Measure, Part, and Multiple

Theorem 41 (Euclid VII.1). Two unequal numbers being set out, and the less being continually subtracted in turn from the greater, if the number which is left never measures the one before it until a unit is left, the original numbers will be prime to one another.

Proof: Let the numbers be AB, CD. If they are not prime to one another, some number E measures them. Let CD, measuring BF, leave FA less than itself; let AF, measuring DG, leave GC less than itself; and let GC, measuring FH, leave the unit HA. E measures CD and CD measures BF, so E measures BF. It measures the whole BA, so it measures the remainder AF. AF measures DG, so E measures DG. It measures the whole DC, so it measures the remainder CG. CG measures FH, so E measures FH. It measures the whole FA, so it measures the remaining unit AH, though it is a number: which is impossible.

The proof supposes: a number that measures two numbers measures their difference, and measures whatever either of them measures (Theorem 32). Three subtractions stand for however many there are (Axiom 3).

Theorem 42 (Euclid VII.2). Given two numbers not prime to one another, to find their greatest common measure.

Proof: Let them be AB, CD, with CD the less. If CD measures AB, it is a common measure, since it measures itself, and the greatest, since nothing greater than CD measures CD.

If not, then, the less being continually subtracted from the greater, some number will be left which measures the one before it. For a unit will not be left; otherwise AB, CD would be prime to one another (Theorem 41). Let CD, measuring BE, leave EA less than itself; let EA, measuring DF, leave FC less than itself; and let CF measure AE. CF measures AE and AE measures DF, so CF measures DF. It measures itself, so it measures the whole CD. CD measures BE, so CF measures BE. It measures EA, so it measures the whole BA. So CF is a common measure.

It is the greatest. If not, let G, greater than CF, measure AB, CD. G measures CD and so BE; it measures the whole BA, so the remainder AE; so DF; it measures the whole DC, so the remainder CF, the greater the less: which is impossible.

Porism. If a number measures two numbers, it also measures their greatest common measure.

The proof supposes: that the subtraction comes to an end. The remainders are each less than the one before, and among numbers one is least (Theorem 22).

Theorem 43 (Euclid VII.3). Given three numbers not prime to one another, to find their greatest common measure.

Proof: Let them be A, B, C. Let D be the greatest common measure of A, B (Theorem 42).

If D measures C, then D measures A, B, C. It is the greatest: if E, greater than D, measured A, B, C, it would measure D (Theorem 42, Porism), the greater the less: which is impossible.

If D does not measure C: C and D are not prime to one another. For some number measures A, B, C; measuring A, B, it measures D (Theorem 42, Porism); and it measures C. Let E be the greatest common measure of C, D. E measures D and D measures A, B, so E measures A, B; and it measures C. It is the greatest: if F, greater than E, measured A, B, C, it would measure D (Theorem 42, Porism), and C, and so E (Theorem 42, Porism), the greater the less: which is impossible.

Theorem 44 (Euclid VII.4). Any number is either a part or parts of any number, the less of the greater.

Proof: Let A, BC be two numbers, BC the less. They are either prime to one another or not.

If prime: divide BC into the units in it. Each unit is a part of A, so BC is parts of A.

If not prime: BC either measures A or does not. If it measures A, it is a part of A. If not, let D be the greatest common measure of A, BC (Theorem 42), and divide BC into the numbers BE, EF, FC equal to D. D measures A, so D is a part of A. So each of BE, EF, FC is a part of A, and BC is parts of A.

Theorem 45 (Euclid VII.5). If a number be a part of a number, and another be the same part of another, the sum will also be the same part of the sum that the one is of the one.

Proof: Let A be a part of BC, and D the same part of EF. As many numbers as there are in BC equal to A, so many are there in EF equal to D. Divide BC into BG, GC equal to A, and EF into EH, HF equal to D. BG, EH together are equal to A, D together, and likewise GC, HF. So as many numbers as there are in BC equal to A, so many are there in BC, EF together equal to A, D together.

Theorem 46 (Euclid VII.6). If a number be parts of a number, and another be the same parts of another, the sum will also be the same parts of the sum that the one is of the one.

Proof: Let AB be parts of C, and DE the same parts of F. Divide AB into the parts of C, namely AG, GB, and DE into the parts of F, namely DH, HE. AG is the same part of C that DH is of F, so AG, DH together are the same part of C, F together (Theorem 45). Likewise GB, HE. So AB, DE together are the same parts of C, F together that AB is of C.

Theorem 47 (Euclid VII.7). If a number be that part of a number, which a number subtracted is of a number subtracted, the remainder will also be the same part of the remainder that the whole is of the whole.

Proof: Let AB be that part of CD which AE subtracted is of CF subtracted. Let EB be the same part of CG that AE is of CF. Then AB is the same part of GF that AE is of CF (Theorem 45). But AB is that same part of CD. So GF is equal to CD. Subtract CF from each: GC is equal to FD. So EB is the same part of FD that AE is of CF, that is, that AB is of CD.

Theorem 48 (Euclid VII.8). If a number be the same parts of a number that a number subtracted is of a number subtracted, the remainder will also be the same parts of the remainder that the whole is of the whole.

Proof: Let AB be the same parts of CD that AE subtracted is of CF subtracted. Make GH equal to AB. Divide GH into the parts of CD, namely GK, KH, and AE into the parts of CF, namely AL, LE. GK is the same part of CD that AL is of CF, and CD is greater than CF, so GK is greater than AL. Make GM equal to AL. Then the remainder MK is the same part of the remainder FD that GK is of CD (Theorem 47). Likewise, making KN equal to LE, the remainder NH is the same part of FD that KH is of CD. So MK, NH together are the same parts of FD that GH is of CD. And MK, NH together are equal to EB.

Theorem 49 (Euclid VII.9). If a number be a part of a number, and another be the same part of another, alternately also, whatever part or parts the first is of the third, the same part, or the same parts, will the second also be of the fourth.

Proof: Let A be a part of BC, and D the same part of EF. Divide BC into BG, GC equal to A, and EF into EH, HF equal to D. Since BG is equal to GC and EH to HF, whatever part or parts BG is of EH, the same is GC of HF. So the same also is the sum BC of the sum EF (Theorem 45, Theorem 46). And BG is equal to A, EH to D.

Theorem 50 (Euclid VII.10). If a number be parts of a number, and another be the same parts of another, alternately also, whatever parts or part the first is of the third, the same parts or the same part will the second also be of the fourth.

Proof: Let AB be parts of C, and DE the same parts of F. Divide AB into the parts of C, namely AG, GB, and DE into the parts of F, namely DH, HE. AG is the same part of C that DH is of F; so, alternately, whatever part or parts AG is of DH, the same is C of F (Theorem 49). Likewise GB of HE. So whatever part or parts AG is of DH, the same is GB of HE, and so the same is the sum AB of the sum DE (Theorem 45, Theorem 46). So whatever parts or part AB is of DE, the same is C of F.