BE to the square on BF; therefore the square on AF is incommensurable with the square on FB; therefore AF, FB are incommensurable in square.

10.prop.1.33.p3602

And, since AB is rational, therefore the square on AB is also rational; so that the sum of the squares on AF, FB is also rational. [I. 47]

10.prop.1.33.p3603

And since, again, the rectangle AE, EB is equal to the square on EF, and, by hypothesis, the rectangle AE, EB is also equal to the square on BD, therefore FE is equal to BD; therefore BC is double of FE, so that the rectangle AB, BC is also commensurable with the rectangle AB, EF.

10.prop.1.33.p3604

But the rectangle AB, BC is medial; [X. 21] therefore the rectangle AB, EF is also medial. [X. 23, Por.]

10.prop.1.33.p3605

But the rectangle AB, EF is equal to the rectangle AF, FB; [Lemma] therefore the rectangle AF, FB is also medial.

10.prop.1.33.p3606

But it was also proved that the sum of the squares on these straight lines is rational.

10.prop.1.33.p3607

Therefore two straight lines AF, FB incommensurable in square have been found which make the sum of the squares on them rational, but the rectangle contained by them medial. Q. E. D.

10.prop.1.34.p3608

To find two straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational.

10.prop.1.34.p3609

Let there be set out two medial straight lines AB, BC, commensurable in square only, such that the rectangle which they contain is rational, and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB; [X. 31, ad fin.] let the semicircle ADB be described on AB, let BC be bisected at E, let there be applied to AB a parallelogram equal to the square on BE and deficient by a square figure, namely the rectangle AF, FB; [VI. 28] therefore AF is incommensurable in length with FB. [X. 18]

10.prop.1.34.p3610

Let FD be drawn from F at right angles to AB, and let AD, DB be joined.

10.prop.1.34.p3611

Since AF is incommensurable in length with FB, therefore the rectangle BA, AF is also incommensurable with the rectangle AB, BF. [X. 11]

10.prop.1.34.p3612

But the rectangle BA, AF is equal to the square on AD, and the rectangle AB, BF to the square on DB; therefore the square on AD is also incommensurable with the square on DB.

10.prop.1.34.p3613

And, since the square on AB is medial, therefore the sum of the squares on AD, DB is also medial. [III. 31, I. 47]

10.prop.1.34.p3614

And, since BC is double of DF, therefore the rectangle AB, BC is also double of the rectangle AB, FD.

10.prop.1.34.p3615

But the rectangle AB, BC is rational; therefore the rectangle AB, FD is also rational. [X. 6]

10.prop.1.34.p3616

But the rectangle AB, FD is equal to the rectangle AD, DB; [Lemma] so that the rectangle AD, DB is also rational.

10.prop.1.34.p3617

Therefore two straight lines AD, DB incommensurable in square have been found which make the sum of the squares on them medial, but the rectangle contained by them rational. Q. E. D.

10.prop.1.35.p3618

To find two straight lines incommensurable in square which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them.

10.prop.1.35.p3619

Let there be set out two medial straight lines AB, BC commensurable in square only, containing a medial rectangle, and such that the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB; [X. 32 , ad fin.] let the semicircle ADB be described on AB, and let the rest of the construction be as above.

10.prop.1.35.p3620

Then, since AF is incommensurable in length with FB, [X. 18 ] AD is also incommensurable in square with DB. [X. 11 ]

10.prop.1.35.p3621

And, since the square on AB is medial, therefore the sum of the squares on AD, DB is also medial. [III. 31 , I. 47 ]

10.prop.1.35.p3622

And, since the rectangle AF, FB is equal to the square on each of the straight lines BE, DF, therefore BE is equal to DF; therefore BC is double of FD, so that the rectangle AB, BC is also double of the rectangle AB, FD.

10.prop.1.35.p3623

But the rectangle AB, BC is medial; therefore the rectangle AB, FD is also medial. [X. 32, Por.]

10.prop.1.35.p3624

And it is equal to the rectangle AD, DB; [Lemma after X. 32 ] therefore the rectangle AD, DB is also medial.

10.prop.1.35.p3625

And, since AB is incommensurable in length with BC, while CB is commensurable with BE, therefore AB is also incommensurable in length with BE, [X. 13 ] so that the square on AB is also incommensurable with the rectangle AB, BE. [X. 11 ]

10.prop.1.35.p3626

But the squares on AD, DB are equal to the square on AB, [I. 47 ] and the rectangle AB, FD, that is, the rectangle AD, DB, is equal to the rectangle AB, BE; therefore the sum of the squares on AD, DB is incommensurable with the rectangle AD, DB.

10.prop.1.35.p3627

Therefore two straight lines AD, DB incommensurable in square have been found which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them. Q. E. D.

10.prop.1.36.p3628

If two rational straight lines commensurable in square only be added together, the whole is irrational; and let it be called binomial.

10.prop.1.36.p3629

For let two rational straight lines AB, BC commensurable in square only be added together; I say that the whole AC is irrational.

10.prop.1.36.p3630

For, since AB is incommensurable in length with BC— for they are commensurable in square only— and, as AB is to BC, so is the rectangle AB, BC to the square on BC, therefore the rectangle AB, BC is incommensurable with the square on BC. [X. 11 ]

10.prop.1.36.p3631

But twice the rectangle AB, BC is commensurable with the rectangle AB, BC [X. 6 ], and the squares on AB, BC are commensurable with the square on BC—for AB, BC are rational straight lines commensurable in square only— [X. 15 ] therefore twice the rectangle AB, BC is incommensurable with the squares on AB, BC. [X. 13 ]

10.prop.1.36.p3632

And, componendo, twice the rectangle AB, BC together with the squares on AB, BC, that is, the square on AC [II. 4 ], is incommensurable with the sum of the squares on AB, BC. [X. 16 ]

10.prop.1.36.p3633

But the sum of the squares on AB, BC is rational; therefore the square on AC is irrational, so that AC is also irrational. [X. Def. 4 ]

10.prop.1.36.p3634

And let it be called binomial.

10.prop.1.37.p3635

If two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational; and let it be called a first bimedial straight line.

10.prop.1.37.p3636

For let two medial straight lines AB, BC commensurable in square only and containing a rational rectangle be added together; I say that the whole AC is irrational.

10.prop.1.37.p3637

For, since AB is incommensurable in length with BC, therefore the squares on AB, BC are also incommensurable with twice the rectangle AB, BC; [cf. X. 36, ll. 9-20] and, componendo, the squares on AB, BC together with twice the rectangle AB, BC, that is, the square on AC [II. 4], is incommensurable with the rectangle AB, BC. [X. 16 ]

10.prop.1.37.p3638

But the rectangle AB, BC is rational, for, by hypothesis, AB, BC are straight lines containing a rational rectangle; therefore the square on AC is irrational; therefore AC is irrational. [X. Def. 4 ]

10.prop.1.37.p3639

And let it be called a first bimedial straight line. Q. E. D.

10.prop.1.38.p3640

If two medial straight lines commensurable in square only and containing a medial rectangle be added together, the whole is irrational; and let it be called a second bimedial straight line.

10.prop.1.38.p3641

For let two medial straight lines AB, BC commensurable in square only and containing a medial rectangle be added together; I say that AC is irrational.

10.prop.1.38.p3642

For let a rational straight line DE be set out, and let the parallelogram DF equal to the square on AC be applied to DE, producing DG as breadth. [I. 44 ]

10.prop.1.38.p3643

Then, since the square on AC is equal to the squares on AB, BC and twice the rectangle AB, BC, [II. 4 ] let EH, equal to the squares on AB, BC, be applied to DE; therefore the remainder HF is equal to twice the rectangle AB, BC.

10.prop.1.38.p3644

And, since each of the straight lines AB, BC is medial, therefore the squares on AB, BC are also medial.

10.prop.1.38.p3645

But, by hypothesis, twice the rectangle AB, BC is also medial.

10.prop.1.38.p3646

And EH is equal to the squares on AB, BC, while FH is equal to twice the rectangle AB, BC; therefore each of the rectangle EH, HF is medial.

10.prop.1.38.p3647

And they are applied to the rational straight line DE; therefore each of the straight lines DH, HG is rational and incommensurable in length with DE. [X. 22 ]

10.prop.1.38.p3648

Since then AB is incommensurable in length with BC, and, as AB is to BC, so is the square on AB to the rectangle AB, BC, therefore the square on AB is incommensurable with the rectangle AB, BC. [X. 11 ]

10.prop.1.38.p3649

But the sum of the squares on AB, BC is commensurable with the square on AB, [X. 15 ] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC. [X. 6 ]

10.prop.1.38.p3650

Therefore the sum of the squares on AB, BC is incommensurable with twice the rectangle AB, BC. [X. 13 ]

10.prop.1.38.p3651

But EH is equal to the squares on AB, BC, and HF is equal to twice the rectangle AB, BC.

10.prop.1.38.p3652

Therefore EH is incommensurable with HF, so that DH is also incommensurable in length with HG. [VI. 1 , X. 11 ]

10.prop.1.38.p3653

Therefore DH, HG are rational straight lines commensurable in square only; so that DG is irrational. [X. 36 ]

10.prop.1.38.p3654

But DE is rational; and the rectangle contained by an irrational and a rational straight line is irrational; [cf. X. 20 ] therefore the area DF is irrational, and the side of the square equal to it is irrational. [X. Def. 4 ]

10.prop.1.38.p3655

But AC is the side of the square equal to DF; therefore AC is irrational.

10.prop.1.38.p3656

And let it be called a second bimedial straight line. Q. E. D.

10.prop.1.39.p3657

If two straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial, be added together, the whole straight line is irrational : and let it be called major.

10.prop.1.39.p3658

For let two straight lines AB, BC incommensurable in square, and fulfilling the given conditions [X. 33 ], be added together; I say that AC is irrational.

10.prop.1.39.p3659

For, since the rectangle AB, BC is medial, twice the rectangle AB, BC is also medial. [X. 6 and 23, Por.]

10.prop.1.39.p3660

But the sum of the squares on AB, BC is rational; therefore twice the rectangle AB, BC is incommensurable with the sum of the squares on AB, BC, so that the squares on AB, BC together with twice the rectangle AB, BC that is, the square on AC, is also incommensurable with the sum of the squares on AB, BC; [X. 16 ] therefore the square on AC is irrational, so that AC is also irrational. [X. Def. 4 ]

10.prop.1.39.p3661

And let it be called major. Q. E. D.

10.prop.1.40.p3662

If two straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational, be added together, the whole straight line is irrational; and let it be called the side of a rational plus a medial area.

10.prop.1.40.p3663

For let two straight lines AB, BC incommensurable in square, and fulfilling the given conditions [X. 34 ], be added together; I say that AC is irrational.

10.prop.1.40.p3664

For, since the sum of the squares on AB, BC is medial, while twice the rectangle AB, BC is rational, therefore the sum of the squares on AB, BC is incommensurable with twice the rectangle AB, BC; so that the square on AC is also incommensurable with twice the rectangle AB, BC. [X. 16 ]

10.prop.1.40.p3665

But twice the rectangle AB, BC is rational; therefore the square on AC is irrational.

10.prop.1.40.p3666

Therefore AC is irrational. [X. Def. 4 ]

10.prop.1.40.p3667

And let it be called the side of a rational plus a medial area. Q. E. D.

10.prop.1.41.p3668

If two straight lines incommensurable in square which make the sum of the squares on them medial, and the rectangle contained by them medial and also incommensurable with the sum of the squares on them, be added together, the whole straight line is irrational; and let it be called the side of the sum of two medial areas.

10.prop.1.41.p3669

For let two straight lines AB, BC incommensurable in square and satisfying the given conditions [X. 35 ] be added together; I say that AC is irrational.

10.prop.1.41.p3670

Let a rational straight line DE be set out, and let there be applied to DE the rectangle DF equal to the squares on AB, BC, and the rectangle GH equal to twice the rectangle AB, BC; therefore the whole DH is equal to the square on AC. [II. 4 ]

10.prop.1.41.p3671

Now, since the sum of the squares on AB, BC is medial, and is equal to DF, therefore DF is also medial.

10.prop.1.41.p3672

And it is applied to the rational straight line DE; therefore DG is rational and incommensurable in length with DE. [X. 22 ]

10.prop.1.41.p3673

For the same reason GK is also rational and incommensurable in length with GF, that is, DE.

10.prop.1.41.p3674

And, since the squares on AB, BC are incommensurable with twice the rectangle AB, BC, DF is incommensurable with GH; so that DG is also incommensurable with GK. [VI. 1 , X. 11 ]

10.prop.1.41.p3675

And they are rational; therefore DG, GK are rational straight lines commensurable in square only; therefore DK is irrational and what is called binomial. [X. 36 ]

10.prop.1.41.p3676

But DE is rational; therefore DH is irrational, and the side of the square which is equal to it is irrational. [X. Def. 4 ]

10.prop.1.41.p3677

But AC is the side of the square equal to HD; therefore AC is irrational.

10.prop.1.41.p3678

And let it be called the side of the sum of two medial areas. Q. E. D.

10.prop.1.41.p3679

Lemma. And that the aforesaid irrational straight lines are divided only in one way into the straight lines of which they are the sum and which produce the types in question, we will now prove after premising the following lemma.

10.prop.1.41.p3680

Let the straight line AB be set out, let the whole be cut into unequal parts at each of the points C, D, and let AC be supposed greater than DB; I say that the squares on AC, CB are greater than the squares on AD, DB.

10.prop.1.41.p3681

For let AB be bisected at E.

10.prop.1.41.p3682

Then, since AC is greater than DB, let DC be subtracted from each; therefore the remainder AD is greater than the remainder CB.

10.prop.1.41.p3683

But AE is equal to EB; therefore DE is less than EC; therefore the points C, D are not equidistant from the point of bisection.

10.prop.1.41.p3684

And, since the rectangle AC, CB together with the square on EC is equal to the square on EB, [II. 5 ] and, further, the rectangle AD, DB together with the square on DE is equal to the square on EB, [id.] therefore the rectangle AC, CB together with the square on EC is equal to the rectangle AD, DB together with the square on DE.

10.prop.1.41.p3685

And of these the square on DE is less than the square on EC; therefore the remainder, the rectangle AC, CB, is also less than the rectangle AD, DB, so that twice the rectangle AC, CB is also less than twice the rectangle AD, DB.

10.prop.1.41.p3686

Therefore also the remainder, the sum of the squares on

10.prop.1.41.p3686

AC

10.prop.1.41.p3686

,

10.prop.1.41.p3686

CB

10.prop.1.41.p3686

, is greater than the sum of the squares on

10.prop.1.41.p3686

AD

10.prop.1.41.p3686

,

10.prop.1.41.p3686

DB

10.prop.1.41.p3686

. Q. E. D.

10.prop.1.41.p3686

1

10.prop.1.42.p3687

Let AB be a binomial straight line divided into its terms at C; therefore AC, CB are rational straight lines commensurable in square only. [X. 36 ]

10.prop.1.42.p3688

I say that AB is not divided at another point into two rational straight lines commensurable in square only.

10.prop.1.42.p3689

For, if possible, let it be divided at D also, so that AD, DB are also rational straight lines commensurable in square only.

10.prop.1.42.p3690

It is then manifest that AC is not the same with DB.

10.prop.1.42.p3691

For, if possible, let it be so.

10.prop.1.42.p3692

Then AD will also be the same as CB, and, as AC is to CB, so will BD be to DA; thus AB will be divided at D also in the same way as by the division at C: which is contrary to the hypothesis.

10.prop.1.42.p3693

Therefore AC is not the same with DB.

10.prop.1.42.p3694

For this reason also the points C, D are not equidistant from the point of bisection.

10.prop.1.42.p3695

Therefore that by which the squares on AC, CB differ from the squares on AD, DB is also that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB, because both the squares on AC, CB together with twice the rectangle AC, CB, and the squares on AD, DB together with twice the rectangle AD, DB, are equal to the square on AB. [II. 4 ]

10.prop.1.42.p3696

But the squares on AC, CB differ from the squares on AD, DB by a rational area, for both are rational; therefore twice the rectangle AD, DB also differs from twice the rectangle AC, CB by a rational area, though they are medial [X. 21 ]: which is absurd, for a medial area does not exceed a medial by a rational area. [x. 26 ]

10.prop.1.42.p3697

Therefore a binomial straight line is not divided at different points; therefore it is divided at one point only. Q. E. D.

10.prop.1.43.p3698

A first bimedial straight line is divided at one point only.

10.prop.1.43.p3699

Let AB be a first bimedial straight line divided at C, so that AC, CB are medial straight lines commensurable in square only and containing a rational rectangle; I say that AB is not so divided at another point.

10.prop.1.43.p3700

For, if possible, let it be divided at D also, so that AD, DB are also medial straight lines commensurable in square only and containing a rational rectangle.

10.prop.1.43.p3701

Since, then, that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB is that by which the squares on AC, CB differ from the squares on AD, DB, while twice the rectangle AD, DB differs from twice the rectangle AC, CB by a rational area—for both are rational— therefore the squares on AC, CB also differ from the squares on AD, DB by a rational area, though they are medial: which is absurd. [x. 26 ]

10.prop.1.43.p3702

Therefore a first bimedial straight line is not divided into its terms at different points; therefore it is so divided at one point only.

10.prop.1.44.p3703

A second bimedial straight line is divided at one point only.

10.prop.1.44.p3704

Let AB be a second bimedial straight line divided at C, so that AC, CB are medial straight lines commensurable in square only and containing a medial rectangle; [X. 38 ] it is then manifest that C is not at the point of bisection, because the segments are not commensurable in length.

10.prop.1.44.p3705

I say that AB is not so divided at another point.

10.prop.1.44.p3706

For, if possible, let it be divided at D also, so that AC is not the same with DB, but AC is supposed greater; it is then clear that the squares on AD, DB are also, as we proved above [Lemma], less than the squares on AC, CB; and suppose that AD, DB are medial straight lines commensurable in square only and containing a medial rectangle.

10.prop.1.44.p3707

Now let a rational straight line EF be set out, let there be applied to EF the rectangular parallelogram EK equal to the square on AB, and let EG equal to the squares on AC, CB be subtracted; therefore the remainder HK is equal to twice the rectangle AC, CB. [II. 4 ]

10.prop.1.44.p3708

Again, let there be subtracted EL, equal to the squares on AD, DB, which were proved less than the squares on AC, CB [Lemma ]; therefore the remainder MK is also equal to twice the rectangle AD, DB.

10.prop.1.44.p3709

Now, since the squares on AC, CB are medial, therefore EG is medial.

10.prop.1.44.p3710

And it is applied to the rational straight line EF; therefore EH is rational and incommensurable in length with EF. [X. 22 ]

10.prop.1.44.p3711

For the same reason HN is also rational and incommensurable in length with EF.

10.prop.1.44.p3712

And, since AC, CB are medial straight lines commensurable in square only, therefore AC is incommensurable in length with CB.

10.prop.1.44.p3713

But, as AC is to CB, so is the square on AC to the rectangle AC, CB; therefore the square on AC is incommensurable with the rectangle AC, CB. [X. 11 ]

10.prop.1.44.p3714

But the squares on AC, CB are commensurable with the square on AC; for AC, CB are commensurable in square. [x. 15 ]

10.prop.1.44.p3715

And twice the rectangle AC, CB is commensurable with the rectangle AC, CB. [X. 6 ]

10.prop.1.44.p3716

Therefore the squares on AC, CB are also incommensurable with twice the rectangle AC, CB. [X. 13 ]

10.prop.1.44.p3717

But EG is equal to the squares on AC, CB, and HK is equal to twice the rectangle AC, CB; therefore EG is incommensurable with HK, so that EH is also incommensurable in length with HN. [VI. 1 , X. 11 ]

10.prop.1.44.p3718

And they are rational; therefore EH, HN are rational straight lines commensurable in square only.

10.prop.1.44.p3719

But, if two rational straight lines commensurable in square only be added together, the whole is the irrational which is called binomial. [X. 36 ]

10.prop.1.44.p3720

Therefore EN is a binomial straight line divided at H.

10.prop.1.44.p3721

In the same way EM, MN will also be proved to be rational straight lines commensurable in square only; and EN will be a binomial straight line divided at different points, H and M.

10.prop.1.44.p3722

And EH is not the same with MN.

10.prop.1.44.p3723

For the squares on AC, CB are greater than the squares on AD, DB.

10.prop.1.44.p3724

But the squares on AD, DB are greater than twice the rectangle AD, DB; therefore also the squares on AC, CB, that is, EG, are much greater than twice the rectangle AD, DB, that is, MK, so that EH is also greater than MN.

10.prop.1.44.p3725

Therefore EH is not the same with MN. Q. E. D.

10.prop.1.45.p3726

A major straight line is divided at one and the same point only.

10.prop.1.45.p3727

Let AB be a major straight line divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB rational, but the rectangle AC, CB medial; [X. 39 ] I say that AB is not so divided at another point.

10.prop.1.45.p3728

For, if possible, let it be divided at D also, so that AD, DB are also incommensurable in square and make the sum of the squares on AD, DB rational, but the rectangle contained by them medial.

10.prop.1.45.p3729

Then, since that by which the squares on AC, CB differ from the squares on AD, DB is also that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB, while the squares on AC, CB exceed the squares on AD, DB by a rational area—for both are rational— therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area, though they are medial: which is impossible. [X. 26 ]

10.prop.1.45.p3730

Therefore a major straight line is not divided at different points; therefore it is only divided at one and the same point. Q. E. D.

10.prop.1.46.p3731

The side of a rational plus a medial area is divided at one point only.

10.prop.1.46.p3732

Let AB be the side of a rational plus a medial area divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB medial, but twice the rectangle AC, CB rational; [X. 40 ] I say that AB is not so divided at another point.

10.prop.1.46.p3733

For, if possible, let it be divided at D also, so that AD, DB are also incommensurable in square and make the sum of the squares on AD, DB medial, but twice the rectangle AD, DB rational.

10.prop.1.46.p3734

Since then that by which twice the rectangle AC, CB differs from twice the rectangle AD, DB is also that by which the squares on AD, DB differ from the squares on AC, CB, while twice the rectangle AC, CB exceeds twice the rectangle AD, DB by a rational area, therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area, though they are medial: which is impossible. [X. 26 ]

10.prop.1.46.p3735

Therefore the side of a rational plus a medial area is not divided at different points; therefore it is divided at one point only. Q. E. D.

10.prop.1.47.p3736

The side of the sum of two medial areas is divided at one point only.

10.prop.1.47.p3737

Let AB be divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB medial, and the rectangle AC, CB medial and also incommensurable with the sum of the squares on them; I say that AB is not divided at another point so as to fulfil the given conditions.

10.prop.1.47.p3738

For, if possible, let it be divided at D, so that again AC is of course not the same as BD, but AC is supposed greater; let a rational straight line EF be set out, and let there be applied to EF the rectangle EG equal to the squares on AC, CB, and the rectangle HK equal to twice the rectangle AC, CB; therefore the whole EK is equal to the square on AB. [II. 4 ]

10.prop.1.47.p3739

Again, let EL, equal to the squares on AD, DB, be applied to EF; therefore the remainder, twice the rectangle AD, DB, is equal to the remainder MK.

10.prop.1.47.p3740

And since, by hypothesis, the sum of the squares on AC, CB is medial, therefore EG is also medial.

10.prop.1.47.p3741

And it is applied to the rational straight line EF; therefore HE is rational and incommensurable in length with EF. [X. 22 ]

10.prop.1.47.p3742

For the same reason HN is also rational and incommensurable in length with EF.

10.prop.1.47.p3743

And, since the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, therefore EG is also incommensurable with GN, so that EH is also incommensurable with HN. [VI. 1 , X. 11 ]

10.prop.1.47.p3744

And they are rational; therefore EH, HN are rational straight lines commensurable in square only; therefore EN is a binomial straight line divided at H. [X. 36 ]

10.prop.1.47.p3745

Similarly we can prove that it is also divided at M.

10.prop.1.47.p3746

And EH is not the same with MN; therefore a binomial has been divided at different points: which is absurd. [X. 42 ]

10.prop.1.47.p3747

Therefore a side of the sum of two medial areas is not divided at different points; therefore it is divided at one point only.

10.def.2.1.p3748

Given a rational straight line and a binomial, divided into its terms, such that the square on the greater term is greater than the square on the lesser by the square on a straight line commensurable in length with the greater, then, if the greater term be commensurable in length with the rational straight line set out, let the whole be called a first binomial straight line;

10.def.2.2.p3749

but if the lesser term be commensurable in length with the rational straight line set out, let the whole be called a second binomial;

10.def.2.3.p3750

and if neither of the terms be commensurable in length with the rational straight line set out, let the whole be called a third binomial.

10.def.2.4.p3751

Again, if the square on the greater term be greater than the square on the lesser by the square on a straight line incommensurable in length with the greater, then, if the greater term be commensurable in length with the rational straight line set out, let the whole be called a fourth binomial;

10.def.2.5.p3752

if the lesser, a fifth binomial;

10.def.2.6.p3753

and if neither, a sixth binomial.

10.prop.2.48.p3754

To find the first binomial straight line.

10.prop.2.48.p3755

Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to CA the ratio which a square number has to a square number; [Lemma I after X. 28] let any rational straight line D be set out, and let EF be commensurable in length with D.

10.prop.2.48.p3756

Therefore EF is also rational.

10.prop.2.48.p3757

Let it be contrived that, as the number BA is to AC, so is the square on EF to the square on FG. [X. 6, Por.]

10.prop.2.48.p3758

But AB has to AC the ratio which a number has to a number; therefore the square on EF also has to the square on FG the ratio which a number has to a number, so that the square on EF is commensurable with the square on FG. [X. 6]

10.prop.2.48.p3759

And EF is rational; therefore FG is also rational.

10.prop.2.48.p3760

And, since BA has not to AC the ratio which a square number has to a square number. neither, therefore, has the square on EF to the square on FG the ratio which a square number has to a square number; therefore EF is incommensurable in length with FG. [X. 9]

10.prop.2.48.p3761

Therefore EF, FG are rational straight lines commensurable in square only; therefore EG is binomial. [X. 36]

10.prop.2.48.p3762

I say that it is also a first binomial straight line.

10.prop.2.48.p3763

For since, as the number BA is to AC, so is the square on EF to the square on FG, while BA is greater than AC, therefore the square on EF is also greater than the square on FG.

10.prop.2.48.p3764

Let then the squares on FG, H be equal to the square on EF.

10.prop.2.48.p3765

Now since, as BA is to AC, so is the square on EF to the square on FG, therefore, convertendo, as AB is to BC, so is the square on EF to the square on H. [V. 19, Por.]

10.prop.2.48.p3766

But AB has to BC the ratio which a square number has to a square number; therefore the square on EF also has to the square on H the ratio which a square number has to a square number.

10.prop.2.48.p3767

Therefore EF is commensurable in length with H; [X. 9] therefore the square on EF is greater than the square on FG by the square on a straight line commensurable with EF.

10.prop.2.48.p3768

And EF, FG are rational, and EF is commensurable in length with D.

10.prop.2.48.p3769

Therefore EF is a first binomial straight line. Q. E. D.

10.prop.2.49.p3770

To find the second binomial straight line.

10.prop.2.49.p3771

Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to AC the ratio which a square number has to a square number; let a rational straight line D be set out, and let EF be commensurable in length with D; therefore EF is rational.

10.prop.2.49.p3772

Let it be contrived then that, as the number CA is to AB, so also is the square on EF to the square on FG; [X. 6, Por.] therefore the square on EF is commensurable with the square on FG. [X. 6]

10.prop.2.49.p3773

Therefore FG is also rational.

10.prop.2.49.p3774

Now, since the number CA has not to AB the ratio which a square number has to a square number, neither has the square on EF to the square on FG the ratio which a square number has to a square number.

10.prop.2.49.p3775

Therefore EF is incommensurable in length with FG; [X. 9] therefore EF, FG are rational straight lines commensurable in square only; therefore EG is binomial. [X. 36]

10.prop.2.49.p3776

It is next to be proved that it is also a second binomial straight line.

10.prop.2.49.p3777

For since, inversely, as the number BA is to AC, so is the square on GF to the square on FE, while BA is greater than AC, therefore the square on GF is greater than the square on FE.

10.prop.2.49.p3778

Let the squares on EF, H be equal to the square on GF; therefore, convertendo, as AB is to BC, so is the square on FG to the square on H. [V. 19, Por.]

10.prop.2.49.p3779

But AB has to BC the ratio which a square number has to a square number; therefore the square on FG also has to the square on H the ratio which a square number has to a square number.

10.prop.2.49.p3780

Therefore FG is commensurable in length with H; [X. 9] so that the square on FG is greater than the square on FE by the square on a straight line commensurable with FG.

10.prop.2.49.p3781

And FG, FE are rational straight lines commensurable in square only, and EF, the lesser term, is commensurable in length with the rational straight line D set out.

10.prop.2.49.p3782

Therefore EG is a second binomial straight line. Q. E. D.

10.prop.2.50.p3783

To find the third binomial straight line.

10.prop.2.50.p3784

Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to AC the ratio which a square number has to a square number.

10.prop.2.50.p3785

Let any other number D, not square, be set out also, and let it not have to either of the numbers BA. AC the ratio which a square number has to a square number.

10.prop.2.50.p3786

Let any rational straight line E be set out, and let it be contrived that, as D is to AB, so is the square on E to the square on FG; [X. 6, Por.] therefore the square on E is commensurable with the square on FG. [X. 6]

10.prop.2.50.p3787

And E is rational; therefore FG is also rational.

10.prop.2.50.p3788

And, since D has not to AB the ratio which a square number has to a square number, neither has the square on E to the square on FG the ratio which a square number has to a square number; therefore E is incommensurable in length with FG. [X. 9]

10.prop.2.50.p3789

Next let it be contrived that, as the number BA is to AC, so is the square on FG to the square on GH; [X. 6, Por.] therefore the square on FG is commensurable with the square on GH. [X. 6]

10.prop.2.50.p3790

But FG is rational; therefore GH is also rational.

10.prop.2.50.p3791

And, since BA has not to AC the ratio which a square number has to a square number, neither has the square on FG to the square on HG the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9]

10.prop.2.50.p3792

Therefore FG, GH are rational straight lines commensurable in square only; therefore FH is binomial. [X. 36]

10.prop.2.50.p3793

I say next that it is also a third binomial straight line.

10.prop.2.50.p3794

For since, as D is to AB, so is the square on E to the square on FG, and, as BA is to AC, so is the square on FG to the square on GH, therefore, ex aequali, as D is to AC, so is the square on E to the square on GH. [V. 22]

10.prop.2.50.p3795

But D has not to AC the ratio which a square number has to a square number; therefore neither has the square on E to the square on GH the ratio which a square number has to a square number; therefore E is incommensurable in length with GH. [X. 9]

10.prop.2.50.p3796

And since, as BA is to AC, so is the square on FG to the square on GH, therefore the square on FG is greater than the square on GH.

10.prop.2.50.p3797

Let then the squares on GH, K be equal to the square on FG; therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19, Por.]

10.prop.2.50.p3798

But AB has to BC the ratio which a square number has to a square number; therefore the square on FG also has to the square on K the ratio which a square number has to a square number; therefore FG is commensurable in length with K. [X. 9]

10.prop.2.50.p3799

Therefore the square on FG is greater than the square on GH by the square on a straight line commensurable with FG.

10.prop.2.50.p3800

And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with E.

10.prop.2.50.p3801

Therefore FH is a third binomial straight line. Q. E. D.

10.prop.2.51.p3802

To find the fourth binomial straight line.

10.prop.2.51.p3803

Let two numbers AC, CB be set out such that AB neither has to BC, nor yet to AC, the ratio which a square number has to a square number.

10.prop.2.51.p3804

Let a rational straight line D be set out, and let EF be commensurable in length with D; therefore EF is also rational.

10.prop.2.51.p3805

Let it be contrived that, as the number BA is to AC, so is the square on EF to the square on FG; [X. 6, Por.] therefore the square on EF is commensurable with the square on FG; [X. 6] therefore FG is also rational.

10.prop.2.51.p3806

Now, since BA has not to AC the ratio which a square number has to a square number, neither has the square on EF to the square on FG the ratio which a square number has to a square number; therefore EF is incommensurable in length with FG. [X. 9]

10.prop.2.51.p3807

Therefore EF, FG are rational straight lines commensurable in square only; so that EG is binomial.

10.prop.2.51.p3808

I say next that it is also a fourth binomial straight line.

10.prop.2.51.p3809

For since, as BA is to AC, so is the square on EF to the square on FG, therefore the square on EF is greater than the square on FG.

10.prop.2.51.p3810

Let then the squares on FG, H be equal to the square on EF; therefore, convertendo, as the number AB is to BC, so is the square on EF to the square on H. [V. 19, Por.]

10.prop.2.51.p3811

But AB has not to BC the ratio which a square number has to a square number; therefore neither has the square on EF to the square on H the ratio which a square number has to a square number.

10.prop.2.51.p3812

Therefore EF is incommensurable in length with H; [X. 9] therefore the square on EF is greater than the square on GF by the square on a straight line incommensurable with EF.

10.prop.2.51.p3813

And EF, FG are rational straight lines commensurable in square only, and EF is commensurable in length with D.

10.prop.2.51.p3814

Therefore EG is a fourth binomial straight line. Q. E. D.

10.prop.2.52.p3815

To find the fifth binomial straight line.

10.prop.2.52.p3816

Let two numbers AC, CB be set out such that AB has not to either of them the ratio which a square number has to a square number; let any rational straight line D be set out, and let EF be commensurable with D; therefore EF is rational.

10.prop.2.52.p3817

Let it be contrived that, as CA is to AB, so is the square on EF to the square on FG. [X. 6, Por.]

10.prop.2.52.p3818

But CA has not to AB the ratio which a square number has to a square number; therefore neither has the square on EF to the square on FG the ratio which a square number has to a square number.

10.prop.2.52.p3819

Therefore EF, FG are rational straight lines commensurable in square only; [X. 9] therefore EG is binomial. [X. 36]

10.prop.2.52.p3820

I say next that it is also a fifth binomial straight line.

10.prop.2.52.p3821

For since, as CA is to AB, so is the square on EF to the square on FG, inversely, as BA is to AC, so is the square on FG to the square on FE; therefore the square on GF is greater than the square on FE.

10.prop.2.52.p3822

Let then the squares on EF, H be equal to the square on GF; therefore, convertendo, as the number AB is to BC, so is the square on GF to the square on H. [V. 19, Por.]

10.prop.2.52.p3823

But AB has not to BC the ratio which a square number has to a square number; therefore neither has the square on FG to the square on H the ratio which a square number has to a square number.

10.prop.2.52.p3824

Therefore FG is incommensurable in length with H; [X. 9] so that the square on FG is greater than the square on FE by the square on a straight line incommensurable with FG.

10.prop.2.52.p3825

And GF, FE are rational straight lines commensurable in square only, and the lesser term EF is commensurable in length with the rational straight line D set out.

10.prop.2.52.p3826

Therefore EG is a fifth binomial straight line. Q. E. D.

10.prop.2.53.p3827

To find the sixth binomial straight line.

10.prop.2.53.p3828

Let two numbers AC, CB be set out such that AB has not to either of them the ratio which a square number has to a square number; and let there also be another number D which is not square and which has not to either of the numbers BA, AC the ratio which a square number has to a square number.

10.prop.2.53.p3829

Let any rational straight line E be set out, and let it be contrived that, as D is to AB, so is the square on E to the square on FG; [X. 6, Por.] therefore the square on E is commensurable with the square on FG. [X. 6]

10.prop.2.53.p3830

And E is rational; therefore FG is also rational.

10.prop.2.53.p3831

Now, since D has not to AB the ratio which a square number has to a square number, neither has the square on E to the square on FG the ratio which a square number has to a square number; therefore E is incommensurable in length with FG. [X. 9]

10.prop.2.53.p3832

Again, let it be contrived that, as BA is to AC, so is the square on FG to the square on GH. [X. 6, Por.]

10.prop.2.53.p3833

Therefore the square on FG is commensurable with the square on HG. [X. 6]

10.prop.2.53.p3834

Therefore the square on HG is rational; therefore HG is rational.

10.prop.2.53.p3835

And, since BA has not to AC the ratio which a square number has to a square number, neither has the square on FG to the square on GH the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9]

10.prop.2.53.p3836

Therefore FG, GH are rational straight lines commensurable in square only; therefore FH is binomial. [X. 36]

10.prop.2.53.p3837

It is next to be proved that it is also a sixth binomial straight line.

10.prop.2.53.p3838

For since, as D is to AB, so is the square on E to the square on FG, and also, as BA is to AC, so is the square on FG to the square on GH, therefore, ex aequali, as D is to AC, so is the square on E to the square on GH. [V. 22]

10.prop.2.53.p3839

But D has not to AC the ratio which a square number has to a square number; therefore neither has the square on E to the square on GH the ratio which a square number has to a square number; therefore E is incommensurable in length with GH. [X. 9]

10.prop.2.53.p3840

But it was also proved incommensurable with FG; therefore each of the straight lines FG, GH is incommensurable in length with E.

10.prop.2.53.p3841

And, since, as BA is to AC, so is the square on FG to the square on GH, therefore the square on FG is greater than the square on GH.

10.prop.2.53.p3842

Let then the squares on GH, K be equal to the square on FG; therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19, Por.]

10.prop.2.53.p3843

But AB has not to BC the ratio which a square number has to a square number; so that neither has the square on FG to the square on K the ratio which a square number has to a square number.

10.prop.2.53.p3844

Therefore FG is incommensurable in length with K; [X. 9] therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable with FG.

10.prop.2.53.p3845

And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with the rational straight line E set out.

10.prop.2.53.p3846

Therefore FH is a sixth binomial straight line. Q. E. D.

10.prop.2.53.p3847

Lemma. Let there be two squares AB, BC, and let them be placed so that DB is in a straight line with BE; therefore FB is also in a straight line with BG.

10.prop.2.53.p3848

Let the parallelogram AC be completed; I say that AC is a square, that DG is a mean proportional between AB, BC, and further that DC is a mean proportional between AC, CB.

10.prop.2.53.p3849

For, since DB is equal to BF, and BE to BG, therefore the whole DE is equal to the whole FG.

10.prop.2.53.p3850

But DE is equal to each of the straight lines AH, KC, and FG is equal to each of the straight lines AK, HC; [I. 34] therefore each of the straight lines AH, KC is also equal to each of the straight lines AK, HC.

10.prop.2.53.p3851

Therefore the parallelogram AC is equilateral.

10.prop.2.53.p3852

And it is also rectangular; therefore AC is a square.

10.prop.2.53.p3853

And since, as FB is to BG, so is DB to BE, while, as FB is to BG, so is AB to DG, and, as DB is to BE, so is DG to BC, [VI. 1] therefore also, as AB is to DG, so is DG to BC. [V. 11]

10.prop.2.53.p3854

Therefore DG is a mean proportional between AB, BC.

10.prop.2.53.p3855

I say next that DC is also a mean proportional between AC, CB.

10.prop.2.53.p3856

For since, as AD is to DK, so is KG to GC— for they are equal respectively— and, componendo, as AK is to KD, so is KC to CG, [V. 18] while, as AK is to KD, so is AC to CD, and, as KC is to CG, so is DC to CB, [VI. 1] therefore also, as AC is to DC, so is DC to BC. [V. 11]

10.prop.2.53.p3857

Therefore DC is a mean proportional between AC, CB. Being what it was proposed to prove.

10.prop.2.54.p3858

If an area be contained by a rational straight line and the first binomial, the side of the area is the irrational straight line which is called binomial.

10.prop.2.54.p3859

For let the area AC be contained by the rational straight line AB and the first binomial AD; I say that the side of the area AC is the irrational straight line which is called binomial.

10.prop.2.54.p3860

For, since AD is a first binomial straight line, let it be divided into its terms at E, and let AE be the greater term.

10.prop.2.54.p3861

It is then manifest that AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and AE is commensurable in length with the rational straight line AB set out. [X. Deff. II. 1]

10.prop.2.54.p3862

Let ED be bisected at the point F.

10.prop.2.54.p3863

Then, since the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, therefore, if there be applied to the greater AE a parallelogram equal to the fourth part of the square on the less, that is, to the square on EF, and deficient by a square figure, it divides it into commensurable parts. [X. 17]

10.prop.2.54.p3864

Let then the rectangle AG, GE equal to the square on EF be applied to AE; therefore AG is commensurable in length with EG.

10.prop.2.54.p3865

Let GH, EK, FL be drawn from G, E, F parallel to either of the straight lines AB, CD; let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, [II. 14] and let them be placed so that MN is in a straight line with NO; therefore RN is also in a straight line with NP.

10.prop.2.54.p3866

And let the parallelogram SQ be completed; therefore SQ is a square. [Lemma]

10.prop.2.54.p3867

Now, since the rectangle AG, GE is equal to the square on EF, therefore, as AG is to EF, so is FE to EG; [VI. 17] therefore also, as AH is to EL, so is EL to KG; [VI. 1] therefore EL is a mean proportional between AH, GK.

10.prop.2.54.p3868

But AH is equal to SN, and GK to NQ; therefore EL is a mean proportional between SN, NQ.

10.prop.2.54.p3869

But MR is also a mean proportional between the same SN, NQ; [Lemma] therefore EL is equal to MR, so that it is also equal to PO.

10.prop.2.54.p3870

But AH, GK are also equal to SN, NQ; therefore the whole AC is equal to the whole SQ, that is, to the square on MO; therefore MO is the side of AC.

10.prop.2.54.p3871

I say next that MO is binomial.

10.prop.2.54.p3872

For, since AG is commensurable with GE, therefore AE is also commensurable with each of the straight lines AG, GE. [X. 15]

10.prop.2.54.p3873

But AE is also, by hypothesis, commensurable with AB; therefore AG, GE are also commensurable with AB. [X. 12]

10.prop.2.54.p3874

And AB is rational; therefore each of the straight lines AG, GE is also rational; therefore each of the rectangles AH, GK is rational, [X. 19] and AH is commensurable with GK.

10.prop.2.54.p3875

But AH is equal to SN, and GK to NQ; therefore SN, NQ, that is, the squares on MN, NO, are rational and commensurable.

10.prop.2.54.p3876

And, since AE is incommensurable in length with ED, while AE is commensurable with AG, and DE is commensurable with EF, therefore AG is also incommensurable with EF, [X. 13] so that AH is also incommensurable with EL. [VI. 1, X. 11]

10.prop.2.54.p3877

But AH is equal to SN, and EL to MR; therefore SN is also incommensurable with MR.

10.prop.2.54.p3878

But, as SN is to MR, so is PN to NR; [VI. 1] therefore PN is incommensurable with NR. [X. 11]

10.prop.2.54.p3879

But PN is equal to MN, and NR to NO; therefore MN is incommensurable with NO.

10.prop.2.54.p3880

And the square on MN is commensurable with the square on NO, and each is rational; therefore MN, NO are rational straight lines commensurable in square only.

10.prop.2.54.p3881

Therefore

10.prop.2.54.p3881

MO

10.prop.2.54.p3881

is binomial [

10.prop.2.54.p3881

X. 36

10.prop.2.54.p3881

] and the

10.prop.2.54.p3881

side

10.prop.2.54.p3881

of

10.prop.2.54.p3881

AC

10.prop.2.54.p3881

. Q. E. D.

10.prop.2.54.p3881

1

10.prop.2.55.p3882

If an area be contained by a rational straight line and the second binomial, the side of the area is the irrational straight line which is called a first bimedial.

10.prop.2.55.p3883

For let the area ABCD be contained by the rational straight line AB and the second binomial AD; I say that the side of the area AC is a first bimedial straight line.

10.prop.2.55.p3884

For, since AD is a second binomial straight line, let it be divided into its terms at E, so that AE is the greater term; therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and the lesser term ED is commensurable in length with AB. [X. Deff. II. 2]

10.prop.2.55.p3885

Let ED be bisected at F, and let there be applied to AE the rectangle AG, GE equal to the square on EF and deficient by a square figure; therefore AG is commensurable in length with GE. [X. 17]

10.prop.2.55.p3886

Through G, E, F let GH, EK, FL be drawn parallel to AB, CD, let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, and let them be placed so that MN is in a straight line with NO; therefore RN is also in a straight line with NP.

10.prop.2.55.p3887

Let the square SQ be completed.

10.prop.2.55.p3888

It is then manifest from what was proved before that MR is a mean proportional between SN, NQ and is equal to EL, and that MO is the side of the area AC.

10.prop.2.55.p3889

It is now to be proved that MO is a first bimedial straight line.

10.prop.2.55.p3890

Since AE is incommensurable in length with ED, while ED is commensurable with AB, therefore AE is incommensurable with AB. [X. 13]

10.prop.2.55.p3891

And, since AG is commensurable with EG, AE is also commensurable with each of the straight lines AG, GE. [X. 15]

10.prop.2.55.p3892

But AE is incommensurable in length with AB; therefore AG, GE are also incommensurable with AB. [X. 13]

10.prop.2.55.p3893

Therefore BA, AG and BA, GE are pairs of rational straight lines commensurable in square only; so that each of the rectangles AH, GK is medial. [X. 21]

10.prop.2.55.p3894

Hence each of the squares SN, NQ is medial.

10.prop.2.55.p3895

Therefore MN, NO are also medial.

10.prop.2.55.p3896

And, since AG is commensurable in length with GE, AH is also commensurable with GK, [VI. 1. X. 11] that is, SN is commensurable with NQ, that is, the square on MN with the square on NO.

10.prop.2.55.p3897

And, since AE is incommensurable in length with ED, while AE is commensurable with AG, and ED is commensurable with EF, therefore AG is incommensurable with EF; [X. 13] so that AH is also incommensurable with EL, that is, SN is incommensurable with MR, that is, PN with NR, [VI. 1, X. 11] that is, MN is incommensurable in length with NO.

10.prop.2.55.p3898

But MN, NO were proved to be both medial and commensurable in square; therefore MN, NO are medial straight lines commensurable in square only.

10.prop.2.55.p3899

I say next that they also contain a rational rectangle.

10.prop.2.55.p3900

For, since DE is, by hypothesis, commensurable with each of the straight lines AB, EF, therefore EF is also commensurable with EK. [X. 12]

10.prop.2.55.p3901

And each of them is rational; therefore EL, that is, MR is rational, [X. 19] and MR is the rectangle MN, NO.

10.prop.2.55.p3902

But, if two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational and is called a first bimedial straight line. [X. 37]

10.prop.2.55.p3903

Therefore

10.prop.2.55.p3903

MO

10.prop.2.55.p3903

is a first bimedial straight line. Q. E. D.

10.prop.2.55.p3903

1

10.prop.2.56.p3904

If an area be contained by a rational straight line and the third binomial, the side of the area is the irrational straight line called a second bimedial.

10.prop.2.56.p3905

For let the area ABCD be contained by the rational straight line AB and the third binomial AD divided into its terms at E, of which terms AE is the greater; I say that the side of the area AC is the irrational straight line called a second bimedial.

10.prop.2.56.p3906

For let the same construction be made as before.

10.prop.2.56.p3907

Now, since AD is a third binomial straight line, therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and neither of the terms AE, ED is commensurable in length with AB. [X. Deff. II. 3]

10.prop.2.56.p3908

Then, in manner similar to the foregoing, we shall prove that MO is the side of the area AC, and MN, NO are medial straight lines commensurable in square only; so that MO is bimedial.

10.prop.2.56.p3909

It is next to be proved that it is also a second bimedial straight line.

10.prop.2.56.p3910

Since DE is incommensurable in length with AB, that is, with EK, and DE is commensurable with EF, therefore EF is incommensurable in length with EK. [X. 13]

10.prop.2.56.p3911

And they are rational; therefore FE, EK are rational straight lines commensurable in square only.

10.prop.2.56.p3912

Therefore EL, that is, MR, is medial. [X. 21]

10.prop.2.56.p3913

And it is contained by MN, NO; therefore the rectangle MN, NO is medial.

10.prop.2.56.p3914

Therefore MO is a second bimedial straight line. [X. 38] Q. E. D.

10.prop.2.57.p3915

If an area be contained by a rational straight line and the fourth binomial, the side of the area is the irrational straight line called major.

10.prop.2.57.p3916

For let the area AC be contained by the rational straight line AB and the fourth binomial AD divided into its terms at E, of which terms let AE be the greater; I say that the side of the area AC is the irrational straight line called major.

10.prop.2.57.p3917

For, since AD is a fourth binomial straight line, therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line incommensurable with AE, and AE is commensurable in length with AB. [X. Deff. II. 4]

10.prop.2.57.p3918

Let DE be bisected at F, and let there be applied to AE a parallelogram, the rectangle AG, GE, equal to the square on EF; therefore AG is incommensurable in length with GE. [X. 18]

10.prop.2.57.p3919

Let GH, EK, FL be drawn parallel to AB, and let the rest of the construction be as before; it is then manifest that MO is the side of the area AC.

10.prop.2.57.p3920

It is next to be proved that MO is the irrational straight line called major.

10.prop.2.57.p3921

Since AG is incommensurable with EG, AH is also incommensurable with GK, that is, SN with NQ; [VI. 1, X. 11] therefore MN, NO are incommensurable in square.

10.prop.2.57.p3922

And, since AE is commensurable with AB, AK is rational; [X. 19] and it is equal to the squares on MN, NO; therefore the sum of the squares on MN, NO is also rational.

10.prop.2.57.p3923

And, since DE is incommensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is incommensurable in length with EK. [X. 13]

10.prop.2.57.p3924

Therefore EK, EF are rational straight lines commensurable in square only; therefore LE, that is, MR, is medial. [X. 21]

10.prop.2.57.p3925

And it is contained by MN, NO; therefore the rectangle MN, NO is medial.

10.prop.2.57.p3926

And the [sum] of the squares on MN, NO is rational, and MN, NO are incommensurable in square.

10.prop.2.57.p3927

But, if two straight lines incommensurable in square and making the sum of the squares on them rational, but the rectangle contained by them medial, be added together, the whole is irrational and is called major. [X. 39]

10.prop.2.57.p3928

Therefore MO is the irrational straight line called major and is the side of the area AC. Q. E. D.

10.prop.2.58.p3929

If an area be contained by a rational straight line and the fifth binomial, the side of the area is the irrational straight line called the side of a rational plus a medial area.

10.prop.2.58.p3930

For let the area AC be contained by the rational straight line AB and the fifth binomial AD divided into its terms at E, so that AE is the greater term; I say that the side of the area AC is the irrational straight line called the side of a rational plus a medial area.

10.prop.2.58.p3931

For let the same construction be made as before shown; it is then manifest that MO is the side of the area AC.

10.prop.2.58.p3932

It is then to be proved that MO is the side of a rational plus a medial area.

10.prop.2.58.p3933

For, since AG is incommensurable with GE, [X. 18] therefore AH is also commensurable with HE, [VI. 1, X. 11] that is, the square on MN with the square on NO; therefore MN, NO are incommensurable in square.

10.prop.2.58.p3934

And, since AD is a fifth binomial straight line, and ED the lesser segment, therefore ED is commensurable in length with AB. [X. Deff. II. 5]

10.prop.2.58.p3935

But AE is incommensurable with ED; therefore AB is also incommensurable in length with AE. [X. 13]

10.prop.2.58.p3936

Therefore AK, that is, the sum of the squares on MN, NO, is medial. [X. 21]

10.prop.2.58.p3937

And, since DE is commensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is also commensurable with EK. [X. 12]

10.prop.2.58.p3938

And EK is rational; therefore EL, that is, MR, that is, the rectangle MN, NO, is also rational. [X. 19]

10.prop.2.58.p3939

Therefore MN, NO are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational.

10.prop.2.58.p3940

Therefore MO is the side of a rational plus a medial area [X. 40] and is the side of the area AC. Q. E. D.

10.prop.2.59.p3941

If an area be contained by a rational straight line and the sixth binomial, the side of the area is the irrational straight line called the side of the sum of two medial areas.

10.prop.2.59.p3942

For let the area ABCD be contained by the rational straight line AB and the sixth binomial AD, divided into its terms at E, so that AE is the greater term; I say that the side of AC is the side of the sum of two medial areas.

10.prop.2.59.p3943

Let the same construction be made as before shown.

10.prop.2.59.p3944

It is then manifest that MO is the side of AC, and that MN is incommensurable in square with NO.

10.prop.2.59.p3945

Now, since EA is incommensurable in length with AB, therefore EA, AB are rational straight lines commensurable in square only; therefore AK, that is, the sum of the squares on MN, NO, is medial. [X. 21]

10.prop.2.59.p3946

Again, since ED is incommensurable in length with AB, therefore FE is also incommensurable with EK; [X. 13] therefore FE, EK are rational straight lines commensurable in square only; therefore EL, that is, MR, that is, the rectangle MN, NO, is medial. [X. 21]

10.prop.2.59.p3947

And, since AE is incommensurable with EF, AK is also incommensurable with EL. [VI. 1, X. 11]

10.prop.2.59.p3948

But AK is the sum of the squares on MN, NO, and EL is the rectangle MN, NO; therefore the sum of the squares on MN, NO is incommensurable with the rectangle MN, NO.

10.prop.2.59.p3949

And each of them is medial, and MN, NO are incommensurable in square.

10.prop.2.59.p3950

Therefore MO is the side of the sum of two medial areas [X. 41], and is the side of AC. Q. E. D.

10.prop.2.59.p3951

[LEMMA. If a straight line be cut into unequal parts, the squares on the unequal parts are greater than twice the rectangle contained by the unequal parts.

10.prop.2.59.p3952

Let AB be a straight line, and let it be cut into unequal parts at C, and let AC be the greater; I say that the squares on AC, CB are greater than twice the rectangle AC, CB.

10.prop.2.59.p3953

For let AB be bisected at D.

10.prop.2.59.p3954

Since then a straight line has been cut into equal parts at D, and into unequal parts at C, therefore the rectangle AC, CB together with the square on CD is equal to the square on AD, [II. 5] so that the rectangle AC, CB is less than double of the square on AD.

10.prop.2.59.p3955

But the squares on AC, CB are double of the squares on AD, DC; [II. 9] therefore the squares on AC, CB are greater than twice the rectangle AC, CB. Q. E. D.]

10.prop.2.60.p3956

The square on the binomial straight line applied to a rational straight line produces as breadth the first binomial.

10.prop.2.60.p3957

Let AB be a binomial straight line divided into its terms at C, so that AC is the greater term; let a rational straight line DE be set out, and let DEFG equal to the square on AB be applied to DE producing DG as its breadth; I say that DG is a first binomial straight line.

10.prop.2.60.p3958

For let there be applied to DE the rectangle DH equal to the square on AC, and KL equal to the square on BC; therefore the remainder, twice the rectangle AC, CB, is equal to MF.

10.prop.2.60.p3959

Let MG be bisected at N, and let NO be drawn parallel [to ML or GF].

10.prop.2.60.p3960

Therefore each of the rectangles MO, NF is equal to once the rectangle AC, CB.

10.prop.2.60.p3961

Now, since AB is a binomial divided into its terms at C, therefore AC, CB are rational straight lines commensurable in square only; [X. 36] therefore the squares on AC, CB are rational and commensurable with one another, so that the sum of the squares on AC, CB is also rational. [X. 15]

10.prop.2.60.p3962

And it is equal to DL; therefore DL is rational.

10.prop.2.60.p3963

And it is applied to the rational straight line DE; therefore DM is rational and commensurable in length with DE. [X. 20]

10.prop.2.60.p3964

Again, since AC, CB are rational straight lines commensurable in square only, therefore twice the rectangle AC, CB, that is MF, is medial. [X. 21]

10.prop.2.60.p3965

And it is applied to the rational straight line ML; therefore MG is also rational and incommensurable in length with ML, that is, DE. [X. 22]

10.prop.2.60.p3966

But MD is also rational and is commensurable in length with DE; therefore DM is incommensurable in length with MG. [X. 13]

10.prop.2.60.p3967

And they are rational; therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36]

10.prop.2.60.p3968

It is next to be proved that it is also a first binomial straight line.

10.prop.2.60.p3969

Since the rectangle AC, CB is a mean proportional between the squares on AC, CB, [cf. Lemma after X. 53] therefore MO is also a mean proportional between DH, KL.

10.prop.2.60.p3970

Therefore, as DH is to MO, so is MO to KL, that is, as DK is to MN, so is MN to MK; [VI. 1] therefore the rectangle DK, KM is equal to the square on MN. [VI. 17]

10.prop.2.60.p3971

And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KM. [VI. 1, X. 11]

10.prop.2.60.p3972

And, since the squares on AC, CB are greater than twice the rectangle AC, CB, [Lemma] therefore DL is also greater than MF, so that DM is also greater than MG. [VI. 1]

10.prop.2.60.p3973

And the rectangle DK, KM is equal to the square on MN, that is, to the fourth part of the square on MG, and DK is commensurable with KM.

10.prop.2.60.p3974

But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into commensurable parts, the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater; [X. 17] therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM.

10.prop.2.60.p3975

And DM, MG are rational, and DM, which is the greater term, is commensurable in length with the rational straight line DE set out.

10.prop.2.60.p3976

Therefore DG is a first binomial straight line. [X. Deff. II. 1] Q. E. D.

10.prop.2.61.p3977

The square on the first bimedial straight line applied to a rational straight line produces as breadth the second binomial.

10.prop.2.61.p3978

Let AB be a first bimedial straight line divided into its medials at C, of which medials AC is the greater; let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a second binominal straight line.

10.prop.2.61.p3979

For let the same construction as before be made.

10.prop.2.61.p3980

Then, since AB is a first bimedial divided at C, therefore AC, CB are medial straight lines commensurable in square only, and containing a rational rectangle, [X. 37] so that the squares on AC, CB are also medial. [X. 21]

10.prop.2.61.p3981

Therefore DL is medial. [X. 15 and 23, Por.]

10.prop.2.61.p3982

And it has been applied to the rational straight line DE; therefore MD is rational and incommensurable in length with DE. [X. 22]

10.prop.2.61.p3983

Again, since twice the rectangle AC, CB is rational, MF is also rational.

10.prop.2.61.p3984

And it is applied to the rational straight line ML; therefore MG is also rational and commensurable in length with ML, that is, DE; [X. 20] therefore DM is incommensurable in length with MG. [X. 13]

10.prop.2.61.p3985

And they are rational; therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36]

10.prop.2.61.p3986

It is next to be proved that it is also a second binomial straight line.

10.prop.2.61.p3987

For, since the squares on AC, CB are greater than twice the rectangle AC, CB, therefore DL is also greater than MF, so that DM is also greater than MG. [VI. 1]

10.prop.2.61.p3988

And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KM. [VI. 1, X. 11]

10.prop.2.61.p3989

And the rectangle DK, KM is equal to the square on MN; therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM. [X. 17]

10.prop.2.61.p3990

And MG is commensurable is length with DE.

10.prop.2.61.p3991

Therefore DG is a second binomial straight line. [X. Deff. II. 2]

10.prop.2.62.p3992

The square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial.

10.prop.2.62.p3993

Let AB be a second bimedial straight line divided into its medials at C, so that AC is the greater segment; let DE be any rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth; I say that DG is a third binomial straight line.

10.prop.2.62.p3994

Let the same construction be made as before shown.

10.prop.2.62.p3995

Then, since AB is a second bimedial divided at C, therefore AC, CB are medial straight lines commensurable in square only and containing a medial rectangle, [X. 38] so that the sum of the squares on AC, CB is also medial. [X. 15 and 23 Por.]

10.prop.2.62.p3996

And it is equal to DL; therefore DL is also medial.

10.prop.2.62.p3997

And it is applied to the rational straight line DE; therefore MD is also rational and incommensurable in length with DE. [X. 22]

10.prop.2.62.p3998

For the same reason, MG is also rational and incommensurable in length with ML, that is, with DE; therefore each of the straight lines DM, MG is rational and incommensurable in length with DE.

10.prop.2.62.p3999

And, since AC is incommensurable in length with CB, and, as AC is to CB, so is the square on AC to the rectangle AC, CB, therefore the square on AC is also incommensurable with the rectangle AC, CB. [X. 11]

10.prop.2.62.p4000

Hence the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, [X. 12, 13] that is, DL is incommensurable with MF, so that DM is also incommensurable with MG. [VI. 1, X. 11]

10.prop.2.62.p4001

And they are rational; therefore DG is binomial. [X. 36]

10.prop.2.62.p4002

It is to be proved that it is also a third binomial straight line.

10.prop.2.62.p4003

In manner similar to the foregoing we may conclude that DM is greater than MG, and that DK is commensurable with KM.

10.prop.2.62.p4004

And the rectangle DK, KM is equal to the square on MN; therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM.

10.prop.2.62.p4005

And neither of the straight lines DM, MG is commensurable in length with DE.

10.prop.2.62.p4006

Therefore DG is a third binomial straight line. [X. Deff. II. 3] Q. E. D.

10.prop.2.63.p4007

The square on the major straight line applied to a rational straight line produces as breadth the fourth binomial.

10.prop.2.63.p4008

Let AB be a major straight line divided at C, so that AC is greater than CB; let DE be a rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth; I say that DG is a fourth binomial straight line.

10.prop.2.63.p4009

Let the same construction be made as before shown.

10.prop.2.63.p4010

Then, since AB is a major straight line divided at C, AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. [X. 39]

10.prop.2.63.p4011

Since then the sum of the squares on AC, CB is rational, therefore DL is rational; therefore DM is also rational and commensurable in length with DE. [X. 20]

10.prop.2.63.p4012

Again, since twice the rectangle AC, CB, that is, MF, is medial, and it is applied to the rational straight line ML, therefore MG is also rational and incommensurable in length with DE; [X. 22] therefore DM is also incommensurable in length with MG. [X. 13]

10.prop.2.63.p4013

Therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36]

10.prop.2.63.p4014

It is to be proved that it is also a fourth binomial straight line.

10.prop.2.63.p4015

In manner similar to the foregoing we can prove that DM is greater than MG, and that the rectangle DK, KM is equal to the square on MN.

10.prop.2.63.p4016

Since then the square on AC is incommensurable with the square on CB, therefore DH is also incommensurable with KL, so that DK is also incommensurable with KM. [VI. 1, X. 11]

10.prop.2.63.p4017

But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into incommensurable parts, then the square on the greater will be greater than the square on the less by the square on a straight line incommensurable in length with the greater; [X. 18] therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM.

10.prop.2.63.p4018

And DM, MG are rational straight lines commensurable in square only, and DM is commensurable with the rational straight line DE set out.

10.prop.2.63.p4019

Therefore DG is a fourth binomial straight line. [X. Deff. II. 4] Q. E. D.

10.prop.2.64.p4020

The square on the side of a rational plus a medial area applied to a rational straight line produces as breadth the fifth binomial.

10.prop.2.64.p4021

Let AB be the side of a rational plus a medial area, divided into its straight lines at C, so that AC is the greater; let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a fifth binomial straight line.

10.prop.2.64.p4022

Let the same construction as before be made.

10.prop.2.64.p4023

Since then AB is the side of a rational plus a medial area, divided at C, therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. [X. 40]

10.prop.2.64.p4024

Since then the sum of the squares on AC, CB is medial, therefore DL is medial, so that DM is rational and incommensurable in length with DE. [X. 22]

10.prop.2.64.p4025

Again, since twice the rectangle AC, CB, that is MF, is rational, therefore MG is rational and commensurable with DE. [X. 20]

10.prop.2.64.p4026

Therefore DM is incommensurable with MG; [X. 13] therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36]

10.prop.2.64.p4027

I say next that it is also a fifth binomial straight line.

10.prop.2.64.p4028

For it can be proved similarly that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM. [X. 18]

10.prop.2.64.p4029

And DM, MG are commensurable in square only, and the less, MG, is commensurable in length with DE.

10.prop.2.64.p4030

Therefore DG is a fifth binomial. Q. E. D.

10.prop.2.65.p4031

The square on the side of the sum of two medial areas applied to a rational straight line produces as breadth the sixth binomial.

10.prop.2.65.p4032

Let AB be the side of the sum of two medial areas, divided at C, let DE be a rational straight line, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a sixth binomial straight line.

10.prop.2.65.p4033

For let the same construction be made as before.

10.prop.2.65.p4034

Then, since AB is the side of the sum of two medial areas, divided at C, therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and moreover the sum of the squares on them incommensurable with the rectangle contained by them, [X. 41] so that, in accordance with what was before proved, each of the rectangles DL, MF is medial.

10.prop.2.65.p4035

And they are applied to the rational straight line DE; therefore each of the straight lines DM, MG is rational and incommensurable in length with DE. [X. 22]

10.prop.2.65.p4036

And, since the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, therefore DL is incommensurable with MF.

10.prop.2.65.p4037

Therefore DM is also incommensurable with MG; [VI. 1, X. 11] therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36]

10.prop.2.65.p4038

I say next that it is also a sixth binomial straight line.

10.prop.2.65.p4039

Similarly again we can prove that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; and, for the same reason, the square on DM is greater than the square on MG by the square on a straight line incommensurable in length with DM.

10.prop.2.65.p4040

And neither of the straight lines DM, MG is commensurable in length with the rational straight line DE set out.

10.prop.2.65.p4041

Therefore DG is a sixth binomial straight line. Q. E. D.

10.prop.2.66.p4042

A straight line commensurable in length with a binomial straight line is itself also binomial and the same in order.

10.prop.2.66.p4043

Let AB be binomial, and let CD be commensurable in length with AB; I say that CD is binomial and the same in order with AB.

10.prop.2.66.p4044

For, since AB is binomial, let it be divided into its terms at E, and let AE be the greater term; therefore AE, EB are rational straight lines commensurable in square only. [X. 36]

10.prop.2.66.p4045

Let it be contrived that, as AB is to CD, so is AE to CF; [VI. 12] therefore also the remainder EB is to the remainder FD as AB is to CD. [V. 19]

10.prop.2.66.p4046

But AB is commensurable in length with CD; therefore AE is also commensurable with CF, and EB with FD. [X. 11]

10.prop.2.66.p4047

And AE, EB are rational; therefore CF, FD are also rational.

10.prop.2.66.p4048

And, as AE is to CF, so is EB to FD. [V. 11]

10.prop.2.66.p4049

Therefore, alternately, as AE is to EB, so is CF to FD. [V. 16]

10.prop.2.66.p4050

But AE, EB are commensurable in square only; therefore CF, FD are also commensurable in square only. [X. 11]

10.prop.2.66.p4051

And they are rational; therefore CD is binomial. [X. 36]

10.prop.2.66.p4052

I say next that it is the same in order with AB.

10.prop.2.66.p4053

For the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it.

10.prop.2.66.p4054

If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line commensurable with CF. [X. 14]

10.prop.2.66.p4055

And, if AE is commensurable with the rational straight line set out, CF will also be commensurable with it, [X. 12] and for this reason each of the straight lines AB, CD is a first binomial, that is, the same in order. [X. Deff. II. 1]

10.prop.2.66.p4056

But, if EB is commensurable with the rational straight line set out, FD is also commensurable with it, [X. 12] and for this reason again CD will be the same in order with AB, for each of them will be a second binomial. [X. Deff. II. 2]

10.prop.2.66.p4057

But, if neither of the straight lines AE, EB is commensurable with the rational straight line set out, neither of the straight lines CF, FD will be commensurable with it, [X. 13] and each of the straight lines AB, CD is a third binomial. [X. Deff. II. 3]

10.prop.2.66.p4058

But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, the square on CF is also greater than the square on FD by the square on a straight line incommensurable with CF. [X. 14]

10.prop.2.66.p4059

And, if AE is commensurable with the rational straight line set out, CF is also commensurable with it, and each of the straight lines AB, CD is a fourth binomial. [X. Deff. II. 4]

10.prop.2.66.p4060

But, if EB is so commensurable, so is FD also, and each of the straight lines AB, CD will be a fifth binomial. [X. Deff. II. 5]

10.prop.2.66.p4061

But, if neither of the straight lines AE, EB is so commensurable, neither of the straight lines CF, FD is commensurable with the rational straight line set out, and each of the straight lines AB, CD will be a sixth binomial. [X. Deff. II. 6]

10.prop.2.66.p4062

Hence a straight line commensurable in length with a binomial straight line is binomial and the same in order. Q. E. D.

10.prop.2.67.p4063

A straight line commensurable in length with a bimedial straight line is itself also bimedial and the same in order.

10.prop.2.67.p4064

Let AB be bimedial, and let CD be commensurable in length with AB; I say that CD is bimedial and the same in order with AB.

10.prop.2.67.p4065

For, since AB is bimedial, let it be divided into its medials at E; therefore AE, EB are medial straight lines commensurable in square only. [X. 37, 38]

10.prop.2.67.p4066

And let it be contrived that, as AB is to CD, so is AE to CF; therefore also the remainder EB is to the remainder FD as AB is to CD. [V. 19]

10.prop.2.67.p4067

But AB is commensurable in length with CD; therefore AE, EB are also commensurable with CF, FD respectively. [X. 11]

10.prop.2.67.p4068

But AE, EB are medial; therefore CF, FD are also medial. [X. 23]

10.prop.2.67.p4069

And since, as AE is to EB, so is CF to FD, [V. 11] and AE, EB are commensurable in square only, CF, FD are also commensurable in square only. [X. 11]

10.prop.2.67.p4070

But they were also proved medial; therefore CD is bimedial.

10.prop.2.67.p4071

I say next that it is also the same in order with AB.

10.prop.2.67.p4072

For since, as AE is to EB, so is CF to FD, therefore also, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD; therefore, alternately, as the square on AE is to the square on CF, so is the rectangle AE, EB to the rectangle CF, FD. [V. 16]

10.prop.2.67.p4073

But the square on AE is commensurable with the square on CF; therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD.

10.prop.2.67.p4074

If therefore the rectangle AE, EB is rational, the rectangle CF, FD is also rational, [and for this reason CD is a first bimedial]; [X. 37] but if medial, medial, [X. 23, Por.] and each of the straight lines AB, CD is a second bimedial. [X. 38]

10.prop.2.67.p4075

And for this reason CD will be the same in order with AB. Q. E. D.

10.prop.2.68.p4076

A straight line commensurable with a major straight line is itself also major.

10.prop.2.68.p4077

Let AB be major, and let CD be commensurable with AB; I say that CD is major.

10.prop.2.68.p4078

Let AB be divided at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. [X. 39]

10.prop.2.68.p4079

Let the same construction be made as before.

10.prop.2.68.p4080

Then since, as AB is to CD, so is AE to CF, and EB to FD, therefore also, as AE is to CF, so is EB to FD. [V. 11]

10.prop.2.68.p4081

But AB is commensurable with CD; therefore AE, EB are also commensurable with CF, FD respectively. [X. 11]

10.prop.2.68.p4082

And since, as AE is to CF, so is EB to FD, alternately also, as AE is to EB, so is CF to FD; [V. 16] therefore also, componendo, as AB is to BE, so is CD to DF; [V. 18] therefore also, as the square on AB is to the square on BE, so is the square on CD to the square on DF. [VI. 20]

10.prop.2.68.p4083

Similarly we can prove that, as the square on AB is to the square on AE, so also is the square on CD to the square on CF.

10.prop.2.68.p4084

Therefore also, as the square on AB is to the squares on AE, EB, so is the square on CD to the squares on CF, FD; therefore also, alternately, as the square on AB is to the square on CD, so are the squares on AE, EB to the squares on CF, FD. [V. 16]

10.prop.2.68.p4085

But the square on AB is commensurable with the square on CD; therefore the squares on AE, EB are also commensurable with the squares on CF, FD.

10.prop.2.68.p4086

And the squares on AE, EB together are rational; therefore the squares on CF, FD together are rational.

10.prop.2.68.p4087

Similarly also twice the rectangle AE, EB is commensurable with twice the rectangle CF, FD.

10.prop.2.68.p4088

And twice the rectangle AE, EB is medial; therefore twice the rectangle CF, FD is also medial. [X. 23, Por.]

10.prop.2.68.p4089

Therefore CF, FD are straight lines incommensurable in square which make, at the same time, the sum of the squares on them rational, but the rectangle contained by them medial; therefore the whole CD is the irrational straight line called major. [X. 39]

10.prop.2.68.p4090

Therefore a straight line commensurable with the major straight line is major. Q. E. D.

10.prop.2.69.p4091

A straight line commensurable with the side of a rational plus a medial area is itself also the side of a rational plus a medial area.

10.prop.2.69.p4092

Let AB be the side of a rational plus a medial area, and let CD be commensurable with AB; it is to be proved that CD is also the side of a rational plus a medial area.

10.prop.2.69.p4093

Let AB be divided into its straight lines at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. [X. 40]

10.prop.2.69.p4094

Let the same construction be made as before.

10.prop.2.69.p4095

We can then prove similarly that CF, FD are incommensurable in square, and the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that the sum of the squares on CF, FD is also medial, and the rectangle CF, FD rational.

10.prop.2.69.p4096

Therefore CD is the side of a rational plus a medial area. Q. E. D.

10.prop.2.70.p4097

A straight line commensurable with the side of the sum of two medial areas is the side of the sum of two medial areas.

10.prop.2.70.p4098

Let AB be the side of the sum of two medial areas, and CD commensurable with AB; it is to be proved that CD is also the side of the sum of two medial areas.

10.prop.2.70.p4099

For, since AB is the side of the sum of two medial areas, let it be divided into its straight lines at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and furthermore the sum of the squares on AE, EB incommensurable with the rectangle AE, EB. [X. 41]

10.prop.2.70.p4100

Let the same construction be made as before.

10.prop.2.70.p4101

We can then prove similarly that CF, FD are also incommensurable in square, the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that the sum of the squares on CF, FD is also medial, the rectangle CF, FD is medial, and moreover the sum of the squares on CF, FD is incommensurable with the rectangle CF, FD.

10.prop.2.70.p4102

Therefore CD is the side of the sum of two medial areas. Q. E. D.

10.prop.2.71.p4103

If a rational and a medial area be added together, four irrational straight lines arise, namely a binomial or a first bimedial or a major or a side of a rational plus a medial area.

10.prop.2.71.p4104

Let. AB be rational, and CD medial; I say that the side of the area AD is a binomial or a first bimedial or a major or a side of a rational plus a medial area.

10.prop.2.71.p4105

For AB is either greater or less than CD.

10.prop.2.71.p4106

First, let it be greater; let a rational straight line EF be set out, let there be applied to EF the rectangle EG equal to AB, producing EH as breadth, and let HI, equal to DC, be applied to EF, producing HK as breadth.

10.prop.2.71.p4107

Then, since AB is rational and is equal to EG, therefore EG is also rational.

10.prop.2.71.p4108

And it has been applied to EF, producing EH as breadth; therefore EH is rational and commensurable in length with EF. [X. 20]

10.prop.2.71.p4109

Again, since CD is medial and is equal to HI, therefore HI is also medial.

10.prop.2.71.p4110

And it is applied to the rational straight line EF, producing HK as breadth; therefore HK is rational and incommensurable in length with EF [X. 22]

10.prop.2.71.p4111

And, since CD is medial, while AB is rational, therefore AB is incommensurable with CD, so that EG is also incommensurable with HI.

10.prop.2.71.p4112

But, as EG is to HI, so is EH to HK; [VI. 1] therefore EH is also incommensurable in length with HK. [X. 11]

10.prop.2.71.p4113

And both are rational; therefore EH, HK are rational straight lines commensurable in square only; therefore EK is a binomial straight line, divided at H. [X. 36]

10.prop.2.71.p4114

And, since AB is greater than CD, while AB is equal to EG and CD to HI, therefore EG is also greater than HI; therefore EH is also greater than HK.

10.prop.2.71.p4115

The square, then, on EH is greater than the square on HK either by the square on a straight line commensurable in length with EH or by the square on a straight line incommensurable with it.

10.prop.2.71.p4116

First, let the square on it be greater by the square on a straight line commensurable with itself.

10.prop.2.71.p4117

Now the greater straight line HE is commensurable in length with the rational straight line EF set out; therefore EK is a first binomial. [X. Deff. II. 1]

10.prop.2.71.p4118

But EF is rational; and, if an area be contained by a rational straight line and the first binomial, the side of the square equal to the area is binomial. [X. 54]

10.prop.2.71.p4119

Therefore the side of EI is binomial; so that the side of AD is also binomial.

10.prop.2.71.p4120

Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable with EH.

10.prop.2.71.p4121

Now the greater straight line EH is commensurable in length with the rational straight line EF set out; therefore EK is a fourth binomial. [X. Deff. II. 4]

10.prop.2.71.p4122

But EF is rational; and, if an area be contained by a rational straight line and the fourth binomial, the side of the area is the irrational straight line called major. [X. 57]

10.prop.2.71.p4123

Therefore the side of the area EI is major; so that the side of the area AD is also major.

10.prop.2.71.p4124

Next, let AB be less than CD; therefore EG is also less than HI, so that EH is also less than HK.

10.prop.2.71.p4125

Now the square on HK is greater than the square on EH either by the square on a straight line commensurable with HK or by the square on a straight line incommensurable with it.

10.prop.2.71.p4126

First, let the square on it be greater by the square on a straight line commensurable in length with itself.

10.prop.2.71.p4127

Now the lesser straight line EH is commensurable in length with the rational straight line EF set out; therefore EK is a second binomial. [X. Deff. II. 2]

10.prop.2.71.p4128

But EF is rational, and, if an area be contained by a rational straight line and the second binomial, the side of the square equal to it is a first bimedial; [X. 55] therefore the side of the area EI is a first bimedial, so that the side of AD is also a first bimedial.

10.prop.2.71.p4129

Next, let the square on HK be greater than the square on HE by the square on a straight line incommensurable with HK.

10.prop.2.71.p4130

Now the lesser straight line EH is commensurable with the rational straight line EF set out; therefore EK is a fifth binomial. [X. Deff. II. 5]

10.prop.2.71.p4131

But EF is rational; and, if an area be contained by a rational straight line and the fifth binomial, the side of the square equal to the area is a side of a rational plus a medial area. [X. 58]

10.prop.2.71.p4132

Therefore the side of the area EI is a side of a rational plus a medial area, so that the side of the area AD is also a side of a rational plus a medial area.

10.prop.2.71.p4133

Therefore etc. Q. E. D.

10.prop.2.72.p4134

If two medial areas incommensurable with one another be added together, the remaining two irrational straight lines arise, namely either a second bimedial or a side of the sum of two medial areas.

10.prop.2.72.p4135

For let two medial areas AB, CD incommensurable with one another be added together; I say that the side of the area AD is either a second bimedial or a side of the sum of two medial areas.

10.prop.2.72.p4136

For AB is either greater or less than CD.

10.prop.2.72.p4137

First, if it so chance, let AB be greater than CD.

10.prop.2.72.p4138

Let the rational straight line EF be set out, and to EF let there be applied the rectangle EG equal to AB and producing EH as breadth, and the rectangle HI equal to CD and producing HK as breadth.

10.prop.2.72.p4139

Now, since each of the areas AB, CD is medial, therefore each of the areas EG, HI is also medial.

10.prop.2.72.p4140

And they are applied to the rational straight line FE, producing EH, HK as breadth; therefore each of the straight lines EH, HK is rational and incommensurable in length with EF. [X. 22]

10.prop.2.72.p4141

And, since AB is incommensurable with CD, and AB is equal to EG, and CD to HI, therefore EG is also incommensurable with HI.

10.prop.2.72.p4142

But, as EG is to HI, so is EH to HK; [VI. 1] therefore EH is incommensurable in length with HK. [X. 11]

10.prop.2.72.p4143

Therefore EH, HK are rational straight lines commensurable in square only; therefore EK is binomial. [X. 36]

10.prop.2.72.p4144

But the square on EH is greater than the square on HK either by the square on a straight line commensurable with EH or by the square on a straight line incommensurable with it.

10.prop.2.72.p4145

First, let the square on it be greater by the square on a straight line commensurable in length with itself.

10.prop.2.72.p4146

Now neither of the straight lines EH, HK is commensurable in length with the rational straight line EF set out; therefore EK is a third binomial. [X. Deff. II. 3]

10.prop.2.72.p4147

But EF is rational; and, if an area be contained by a rational straight line and the third binomial, the side of the area is a second bimedial; [X. 56] therefore the side of EI, that is, of AD, is a second bimedial.

10.prop.2.72.p4148

Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable in length with EH.

10.prop.2.72.p4149

Now each of the straight lines EH, HK is incommensurable in length with EF; therefore EK is a sixth binomial. [X. Deff. II. 6]

10.prop.2.72.p4150

But, if an area be contained by a rational straight line and the sixth binomial, the side of the area is the side of the sum of two medial areas; [X. 59] so that the side of the area AD is also the side of the sum of two medial areas.

10.prop.2.72.p4151

Therefore etc. Q. E. D.

10.prop.2.73.p4152

If from a rational straight line there be subtracted a rational straight line commensurable with the whole in square only, the remainder is irrational; and let it be called an apotome.

10.prop.2.73.p4153

For from the rational straight line AB let the rational straight line BC, commensurable with the whole in square only, be subtracted; I say that the remainder AC is the irrational straight line called apotome.

10.prop.2.73.p4154

For, since AB is incommensurable in length with BC, and, as AB is to BC, so is the square on AB to the rectangle AB, BC, therefore the square on AB is incommensurable with the rectangle AB, BC. [X. 11]

10.prop.2.73.p4155

But the squares on AB, BC are commensurable with the square on AB, [X. 15] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC. [X. 6]

10.prop.2.73.p4156

And, inasmuch as the squares on AB, BC are equal to twice the rectangle AB, BC together with the square on CA, [II. 7] therefore the squares on AB, BC are also incommensurable with the remainder, the square on AC. [X. 13, 16]

10.prop.2.73.p4157

But the squares on AB, BC are rational; therefore AC is irrational. [X. Def. 4]

10.prop.2.73.p4158

And let it be called an apotome. Q. E. D.

10.prop.2.74.p4159

If from a medial straight line there be subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a rational rectangle, the remainder is irrational. And let it be called a first apotome of a medial straight line.

10.prop.2.74.p4160

For from the medial straight line AB let there be subtracted the medial straight line BC which is commensurable with AB in square only and with AB makes the rectangle AB, BC rational; I say that the remainder AC is irrational; and let it be called a first apotome of a medial straight line.

10.prop.2.74.p4161

For, since AB, BC are medial, the squares on AB, BC are also medial.

10.prop.2.74.p4162

But twice the rectangle AB, BC is rational; therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; therefore twice the rectangle AB, BC is also incommensurable with the remainder, the square on AC, [Cf. II. 7] since, if the whole is incommensurable with one of the magnitudes, the original magnitudes will also be incommensurable. [X. 16]

10.prop.2.74.p4163

But twice the rectangle AB, BC is rational; therefore the square on AC is irrational; therefore AC is irrational. [X. Def. 4]

10.prop.2.74.p4164

And let it be called a first apotome of a medial straight line.

10.prop.2.75.p4165

If from a medial straight line there be subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a medial rectangle, the remainder is irrational; and let it be called a second apotome of a medial straight line.

10.prop.2.75.p4166

For from the medial straight line AB let there be subtracted the medial straight line CB which is commensurable with the whole AB in square only and such that the rectangle AB, BC, which it contains with the whole AB, is medial; [X. 28] I say that the remainder AC is irrational; and let it be called a second apotome of a medial straight line.

10.prop.2.75.p4167

For let a rational straight line DI be set out, let DE equal to the squares on AB, BC be applied to DI, producing DG as breadth, and let DH equal to twice the rectangle AB, BC be applied to DI, producing DF as breadth; therefore the remainder FE is equal to the square on AC. [II. 7]

10.prop.2.75.p4168

Now, since the squares on AB, BC are medial and commensurable, therefore DE is also medial. [X. 15 and 23, Por.]

10.prop.2.75.p4169

And it is applied to the rational straight line DI, producing DG as breadth; therefore DG is rational and incommensurable in length with DI. [X. 22]

10.prop.2.75.p4170

Again, since the rectangle AB, BC is medial, therefore twice the rectangle AB, BC is also medial. [X. 23, Por.]

10.prop.2.75.p4171

And it is equal to DH; therefore DH is also medial.

10.prop.2.75.p4172

And it has been applied to the rational straight line DI, producing DF as breadth; therefore DF is rational and incommensurable in length with DI. [X. 22]

10.prop.2.75.p4173

And, since AB, BC are commensurable in square only, therefore AB is incommensurable in length with BC; therefore the square on AB is also incommensurable with the rectangle AB, BC. [X. 11]

10.prop.2.75.p4174

But the squares on AB, BC are commensurable with the square on AB, [X. 15] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC; [X. 6] therefore twice the rectangle AB, BC is incommensurable with the squares on AB, BC. [X. 13]

10.prop.2.75.p4175

But DE is equal to the squares on AB, BC, and DH to twice the rectangle AB, BC; therefore DE is incommensurable with DH.

10.prop.2.75.p4176

But, as DE is to DH, so is GD to DF; [VI. 1] therefore GD is incommensurable with DF. [X. 11]

10.prop.2.75.p4177

And both are rational; therefore GD, DF are rational straight lines commensurable in square only; therefore FG is an apotome. [X. 73]

10.prop.2.75.p4178

But DI is rational, and the rectangle contained by a rational and an irrational straight line is irrational, [deduction from X. 20] and its ’side’ is irrational.

10.prop.2.75.p4179

And AC is the ’side’ of FE; therefore AC is irrational.

10.prop.2.75.p4180

And let it be called a second apotome of a medial straight line. Q. E. D.

10.prop.2.76.p4181

If from a straight line there be subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the squares on them added together rational, but the rectangle contained by them medial, the remainder is irrational; and let it be called minor.

10.prop.2.76.p4182

For from the straight line AB let there be subtracted the straight line BC which is incommensurable in square with the whole and fulfils the given conditions. [X. 33]

10.prop.2.76.p4183

I say that the remainder AC is the irrational straight line called minor.

10.prop.2.76.p4184

For, since the sum of the squares on AB, BC is rational, while twice the rectangle AB, BC is medial, therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; and, convertendo, the squares on AB, BC are incommensurable with the remainder, the square on AC. [II. 7, X. 16]

10.prop.2.76.p4185

But the squares on AB, BC are rational; therefore the square on AC is irrational; therefore AC is irrational.

10.prop.2.76.p4186

And let it be called minor.

10.prop.2.77.p4187

If from a straight line there be subtracted a straight line which is incommensurable in square with the whole, and which with the whole makes the sum of the squares on them medial, but twice the rectangle contained by them rational, the remainder is irrational: and let it be called that which produces with a rational area a medial whole.

10.prop.2.77.p4188

For from the straight line AB let there be subtracted the straight line BC which is incommensurable in square with AB and fulfils the given conditions; [X. 34] I say that the remainder AC is the irrational straight line aforesaid.

10.prop.2.77.p4189

For, since the sum of the squares on AB, BC is medial, while twice the rectangle AB, BC is rational, therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; therefore the remainder also, the square on AC, is incommensurable with twice the rectangle AB, BC. [II. 7, X. 16]

10.prop.2.77.p4190

And twice the rectangle AB, BC is rational; therefore the square on AC is irrational; therefore AC is irrational.

10.prop.2.77.p4191

And let it be called that which produces with a rational area a medial whole. Q. E. D.

10.prop.2.78.p4192

If from a straight line there be subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the sum of the squares on them medial, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them, the remainder is irrational; and let it be called that which produces with a medial area a medial whole.

10.prop.2.78.p4193

For from the straight line AB let there be subtracted the straight line BC incommensurable in square with AB and fulfilling the given conditions; [X. 35] I say that the remainder AC is the irrational straight line called that which produces with a medial area a medial whole.

10.prop.2.78.p4194

For let a rational straight line DI be set out, to DI let there be applied DE equal to the squares on AB, BC, producing DG as breadth, and let DH equal to twice the rectangle AB, BC be subtracted.

10.prop.2.78.p4195

Therefore the remainder FE is equal to the square on AC, [II. 7] so that AC is the side of FE.

10.prop.2.78.p4196

Now, since the sum of the squares on AB, BC is medial and is equal to DE, therefore DE is medial.

10.prop.2.78.p4197

And it is applied to the rational straight line DI, producing DG as breadth; therefore DG is rational and incommensurable in length with DI. [X. 22]

10.prop.2.78.p4198

Again, since twice the rectangle AB, BC is medial and is equal to DH, therefore DH is medial.

10.prop.2.78.p4199

And it is applied to the rational straight line DI, producing DF as breadth; therefore DF is also rational and incommensurable in length with DI. [X. 22]

10.prop.2.78.p4200

And, since the squares on AB, BC are incommensurable with twice the rectangle AB, BC, therefore DE is also incommensurable with DH.

10.prop.2.78.p4201

But, as DE is to DH, so also is DG to DF; [VI. 1] therefore DG is incommensurable with DF. [X. 11]

10.prop.2.78.p4202

And both are rational; therefore GD, DF are rational straight lines commensurable in square only.

10.prop.2.78.p4203

Therefore FG is an apotome. [X. 73]

10.prop.2.78.p4204

And FH is rational; but the rectangle contained by a rational straight line and an apotome is irrational, [deduction from X. 20] and its side is irrational.

10.prop.2.78.p4205

And AC is the side of FE; therefore AC is irrational.

10.prop.2.78.p4206

And let it be called that which produces with a medial area a medial whole. Q. E. D.

10.prop.2.79.p4207

To an apotome only one rational straight line can be annexed which is commensurable with the whole in square only.

10.prop.2.79.p4208

Let AB be an apotome, and BC an annex to it; therefore AC, CB are rational straight lines commensurable in square only. [X. 73]

10.prop.2.79.p4209

I say that no other rational straight line can be annexed to AB which is commensurable with the whole in square only.

10.prop.2.79.p4210

For, if possible, let BD be so annexed; therefore AD, DB are also rational straight lines commensurable in square only. [X. 73]

10.prop.2.79.p4211

Now, since the excess of the squares on AD, DB over twice the rectangle AD, DB is also the excess of the squares on AC, CB over twice the rectangle AC, CB, for both exceed by the same, the square on AB, [II. 7] therefore, alternately, the excess of the squares on AD, DB over the squares on AC, CB is the excess of twice the rectangle AD, DB over twice the rectangle AC, CB.

10.prop.2.79.p4212

But the squares on AD, DB exceed the squares on AC, CB by a rational area, for both are rational; therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area: which is impossible, for both are medial [X. 21], and a medial area does not exceed a medial by a rational area. [X. 26]

10.prop.2.79.p4213

Therefore no other rational straight line can be annexed to AB which is commensurable with the whole in square only.

10.prop.2.79.p4214

Therefore only one rational straight line can be annexed to an apotome which is commensurable with the whole in square only. Q. E. D.

10.prop.2.80.p4215

To a first apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a rational rectangle.

10.prop.2.80.p4216

For let AB be a first apotome of a medial straight line, and let BC be an annex to AB; therefore AC, CB are medial straight lines commensurable in square only and such that the rectangle AC, CB which they contain is rational; [X. 74] I say that no other medial straight line can be annexed to AB which is commensurable with the whole in square only and which contains with the whole a rational area.

10.prop.2.80.p4217

For, if possible, let DB also be so annexed; therefore AD, DB are medial straight lines commensurable in square only and such that the rectangle AD, DB which they contain is rational. [X. 74]

10.prop.2.80.p4218

Now, since the excess of the squares on AD, DB over twice the rectangle AD, DB is also the excess of the squares on AC, CB over twice the rectangle AC, CB, for they exceed by the same, the square on AB, [II. 7] therefore, alternately, the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB.

10.prop.2.80.p4219

But twice the rectangle AD, DB exceeds twice the rectangle AC, CB by a rational area, for both are rational.

10.prop.2.80.p4220

Therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area. which is impossible, for both are medial [X. 15 and 23, Por.], and a medial area does not exceed a medial by a rational area. [X. 26]

10.prop.2.80.p4221

Therefore etc. Q. E. D.

10.prop.2.81.p4222

To a second apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a medial rectangle.

10.prop.2.81.p4223

Let AB be a second apotome of a medial straight line and BC an annex to AB; therefore AC, CB are medial straight lines commensurable in square only and such that the rectangle AC, CB which they contain is medial. [X. 75]

10.prop.2.81.p4224

I say that no other medial straight line can be annexed to AB which is commensurable with the whole in square only and which contains with the whole a medial rectangle.

10.prop.2.81.p4225

For, if possible, let BD also be so annexed; therefore AD, DB are also medial straight lines commensurable in square only and such that the rectangle AD, DB which they contain is medial. [X. 75]

10.prop.2.81.p4226

Let a rational straight line EF be set out, let EG equal to the squares on AC, CB be applied to EF, producing EM as breadth, and let HG equal to twice the rectangle AC, CB be subtracted, producing HM as breadth; therefore the remainder EL is equal to the square on AB, [II. 7] so that AB is the side of EL.

10.prop.2.81.p4227

Again, let EI equal to the squares on AD, DB be applied to EF, producing EN as breadth.

10.prop.2.81.p4228

But EL is also equal to the square on AB; therefore the remainder HI is equal to twice the rectangle AD, DB. [II. 7]

10.prop.2.81.p4229

Now, since AC, CB are medial straight lines, therefore the squares on AC, CB are also medial.

10.prop.2.81.p4230

And they are equal to EG; therefore EG is also medial. [X. 15 and 23, Por.]

10.prop.2.81.p4231

And it is applied to the rational straight line EF, producing EM as breadth; therefore EM is rational and incommensurable in length with EF. [X. 22]

10.prop.2.81.p4232

Again, since the rectangle AC, CB is medial, twice the rectangle AC, CB is also medial. [X. 23, Por.]

10.prop.2.81.p4233

And it is equal to HG; therefore HG is also medial.

10.prop.2.81.p4234

And it is applied to the rational straight line EF, producing HM as breadth; therefore HM is also rational and incommensurable in length with EF. [X. 22]

10.prop.2.81.p4235

And, since AC, CB are commensurable in square only, therefore AC is incommensurable in length with CB.

10.prop.2.81.p4236

But, as AC is to CB, so is the square on AC to the rectangle AC, CB; therefore the square on AC is incommensurable with the rectangle AC, CB. [X. 11]

10.prop.2.81.p4237

But the squares on AC, CB are commensurable with the square on AC, while twice the rectangle AC, CB is commensurable with the rectangle AC, CB; [X. 6] therefore the squares on AC, CB are incommensurable with twice the rectangle AC, CB. [X. 13]

10.prop.2.81.p4238

And EG is equal to the squares on AC, CB, while GH is equal to twice the rectangle AC, CB; therefore EG is incommensurable with HG.

10.prop.2.81.p4239

But, as EG is to HG, so is EM to HM; [VI. 1] therefore EM is incommensurable in length with MH. [X. 11]

10.prop.2.81.p4240

And both are rational; therefore EM, MH are rational straight lines commensurable in square only; therefore EH is an apotome, and HM an annex to it. [X. 73]

10.prop.2.81.p4241

Similarly we can prove that HN is also an annex to it; therefore to an apotome different straight lines are annexed which are commensurable with the wholes in square only: which is impossible. [X. 79]

10.prop.2.81.p4242

Therefore etc. Q. E. D.

10.prop.2.82.p4243

To a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes, with the whole, the sum of the squares on them rational but twice the rectangle contained by them medial.

10.prop.2.82.p4244

Let AB be the minor straight line, and let BC be an annex to AB; therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial. [X. 76]

10.prop.2.82.p4245

I say that no other straight line can be annexed to AB fulfilling the same conditions.

10.prop.2.82.p4246

For, if possible, let BD be so annexed; therefore AD, DB are also straight lines incommensurable in square which fulfil the aforesaid conditions. [X. 76]

10.prop.2.82.p4247

Now, since the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB, while the squares on AD, DB exceed the squares on AC, CB by a rational area, for both are rational, therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area: which is impossible, for both are medial. [X. 26]

10.prop.2.82.p4248

Therefore to a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes the squares on them added together rational, but twice the rectangle contained by them medial. Q. E. D.

10.prop.2.83.p4249

To a straight line which produces with a rational area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial, but twice the rectangle contained by them rational.

10.prop.2.83.p4250

Let AB be the straight line which produces with a rational area a medial whole, and let BC be an annex to AB; therefore AC, CB are straight lines incommensurable in square which fulfil the given conditions. [X. 77]

10.prop.2.83.p4251

I say that no other straight line can be annexed to AB which fulfils the same conditions.

10.prop.2.83.p4252

For, if possible, let BD be so annexed; therefore AD, DB are also straight lines incommensurable in square which fulfil the given conditions. [X. 77]

10.prop.2.83.p4253

Since then, as in the preceding cases, the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB, while twice the rectangle AD, DB exceeds twice the rectangle AC, CB by a rational area, for both are rational, therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area: which is impossible, for both are medial. [X. 26]

10.prop.2.83.p4254

Therefore no other straight line can be annexed to AB which is incommensurable in square with the whole and which with the whole fulfils the aforesaid conditions; therefore only one straight line can be so annexed. Q. E. D.

10.prop.2.84.p4255

To a straight line which produces with a medial area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial and twice the rectangle contained by them both medial and also incommensurable with the sum of the squares on them.

10.prop.2.84.p4256

Let AB be the straight line which produces with a medial area a medial whole, and BC an annex to it; therefore AC, CB are straight lines incommensurable in square which fulfil the aforesaid conditions. [X. 78]

10.prop.2.84.p4257

I say that no other straight line can be annexed to AB which fulfils the aforesaid conditions.

10.prop.2.84.p4258

For, if possible, let BD be so annexed, so that AD, DB are also straight lines incommensurable in square which make the squares on AD, DB added together medial, twice the rectangle AD, DB medial, and also the squares on AD, DB incommensurable with twice the rectangle AD, DB. [X. 78]

10.prop.2.84.p4259

Let a rational straight line EF be set out, let EG equal to the squares on AC, CB be applied to EF, producing EM as breadth, and let HG equal to twice the rectangle AC, CB be applied to EF, producing HM as breadth; therefore the remainder, the square on AB [II. 7], is equal to EL; therefore AB is the side of EL.

10.prop.2.84.p4260

Again, let EI equal to the squares on AD, DB be applied to EF, producing EN as breadth.

10.prop.2.84.p4261

But the square on AB is also equal to EL; therefore the remainder, twice the rectangle AD, DB [II. 7], is equal to HI.

10.prop.2.84.p4262

Now, since the sum of the squares on AC, CB is medial and is equal to EG, therefore EG is also medial.

10.prop.2.84.p4263

And it is applied to the rational straight line EF, producing EM as breadth; therefore EM is rational and incommensurable in length with EF. [X. 22]

10.prop.2.84.p4264

Again, since twice the rectangle AC, CB is medial and is equal to HG, therefore HG is also medial.

10.prop.2.84.p4265

And it is applied to the rational straight line EF, producing HM as breadth; therefore HM is rational and incommensurable in length with EF. [X. 22]

10.prop.2.84.p4266

And, since the squares on AC, CB are incommensurable with twice the rectangle AC, CB, EG is also incommensurable with HG; therefore EM is also incommensurable in length with MH. [VI. 1, X. 11]

10.prop.2.84.p4267

And both are rational; therefore EM, MH are rational straight lines commensurable in square only; therefore EH is an apotome, and HM an annex to it. [X. 73]

10.prop.2.84.p4268

Similarly we can prove that EH is again an apotome and HN an annex to it.

10.prop.2.84.p4269

Therefore to an apotome different rational straight lines are annexed which are commensurable with the wholes in square only: which was proved impossible. [X. 79]

10.prop.2.84.p4270

Therefore no other straight line can be so annexed to AB.

10.prop.2.84.p4271

Therefore to AB only one straight line can be annexed which is incommensurable in square with the whole and which with the whole makes the squares on them added together medial, twice the rectangle contained by them medial, and also the squares on them incommensurable with twice the rectangle contained by them. Q. E. D.

10.def.3.1.p4272

Given a rational straight line and an apotome, if the square on the whole be greater than the square on the annex by the square on a straight line commensurable in length with the whole, and the whole be commensurable in length with the rational straight line set out, let the apotome be called a first apotome.

10.def.3.2.p4273

But if the annex be commensurable in length with the rational straight line set out, and the square on the whole be greater than that on the annex by the square on a straight line commensurable with the whole, let the apotome be called a second apotome.

10.def.3.3.p4274

But if neither be commensurable in length with the rational straight line set out, and the square on the whole be greater than the square on the annex by the square on a straight line commensurable with the whole, let the apotome be called a third apotome.

10.def.3.4.p4275

Again, if the square on the whole be greater than the square on the annex by the square on a straight line incommensurable with the whole, then, if the whole be commensurable in length with the rational straight line set out, let the apotome be called a fourth apotome;

10.def.3.5.p4276

if the annex be so commensurable, a fifth;

10.def.3.6.p4277

and, if neither, a sixth.

10.prop.3.85.p4278

To find the first apotome.

10.prop.3.85.p4279

Let a rational straight line A be set out, and let BG be commensurable in length with A; therefore BG is also rational.

10.prop.3.85.p4280

Let two square numbers DE, EF be set out, and let their difference FD not be square; therefore neither has ED to DF the ratio which a square number has to a square number.

10.prop.3.85.p4281

Let it be contrived that, as ED is to DF, so is the square on BG to the square on GC; [X. 6, Por.] therefore the square on BG is commensurable with the square on GC. [X. 6]

10.prop.3.85.p4282

But the square on BG is rational; therefore the square on GC is also rational; therefore GC is also rational.

10.prop.3.85.p4283

And, since ED has not to DF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9]

10.prop.3.85.p4284

And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73]

10.prop.3.85.p4285

I say next that it is also a first apotome.

10.prop.3.85.p4286

For let the square on H be that by which the square on BG is greater than the square on GC.

10.prop.3.85.p4287

Now since. as ED is to FD, so is the square on BG to the square on GC, therefore also, convertendo, [v. 19, Por.] as DE is to EF, so is the square on GB to the square on H.

10.prop.3.85.p4288

But DE has to EF the ratio which a square number has to a square number, for each is square; therefore the square on GB also has to the square on H the ratio which a square number has to a square number; therefore BG is commensurable in length with H. [X. 9]

10.prop.3.85.p4289

And the square on BG is greater than the square on GC by the square on a straight line commensurable in length with BG.

10.prop.3.85.p4290

And the whole BG is commensurable in length with the rational straight line A set out.

10.prop.3.85.p4291

Therefore BC is a first apotome. [X. Deff. III. 1]

10.prop.3.85.p4292

Therefore the first apotome BC has been found. (Being) that which it was required to find.

10.prop.3.86.p4293

To find the second apotome.

10.prop.3.86.p4294

Let a rational straight line A be set out, and GC commensurable in length with A; therefore GC is rational.

10.prop.3.86.p4295

Let two square numbers DE, EF be set out, and let their difference DF not be square.

10.prop.3.86.p4296

Now let it be contrived that, as FD is to DE, so is the square on CG to the square on GB. [X. 6, Por.]

10.prop.3.86.p4297

Therefore the square on CG is commensurable with the square on GB. [X. 6]

10.prop.3.86.p4298

But the square on CG is rational; therefore the square on GB is also rational; therefore BG is rational.

10.prop.3.86.p4299

And, since the square on GC has not to the square on GB the ratio which a square number has to a square number, CG is incommensurable in length with GB. [X. 9]

10.prop.3.86.p4300

And both are rational; therefore CG, GB are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73]

10.prop.3.86.p4301

I say next that it is also a second apotome.

10.prop.3.86.p4302

For let the square on H be that by which the square on BG is greater than the square on GC.

10.prop.3.86.p4303

Since then, as the square on BG is to the square on GC, so is the number ED to the number DF, therefore, convertendo, as the square on BG is to the square on H, so is DE to EF. [V. 19, Por.]

10.prop.3.86.p4304

And each of the numbers DE, EF is square; therefore the square on BG has to the square on H the ratio which a square number has to a square number; therefore BG is commensurable in length with H. [X. 9]

10.prop.3.86.p4305

And the square on BG is greater than the square on GC by the square on H; therefore the square on BG is greater than the square on GC by the square on a straight line commensurable in length with BG.

10.prop.3.86.p4306

And CG, the annex, is commensurable with the rational straight line A set out.

10.prop.3.86.p4307

Therefore BC is a second apotome. [X. Deff. III. 2]

10.prop.3.86.p4308

Therefore the second apotome BC has been found. Q. E. D.

10.prop.3.87.p4309

To find the third apotome.

10.prop.3.87.p4310

Let a rational straight line A be set out, let three numbers E, BC, CD be set out which have not to one another the ratio which a square number has to a square number, but let CB have to BD the ratio which a square number has to a square number.

10.prop.3.87.p4311

Let it be contrived that, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH. [X. 6, Por.]

10.prop.3.87.p4312

Since then, as E is to BC, so is the square on A to the square on FG, therefore the square on A is commensurable with the square on FG. [X. 6]

10.prop.3.87.p4313

But the square on A is rational; therefore the square on FG is also rational; therefore FG is rational.

10.prop.3.87.p4314

And, since E has not to BC the ratio which a square number has to a square number, therefore neither has the square on A to the square on FG the ratio which a square number has to a square number; therefore A is incommensurable in length with FG. [X. 9]

10.prop.3.87.p4315

Again, since, as BC is to CD, so is the square on FG to the square on GH, therefore the square on FG is commensurable with the square on GH. [X. 6]

10.prop.3.87.p4316

But the square on FG is rational; therefore the square on GH is also rational; therefore GH is rational.

10.prop.3.87.p4317

And, since BC has not to CD the ratio which a square number has to a square number, therefore neither has the square on FG to the square on GH the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9]

10.prop.3.87.p4318

And both are rational; therefore FG, GH are rational straight lines commensurable in square only; therefore FH is an apotome. [X. 73]

10.prop.3.87.p4319

I say next that it is also a third apotome.

10.prop.3.87.p4320

For since, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on HG, therefore, ex aequali, as E is to CD, so is the square on A to the square on HG. [V. 22]

10.prop.3.87.p4321

But E has not to CD the ratio which a square number has to a square number; therefore neither has the square on A to the square on GH the ratio which a square number has to a square number; therefore A is incommensurable in length with GH. [X. 9]

10.prop.3.87.p4322

Therefore neither of the straight lines FG, GH is commensurable in length with the rational straight line A set out.

10.prop.3.87.p4323

Now let the square on K be that by which the square on FG is greater than the square on GH.

10.prop.3.87.p4324

Since then, as BC is to CD, so is the square on FG to the square on GH, therefore, convertendo, as BC is to BD, so is the square on FG to the square on K. [V. 19, Por.]

10.prop.3.87.p4325

But BC has to BD the ratio which a square number has to a square number; therefore the square on FG also has to the square on K the ratio which a square number has to a square number.

10.prop.3.87.p4326

Therefore FG is commensurable in length with K, [X. 9] and the square on FG is greater than the square on GH by the square on a straight line commensurable with FG.

10.prop.3.87.p4327

And neither of the straight lines FG, GH is commensurable in length with the rational straight line A set out; therefore FH is a third apotome. [X. Deff. III. 3]

10.prop.3.87.p4328

Therefore the third apotome FH has been found. Q. E. D.

10.prop.3.88.p4329

To find the fourth apotome.

10.prop.3.88.p4330

Let a rational straight line A be set out, and BG commensurable in length with it; therefore BG is also rational.

10.prop.3.88.p4331

Let two numbers DF, FE be set out such that the whole DE has not to either of the numbers DF, EF the ratio which a square number has to a square number.

10.prop.3.88.p4332

Let it be contrived that, as DE is to EF, so is the square on BG to the square on GC; [X. 6, Por.] therefore the square on BG is commensurable with the square on GC. [X. 6]

10.prop.3.88.p4333

But the square on BG is rational; therefore the square on GC is also rational; therefore GC is rational.

10.prop.3.88.p4334

Now, since DE has not to EF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9]

10.prop.3.88.p4335

And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73]

10.prop.3.88.p4336

Now let the square on H be that by which the square on BG is greater than the square on GC.

10.prop.3.88.p4337

Since then, as DE is to EF, so is the square on BG to the square on GC, therefore also, convertendo, as ED is to DF, so is the square on GB to the square on H. [v. 19, Por.]

10.prop.3.88.p4338

But ED has not to DF the ratio which a square number has to a square number; therefore neither has the square on GB to the square on H the ratio which a square number has to a square number; therefore BG is incommensurable in length with H. [X. 9]

10.prop.3.88.p4339

And the square on BG is greater than the square on GC by the square on H; therefore the square on BG is greater than the square on GC by the square on a straight line incommensurable with BG.

10.prop.3.88.p4340

And the whole BG is commensurable in length with the rational straight line A set out.

10.prop.3.88.p4341

Therefore BC is a fourth apotome. [X. Deff. III. 4]

10.prop.3.88.p4342

Therefore the fourth apotome has been found. Q. E. D.

10.prop.3.89.p4343

To find the fifth apotome.

10.prop.3.89.p4344

Let a rational straight line A be set out, and let CG be commensurable in length with A; therefore CG is rational.

10.prop.3.89.p4345

Let two numbers DF, FE be set out such that DE again has not to either of the numbers DF, FE the ratio which a square number has to a square number; and let it be contrived that, as FE is to ED, so is the square on CG to the square on GB.

10.prop.3.89.p4346

Therefore the square on GB is also rational; [X. 6] therefore BG is also rational.

10.prop.3.89.p4347

Now since, as DE is to EF, so is the square on BG to the square on GC, while DE has not to EF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9]

10.prop.3.89.p4348

And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73]

10.prop.3.89.p4349

I say next that it is also a fifth apotome.

10.prop.3.89.p4350

For let the square on H be that by which the square on BG is greater than the square on GC.

10.prop.3.89.p4351

Since then, as the square on BG is to the square on GC, so is DE to EF, therefore, convertendo, as ED is to DF, so is the square on BG to the square on H. [V. 19, Por.]

10.prop.3.89.p4352

But ED has not to DF the ratio which a square number has to a square number; therefore neither has the square on BG to the square on H the ratio which a square number has to a square number; therefore BG is incommensurable in length with H. [X. 9]

10.prop.3.89.p4353

And the square on BG is greater than the square on GC by the square on H; therefore the square on GB is greater than the square on GC by the square on a straight line incommensurable in length with GB.

10.prop.3.89.p4354

And the annex CG is commensurable in length with the rational straight line A set out; therefore BC is a fifth apotome. [X. Deff. III. 5]

10.prop.3.89.p4355

Therefore the fifth apotome BC has been found. Q. E. D.

10.prop.3.90.p4356

To find the sixth apotome.

10.prop.3.90.p4357

Let a rational straight line A be set out, and three numbers E, BC, CD not having to one another the ratio which a square number has to a square number; and further let CB also not have to BD the ratio which a square number has to a square number.

10.prop.3.90.p4358

Let it be contrived that, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH. [X. 6, Por.]

10.prop.3.90.p4359

Now since, as E is to BC, so is the square on A to the square on FG, therefore the square on A is commensurable with the square on FG. [X. 6]

10.prop.3.90.p4360

But the square on A is rational; therefore the square on FG is also rational; therefore FG is also rational.

10.prop.3.90.p4361

And, since E has not to BC the ratio which a square number has to a square number, therefore neither has the square on A to the square on FG the ratio which a square number has to a square number; therefore A is incommensurable in length with FG. [X. 9]

10.prop.3.90.p4362

Again, since, as BC is to CD, so is the square on FG to the square on GH, therefore the square on FG is commensurable with the square on GH. [X. 6]

10.prop.3.90.p4363

But the square on FG is rational; therefore the square on GH is also rational; therefore GH is also rational.

10.prop.3.90.p4364

And, since BC has not to CD the ratio which a square number has to a square number, therefore neither has the square on FG to the square on GH the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9]

10.prop.3.90.p4365

And both are rational; therefore FG, GH are rational straight lines commensurable in square only; therefore FH is an apotome. [X. 73]

10.prop.3.90.p4366

I say next that it is also a sixth apotome.

10.prop.3.90.p4367

For since, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH, therefore, ex aequali, as E is to CD, so is the square on A to the square on GH. [v. 22]

10.prop.3.90.p4368

But E has not to CD the ratio which a square number has to a square number; therefore neither has the square on A to the square on GH the ratio which a square number has to a square number; therefore A is incommensurable in length with GH; [X. 9] therefore neither of the straight lines FG, GH is commensurable in length with the rational straight line A.

10.prop.3.90.p4369

Now let the square on K be that by which the square on FG is greater than the square on GH.

10.prop.3.90.p4370

Since then, as BC is to CD, so is the square on FG to the square on GH, therefore, convertendo, as CB is to BD, so is the square on FG to the square on K. [v. 19, Por.]

10.prop.3.90.p4371

But CB has not to BD the ratio which a square number has to a square number; therefore neither has the square on FG to the square on K the ratio which a square number has to a square number; therefore FG is incommensurable in length with K. [X. 9]

10.prop.3.90.p4372

And the square on FG is greater than the square on GH by the square on K; therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable in length with FG.

10.prop.3.90.p4373

And neither of the straight lines FG, GH is commensurable with the rational straight line A set out.

10.prop.3.90.p4374

Therefore FH is a sixth apotome. [X. Deff. III. 6]

10.prop.3.90.p4375

Therefore the sixth apotome FH has been found. Q. E. D.

10.prop.3.91.p4376

If an area be contained by a rational straight line and a first apotome, the side of the area is an apotome.

10.prop.3.91.p4377

For let the area AB be contained by the rational straight line AC and the first apotome AD;

10.prop.3.91.p4378

I say that the side of the area AB is an apotome.

10.prop.3.91.p4379

For, since AD is a first apotome, let DG be its annex; therefore AG, GD are rational straight lines commensurable in square only. [X. 73]

10.prop.3.91.p4380

And the whole AG is commensurable with the rational straight line AC set out, and the square on AG is greater than the square on GD by the square on a straight line commensurable in length with AG; [X. Deff. III. 1] if therefore there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it divides it into commensurable parts. [X. 17]

10.prop.3.91.p4381

Let DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is commensurable with FG.

10.prop.3.91.p4382

And through the points E, F, G let EH, FI, GK be drawn parallel to AC.

10.prop.3.91.p4383

Now, since AF is commensurable in length with FG, therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]

10.prop.3.91.p4384

But AG is commensurable with AC; therefore each of the straight lines AF, FG is commensurable in length with AC. [X. 12]

10.prop.3.91.p4385

And AC is rational; therefore each of the straight lines AF, FG is also rational, so that each of the rectangles AI, FK is also rational. [X. 19]

10.prop.3.91.p4386

Now, since DE is commensurable in length with EG, therefore DG is also commensurable in length with each of the straight lines DE, EG. [X. 15]

10.prop.3.91.p4387

But DG is rational and incommensurable in length with AC; therefore each of the straight lines DE, EG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles DH, EK is medial. [X. 21]

10.prop.3.91.p4388

Now let the square LM be made equal to AI, and let there be subtracted the square NO having a common angle with it, the angle LPM, and equal to FK; therefore the squares LM, NO are about the same diameter. [VI. 26]

10.prop.3.91.p4389

Let PR be their diameter, and let the figure be drawn.

10.prop.3.91.p4390

Since then the rectangle contained by AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG. [VI. 17]

10.prop.3.91.p4391

But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to KF; [VI. 1] therefore EK is a mean proportional between AI, KF. [V. 11]

10.prop.3.91.p4392

But MN is also a mean proportional between LM, NO, as was before proved, [Lemma after X. 53] and AI is equal to the square LM, and KF to NO; therefore MN is also equal to EK.

10.prop.3.91.p4393

But EK is equal to DH, and MN to LO; therefore DK is equal to the gnomon UVW and NO.

10.prop.3.91.p4394

But AK is also equal to the squares LM, NO; therefore the remainder AB is equal to ST.

10.prop.3.91.p4395

But ST is the square on LN; therefore the square on LN is equal to AB; therefore LN is the side of AB.

10.prop.3.91.p4396

I say next that LN is an apotome.

10.prop.3.91.p4397

For, since each of the rectangles AI, FK is rational, and they are equal to LM, NO, therefore each of the squares LM, NO, that is, the squares on LP, PN respectively, is also rational; therefore each of the straight lines LP, PN is also rational.

10.prop.3.91.p4398

Again, since DH is medial and is equal to LO, therefore LO is also medial.

10.prop.3.91.p4399

Since then LO is medial, while NO is rational, therefore LO is incommensurable with NO.

10.prop.3.91.p4400

But, as LO is to NO, so is LP to PN; [VI. 1] therefore LP is incommensurable in length with PN. [X. 11]

10.prop.3.91.p4401

And both are rational; therefore LP, PN are rational straight lines commensurable in square only; therefore LN is an apotome. [X. 73]

10.prop.3.91.p4402

And it is the side of the area AB; therefore the side of the area AB is an apotome.

10.prop.3.91.p4403

Therefore etc.

10.prop.3.92.p4404

If an area be contained by a rational straight line and a second apotome, the side of the area is a first apotome of a medial straight line.

10.prop.3.92.p4405

For let the area AB be contained by the rational straight line AC and the second apotome AD; I say that the side of the area AB is a first apotome of a medial straight line.

10.prop.3.92.p4406

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, [X. 73] and the annex DG is commensurable with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex GD by the square on a straight line commensurable in length with AG. [X. Deff. III. 2]

10.prop.3.92.p4407

Since then the square on AG is greater than the square on GD by the square on a straight line commensurable with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on GD and deficient by a square figure, it divides it into commensurable parts. [X. 17]

10.prop.3.92.p4408

Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is commensurable in length with FG.

10.prop.3.92.p4409

Therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]

10.prop.3.92.p4410

But AG is rational and incommensurable in length with AC; therefore each of the straight lines AF, FG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles AI, FK is medial. [X. 21]

10.prop.3.92.p4411

Again, since DE is commensurable with EG, therefore DG is also commensurable with each of the straight lines DE, EG. [X. 15]

10.prop.3.92.p4412

But DG is commensurable in length with AC.

10.prop.3.92.p4413

Therefore each of the rectangles DH, EK is rational. [X. 19]

10.prop.3.92.p4414

Let then the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and being about the same angle with LM, namely the angle LPM; therefore the squares LM, NO are about the same diameter. [VI. 26]

10.prop.3.92.p4415

Let PR be their diameter, and let the figure be drawn.

10.prop.3.92.p4416

Since then AI, FK are medial and are equal to the squares on LP, PN, the squares on LP, PN are also medial; therefore LP, PN are also medial straight lines commensurable in square only.

10.prop.3.92.p4417

And, since the rectangle AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG, [VI. 17] while, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore EK is a mean proportional between AI, FK. [V. 11]

10.prop.3.92.p4418

But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore MN is also equal to EK.

10.prop.3.92.p4419

But DH is equal to EK, and LO equal to MN; therefore the whole DK is equal to the gnomon UVW and NO.

10.prop.3.92.p4420

Since then the whole AK is equal to LM, NO, and, in these, DK is equal to the gnomon UVW and NO, therefore the remainder AB is equal to TS.

10.prop.3.92.p4421

But TS is the square on LN; therefore the square on LN is equal to the area AB; therefore LN is the side of the area AB.

10.prop.3.92.p4422

I say that LN is a first apotome of a medial straight line.

10.prop.3.92.p4423

For, since EK is rational and is equal to LO, therefore LO, that is, the rectangle LP, PN, is rational.

10.prop.3.92.p4424

But NO was proved medial; therefore LO is incommensurable with NO.

10.prop.3.92.p4425

But, as LO is to NO, so is LP to PN; [VI. 1] therefore LP, PN are incommensurable in length. [X. 11]

10.prop.3.92.p4426

Therefore LP, PN are medial straight lines commensurable in square only which contain a rational rectangle; therefore LN is a first apotome of a medial straight line. [X. 74]

10.prop.3.92.p4427

And it is the side of the area AB.

10.prop.3.92.p4428

Therefore the side of the area AB is a first apotome of a medial straight line. Q. E. D.

10.prop.3.93.p4429

If an area be contained by a rational straight line and a third apotome, the side of the area is a second apotome of a medial straight line.

10.prop.3.93.p4430

For let the area AB be contained by the rational straight line AC and the third apotome AD; I say that the side of the area AB is a second apotome of a medial straight line.

10.prop.3.93.p4431

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, and neither of the straight lines AG, GD is commensurable in length with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex DG by the square on a straight line commensurable with AG. [X. Deff. III. 3]

10.prop.3.93.p4432

Since then the square on AG is greater than the square on GD by the square on a straight line commensurable with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into commensurable parts. [X. 17]

10.prop.3.93.p4433

Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG.

10.prop.3.93.p4434

Let EH, FI, GK be drawn through the points E, F, G parallel to AC.

10.prop.3.93.p4435

Therefore AF, FG are commensurable; therefore AI is also commensurable with FK. [VI. 1, X. 11]

10.prop.3.93.p4436

And, since AF, FG are commensurable in length, therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]

10.prop.3.93.p4437

But AG is rational and incommensurable in length with AC; so that AF, FG are so also. [X. 13]

10.prop.3.93.p4438

Therefore each of the rectangles AI, FK is medial. [X. 21]

10.prop.3.93.p4439

Again, since DE is commensurable in length with EG, therefore DG is also commensurable in length with each of the straight lines DE, EG. [X. 15]

10.prop.3.93.p4440

But GD is rational and incommensurable in length with AC; therefore each of the straight lines DE, EG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles DH, EK is medial. [X. 21]

10.prop.3.93.p4441

And, since AG, GD are commensurable in square only, therefore AG is incommensurable in length with GD.

10.prop.3.93.p4442

But AG is commensurable in length with AF, and DG with EG; therefore AF is incommensurable in length with EG. [X. 13]

10.prop.3.93.p4443

But, as AF is to EG, so is AI to EK; [VI. 1] therefore AI is incommensurable with EK. [X. 11]

10.prop.3.93.p4444

Now let the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and being about the same angle with LM; therefore LM, NO are about the same diameter. [VI. 26]

10.prop.3.93.p4445

Let PR be their diameter, and let the figure be drawn.

10.prop.3.93.p4446

Now, since the rectangle AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG. [VI. 17]

10.prop.3.93.p4447

But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore also, as AI is to EK, so is EK to FK; [V. 11] therefore EK is a mean proportional between AI, FK.

10.prop.3.93.p4448

But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore EK is also equal to MN.

10.prop.3.93.p4449

But MN is equal to LO, and EK equal to DH; therefore the whole DK is also equal to the gnomon UVW and NO.

10.prop.3.93.p4450

But AK is also equal to LM, NO; therefore the remainder AB is equal to ST, that is, to the square on LN; therefore LN is the side of the area AB.

10.prop.3.93.p4451

I say that LN is a second apotome of a medial straight line.

10.prop.3.93.p4452

For, since AI, FK were proved medial, and are equal to the squares on LP, PN, therefore each of the squares on LP, PN is also medial; therefore each of the straight lines LP, PN is medial.

10.prop.3.93.p4453

And, since AI is commensurable with FK, [VI. 1, X. 11] therefore the square on LP is also commensurable with the square on PN.

10.prop.3.93.p4454

Again, since AI was proved incommensurable with EK, therefore LM is also incommensurable with MN, that is, the square on LP with the rectangle LP, PN; so that LP is also incommensurable in length with PN; [VI. 1, X. 11] therefore LP, PN are medial straight lines commensurable in square only.

10.prop.3.93.p4455

I say next that they also contain a medial rectangle.

10.prop.3.93.p4456

For, since EK was proved medial, and is equal to the rectangle LP, PN, therefore the rectangle LP, PN is also medial, so that LP, PN are medial straight lines commensurable in square only which contain a medial rectangle.

10.prop.3.93.p4457

Therefore LN is a second apotome of a medial straight line; [X. 75] and it is the side of the area AB.

10.prop.3.93.p4458

Therefore the side of the area AB is a second apotome of a medial straight line. Q. E. D.

10.prop.3.94.p4459

If an area be contained by a rational straight line and a fourth apotome, the side of the area is minor.

10.prop.3.94.p4460

For let the area AB be contained by the rational straight line AC and the fourth apotome AD; I say that the side of the area AB is minor.

10.prop.3.94.p4461

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, AG is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG, [X. Deff. III. 4]

10.prop.3.94.p4462

Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18]

10.prop.3.94.p4463

Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG.

10.prop.3.94.p4464

Let EH, FI, GK be drawn through E, F, G parallel to AC, BD.

10.prop.3.94.p4465

Since then AG is rational and commensurable in length with AC, therefore the whole AK is rational. [X. 19]

10.prop.3.94.p4466

Again, since DG is incommensurable in length with AC, and both are rational, therefore DK is medial. [X. 21]

10.prop.3.94.p4467

Again, since AF is incommensurable in length with FG, therefore AI is also incommensurable with FK. [VI. 1, X. 11]

10.prop.3.94.p4468

Now let the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and about the same angle, the angle LPM.

10.prop.3.94.p4469

Therefore the squares LM, NO are about the same diameter. [VI. 26]

10.prop.3.94.p4470

Let PR be their diameter, and let the figure be drawn.

10.prop.3.94.p4471

Since then the rectangle AF, FG is equal to the square on EG, therefore, proportionally, as AF is to EG, so is EG to FG. [VI. 17]

10.prop.3.94.p4472

But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore EK is a mean proportional between AI, FK. [V. 11]

10.prop.3.94.p4473

But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore EK is also equal to MN.

10.prop.3.94.p4474

But DH is equal to EK, and LO is equal to MN; therefore the whole DK is equal to the gnomon UVW and NO.

10.prop.3.94.p4475

Since, then, the whole AK is equal to the squares LM, NO, and, in these, DK is equal to the gnomon UVW and the square NO, therefore the remainder AB is equal to ST, that is, to the square on LN; therefore LN is the side of the area AB.

10.prop.3.94.p4476

I say that LN is the irrational straight line called minor.

10.prop.3.94.p4477

For, since AK is rational and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is rational.

10.prop.3.94.p4478

Again, since DK is medial, and DK is equal to twice the rectangle LP, PN, therefore twice the rectangle LP, PN is medial.

10.prop.3.94.p4479

And, since AI was proved incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN.

10.prop.3.94.p4480

Therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial.

10.prop.3.94.p4481

Therefore LN is the irrational straight line called minor; [X. 76] and it is the side of the area AB.

10.prop.3.94.p4482

Therefore the side of the area AB is minor. Q. E. D.

10.prop.3.95.p4483

If an area be contained by a rational straight line and a fifth apotome, the side of the area is a straight line which produces with a rational area a medial whole.

10.prop.3.95.p4484

For let the area AB be contained by the rational straight line AC and the fifth apotome AD; I say that the side of the area AB is a straight line which produces with a rational area a medial whole.

10.prop.3.95.p4485

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, the annex GD is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable with AG. [X. Deff. III. 5]

10.prop.3.95.p4486

Therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18]

10.prop.3.95.p4487

Let then DG be bisected at the point E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG.

10.prop.3.95.p4488

Now, since AG is incommensurable in length with CA, and both are rational, therefore AK is medial. [X. 21]

10.prop.3.95.p4489

Again, since DG is rational and commensurable in length with AC, DK is rational. [X. 19]

10.prop.3.95.p4490

Now let the square LM be constructed equal to AI, and let the square NO equal to FK and about the same angle, the angle LPM, be subtracted; therefore the squares LM, NO are about the same diameter. [VI. 26]

10.prop.3.95.p4491

Let PR be their diameter, and let the figure be drawn.

10.prop.3.95.p4492

Similarly then we can prove that LN is the side of the area AB.

10.prop.3.95.p4493

I say that LN is the straight line which produces with a rational area a medial whole.

10.prop.3.95.p4494

For, since AK was proved medial and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is medial.

10.prop.3.95.p4495

Again, since DK is rational and is equal to twice the rectangle LP, PN, the latter is itself also rational.

10.prop.3.95.p4496

And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN; therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational.

10.prop.3.95.p4497

Therefore the remainder LN is the irrational straight line called that which produces with a rational area a medial whole; [X. 77] and it is the side of the area AB.

10.prop.3.95.p4498

Therefore the side of the area AB is a straight line which produces with a rational area a medial whole. Q. E. D.

10.prop.3.96.p4499

If an area be contained by a rational straight line and a sixth apotome, the side of the area is a straight line which produces with a medial area a medial whole.

10.prop.3.96.p4500

For let the area AB be contained by the rational straight line AC and the sixth apotome AD; I say that the side of the area AB is a straight line which produces with a medial area a medial whole.

10.prop.3.96.p4501

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, neither of them is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG. [X. Deff. III. 6]

10.prop.3.96.p4502

Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18]

10.prop.3.96.p4503

Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG.

10.prop.3.96.p4504

But, as AF is to FG, so is AI to FK. [VI. 1] therefore AI is incommensurable with FK. [X. 11]

10.prop.3.96.p4505

And, since AG, AC are rational straight lines commensurable in square only, AK is medial. [X. 21]

10.prop.3.96.p4506

Again, since AC, DG are rational straight lines and incommensurable in length, DK is also medial. [X. 21]

10.prop.3.96.p4507

Now, since AG, GD are commensurable in square only, therefore AG is incommensurable in length with GD.

10.prop.3.96.p4508

But, as AG is to GD, so is AK to KD; [VI. 1] therefore AK is incommensurable with KD. [X. 11]

10.prop.3.96.p4509

Now let the square LM be constructed equal to AI, and let NO equal to FK, and about the same angle, be subtracted; therefore the squares LM, NO are about the same diameter. [VI. 26]

10.prop.3.96.p4510

Let PR be their diameter, and let the figure be drawn.

10.prop.3.96.p4511

Then in manner similar to the above we can prove that LN is the side of the area AB.

10.prop.3.96.p4512

I say that LN is a straight line which produces with a medial area a medial whole.

10.prop.3.96.p4513

For, since AK was proved medial and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is medial.

10.prop.3.96.p4514

Again, since DK was proved medial and is equal to twice the rectangle LP, PN, twice the rectangle LP, PN is also medial.

10.prop.3.96.p4515

And, since AK was proved incommensurable with DK, the squares on LP, PN are also incommensurable with twice the rectangle LP, PN.

10.prop.3.96.p4516

And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN; therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them.

10.prop.3.96.p4517

Therefore LN is the irrational straight line called that which produces with a medial area a medial whole; [X. 78] and it is the side of the area AB.

10.prop.3.96.p4518

Therefore the side of the area is a straight line which produces with a medial area a medial whole. Q. E. D.

10.prop.3.97.p4519

The square on an apotome applied to a rational straight line produces as breadth a first apotome.

10.prop.3.97.p4520

Let AB be an apotome, and CD rational, and to CD let there be applied CE equal to the square on AB and producing CF as breadth; I say that CF is a first apotome.

10.prop.3.97.p4521

For let BG be the annex to AB; therefore AG, GB are rational straight lines commensurable in square only. [X. 73]

10.prop.3.97.p4522

To CD let there be applied CH equal to the square on AG, and KL equal to the square on BG.

10.prop.3.97.p4523

Therefore the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB; therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]

10.prop.3.97.p4524

Let FM be bisected at the point N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, LN is equal to the rectangle AG, GB.

10.prop.3.97.p4525

Now, since the squares on AG, GB are rational, and DM is equal to the squares on AG, GB,. therefore DM is rational.

10.prop.3.97.p4526

And it has been applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and commensurable in length with CD. [X. 20]

10.prop.3.97.p4527

Again, since twice the rectangle AG, GB is medial, and FL is equal to twice the rectangle AG, GB, therefore FL is medial.

10.prop.3.97.p4528

And it is applied to the rational straight line CD, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22]

10.prop.3.97.p4529

And, since the squares on AG, GB are rational, while twice the rectangle AG, GB is medial, therefore the squares on AG, GB are incommensurable with twice the rectangle AG, GB.

10.prop.3.97.p4530

And CL is equal to the squares on AG, GB, and FL to twice the rectangle AG, GB; therefore DM is incommensurable with FL.

10.prop.3.97.p4531

But, as DM is to FL, so is CM to FM; [VI. 1] therefore CM is incommensurable in length with FM. [X. 11]

10.prop.3.97.p4532

And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]

10.prop.3.97.p4533

I say next that it is also a first apotome.

10.prop.3.97.p4534

For, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on BG, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

10.prop.3.97.p4535

But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore the rectangle CK, KM is equal to the square on NM [VI. 17], that is, to the fourth part of the square on FM.

10.prop.3.97.p4536

And, since the square on AG is commensurable with the square on GB, CH is also commensurable with KL.

10.prop.3.97.p4537

But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is commensurable with KM. [X. 11]

10.prop.3.97.p4538

Since then CM, MF are two unequal straight lines, and to CM there has been applied the rectangle CK, KM equal to the fourth part of the square on FM and deficient by a square figure, while CK is commensurable with KM, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. [X. 17]

10.prop.3.97.p4539

And CM is commensurable in length with the rational straight line CD set out; therefore CF is a first apotome. [X. Deff. III. 1]

10.prop.3.97.p4540

Therefore etc. Q. E. D.

10.prop.3.98.p4541

The square on a first apotome of a medial straight line applied to a rational straight line produces as breadth a second apotome.

10.prop.3.98.p4542

Let AB be a first apotome of a medial straight line and CD a rational straight line, and to CD let there be applied CE equal to the square on AB, producing CF as breadth; I say that CF is a second apotome.

10.prop.3.98.p4543

For let BG be the annex to AB;. therefore AG, GB are medial straight lines commensurable in square only which contain a rational rectangle. [X. 74]

10.prop.3.98.p4544

To CD let there be applied CH equal to the square on AG, producing CK as breadth, and KL equal to the square on GB, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. [X. 15 and 23, Por.]

10.prop.3.98.p4545

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22]

10.prop.3.98.p4546

Now, since CL is equal to the squares on AG, GB, and, in these, the square on AB is equal to CE, therefore the remainder, twice the rectangle AG, GB, is equal to FL. [II. 7]

10.prop.3.98.p4547

But twice the rectangle AG, GB is rational; therefore FL is rational.

10.prop.3.98.p4548

And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is also rational and commensurable in length with CD. [X. 20]

10.prop.3.98.p4549

Now, since the sum of the squares on AG, GB, that is, CL, is medial, while twice the rectangle AG, GB, that is, FL, is rational, therefore CL is incommensurable with FL.

10.prop.3.98.p4550

But, as CL is to FL, so is CM to FM; [VI. 1] therefore CM is incommensurable in length with FM. [X. 11]

10.prop.3.98.p4551

And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]

10.prop.3.98.p4552

I say next that it is also a second apotome.

10.prop.3.98.p4553

For let FM be bisected at N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.

10.prop.3.98.p4554

Now, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and the square on AG is equal to CH, the rectangle AG, GB to NL, and the square on BG to KL, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

10.prop.3.98.p4555

But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to MK; [VI. 1] therefore, as CK is to NM, so is NM, so is KM; [V. 11] therefore the rectangle CK, KM is equal to the square on NM [VI. 17], that is, to the fourth part of the square on FM.

10.prop.3.98.p4556

Since the CM, MF are two unequal straight lines, and the rectangle CK, KM equal to the fourth part of the square on MF and deficient by a square figure has been applied to the greater, CM, and divides it into commensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. [X. 17]

10.prop.3.98.p4557

And the annex FM is commensurable in length with the rational straight line CD set out; therefore CF is a second apotome. [X. Deff. III. 2]

10.prop.3.98.p4558

Therefore etc. Q. E. D.

10.prop.3.99.p4559

The square on a second apotome of a medial straight line applied to a rational straight line produces as breadth a third apotome.

10.prop.3.99.p4560

Let AB be a second apotome of a medial straight line, and CD rational, and to CD let there be applied CE equal to the square on AB, producing CF as breadth; I say that CF is a third apotome.

10.prop.3.99.p4561

For let BG be the annex to AB; therefore AG, GB are medial straight lines commensurable in square only which contain a medial rectangle. [X. 75]

10.prop.3.99.p4562

Let CH equal to the square on AG be applied to CD, producing CK as breadth, and let KL equal to the square on BG be applied to KH, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. [X. 15 and 23, Por.]

10.prop.3.99.p4563

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22]

10.prop.3.99.p4564

Now, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder LF is equal to twice the rectangle AG, GB. [II. 7]

10.prop.3.99.p4565

Let then FM be bisected at the point N, and let NO be drawn parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.

10.prop.3.99.p4566

But the rectangle AG, GB is medial; therefore FL is also medial.

10.prop.3.99.p4567

And it is applied to the rational straight line EF, producing FM as breadth; therefore FM is also rational and incommensurable in length with CD. [X. 22]

10.prop.3.99.p4568

And, since AG, GB are commensurable in square only, therefore AG is incommensurable in length with GB; therefore the square on AG is also incommensurable with the rectangle AG, GB. [VI. 1, X. 11]

10.prop.3.99.p4569

But the squares on AG, GB are commensurable with the square on AG, and twice the rectangle AG, GB with the rectangle AG, GB; therefore the squares on AG, GB are incommensurable with twice the rectangle AG, GB. [X. 13]

10.prop.3.99.p4570

But CL is equal to the squares on AG, GB, and FL is equal to twice the rectangle AG, GB; therefore CL is also incommensurable with FL.

10.prop.3.99.p4571

But, as CL is to FL, so is CM to FM; [VI. 1] therefore CM is incommensurable in length with FM. [X. 11]

10.prop.3.99.p4572

And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]

10.prop.3.99.p4573

I say next that it is also a third apotome.

10.prop.3.99.p4574

For, since the square on AG is commensurable with the square on GB, therefore CH is also commensurable with KL, so that CK is also commensurable with KM. [VI. 1, X. 11]

10.prop.3.99.p4575

And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

10.prop.3.99.p4576

But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore, as CK is to MN, so is MN to KM; [V. 11] therefore the rectangle CK, KM is equal to [the square on MN, that is, to] the fourth part of the square on FM.

10.prop.3.99.p4577

Since then CM, MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to CM, and divides it into commensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable with CM. [X. 17]

10.prop.3.99.p4578

And neither of the straight lines CM, MF is commensurable in length with the rational straight line CD set out; therefore CF is a third apotome. [X. Deff. III. 3]

10.prop.3.99.p4579

Therefore etc. Q. E. D.

10.prop.3.100.p4580

The square on a minor straight line applied to a rational straight line produces as breadth a fourth apotome.

10.prop.3.100.p4581

Let AB be a minor and CD a rational straight line, and to the rational straight line CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a fourth apotome.

10.prop.3.100.p4582

For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on AG, GB rational, but twice the rectangle AG, GB medial. [X. 76]

10.prop.3.100.p4583

To CD let there be applied CH equal to the square on AG and producing CK as breadth, and KL equal to the square on BG, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB.

10.prop.3.100.p4584

And the sum of the squares on AG, GB is rational; therefore CL is also rational.

10.prop.3.100.p4585

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is also rational and commensurable in length with CD. [X. 20]

10.prop.3.100.p4586

And, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]

10.prop.3.100.p4587

Let then FM be bisected at the point N, and let NO be drawn through N parallel to either of the straight lines CD, ML; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.

10.prop.3.100.p4588

And, since twice the rectangle AG, GB is medial and is equal to FL, therefore FL is also medial.

10.prop.3.100.p4589

And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22]

10.prop.3.100.p4590

And, since the sum of the squares on AG, GB is rational, while twice the rectangle AG, GB is medial, the squares on AG, GB are incommensurable with twice the rectangle AG, GB.

10.prop.3.100.p4591

But CL is equal to the squares on AG, GB, and FL equal to twice the rectangle AG, GB; therefore CL is incommensurable with FL.

10.prop.3.100.p4592

But, as CL is to FL, so is CM to MF; [VI. 1] therefore CM is incommensurable in length with MF. [X. 11]

10.prop.3.100.p4593

And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]

10.prop.3.100.p4594

I say that it is also a fourth apotome.

10.prop.3.100.p4595

For, since AG, GB are incommensurable in square, therefore the square on AG is also incommensurable with the square on GB.

10.prop.3.100.p4596

And CH is equal to the square on AG, and KL equal to the square on GB; therefore CH is incommensurable with KL.

10.prop.3.100.p4597

But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable in length with KM. [X. 11]

10.prop.3.100.p4598

And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and the square on AG is equal to CH, the square on GB to KL, and the rectangle AG, GB to NL, therefore NL is a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

10.prop.3.100.p4599

But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore, as CK is to MN, so is MN to KM; [V. 11] therefore the rectangle CK, KM is equal to the square on MN [VI. 17], that is, to the fourth part of the square on FM.

10.prop.3.100.p4600

Since then CM, MF are two unequal straight lines, and the rectangle CK, KM equal to the fourth part of the square on MF and deficient by a square figure has been applied to CM and divides it into incommensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18]

10.prop.3.100.p4601

And the whole CM is commensurable in length with the rational straight line CD set out; therefore CF is a fourth apotome. [X. Deff. III. 4]

10.prop.3.100.p4602

Therefore etc. Q. E. D.

10.prop.3.101.p4603

The square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome.

10.prop.3.101.p4604

Let AB be the straight line which produces with a rational area a medial whole, and CD a rational straight line, and to CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a fifth apotome.

10.prop.3.101.p4605

For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational. [X. 77]

10.prop.3.101.p4606

To CD let there be applied CH equal to the square on AG, and KL equal to the square on GB; therefore the whole CL is equal to the squares on AG, GB.

10.prop.3.101.p4607

But the sum of the squares on AG, GB together is medial; therefore CL is medial.

10.prop.3.101.p4608

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable with CD. [X. 22]

10.prop.3.101.p4609

And, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]

10.prop.3.101.p4610

Let then FM be bisected at N, and through N let NO be drawn parallel to either of the straight lines CD, ML; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB:

10.prop.3.101.p4611

And, since twice the rectangle AG, GB is rational and equal to FL, therefore FL is rational.

10.prop.3.101.p4612

And it is applied to the rational straight line EF, producing FM as breadth; therefore FM is rational and commensurable in length with CD. [X. 20]

10.prop.3.101.p4613

Now, since CL is medial, and FL rational, therefore CL is incommensurable with FL.

10.prop.3.101.p4614

But, as CL is to FL, so is CM to MF; [VI. 1] therefore CM is incommensurable in length with MF. [X. 11]

10.prop.3.101.p4615

And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]

10.prop.3.101.p4616

I say next that it is also a fifth apotome.

10.prop.3.101.p4617

For we can prove similarly that the rectangle CK, KM is equal to the square on NM, that is, to the fourth part of the square on FM.

10.prop.3.101.p4618

And, since the square on AG is incommensurable with the square on GB, while the square on AG is equal to CH, and the square on GB to KL, therefore CH is incommensurable with KL.

10.prop.3.101.p4619

But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable in length with KM. [X. 11]

10.prop.3.101.p4620

Since then CM, MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to CM, and divides it into incommensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18]

10.prop.3.101.p4621

And the annex FM is commensurable with the rational straight line CD set out; therefore CF is a fifth apotome. [X. Deff. III. 5] Q. E. D.

10.prop.3.102.p4622

The square on the straight line which produces with a medial area a medial whole, if applied to a rational straight line, produces as breadth a sixth apotome.

10.prop.3.102.p4623

Let AB be the straight line which produces with a medial area a medial whole, and CD a rational straight line, and to CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a sixth apotome.

10.prop.3.102.p4624

For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on them medial, twice the rectangle AG, GB medial, and the squares on AG, GB incommensurable with twice the rectangle AG, GB. [X. 78]

10.prop.3.102.p4625

Now to CD let there be applied CH equal to the square on AG and producing CK as breadth, and KL equal to the square on BG; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial.

10.prop.3.102.p4626

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22]

10.prop.3.102.p4627

Since now CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]

10.prop.3.102.p4628

And twice the rectangle AG, GB is medial; therefore FL is also medial.

10.prop.3.102.p4629

And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22]

10.prop.3.102.p4630

And, since the squares on AG, GB are incommensurable with twice the rectangle AG, GB, and CL is equal to the squares on AG, GB, and FL equal to twice the rectangle AG, GB, therefore CL is incommensurable with FL.

10.prop.3.102.p4631

But, as CL is to FL, so is CM to MF; [VI. 1] therefore CM is incommensurable in length with MF. [X. 11]

10.prop.3.102.p4632

And both are rational.

10.prop.3.102.p4633

Therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]

10.prop.3.102.p4634

I say next that it is also a sixth apotome.

10.prop.3.102.p4635

For, since FL is equal to twice the rectangle AG, GB, let FM be bisected at N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.

10.prop.3.102.p4636

And, since AG, GB are incommensurable in square, therefore the square on AG is incommensurable with the square on GB.

10.prop.3.102.p4637

But CH is equal to the square on AG, and KL is equal to the square on GB; therefore CH is incommensurable with KL.

10.prop.3.102.p4638

But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable with KM. [X. 11]

10.prop.3.102.p4639

And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

10.prop.3.102.p4640

And for the same reason as before the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18]

10.prop.3.102.p4641

And neither of them is commensurable with the rational straight line CD set out; therefore CF is a sixth apotome. [X. Deff. III. 6] Q. E. D.

10.prop.3.103.p4642

A straight line commensurable in length with an apotome is an apotome and the same in order.

10.prop.3.103.p4643

Let AB be an apotome, and let CD be commensurable in length with AB; I say that CD is also an apotome and the same in order with AB.

10.prop.3.103.p4644

For, since AB is an apotome, let BE be the annex to it; therefore AE, EB are rational straight lines commensurable in square only. [X. 73]

10.prop.3.103.p4645

Let it be contrived that the ratio of BE to DF is the same as the ratio of AB to CD; [VI. 12] therefore also, as one is to one, so are all to all; [V. 12] therefore also, as the whole AE is to the whole CF, so is AB to CD.

10.prop.3.103.p4646

But AB is commensurable in length with CD.

10.prop.3.103.p4647

Therefore AE is also commensurable with CF, and BE with DF. [X. 11]

10.prop.3.103.p4648

And AE, EB are rational straight lines commensurable in square only; therefore CF, FD are also rational straight lines commensurable in square only. [X. 13]

10.prop.3.103.p4649

Now since, as AE is to CF, so is BE to DF, alternately therefore, as AE is to EB, so is CF to FD. [V. 16]

10.prop.3.103.p4650

And the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it.

10.prop.3.103.p4651

If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line commensurable with CF. [X. 14]

10.prop.3.103.p4652

And, if AE is commensurable in length with the rational straight line set out, CF is so also, [X. 12] if BE, then DF also, [id.] and, if neither of the straight lines AE, EB, then neither of the straight lines CF, FD. [X. 13]

10.prop.3.103.p4653

But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line incommensurable with CF. [X. 14]

10.prop.3.103.p4654

And, if AE is commensurable in length with the rational straight line set out, CF is so also, if BE, then DF also, [X. 12] and, if neither of the straight lines AE, EB, then neither of the straight lines CF, FD. [X. 13]

10.prop.3.103.p4655

Therefore CD is an apotome and the same in order with AB. Q. E. D.

10.prop.3.104.p4656

A straight line commensurable with an apotome of a medial straight line is an apotome of a medial straight line and the same in order.

10.prop.3.104.p4657

Let AB be an apotome of a medial straight line, and let CD be commensurable in length with AB; I say that CD is also an apotome of a medial straight line and the same in order with AB.

10.prop.3.104.p4658

For, since AB is an apotome of a medial straight line, let EB be the annex to it.

10.prop.3.104.p4659

Therefore AE, EB are medial straight lines commensurable in square only. [X. 74, 75]

10.prop.3.104.p4660

Let it be contrived that, as AB is to CD, so is BE to DF; [VI. 12] therefore AE is also commensurable with CF, and BE with DF. [V. 12, X. 11]

10.prop.3.104.p4661

But AE, EB are medial straight lines commensurable in square only; therefore CF, FD are also medial straight lines [X. 23] commensurable in square only; [X. 13] therefore CD is an apotome of a medial straight line. [X. 74, 75]

10.prop.3.104.p4662

I say next that it is also the same in order with AB.

10.prop.3.104.p4663

Since, as AE is to EB, so is CF to FD, therefore also, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD.

10.prop.3.104.p4664

But the square on AE is commensurable with the square on CF; therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD. [V. 16, X. 11]

10.prop.3.104.p4665

Therefore, if the rectangle AE, EB is rational, the rectangle CF, FD will also be rational, [X. Def. 4] and if the rectangle AE, EB is medial, the rectangle CF, FD is also medial. [X. 23, Por.]

10.prop.3.104.p4666

Therefore CD is an apotome of a medial straight line and the same in order with AB. [X. 74, 75] Q. E. D.

10.prop.3.105.p4667

A straight line commensurable with a minor straight line is minor.

10.prop.3.105.p4668

Let AB be a minor straight line, and CD commensurable with AB; I say that CD is also minor.

10.prop.3.105.p4669

Let the same construction be made as before; then, since AE, EB are incommensurable in square, [X. 76] therefore CF, FD are also incommensurable in square. [X. 13]

10.prop.3.105.p4670

Now since, as AE is to EB, so is CF to FD, [V. 12, V. 16] therefore also, as the square on AE is to the square on EB, so is the square on CF to the square on FD. [VI. 22]

10.prop.3.105.p4671

Therefore, componendo, as the squares on AE, EB are to the square on EB, so are the squares on CF, FD to the square on FD. [V. 18]

10.prop.3.105.p4672

But the square on BE is commensurable with the square on DF; therefore the sum of the squares on AE, EB is also commensurable with the sum of the squares on CF, FD. [V. 16, X. 11]

10.prop.3.105.p4673

But the sum of the squares on AE, EB is rational; [X. 76] therefore the sum of the squares on CF, FD is also rational. [X. Def. 4]

10.prop.3.105.p4674

Again, since, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD, while the square on AE is commensurable with the square on CF, therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD.

10.prop.3.105.p4675

But the rectangle AE, EB is medial; [X. 76] therefore the rectangle CF, FD is also medial; [X. 23, Por.] therefore CF, FD are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial.

10.prop.3.105.p4676

Therefore CD is minor. [X. 76] Q. E. D.

10.prop.3.106.p4677

A straight line commensurable with that which produces with a rational area a medial whole is a straight line which produces with a rational area a medial whole.

10.prop.3.106.p4678

Let AB be a straight line which produces with a rational area a medial whole, and CD commensurable with AB; I say that CD is also a straight line which produces with a rational area a medial whole.

10.prop.3.106.p4679

For let BE be the annex to AB; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on AE, EB medial, but the rectangle contained by them rational. [X. 77]

10.prop.3.106.p4680

Let the same construction be made.

10.prop.3.106.p4681

Then we can prove, in manner similar to the foregoing, that CF, FD are in the same ratio as AE, EB, the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that CF, FD are also straight lines incommensurable in square which make the sum of the squares on CF, FD medial, but the rectangle contained by them rational.

10.prop.3.106.p4682

Therefore CD is a straight line which produces with a rational area a medial whole. [X. 77] Q. E. D.

10.prop.3.107.p4683

A straight line commensurable with that which produces with a medial area a medial whole is itself also a straight line which produces with a medial area a medial whole.

10.prop.3.107.p4684

Let AB be a straight line which produces with a medial area a medial whole, and let CD be commensurable with AB; I say that CD is also a straight line which produces with a medial area a medial whole.

10.prop.3.107.p4685

For let BE be the annex to AB, and let the same construction be made; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and further the sum of the squares on them incommensurable with the rectangle contained by them. [X. 78]

10.prop.3.107.p4686

Now, as was proved, AE, EB are commensurable with CF, FD, the sum of the squares on AE, EB with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; therefore CF, FD are also straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and further the sum of the squares on them incommensurable with the rectangle contained by them.

10.prop.3.107.p4687

Therefore CD is a straight line which produces with a medial area a medial whole. [X. 78]

10.prop.3.108.p4688

If from a rational area a medial area be subtracted, the side of the remaining area becomes one of two irrational straight lines, either an apotome or a minor straight line.

10.prop.3.108.p4689

For from the rational area BC let the medial area BD be subtracted; I say that the side of the remainder EC becomes one of two irrational straight lines, either an apotome or a minor straight line.

10.prop.3.108.p4690

For let a rational straight line FG be set out, to FG let there be applied the rectangular parallelogram GH equal to BC, and let GK equal to DB be subtracted; therefore the remainder EC is equal to LH.

10.prop.3.108.p4691

Since then BC is rational, and BD medial, while BC is equal to GH, and BD to GK, therefore GH is rational, and GK medial.

10.prop.3.108.p4692

And they are applied to the rational straight line FG; therefore FH is rational and commensurable in length with FG, [X. 20] while FK is rational and incommensurable in length with FG; [X. 22] therefore FH is incommensurable in length with FK. [X. 13]

10.prop.3.108.p4693

Therefore FH, FK are rational straight lines commensurable in square only; therefore KH is an apotome [X. 73], and KF the annex to it.

10.prop.3.108.p4694

Now the square on HF is greater than the square on FK by the square on a straight line either commensurable with HF or not commensurable.

10.prop.3.108.p4695

First, let the square on it be greater by the square on a straight line commensurable with it.

10.prop.3.108.p4696

Now the whole HF is commensurable in length with the rational straight line FG set out; therefore KH is a first apotome. [X. Deff. III. 1]

10.prop.3.108.p4697

But the side of the rectangle contained by a rational straight line and a first apotome is an apotome. [X. 91]

10.prop.3.108.p4698

Therefore the side of LH, that is, of EC, is an apotome.

10.prop.3.108.p4699

But, if the square on HF is greater than the square on FK by the square on a straight line incommensurable with HF, while the whole FH is commensurable in length with the rational straight line FG set out, KH is a fourth apotome. [X. Deff. III. 4]

10.prop.3.108.p4700

But the side of the rectangle contained by a rational straight line and a fourth apotome is minor. [X. 94] Q. E. D.

10.prop.3.109.p4701

If from a medial area a rational area be subtracted, there arise two other irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole.

10.prop.3.109.p4702

For from the medial area BC let the rational area BD be subtracted.

10.prop.3.109.p4703

I say that the side of the remainder EC becomes one of two irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole.

10.prop.3.109.p4704

For let a rational straight line FG be set out, and let the areas be similarly applied.

10.prop.3.109.p4705

It follows then that FH is rational and incommensurable in length with FG, while KF is rational and commensurable in length with FG; therefore FH, FK are rational straight lines commensurable in square only; [X. 13] therefore KH is an apotome, and FK the annex to it. [X. 73]

10.prop.3.109.p4706

Now the square on HF is greater than the square on FK either by the square on a straight line commensurable with HF or by the square on a straight line incommensurable with it.

10.prop.3.109.p4707

If then the square on HF is greater than the square on FK by the square on a straight line commensurable with HF, while the annex FK is commensurable in length with the rational straight line FG set out, KH is a second apotome. [X. Deff. III. 2]

10.prop.3.109.p4708

But FG is rational; so that the side of LH, that is, of EC, is a first apotome of a medial straight line. [X. 92]

10.prop.3.109.p4709

But, if the square on HF is greater than the square on FK by the square on a straight line incommensurable with HF, while the annex FK is commensurable in length with the rational straight line FG set out, KH is a fifth apotome; [X. Deff. III. 5] so that the side of EC is a straight line which produces with a rational area a medial whole. [X. 95]

10.prop.3.110.p4710

If from a medial area there be subtracted a medial area incommensurable with the whole, the two remaining irrational straight lines arise, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole.

10.prop.3.110.p4711

For, as in the foregoing figures, let there be subtracted from the medial area BC the medial area BD incommensurable with the whole; I say that the side of EC is one of two irrational straight lines, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole.

10.prop.3.110.p4712

For, since each of the rectangles BC, BD is medial, and BC is incommensurable with BD, it follows that each of the straight lines FH, FK will be rational and incommensurable in length with FG. [X. 22]

10.prop.3.110.p4713

And, since BC is incommensurable with BD, that is, GH with GK, HF is also incommensurable with FK; [VI. 1, X. 11] therefore FH, FK are rational straight lines commensurable in square only; therefore KH is an apotome. [X. 73]

10.prop.3.110.p4714

If then the square on FH is greater than the square on FK by the square on a straight line commensurable with FH, while neither of the straight lines FH, FK is commensurable in length with the rational straight line FG set out, KH is a third apotome. [X. Deff. III. 3]

10.prop.3.110.p4715

But KL is rational, and the rectangle contained by a rational straight line and a third apotome is irrational, and the side of it is irrational, and is called a second apotome of a medial straight line; [X. 93] so that the side of LH, that is, of EC, is a second apotome of a medial straight line.

10.prop.3.110.p4716

But, if the square on FH is greater than the square on FK by the square on a straight line incommensurable with FH, while neither of the straight lines HF, FK is commensurable in length with FG, KH is a sixth apotome. [X. Deff. III. 6]

10.prop.3.110.p4717

But the side of the rectangle contained by a rational straight line and a sixth apotome is a straight line which produces with a medial area a medial whole. [X. 96]

10.prop.3.110.p4718

Therefore the side of LH, that is, of EC, is a straight line which produces with a medial area a medial whole. Q. E. D.

10.prop.3.111.p4719

The apotome is not the same with the binomial straight line.

10.prop.3.111.p4720

Let AB be an apotome; I say that AB is not the same with the binomial straight line.

10.prop.3.111.p4721

For, if possible, let it be so; let a rational straight line DC be set out, and to CD let there be applied the rectangle CE equal to the square on AB and producing DE as breadth.

10.prop.3.111.p4722

Then, since AB is an apotome, DE is a first apotome. [X. 97]

10.prop.3.111.p4723

Let EF be the annex to it; therefore DF, FE are rational straight lines commensurable in square only, the square on DF is greater than the square on FE by the square on a straight line commensurable with DF, and DF is commensurable in length with the rational straight line DC set out. [X. Deff. III. 1]

10.prop.3.111.p4724

Again, since AB is binomial, therefore DE is a first binomial straight line. [X. 60]

10.prop.3.111.p4725

Let it be divided into its terms at G, and let DG be the greater term; therefore DG, GE are rational straight lines commensurable in square only, the square on DG is greater than the square on GE by the square on a straight line commensurable with DG, and the greater term DG is commensurable in length with the rational straight line DC set out. [X. Deff. II. 1]

10.prop.3.111.p4726

Therefore DF is also commensurable in length with DG; [X. 12] therefore the remainder GF is also commensurable in length with DF. [X. 15]

10.prop.3.111.p4727

But DF is incommensurable in length with EF; therefore FG is also incommensurable in length with EF. [X. 13]

10.prop.3.111.p4728

Therefore GF, FE are rational straight lines commensurable in square only; therefore EG is an apotome. [X. 73]

10.prop.3.111.p4729

But it is also rational: which is impossible.

10.prop.3.111.p4730

Therefore the apotome is not the same with the binomial straight line. Q. E. D.

10.prop.3.112.p4731

The square on a rational straight line applied to the binomial straight line produces as breadth an apotome the terms of which are commensurable with the terms of the binomial and moreover in the same ratio; and further the apotome so arising will have the same order as the binomial straight line.

10.prop.3.112.p4732

Let A be a rational straight line, let BC be a binomial, and let DC be its greater term; let the rectangle BC, EF be equal to the square on A; I say that EF is an apotome the terms of which are commensurable with CD, DB, and in the same ratio, and further EF will have the same order as BC.

10.prop.3.112.p4733

For again let the rectangle BD, G be equal to the square on A.

10.prop.3.112.p4734

Since then the rectangle BC, EF is equal to the rectangle BD, G, therefore, as CB is to BD, so is G to EF. [VI. 16]

10.prop.3.112.p4735

But CB is greater than BD; therefore G is also greater than EF. [V. 16, V. 14]

10.prop.3.112.p4736

Let EH be equal to G; therefore, as CB is to BD, so is HE to EF; therefore, separando, as CD is to BD, so is HF to FE. [V. 17]

10.prop.3.112.p4737

Let it be contrived that, as HF is to FE, so is FK to KE; therefore also the whole HK is to the whole KF as FK is to KE; for, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents. [V. 12]

10.prop.3.112.p4738

But, as FK is to KE, so is CD to DB; [V. 11] therefore also, as HK is to KF, so is CD to DB. [id.]

10.prop.3.112.p4739

But the square on CD is commensurable with the square on DB; [X. 36] therefore the square on HK is also commensurable with the square on KF. [VI. 22, X. 11]

10.prop.3.112.p4740

And, as the square on HK is to the square on KF, so is HK to KE, since the three straight lines HK, KF, KE are proportional. [V. Def. 9]

10.prop.3.112.p4741

Therefore HK is commensurable in length with KE, so that HE is also commensurable in length with EK. [X. 15]

10.prop.3.112.p4742

Now, since the square on A is equal to the rectangle EH, BD, while the square on A is rational, therefore the rectangle EH, BD is also rational.

10.prop.3.112.p4743

And it is applied to the rational straight line BD; therefore EH is rational and commensurable in length with BD; [X. 20] so that EK, being commensurable with it, is also rational and commensurable in length with BD.

10.prop.3.112.p4744

Since, then, as CD is to DB, so is FK to KE, while CD, DB are straight lines commensurable in square only, therefore FK, KE are also commensurable in square only. [X. 11]

10.prop.3.112.p4745

But KE is rational; therefore FK is also rational.

10.prop.3.112.p4746

Therefore FK, KE are rational straight lines commensurable in square only; therefore EF is an apotome. [X. 73]

10.prop.3.112.p4747

Now the square on CD is greater than the square on DB either by the square on a straight line commensurable with CD or by the square on a straight line incommensurable with it.

10.prop.3.112.p4748

If then the square on CD is greater than the square on DB by the square on a straight line commensurable with CD, the square on FK is also greater than the square on KE by the square on a straight line commensurable with FK. [X. 14]

10.prop.3.112.p4749

And, if CD is commensurable in length with the rational straight line set out, so also is FK; [X. 11, 12] if BD is so commensurable, so also is KE; [X. 12] but, if neither of the straight lines CD, DB is so commensurable, neither of the straight lines FK, KE is so.

10.prop.3.112.p4750

But, if the square on CD is greater than the square on DB by the square on a straight line incommensurable with CD, the square on FK is also greater than the square on KE by the square on a straight line incommensurable with FK. [X. 14]

10.prop.3.112.p4751

And, if CD is commensurable with the rational straight line set out, so also is FK; if BD is so commensurable, so also is KE; but, if neither of the straight lines CD, DB is so commensurable, neither of the straight lines FK, KE is so; so that FE is an apotome, the terms of which FK, KE are commensurable with the terms CD, DB of the binomial straight line and in the same ratio, and it has the same order as BC. Q. E. D.

10.prop.3.113.p4752

The square on a rational straight line, if applied to an apotome, produces as, breadth the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio; and further the binomial so arising has the same order as the apotome.

10.prop.3.113.p4753

Let A be a rational straight line and BD an apotome, and let the rectangle BD, KH be equal to the square on A, so that the square on the rational straight line A when applied to the apotome BD produces KH as breadth; I say that KH is a binomial straight line the terms of which are commensurable with the terms of BD and in the same ratio; and further KH has the same order as BD.

10.prop.3.113.p4754

For let DC be the annex to BD; therefore BC, CD are rational straight lines commensurable in square only. [X. 73]

10.prop.3.113.p4755

Let the rectangle BC, G be also equal to the square on A.

10.prop.3.113.p4756

But the square on A is rational; therefore the rectangle BC, G is also rational.

10.prop.3.113.p4757

And it has been applied to the rational straight line BC; therefore G is rational and commensurable in length with BC. [X. 20]

10.prop.3.113.p4758

Since now the rectangle BC, G is equal to the rectangle BD, KH, therefore, proportionally, as CB is to BD, so is KH to G. [VI. 16]

10.prop.3.113.p4759

But BC is greater than BD; therefore KH is also greater than G. [V. 16, V. 14]

10.prop.3.113.p4760

Let KE be made equal to G; therefore KE is commensurable in length with BC.

10.prop.3.113.p4761

And since, as CB is to BD, so is HK to KE, therefore, convertendo, as BC is to CD, so is KH to HE. [V. 19, Por.]

10.prop.3.113.p4762

Let it be contrived that, as KH is to HE, so is HF to FE; therefore also the remainder KF is to FH as KH is to HE, that is, as BC is to CD. [V. 19]

10.prop.3.113.p4763

But BC, CD are commensurable in square only; therefore KF, FH are also commensurable in square only. [X. 11]

10.prop.3.113.p4764

And since, as KH is to HE, so is KF to FH, while, as KH is to HE, so is HF to FE, therefore also, as KF is to FH, so is HF to FE, [V. 11] so that also, as the first is to the third, so is the square on the first to the square on the second; [V. Def. 9] therefore also, as KF is to FE, so is the square on KF to the square on FH.

10.prop.3.113.p4765

But the square on KF is commensurable with the square on FH, for KF, FH are commensurable in square; therefore KF is also commensurable in length with FE, [X. 11] so that KF is also commensurable in length with KE. [X. 15]

10.prop.3.113.p4766

But KE is rational and commensurable in length with BC; therefore KF is also rational and commensurable in length with BC. [X. 12]

10.prop.3.113.p4767

And, since, as BC is to CD, so is KF to FH, alternately, as BC is to KF, so is DC to FH. [V. 16]

10.prop.3.113.p4768

But BC is commensurable with KF; therefore FH is also commensurable in length with CD. [X. 11]

10.prop.3.113.p4769

But BC, CD are rational straight lines commensurable in square only; therefore KF, FH are also rational straight lines [X. Def. 3] commensurable in square only; therefore KH is binomial. [X. 36]

10.prop.3.113.p4770

If now the square on BC is greater than the square on CD by the square on a straight line commensurable with BC, the square on KF will also be greater than the square on FH by the square on a straight line commensurable with KF. [X 14]

10.prop.3.113.p4771

And, if BC is commensurable in length with the rational straight line set out, so also is KF; if CD is commensurable in length with the rational straight line set out, so also is FH, but, if neither of the straight lines BC, CD, then neither of the straight lines KF, FH.

10.prop.3.113.p4772

But, if the square on BC is greater than the square on CD by the square on a straight line incommensurable with BC, the square on KF is also greater than the square on FH by the square on a straight line incommensurable with KF. [X. 14]

10.prop.3.113.p4773

And, if BC is commensurable with the rational straight line set out, so also is KF; if CD is so commensurable, in length with the rational straight line set out, so also is FH; but, if neither of the straight lines BC, CD, then neither of the straight lines KF, FH.

10.prop.3.113.p4774

Therefore KH is a binomial straight line, the terms of which KF, FH are commensurable with the terms BC, CD of the apotome and in the same ratio, and further KH has the same order as BD. Q. E. D.

10.prop.3.114.p4775

If an area be contained by an apotome and the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio, the side of the area is rational.

10.prop.3.114.p4776

For let an area, the rectangle AB, CD, be contained by the apotome AB and the binomial straight line CD, and let CE be the greater term of the latter; let the terms CE, ED of the binomial straight line be commensurable with the terms AF, FB of the apotome and in the same ratio; and let the side of the rectangle AB, CD be G; I say that G is rational.

10.prop.3.114.p4777

For let a rational straight line H be set out, and to CD let there be applied a rectangle equal to the square on H and producing KL as breadth.

10.prop.3.114.p4778

Therefore KL is an apotome.

10.prop.3.114.p4779

Let its terms be KM, ML commensurable with the terms CE, ED of the binomial straight line and in the same ratio. [X. 112]

10.prop.3.114.p4780

But CE, ED are also commensurable with AF, FB and in the same ratio; therefore, as AF is to FB, so is KM to ML.

10.prop.3.114.p4781

Therefore, alternately, as AF is to KM, so is BF to LM; therefore also the remainder AB is to the remainder KL as AF is to KM. [V. 19]

10.prop.3.114.p4782

But AF is commensurable with KM; [X. 12] therefore AB is also commensurable with KL. [X. 11]

10.prop.3.114.p4783

And, as AB is to KL, so is the rectangle CD, AB to the rectangle CD, KL; [VI. 1] therefore the rectangle CD, AB is also commensurable with the rectangle CD, KL. [X. 11]

10.prop.3.114.p4784

But the rectangle CD, KL is equal to the square on H; therefore the rectangle CD, AB is commensurable with the square on H.

10.prop.3.114.p4785

But the square on G is equal to the rectangle CD, AB; therefore the square on G is commensurable with the square on H.

10.prop.3.114.p4786

But the square on H is rational; therefore the square on G is also rational; therefore G is rational.

10.prop.3.114.p4787

And it is the side of the rectangle CD, AB.

10.prop.3.114.p4788

Therefore etc.

10.prop.3.114.p4789

Porism. And it is made manifest to us by this also that it is possible for a rational area to be contained by irrational straight lines. Q. E. D.

10.prop.3.115.p4790

From a medial straight line there arise irrational straight lines infinite in number, and none of them is the same as any of the preceding.

10.prop.3.115.p4791

Let A be a medial straight line; I say that from A there arise irrational straight lines infinite in number, and none of them is the same as any of the preceding.

10.prop.3.115.p4792

Let a rational straight line B be set out, and let the square on C be equal to the rectangle B, A; therefore C is irrational; [X. Def. 4] for that which is contained by an irrational and a rational straight line is irrational. [deduction from X. 20]

10.prop.3.115.p4793

And it is not the same with any of the preceding; for the square on none of the preceding, if applied to a rational straight line produces as breadth a medial straight line.

10.prop.3.115.p4794

Again, let the square on D be equal to the rectangle B, C; therefore the square on D is irrational. [deduction from X. 20]

10.prop.3.115.p4795

Therefore D is irrational; [X. Def. 4] and it is not the same with any of the preceding, for the square on none of the preceding, if applied to a rational straight line, produces C as breadth.

10.prop.3.115.p4796

Similarly, if this arrangement proceeds ad infinitum, it is manifest that from the medial straight line there arise irrational straight lines infinite in number, and none is the same with any of the preceding. Q. E. D.

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