Elementa

Euclid

Elementa

Euclid · Elem 9 · perseus-eng2 · More by Euclid

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9.prop.1.p2731

If two similar plane numbers by multiplying one another make some number, the product will be square.

9.prop.1.p2732

Let A, B be two similar plane numbers, and let A by multiplying B make C; I say that C is square.

9.prop.1.p2733

For let A by multiplying itself make D.

9.prop.1.p2734

Therefore D is square.

9.prop.1.p2735

Since then A by multiplying itself has made D, and by multiplying B has made C, therefore, as A is to B, so is D to C. [VII. 17]

9.prop.1.p2736

And, since A, B are similar plane numbers, therefore one mean proportional number falls between A, B. [VIII. 18]

9.prop.1.p2737

But, if numbers fall between two numbers in continued proportion, as many as fall between them, so many also fall between those which have the same ratio; [VIII. 8] so that one mean proportional number falls between D, C also.

9.prop.1.p2738

And D is square; therefore C is also square. [VIII. 22] Q. E. D.

9.prop.2.p2739

If two numbers by multiplying one another make a square number, they are similar plane numbers.

9.prop.2.p2740

Let A, B be two numbers, and let A by multiplying B make the square number C; I say that A, B are similar plane numbers.

9.prop.2.p2741

For let A by multiplying itself make D; therefore D is square.

9.prop.2.p2742

Now, since A by multiplying itself has made D, and by multiplying B has made C, therefore, as A is to B, so is D to C. [VII. 17]

9.prop.2.p2743

And, since D is square, and C is so also, therefore D, C are similar plane numbers.

9.prop.2.p2744

Therefore one mean proportional number falls between D, C. [VIII. 18]

9.prop.2.p2745

And, as D is to C, so is A to B; therefore one mean proportional number falls between A, B also. [VIII. 8]

9.prop.2.p2746

But, if one mean proportional number fall between two numbers, they are similar plane numbers; [VIII. 20] therefore A, B are similar plane numbers. Q. E. D.

9.prop.3.p2747

If a cube number by multiplying itself make some number, the product will be cube.

9.prop.3.p2748

For let the cube number A by multiplying itself make B; I say that B is cube.

9.prop.3.p2749

For let C, the side of A, be taken, and let C by multiplying itself make D.

9.prop.3.p2750

It is then manifest that C by multiplying D has made A.

9.prop.3.p2751

Now, since C by multiplying itself has made D, therefore C measures D according to the units in itself.

9.prop.3.p2752

But further the unit also measures C according to the units in it; therefore, as the unit is to C, so is C to D. [VII. Def. 20]

9.prop.3.p2753

Again, since C by multiplying D has made A, therefore D measures A according to the units in C.

9.prop.3.p2754

But the unit also measures C according to the units in it; therefore, as the unit is to C, so is D to A.

9.prop.3.p2755

But, as the unit is to C, so is C to D; therefore also, as the unit is to C, so is C to D, and D to A.

9.prop.3.p2756

Therefore between the unit and the number A two mean proportional numbers C, D have fallen in continued proportion.

9.prop.3.p2757

Again, since A by multiplying itself has made B, therefore A measures B according to the units in itself.

9.prop.3.p2758

But the unit also measures A according to the units in it; therefore, as the unit is to A, so is A to B. [VII. Def. 20]

9.prop.3.p2759

But between the unit and A two mean proportional numbers have fallen; therefore two mean proportional numbers will also fall between A, B. [VIII. 8]

9.prop.3.p2760

But, if two mean proportional numbers fall between two numbers, and the first be cube, the second will also be cube. [VIII. 23]

9.prop.3.p2761

And A is cube; therefore B is also cube. Q. E. D.

9.prop.4.p2762

If a cube number by multiplying a cube number make some number, the product will be cube.

9.prop.4.p2763

For let the cube number A by multiplying the cube number B make C; I say that C is cube.

9.prop.4.p2764

For let A by multiplying itself make D; therefore D is cube. [IX. 3]

9.prop.4.p2765

And, since A by multiplying itself has made D, and by multiplying B has made C therefore, as A is to B, so is D to C. [VII. 17]

9.prop.4.p2766

And, since A, B are cube numbers, A, B are similar solid numbers.

9.prop.4.p2767

Therefore two mean proportional numbers fall between A, B; [VIII. 19] so that two mean proportional numbers will fall between D, C also. [VIII. 8]

9.prop.4.p2768

And D is cube; therefore C is also cube [VIII. 23] Q. E. D.

9.prop.5.p2769

If a cube number by multiplying any number make a cube number, the multiplied number will also be cube.

9.prop.5.p2770

For let the cube number A by multiplying any number B make the cube number C; I say that B is cube.

9.prop.5.p2771

For let A by multiplying itself make D; therefore D is cube. [IX. 3]

9.prop.5.p2772

Now, since A by multiplying itself has made D, and by multiplying B has made C, therefore, as A is to B, so is D to C. [VII. 17]

9.prop.5.p2773

And since D, C are cube, they are similar solid numbers.

9.prop.5.p2774

Therefore two mean proportional numbers fall between D, C. [VIII. 19]

9.prop.5.p2775

And, as D is to C, so is A to B; therefore two mean proportional numbers fall between A, B also. [VIII. 8]

9.prop.5.p2776

And A is cube; therefore B is also cube. [VIII. 23]

9.prop.6.p2777

If a number by multiplying itself make a cube number, it will itself also be cube.

9.prop.6.p2778

For let the number A by multiplying itself make the cube number B; I say that A is also cube.

9.prop.6.p2779

For let A by multiplying B make C.

9.prop.6.p2780

Since, then, A by multiplying itself has made B, and by multiplying B has made C, therefore C is cube.

9.prop.6.p2781

And, since A by multiplying itself has made B, therefore A measures B according to the units in itself.

9.prop.6.p2782

But the unit also measures A according to the units in it.

9.prop.6.p2783

Therefore, as the unit is to A, so is A to B. [VII. Def. 20]

9.prop.6.p2784

And, since A by multiplying B has made C, therefore B measures C according to the units in A.

9.prop.6.p2785

But the unit also measures A according to the units in it.

9.prop.6.p2786

Therefore, as the unit is to A, so is B to C. [VII. Def. 20]

9.prop.6.p2787

But, as the unit is to A, so is A to B; therefore also, as A is to B, so is B to C.

9.prop.6.p2788

And, since B, C are cube, they are similar solid numbers.

9.prop.6.p2789

Therefore there are two mean proportional numbers between B, C. [VIII. 19]

9.prop.6.p2790

And, as B is to C, so is A to B.

9.prop.6.p2791

Therefore there are two mean proportional numbers between A, B also. [VIII. 8]

9.prop.6.p2792

And B is cube; therefore A is also cube. [cf. VIII. 23] Q. E. D.

9.prop.7.p2793

If a composite number by multiplying any number make some number, the product will be solid.

9.prop.7.p2794

For let the composite number A by multiplying any number B make C; I say that C is solid.

9.prop.7.p2795

For, since A is composite, it will be measured by some number. [VII. Def. 13]

9.prop.7.p2796

Let it be measured by D; and, as many times as D measures A, so many units let there be in E.

9.prop.7.p2797

Since then D measures A according to the units in E, therefore E by multiplying D has made A. [VII. Def. 15]

9.prop.7.p2798

And, since A by multiplying B has made C, and A is the product of D, E, therefore the product of D, E by multiplying B has made C.

9.prop.7.p2799

Therefore C is solid, and D, E, B are its sides. Q. E. D.

9.prop.8.p2800

If as many numbers as we please beginning from an unit be in continued proportion, the third from the unit will be square, as will also those which successively leave out one; the fourth will be cube, as will also all those which leave out two; and the seventh will be at once cube and square, as will also those which leave out five.

9.prop.8.p2801

Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion; I say that B, the third from the unit, is square, as are also all those which leave out one; C, the fourth, is cube, as are also all those which leave out two; and F, the seventh, is at once cube and square, as are also all those which leave out five.

9.prop.8.p2802

For since, as the unit is to A, so is A to B, therefore the unit measures the number A the same number of times that A measures B. [VII. Def. 20]

9.prop.8.p2803

But the unit measures the number A according to the units in it; therefore A also measures B according to the units in A.

9.prop.8.p2804

Therefore A by multiplying itself has made B; therefore B is square.

9.prop.8.p2805

And, since B, C, D are in continued proportion, and B is square, therefore D is also square. [VIII. 22]

9.prop.8.p2806

For the same reason F is also square.

9.prop.8.p2807

Similarly we can prove that all those which leave out one are square.

9.prop.8.p2808

I say next that C, the fourth from the unit, is cube, as are also all those which leave out two.

9.prop.8.p2809

For since, as the unit is to A, so is B to C, therefore the unit measures the number A the same number of times that B measures C.

9.prop.8.p2810

But the unit measures the number A according to the units in A; therefore B also measures C according to the units in A.

9.prop.8.p2811

Therefore A by multiplying B has made C.

9.prop.8.p2812

Since then A by multiplying itself has made B, and by multiplying B has made C, therefore C is cube.

9.prop.8.p2813

And, since C, D, E, F are in continued proportion, and C is cube, therefore F is also cube. [VIII. 23]

9.prop.8.p2814

But it was also proved square; therefore the seventh from the unit is both cube and square.

9.prop.8.p2815

Similarly we can prove that all the numbers which leave out five are also both cube and square. Q. E. D.

9.prop.9.p2816

If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be square, all the rest will also be square. And, if the number after the unit be cube, all the rest will also be cube.

9.prop.9.p2817

Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion, and let A, the number after the unit, be square; I say that all the rest will also be square.

9.prop.9.p2818

Now it has been proved that B, the third from the unit, is square, as are also all those which leave out one; [IX. 8] I say that all the rest are also square.

9.prop.9.p2819

For, since A, B, C are in continued proportion, and A is square, therefore C is also square. [VIII. 22]

9.prop.9.p2820

Again, since B, C, D are in continued proportion, and B is square, D is also square. [VIII. 22]

9.prop.9.p2821

Similarly we can prove that all the rest are also square.

9.prop.9.p2822

Next, let A be cube; I say that all the rest are also cube.

9.prop.9.p2823

Now it has been proved that C, the fourth from the unit, is cube, as also are all those which leave out two; [IX. 8] I say that all the rest are also cube.

9.prop.9.p2824

For, since, as the unit is to A, so is A to B, therefore the unit measures A the same number of times as A measures B.

9.prop.9.p2825

But the unit measures A according to the units in it; therefore A also measures B according to the units in itself; therefore A by multiplying itself has made B.

9.prop.9.p2826

And A is cube.

9.prop.9.p2827

But, if a cube number by multiplying itself make some number, the product is cube. [IX. 3]

9.prop.9.p2828

Therefore B is also cube.

9.prop.9.p2829

And, since the four numbers A, B, C, D are in continued proportion, and A is cube, D also is cube. [VIII. 23]

9.prop.9.p2830

For the same reason E is also cube, and similarly all the rest are cube. Q. E. D.

9.prop.10.p2831

If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be not square, neither will any other be square except the third from the unit and all those which leave out one. And, if the number after the unit be not cube, neither will any other be cube except the fourth from the unit and all those which leave out two.

9.prop.10.p2832

Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion, and let A, the number after the unit, not be square; I say that neither will any other be square except the third from the unit ltand those which leave out onegt.

9.prop.10.p2833

For, if possible, let C be square.

9.prop.10.p2834

But B is also square; [IX. 8] [therefore B, C have to one another the ratio which a square number has to a square number].

9.prop.10.p2835

And, as B is to C, so is A to B; therefore A, B have to one another the ratio which a square number has to a square number; [so that A, B are similar plane numbers]. [VIII. 26, converse]

9.prop.10.p2836

And B is square; therefore A is also square: which is contrary to the hypothesis.

9.prop.10.p2837

Therefore C is not square.

9.prop.10.p2838

Similarly we can prove that neither is any other of the numbers square except the third from the unit and those which leave out one.

9.prop.10.p2839

Next, let A not be cube.

9.prop.10.p2840

I say that neither will any other be cube except the fourth from the unit and those which leave out two.

9.prop.10.p2841

For, if possible, let D be cube.

9.prop.10.p2842

Now C is also cube; for it is fourth from the unit. [IX. 8]

9.prop.10.p2843

And, as C is to D, so is B to C; therefore B also has to C the ratio which a cube has to a cube.

9.prop.10.p2844

And C is cube; therefore B is also cube. [VIII. 25]

9.prop.10.p2845

And since, as the unit is to A, so is A to B, and the unit measures A according to the units in it, therefore A also measures B according to the units in itself; therefore A by multiplying itself has made the cube number B.

9.prop.10.p2846

But, if a number by multiplying itself make a cube number, it is also itself cube. [IX. 6]

9.prop.10.p2847

Therefore A is also cube: which is contrary to the hypothesis.

9.prop.10.p2848

Therefore D is not cube.

9.prop.10.p2849

Similarly we can prove that neither is any other of the numbers cube except the fourth from the unit and those which leave out two. Q. E. D.

9.prop.11.p2850

If as many numbers as we please beginning from an unit be in continued proportion, the less measures the greater according to some one of the numbers which have place among the proportional numbers.

9.prop.11.p2851

Let there be as many numbers as we please, B, C, D, E, beginning from the unit A and in continued proportion; I say that B, the least of the numbers B, C, D, E, measures E according to some one of the numbers C, D.

9.prop.11.p2852

For since, as the unit A is to B, so is D to E, therefore the unit A measures the number B the same number of times as D measures E; therefore, alternately, the unit A measures D the same number of times as B measures E. [VII. 15]

9.prop.11.p2853

But the unit A measures D according to the units in it; therefore B also measures E according to the units in D; so that B the less measures E the greater according to some number of those which have place among the proportional numbers.—

9.prop.11.p2854

Porism. And it is manifest that, whatever place the measuring number has, reckoned from the unit, the same place also has the number according to which it measures, reckoned from the number measured, in the direction of the number before it.—

9.prop.11.trailer

Q. E. D.

9.prop.12.p2856

If as many numbers as we please beginning from an unit be in continued proportion, by however many prime numbers the last is measured, the next to the unit will also be measured by the same.

9.prop.12.p2857

Let there be as many numbers as we please, A, B, C, D, beginning from an unit, and in continued proportion; I say that, by however many prime numbers D is measured, A will also be measured by the same.

9.prop.12.p2858

For let D be measured by any prime number E; I say that E measures A.

9.prop.12.p2859

For suppose it does not; now E is prime, and any prime number is prime to any which it does not measure; [VII. 29] therefore E, A are prime to one another.

9.prop.12.p2860

And, since E measures D, let it measure it according to F, therefore E by multiplying F has made D.

9.prop.12.p2861

Again, since A measures D according to the units in C, [IX. 11 and Por.] therefore A by multiplying C has made D.

9.prop.12.p2862

But, further, E has also by multiplying F made D; therefore the product of A, C is equal to the product of E, F.

9.prop.12.p2863

Therefore, as A is to E, so is F to C. [VII. 19]

9.prop.12.p2864

But A, E are prime, primes are also least, [VII. 21] and the least measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures C.

9.prop.12.p2865

Let it measure it according to G; therefore E by multiplying G has made C.

9.prop.12.p2866

But, further, by the theorem before this, A has also by multiplying B made C. [IX. 11 and Por.]

9.prop.12.p2867

Therefore the product of A, B is equal to the product of E, G.

9.prop.12.p2868

Therefore, as A is to E, so is G to B. [VII. 19]

9.prop.12.p2869

But A, E are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio with them the same number of times, the antecedent the antecedent and the consequent the consequent: [VII. 20] therefore E measures B.

9.prop.12.p2870

Let it measure it according to H; therefore E by multiplying H has made B.

9.prop.12.p2871

But further A has also by multiplying itself made B; [IX. 8] therefore the product of E, H is equal to the square on A.

9.prop.12.p2872

Therefore, as E is to A, so is A to H. [VII. 19]

9.prop.12.p2873

But A, E are prime, primes are also least, [VII. 21] and the least measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures A, as antecedent antecedent.

9.prop.12.p2874

But, again, it also does not measure it: which is impossible.

9.prop.12.p2875

Therefore E, A are not prime to one another.

9.prop.12.p2876

Therefore they are composite to one another.

9.prop.12.p2877

But numbers composite to one another are measured by some number. [VII. Def. 14]

9.prop.12.p2878

And, since E is by hypothesis prime, and the prime is not measured by any number other than itself, therefore E measures A, E, so that E measures A.

9.prop.12.p2879

[But it also measures D; therefore E measures A, D.]

9.prop.12.p2880

Similarly we can prove that, by however many prime numbers D is measured, A will also be measured by the same. Q. E. D.

9.prop.13.p2881

If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be prime, the greatest will not be measured by any except those which have a place among the proportional numbers.

9.prop.13.p2882

Let there be as many numbers as we please, A, B, C, D, beginning from an unit and in continued proportion, and let A, the number after the unit, be prime; I say that D, the greatest of them, will not be measured by any other number except A, B, C.

9.prop.13.p2883

For, if possible, let it be measured by E, and let E not be the same with any of the numbers A, B, C.

9.prop.13.p2884

It is then manifest that E is not prime.

9.prop.13.p2885

For, if E is prime and measures D, it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible.

9.prop.13.p2886

Therefore E is not prime.

9.prop.13.p2887

Therefore it is composite.

9.prop.13.p2888

But any composite number is measured by some prime number; [VII. 31] therefore E is measured by some prime number.

9.prop.13.p2889

I say next that it will not be measured by any other prime except A.

9.prop.13.p2890

For, if E is measured by another, and E measures D, that other will also measure D; so that it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible.

9.prop.13.p2891

Therefore A measures E.

9.prop.13.p2892

And, since E measures D, let it measure it according to F.

9.prop.13.p2893

I say that F is not the same with any of the numbers A, B, C.

9.prop.13.p2894

For, if F is the same with one of the numbers A, B, C, and measures D according to E, therefore one of the numbers A, B, C also measures D according to E.

9.prop.13.p2895

But one of the numbers A, B, C measures D according to some one of the numbers A, B, C; [IX. 11] therefore E is also the same with one of the numbers A, B, C: which is contrary to the hypothesis.

9.prop.13.p2896

Therefore F is not the same as any one of the numbers A, B, C.

9.prop.13.p2897

Similarly we can prove that F is measured by A, by proving again that F is not prime.

9.prop.13.p2898

For, if it is, and measures D, it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible; therefore F is not prime.

9.prop.13.p2899

Therefore it is composite.

9.prop.13.p2900

But any composite number is measured by some prime number; [VII. 31] therefore F is measured by some prime number.

9.prop.13.p2901

I say next that it will not be measured by any other prime except A.

9.prop.13.p2902

For, if any other prime number measures F, and F measures D, that other will also measure D; so that it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible.

9.prop.13.p2903

Therefore A measures F.

9.prop.13.p2904

And, since E measures D according to F, therefore E by multiplying F has made D.

9.prop.13.p2905

But, further, A has also by multiplying C made D; [IX. 11] therefore the product of A, C is equal to the product of E, F.

9.prop.13.p2906

Therefore, proportionally, as A is to E, so is F to C. [VII. 19]

9.prop.13.p2907

But A measures E; therefore F also measures C.

9.prop.13.p2908

Let it measure it according to G.

9.prop.13.p2909

Similarly, then, we can prove that G is not the same with any of the numbers A, B, and that it is measured by A.

9.prop.13.p2910

And, since F measures C according to G therefore F by multiplying G has made C.

9.prop.13.p2911

But, further, A has also by multiplying B made C; [IX. 11] therefore the product of A, B is equal to the product of F, G.

9.prop.13.p2912

Therefore, proportionally, as A is to F, so is G to B. [VII. 19]

9.prop.13.p2913

But A measures F; therefore G also measures B.

9.prop.13.p2914

Let it measure it according to H.

9.prop.13.p2915

Similarly then we can prove that H is not the same with A.

9.prop.13.p2916

And, since G measures B according to H, therefore G by multiplying H has made B.

9.prop.13.p2917

But further A has also by multiplying itself made B; [IX. 8] therefore the product of H, G is equal to the square on A.

9.prop.13.p2918

Therefore, as H is to A, so is A to G. [VII. 19]

9.prop.13.p2919

But A measures G;lt therefore H also measures A, which is prime, though it is not the same with it: which is absurd.

9.prop.13.p2920

Therefore D the greatest will not be measured by any other number except A, B, C. Q. E. D.

9.prop.14.p2921

If a number be the least that is measured by prime numbers, it will not be measured by any other prime number except those originally measuring it.

9.prop.14.p2922

For let the number A be the least that is measured by the prime numbers B, C, D; I say that A will not be measured by any other prime number except B, C, D.

9.prop.14.p2923

For, if possible, let it be measured by the prime number E, and let E not be the same with any one of the numbers B, C, D.

9.prop.14.p2924

Now, since E measures A, let it measure it according to F; therefore E by multiplying F has made A.

9.prop.14.p2925

And A is measured by the prime numbers B, C, D.

9.prop.14.p2926

But, if two numbers by multiplying one another make some number, and any prime number measure the product, it will also measure one of the original numbers; [VII. 30] therefore B, C, D will measure one of the numbers E, F.

9.prop.14.p2927

Now they will not measure E; for E is prime and not the same with any one of the numbers B, C, D.

9.prop.14.p2928

Therefore they will measure F, which is less than A: which is impossible, for A is by hypothesis the least number measured by B, C, D.

9.prop.14.p2929

Therefore no prime number will measure A except B, C, D. Q. E. D.

9.prop.15.p2930

If three numbers in continued proportion be the least of those which have the same ratio with them, any two whatever added together will be prime to the remaining number.

9.prop.15.p2931

Let A, B, C, three numbers in continued proportion, be the least of those which have the same ratio with them; I say that any two of the numbers A, B, C whatever added together are prime to the remaining number, namely A, B to C; B, C to A; and further A, C to B.

9.prop.15.p2932

For let two numbers DE, EF, the least of those which have the same ratio with A, B, C, be taken. [VIII. 2]

9.prop.15.p2933

It is then manifest that DE by multiplying itself has made A, and by multiplying EF has made B, and, further, EF by multiplying itself has made C. [VIII. 2]

9.prop.15.p2934

Now, since DE, EF are least, they are prime to one another. [VII. 22]

9.prop.15.p2935

But, if two numbers be prime to one another, their sum is also prime to each; [VII. 28] therefore DF is also prime to each of the numbers DE, EF.

9.prop.15.p2936

But further DE is also prime to EF; therefore DF, DE are prime to EF.

9.prop.15.p2937

But, if two numbers be prime to any number, their product is also prime to the other; [VII. 24] so that the product of FD, DE is prime to EF; hence the product of FD, DE is also prime to the square on EF. [VII. 25]

9.prop.15.p2938

But the product of FD, DE is the square on DE together with the product of DE, EF; [II. 3] therefore the square on DE together with the product of DE, EF is prime to the square on EF.

9.prop.15.p2939

And the square on DE is A, the product of DE, EF is B, and the square on EF is C; therefore A, B added together are prime to C.

9.prop.15.p2940

Similarly we can prove that B, C added together are prime to A.

9.prop.15.p2941

I say next that A, C added together are also prime to B.

9.prop.15.p2942

For, since DF is prime to each of the numbers DE, EF, the square on DF is also prime to the product of DE, EF. [VII. 24, 25]

9.prop.15.p2943

But the squares on DE, EF together with twice the product of DE, EF are equal to the square on DF; [II. 4] therefore the squares on DE, EF together with twice the product of DE, EF are prime to the product of DE, EF.

9.prop.15.p2944

Separando, the squares on DE, EF together with once the product of DE, EF are prime to the product of DE, EF.

9.prop.15.p2945

Therefore, separando again, the squares on DE, EF are prime to the product of DE, EF.

9.prop.15.p2946

And the square on DE is A, the product of DE, EF is B, and the square on EF is C.

9.prop.15.p2947

Therefore A, C added together are prime to B. Q. E. D.

9.prop.16.p2948

If two numbers be prime to one another, the second will not be to any other number as the first is to the second.

9.prop.16.p2949

For let the two numbers A, B be prime to one another; I say that B is not to any other number as A is to B.

9.prop.16.p2950

For, if possible, as A is to B, so let B be to C.

9.prop.16.p2951

Now A, B are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore A measures B as antecedent antecedent.

9.prop.16.p2952

But it also measures itself; therefore A measures A, B which are prime to one another: which is absurd.

9.prop.16.p2953

Therefore B will not be to C, as A is to B. Q. E. D.

9.prop.17.p2954

If there be as many numbers as we please in continued proportion, and the extremes of them be prime to one another, the last will not be to any other number as the first to the second.

9.prop.17.p2955

For let there be as many numbers as we please, A, B, C, D, in continued proportion, and let the extremes of them, A, D, be prime to one another; I say that D is not to any other number as A is to B.

9.prop.17.p2956

For, if possible, as A is to B, so let D be to E; therefore, alternately, as A is to D, so is B to E. [VII. 13]

9.prop.17.p2957

But A, D are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent. [VII. 20]

9.prop.17.p2958

Therefore A measures B.

9.prop.17.p2959

And, as A is to B, so is B to C.

9.prop.17.p2960

Therefore B also measures C; so that A also measures C.

9.prop.17.p2961

And since, as B is to C, so is C to D, and B measures C, therefore C also measures D.

9.prop.17.p2962

But A measured C; so that A also measures D.

9.prop.17.p2963

But it also measures itself; therefore A measures A, D which are prime to one another : which is impossible.

9.prop.17.p2964

Therefore D will not be to any other number as A is to B. Q. E. D.

9.prop.18.p2965

Given two numbers, to investigate whether it is possible to find a third proportional to them.

9.prop.18.p2966

Let A, B be the given two numbers, and let it be required to investigate whether it is possible to find a third proportional to them.

9.prop.18.p2967

Now A, B are either prime to one another or not.

9.prop.18.p2968

And, if they are prime to one another, it has been proved that it is impossible to find a third proportional to them. [IX. 16]

9.prop.18.p2969

Next, let A, B not be prime to one another, and let B by multiplying itself make C.

9.prop.18.p2970

Then A either measures C or does not measure it.

9.prop.18.p2971

First, let it measure it according to D; therefore A by multiplying D has made C.

9.prop.18.p2972

But, further, B has also by multiplying itself made C; therefore the product of A, D is equal to the square on B.

9.prop.18.p2973

Therefore, as A is to B, so is B to D; [VII. 19] therefore a third proportional number D has been found to A, B.

9.prop.18.p2974

Next, let A not measure C; I say that it is impossible to find a third proportional number to A, B.

9.prop.18.p2975

For, if possible, let D, such third proportional, have been found.

9.prop.18.p2976

Therefore the product of A, D is equal to the square on B.

9.prop.18.p2977

But the square on B is C; therefore the product of A, D is equal to C.

9.prop.18.p2978

Hence A by multiplying D has made C; therefore A measures C according to D.

9.prop.18.p2979

But, by hypothesis, it also does not measure it: which is absurd.

9.prop.18.p2980

Therefore it is not possible to find a third proportional number to A, B when A does not measure C. Q. E. D.

9.prop.19.p2981

Given three numbers, to investigate when it is possible to find a fourth proportional to them.

9.prop.19.p2982

Let A, B, C be the given three numbers, and let it be required to investigate when it is possible to find a fourth proportional to them.

9.prop.19.p2983

Now either they are not in continued proportion, and the extremes of them are prime to one another; or they are in continued proportion, and the extremes of them are not prime to one another; or they are not in continued proportion, nor are the extremes of them prime to one another; or they are in continued proportion, and the extremes of them are prime to one another.

9.prop.19.p2984

If then A, B, C are in continued proportion, and the extremes of them A, C are prime to one another, it has been proved that it is impossible to find a fourth proportional number to them. [IX. 17]

9.prop.19.p2985

†Next, let A, B, C not be in continued proportion, the extremes being again prime to one another; I say that in this case also it is impossible to find a fourth proportional to them.

9.prop.19.p2986

For, if possible, let D have been found, so that, as A is to B, so is C to D, and let it be contrived that, as B is to C, so is D to E.

9.prop.19.p2987

Now, since, as A is to B, so is C to D, and, as B is to C, so is D to E, therefore, ex aequali, as A is to C, so is C to E. [VII. 14]

9.prop.19.p2988

But A, C are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio, the antecedent the antecedent and the consequent the consequent. [VII. 20]

9.prop.19.p2989

Therefore A measures C as antecedent antecedent.

9.prop.19.p2990

But it also measures itself; therefore A measures A, C which are prime to one another: which is impossible.

9.prop.19.p2991

Therefore it is not possible to find a fourth proportional to A, B, C.†

9.prop.19.p2992

Next, let A, B, C be again in continued proportion, but let A, C not be prime to one another.

9.prop.19.p2993

I say that it is possible to find a fourth proportional to them.

9.prop.19.p2994

For let B by multiplying C make D; therefore A either measures D or does not measure it.

9.prop.19.p2995

First, let it measure it according to E; therefore A by multiplying E has made D.

9.prop.19.p2996

But, further, B has also by multiplying C made D; therefore the product of A, E is equal to the product of B, C; therefore, proportionally, as A is to B, so is C to E; [VII. 19] therefore E has been found a fourth proportional to A, B, C.

9.prop.19.p2997

Next, let A not measure D; I say that it is impossible to find a fourth proportional number to A, B, C.

9.prop.19.p2998

For, if possible, let E have been found; therefore the product of A, E is equal to the product of B, C. [VII. 19]

9.prop.19.p2999

But the product of B, C is D; therefore the product of A, E is also equal to D.

9.prop.19.p3000

Therefore A by multiplying E has made D; therefore A measures D according to E, so that A measures D.

9.prop.19.p3001

But it also does not measure it: which is absurd.

9.prop.19.p3002

Therefore it is not possible to find a fourth proportional number to A, B, C when A does not measure D.

9.prop.19.p3003

Next, let A, B, C not be in continued proportion, nor the extremes prime to one another.

9.prop.19.p3004

And let B by multiplying C make D.

9.prop.19.p3005

Similarly then it can be proved that, if A measures D, it is possible to find a fourth proportional to them, but, if it does not measure it, impossible. Q. E. D.

9.prop.20.p3006

Prime numbers are more than any assigned multitude of prime numbers.

9.prop.20.p3007

Let A, B, C be the assigned prime numbers; I say that there are more prime numbers than A, B, C.

9.prop.20.p3008

For let the least number measured by A, B, C be taken, and let it be DE; let the unit DF be added to DE.

9.prop.20.p3009

Then EF is either prime or not.

9.prop.20.p3010

First, let it be prime; then the prime numbers A, B, C, EF have been found which are more than A, B, C.

9.prop.20.p3011

Next, let EF not be prime; therefore it is measured by some prime number. [VII. 31]

9.prop.20.p3012

Let it be measured by the prime number G.

9.prop.20.p3013

I say that G is not the same with any of the numbers A, B, C.

9.prop.20.p3014

For, if possible, let it be so.

9.prop.20.p3015

Now A, B, C measure DE; therefore G also will measure DE.

9.prop.20.p3016

But it also measures EF.

9.prop.20.p3017

Therefore G, being a number, will measure the remainder, the unit DF: which is absurd.

9.prop.20.p3018

Therefore G is not the same with any one of the numbers A, B, C.

9.prop.20.p3019

And by hypothesis it is prime.

9.prop.20.p3020

Therefore the prime numbers A, B, C, G have been found which are more than the assigned multitude of A, B, C. Q. E. D.

9.prop.21.p3021

If as many even numbers as we please be added together, the whole is even.

9.prop.21.p3022

For let as many even numbers as we please, AB, BC, CD, DE, be added together; I say that the whole AE is even.

9.prop.21.p3023

For, since each of the numbers AB, BC, CD, DE is even, it has a half part; [VII. Def. 6] so that the whole AE also has a half part.

9.prop.21.p3024

But an even number is that which is divisible into two equal parts; [id.] therefore AE is even. Q. E. D.

9.prop.22.p3025

If as many odd numbers as we please be added together, and their multitude be even, the whole will be even.

9.prop.22.p3026

For let as many odd numbers as we please, AB, BC, CD, DE, even in multitude, be added together; I say that the whole AE is even.

9.prop.22.p3027

For, since each of the numbers AB, BC, CD, DE is odd, if an unit be subtracted from each, each of the remainders will be even; [VII. Def. 7] so that the sum of them will be even. [IX. 21]

9.prop.22.p3028

But the multitude of the units is also even.

9.prop.22.p3029

Therefore the whole AE is also even. [IX. 21] Q. E. D.

9.prop.23.p3030

If as many odd numbers as we please be added together, and their multitude be odd, the whole will also be odd.

9.prop.23.p3031

For let as many odd numbers as we please, AB, BC, CD, the multitude of which is odd, be added together; I say that the whole AD is also odd.

9.prop.23.p3032

Let the unit DE be subtracted from CD; therefore the remainder CE is even. [VII. Def. 7]

9.prop.23.p3033

But CA is also even; [IX. 22] therefore the whole AE is also even. [IX. 21]

9.prop.23.p3034

And DE is an unit.

9.prop.23.p3035

Therefore

9.prop.23.p3035

AD

9.prop.23.p3035

is odd. [

9.prop.23.p3035

VII. Def. 7

9.prop.23.p3035

] Q. E. D.

9.prop.23.p3035

1

9.prop.24.p3036

If from an even number an even number be subtracted, the remainder will be even.

9.prop.24.p3037

For from the even number AB let the even number BC be subtracted: I say that the remainder CA is even.

9.prop.24.p3038

For, since AB is even, it has a half part. [VII. Def. 6]

9.prop.24.p3039

For the same reason BC also has a half part; so that the remainder [CA also has a half part, and] AC is therefore even. Q. E. D.

9.prop.25.p3040

If from an even number an odd number be subtracted, the remainder will be odd.

9.prop.25.p3041

For from the even number AB let the odd number BC be subtracted; I say that the remainder CA is odd.

9.prop.25.p3042

For let the unit CD be subtracted from BC; therefore DB is even. [VII. Def. 7]

9.prop.25.p3043

But AB is also even; therefore the remainder AD is also even. [IX. 24]

9.prop.25.p3044

And CD is an unit; therefore CA is odd. [VII. Def. 7] Q. E. D.

9.prop.26.p3045

If from an odd number an odd number be subtracted, the remainder will be even.

9.prop.26.p3046

For from the odd number AB let the odd number BC be subtracted; I say that the remainder CA is even.

9.prop.26.p3047

For, since AB is odd, let the unit BD be subtracted; therefore the remainder AD is even. [VII. Def. 7]

9.prop.26.p3048

For the same reason CD is also even; [VII. Def. 7] so that the remainder CA is also even. [IX. 24] Q. E. D.

9.prop.27.p3049

If from an odd number an even number be subtracted, the remainder will be odd.

9.prop.27.p3050

For from the odd number AB let the even number BC be subtracted; I say that the remainder CA is odd.

9.prop.27.p3051

Let the unit AD be subtracted; therefore DB is even. [VII. Def. 7]

9.prop.27.p3052

But BC is also even; therefore the remainder CD is even. [ IX. 24 ]

9.prop.27.p3053

Therefore CA is odd. [VII. Def. 7] Q. E. D.

9.prop.28.p3054

If an odd number by multiplying an even number make some number, the product will be even.

9.prop.28.p3055

For let the odd number A by multiplying the even number B make C; I say that C is even.

9.prop.28.p3056

For, since A by multiplying B has made C, therefore C is made up of as many numbers equal to B as there are units in A. [VII. Def. 15]

9.prop.28.p3057

And B is even; therefore C is made up of even numbers.

9.prop.28.p3058

But, if as many even numbers as we please be added together, the whole is even. [IX. 21]

9.prop.28.p3059

Therefore C is even. Q. E. D.

9.prop.29.p3060

If an odd number by multiplying an odd number make some number, the product will be odd.

9.prop.29.p3061

For let the odd number A by multiplying the odd number B make C; I say that C is odd.

9.prop.29.p3062

For, since A by multiplying B has made C, therefore C is made up of as many numbers equal to B as there are units in A. [VII. Def. 15]

9.prop.29.p3063

And each of the numbers A, B is odd; therefore C is made up of odd numbers the multitude of which is odd.

9.prop.29.p3064

Thus C is odd. [IX. 23] Q. E. D.

9.prop.30.p3065

If an odd number measure an even number, it will also measure the half of it.

9.prop.30.p3066

For let the odd number A measure the even number B; I say that it will also measure the half of it.

9.prop.30.p3067

For, since A measures B, let it measure it according to C; I say that C is not odd.

9.prop.30.p3068

For, if possible, let it be so.

9.prop.30.p3069

Then, since A measures B according to C, therefore A by multiplying C has made B.

9.prop.30.p3070

Therefore B is made up of odd numbers the multitude of which is odd.

9.prop.30.p3071

Therefore B is odd: [IX. 23] which is absurd, for by hypothesis it is even.

9.prop.30.p3072

Therefore C is not odd; therefore C is even.

9.prop.30.p3073

Thus A measures B an even number of times.

9.prop.30.p3074

For this reason then it also measures the half of it. Q. E. D.

9.prop.31.p3075

If an odd number be prime to any number, it will also be prime to the double of it.

9.prop.31.p3076

For let the odd number A be prime to any number B, and let C be double of B; I say that A is prime to C.

9.prop.31.p3077

For, if they are not prime to one another, some number will measure them.

9.prop.31.p3078

Let a number measure them, and let it be D.

9.prop.31.p3079

Now A is odd; therefore D is also odd.

9.prop.31.p3080

And since D which is odd measures C, and C is even, therefore [D] will measure the half of C also. [IX. 30]

9.prop.31.p3081

But B is half of C; therefore D measures B.

9.prop.31.p3082

But it also measures A; therefore D measures A, B which are prime to one another: which is impossible.

9.prop.31.p3083

Therefore A cannot but be prime to C.

9.prop.31.p3084

Therefore A, C are prime to one another. Q. E. D.

9.prop.32.p3085

Each of the numbers which are continually doubled beginning from a dyad is even-times even only.

9.prop.32.p3086

For let as many numbers as we please, B, C, D, have been continually doubled beginning from the dyad A; I say that B, C, D are eventimes even only.

9.prop.32.p3087

Now that each of the numbers B, C, D is even-times even is manifest; for it is doubled from a dyad.

9.prop.32.p3088

I say that it is also even-times even only.

9.prop.32.p3089

For let an unit be set out.

9.prop.32.p3090

Since then as many numbers as we please beginning from an unit are in continued proportion, and the number A after the unit is prime, therefore D, the greatest of the numbers A, B, C, D, will not be measured by any other number except A, B, C. [IX. 13]

9.prop.32.p3091

And each of the numbers A, B, C is even; therefore D is even-times even only. [VII. Def. 8]

9.prop.32.p3092

Similarly we can prove that each of the numbers B, C is even-times even only. Q. E. D.

9.prop.33.p3093

If a number have its half odd, it is even-times odd only.

9.prop.33.p3094

For let the number A have its half odd; I say that A is even-times odd only.

9.prop.33.p3095

Now that it is even-times odd is manifest; for the half of it, being odd, measures it an even number of times. [VII. Def. 9]

9.prop.33.p3096

I say next that it is also even-times odd only.

9.prop.33.p3097

For, if A is even-times even also, it will be measured by an even number according to an even number; [VII. Def. 8] so that the half of it will also be measured by an even number though it is odd: which is absurd.

9.prop.33.p3098

Therefore A is even-times odd only. Q. E. D.

9.prop.34.p3099

If a number neither be one of those which are continually doubled from a dyad, nor have its half odd, it is both eventimes even and even-times odd.

9.prop.34.p3100

For let the number A neither be one of those doubled from a dyad, nor have its half odd; I say that A is both even-times even and even-times odd.

9.prop.34.p3101

Now that A is even-times even is manifest; for it has not its half odd. [VII. Def. 8]

9.prop.34.p3102

I say next that it is also even-times odd.

9.prop.34.p3103

For, if we bisect A, then bisect its half, and do this continually, we shall come upon some odd number which will measure A according to an even number.

9.prop.34.p3104

For, if not, we shall come upon a dyad, and A will be among those which are doubled from a dyad: which is contrary to the hypothesis.

9.prop.34.p3105

Thus A is even-times odd.

9.prop.34.p3106

But it was also proved even-times even.

9.prop.34.p3107

Therefore A is both even-times even and even-times odd. Q. E. D.

9.prop.35.p3108

If as many numbers as we please be in continued proportion, and there be subtracted from the second and the last numbers equal to the first, then, as the excess of the second is to the first, so will the excess of the last be to all those before it.

9.prop.35.p3109

Let there be as many numbers as we please in continued proportion, A, BC, D, EF, beginning from A as least, and let there be subtracted from BC and EF the numbers BG, FH, each equal to A; I say that, as GC is to A, so is EH to A, BC, D.

9.prop.35.p3110

For let FK be made equal to BC, and FL equal to D.

9.prop.35.p3111

Then, since FK is equal to BC, and of these the part FH is equal to the part BG, therefore the remainder HK is equal to the remainder GC.

9.prop.35.p3112

And since, as EF is to D, so is D to BC, and BC to A, while D is equal to FL, BC to FK, and A to FH, therefore, as EF is to FL, so is LF to FK, and FK to FH.

9.prop.35.p3113

Separando, as EL is to LF, so is LK to FK, and KH to FH. [VII. 11, 13]

9.prop.35.p3114

Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [VII. 12] therefore, as KH is to FH, so are EL, LK. KH to LF, FK, HF.

9.prop.35.p3115

But KH is equal to CG, FH to A, and LF, FK, HF to D, BC, A; therefore, as CG is to A, so is EH to D, BC, A.

9.prop.35.p3116

Therefore, as the excess of the second is to the first, so is the excess of the last to all those before it. Q. E. D.

9.prop.36.p3117

If as many numbers as we please beginning from an unit be set out continuously in double proportion, until the sum of all becomes prime, and if the sum multiplied into the last make some number, the product will be perfect.

9.prop.36.p3118

For let as many numbers as we please, A, B, C, D, beginning from an unit be set out in double proportion, until the sum of all becomes prime, let E be equal to the sum, and let E by multiplying D make FG; I say that FG is perfect.

9.prop.36.p3119

For, however many A, B, C, D are in multitude, let so many E, HK, L, M be taken in double proportion beginning from E; therefore, ex aequali, as A is to D, so is E to M. [VII. 14]

9.prop.36.p3120

Therefore the product of E, D is equal to the product of A, M. [VII. 19]

9.prop.36.p3121

And the product of E, D is FG; therefore the product of A, M is also FG.

9.prop.36.p3122

Therefore A by multiplying M has made FG; therefore M measures FG according to the units in A.

9.prop.36.p3123

And A is a dyad; therefore FG is double of M.

9.prop.36.p3124

But M, L, HK, E are continuously double of each other; therefore E, HK, L, M, FG are continuously proportional in double proportion.

9.prop.36.p3125

Now let there be subtracted from the second HK and the last FG the numbers HN, FO, each equal to the first E; therefore, as the excess of the second is to the first, so is the excess of the last to all those before it. [IX. 35]

9.prop.36.p3126

Therefore, as NK is to E, so is OG to M, L, KH, E.

9.prop.36.p3127

And NK is equal to E; therefore OG is also equal to M, L, HK, E.

9.prop.36.p3128

But FO is also equal to E, and E is equal to A, B, C, D and the unit.

9.prop.36.p3129

Therefore the whole FG is equal to E, HK, L, M and A, B, C, D and the unit; and it is measured by them.

9.prop.36.p3130

I say also that FG will not be measured by any other number except A, B, C, D, E, HK, L, M and the unit.

9.prop.36.p3131

For, if possible, let some number P measure FG, and let P not be the same with any of the numbers A, B, C, D, E, HK, L, M.

9.prop.36.p3132

And, as many times as P measures FG, so many units let there be in Q; therefore Q by multiplying P has made FG.

9.prop.36.p3133

But, further, E has also by multiplying D made FG; therefore, as E is to Q, so is P to D. [VII. 19]

9.prop.36.p3134

And, since A, B, C, D are continuously proportional beginning from an unit, therefore D will not be measured by any other number except A, B, C. [IX. 13]

9.prop.36.p3135

And, by hypothesis, P is not the same with any of the numbers A, B, C; therefore P will not measure D.

9.prop.36.p3136

But, as P is to D, so is E to Q; therefore neither does E measure Q. [VII. Def. 20]

9.prop.36.p3137

And E is prime; and any prime number is prime to any number which it does not measure. [VII. 29]

9.prop.36.p3138

Therefore E, Q are prime to one another.

9.prop.36.p3139

But primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20] and, as E is to Q, so is P to D; therefore E measures P the same number of times that Q measures D.

9.prop.36.p3140

But D is not measured by any other number except A, B, C; therefore Q is the same with one of the numbers A, B, C.

9.prop.36.p3141

Let it be the same with B.

9.prop.36.p3142

And, however many B, C, D are in multitude, let so many E, HK, L be taken beginning from E.

9.prop.36.p3143

Now E, HK, L are in the same ratio with B, C, D; therefore, ex aequali, as B is to D, so is E to L. [VII. 14]

9.prop.36.p3144

Therefore the product of B, L is equal to the product of D, E. [VII. 19]

9.prop.36.p3145

But the product of D, E is equal to the product of Q, P; therefore the product of Q, P is also equal to the product of B, L.

9.prop.36.p3146

Therefore, as Q is to B, so is L to P. [VII. 19]

9.prop.36.p3147

And Q is the same with B; therefore L is also the same with P; which is impossible, for by hypothesis P is not the same with any of the numbers set out.

9.prop.36.p3148

Therefore no number will measure FG except A, B, C, D, E, HK, L, M and the unit.

9.prop.36.p3149

And FG was proved equal to A, B, C, D, E, HK, L, M and the unit; and a perfect number is that which is equal to its own parts; [VII. Def. 22] therefore FG is perfect. Q. E. D.

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