Theorem 119 (Euclid IX.8). 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.
Proof: Let them be the unit, A, B, C, D, E, F. The unit is to A as A to B, so A by multiplying itself makes B: B is square. B, C, D are in continued proportion and B is square, so D is square (Theorem 106); likewise F. Again the unit is to A as B to C, so A by multiplying B makes C: C is cube. C, D, E, F are in continued proportion and C is cube, so F is cube (Theorem 107). And F was shown square.
Theorem 120 (Euclid IX.9). 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.
Proof: Let them be the unit, A, B, C, D, E, F, with A square. B is square, and those leaving out one (Theorem 119). A, B, C are in continued proportion and A is square, so C is square (Theorem 106); likewise the rest. Next let A be cube. C is cube, and those leaving out two (Theorem 119). A by multiplying itself makes B, so B is cube (Theorem 114). A, B, C, D are in continued proportion and A is cube, so D is cube (Theorem 107); likewise the rest.
Theorem 121 (Euclid IX.10). 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.
Proof: Let them be the unit, A, B, C, D, E, F, with A not square. If possible, let C be square. B is square (Theorem 119), and A : B = B : C, so A has to B the ratio of a square to a square; B is square, so A is square (Theorem 108): contrary to the hypothesis. Likewise for the others. Next let A be not cube. If possible, let D be cube. C is cube (Theorem 119), and B : C = C : D, so B is cube (Theorem 109). And A by multiplying itself makes B, so A is cube (Theorem 117): contrary to the hypothesis.
Theorem 122 (Euclid IX.11). 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.
Proof: Let them be the unit A, B, C, D, E. The unit is to B as D to E, so the unit measures B the same number of times that D measures E. Alternately, the unit measures D the same number of times that B measures E (Theorem 55). So B measures E according to the units in D.
Porism. Whatever place the measuring number has from the unit, the same place has the number according to which it measures, counting from the number measured toward the one before it.
Theorem 123 (Euclid IX.12). 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.
Proof: Let them be the unit, A, B, C, D, and let the prime E measure D. If E does not measure A, then E, A are prime to one another (Theorem 68). Let E measure D according to F. A measures D according to C (Theorem 122). So the product of A, C is equal to the product of E, F, and A : E = F : C (Theorem 58). A, E are prime, so least, so E measures C (Theorem 60, Theorem 59); let it be according to G. A by multiplying B makes C (Theorem 122), so A : E = G : B, and E measures B; let it be according to H. A by multiplying itself makes B, so E : A = A : H, and E measures A. But it also does not: which is impossible.
Theorem 124 (Euclid IX.13). 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.
Proof: Let them be the unit, A, B, C, D, with A prime. If possible, let E measure D, E being none of A, B, C. E is not prime: if it were, measuring D it would measure A (Theorem 123), which is prime and not the same as E. So E is composite, and some prime measures it (Theorem 70); and no prime other than A, since such a prime would measure D and so A (Theorem 123). So A measures E. Let E measure D according to F. F is none of A, B, C; otherwise E would be one of them (Theorem 122). In the same way A measures F. A by multiplying C makes D, so A : E = F : C (Theorem 58); A measures E, so F measures C, according to G. G is none of A, B, and A measures G. A by multiplying B makes C, so A : F = G : B; A measures F, so G measures B, according to H. H is not A. A by multiplying itself makes B, so H : A = A : G; A measures G, so H measures A, which is prime, H not being the same as it: which is impossible.
Definition 24 (Euclid VII, Def. 8). An even-times even number is that which is measured by an even number according to an even number.
Definition 25 (Euclid VII, Def. 9). An even-times odd number is that which is measured by an even number according to an odd number.
Definition 26 (Euclid VII, Def. 10). An odd-times odd number is that which is measured by an odd number according to an odd number.
Theorem 125 (Euclid IX.32). Each of the numbers which are continually doubled beginning from a dyad is even-times even only.
Proof: Let A be the dyad and B, C, D be doubled from it. That each is even-times even is manifest. Set out a unit. The unit, A, B, C, D are in continued proportion and A is prime, so D is measured by none except A, B, C (Theorem 124). Each of these is even. So D is even-times even only.
Theorem 126 (Euclid IX.33). If a number have its half odd, it is even-times odd only.
Proof: It is even-times odd, since its half, being odd, measures it an even number of times. If it were also even-times even, its half would be measured by an even number, though odd: which is absurd.
Theorem 127 (Euclid IX.34). If a number neither be one of those which are continually doubled from a dyad, nor have its half odd, it is both even-times even and even-times odd.
Proof: Let A be such. Its half is not odd, so A is even-times even. Bisect A, then its half, and so on continually. We shall come upon some odd number which measures A according to an even number. For if not, we shall come upon a dyad, and A will be among those doubled from a dyad: contrary to the hypothesis. So A is even-times odd also.
The proof supposes: that the bisection ends (Theorem 22).
Theorem 128 (Euclid IX.35). 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.
Proof: Let A, BC, D, EF be in continued proportion beginning from the least A, and let BG, FH, each equal to A, be subtracted from BC, EF. Make FK equal to BC and FL equal to D. Then HK is equal to GC. Since EF : D = D : BC = BC : A, we have EF : FL = LF : FK = FK : FH. Separando, EL : LF = LK : FK = KH : FH. So as one antecedent is to one consequent, so are all to all (Theorem 52): KH : FH = (EL, LK, KH together) : (LF, FK, HF together). KH is equal to CG, FH to A, and LF, FK, HF to D, BC, A. So CG : A = EH : (D, BC, A together).
The proof supposes: separando for numbers, which is Theorem 51 with Theorem 53.
Definition 27 (Euclid VII, Def. 22). A perfect number is that which is equal to its own parts.
Theorem 129 (Euclid IX.36). 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.
Proof: Let the unit, A, B, C, D be set out in double proportion, let their sum E be prime, and let E by multiplying D make FG. Take as many as A, B, C, D in double proportion beginning from E: E, HK, L, M. Ex aequali, A : D = E : M (Theorem 54), so the product of E, D is equal to the product of A, M (Theorem 58). So A by multiplying M makes FG, and A is a dyad, so FG is double of M. So E, HK, L, M, FG are in continued double proportion.
Subtract from the second HK and the last FG the numbers HN, FO, each equal to the first E. Then NK : E = OG : (M, L, KH, E together) (Theorem 128). NK is equal to E, so OG is equal to M, L, HK, E together. And FO is equal to E, that is, to the unit, A, B, C, D together. So the whole FG is equal to E, HK, L, M and A, B, C, D and the unit together. And it is measured by each of them.
No other number measures FG. If possible, let P measure it, P being none of these, according to Q. Then the product of Q, P is equal to the product of E, D, so E : Q = P : D (Theorem 58). The unit, A, B, C, D are in continued proportion with A prime, so D is measured by none except A, B, C (Theorem 124). P is none of these, so P does not measure D, and so E does not measure Q. E is prime, so E, Q are prime to one another (Theorem 68), so least (Theorem 60), and E measures P the same number of times that Q measures D (Theorem 59). So Q is one of A, B, C; let it be B. Take as many as B, C, D beginning from E: E, HK, L. Ex aequali, B : D = E : L, so the product of B, L is equal to the product of D, E, that is, of Q, P. Q is B, so L is equal to P. But P is none of the numbers set out: which is impossible. So FG is equal to all its parts, and is perfect.