# How did Euclid prove Euclid's Lemma

In Elements, Book VII, Proposition 7, Euclid states: If a number is that part of a number which a subtracted number is of a subtracted number, then the remainder is also the same part of the remainder that the whole is of the whole. He then gives a proof, but the proof isn't clear to me.

The modern version of Euclid's Lemma states that if p is prime and p|ab then either p|a or p|b, or both. I am familiar with the proof by Bezout's Identity, but Euclid didn't know Bezout's Identity.

I am looking for a simple, clear proof of the modern form of Euclid's Lemma, but stated in a way that uses the same concepts as Euclid did. Does anyone have such a proof?

• The guide section in the link you give seems to imply you have the modern statement wrong. (of course it's true, but it's not what he meant) – Ben Millwood Sep 18 '12 at 13:36
• I don't know what he used, but this trivially follows from unique factorization of integers which, according to Wikipedia, he knew. – Karolis Juodelė Sep 18 '12 at 13:37
• @KarolisJuodelė: it's often used to prove unique factorisation of integers, though. – Ben Millwood Sep 18 '12 at 13:37
• @KarolisJuodelė: Yes, as BenMillwood suggests, I want to use Euclid's Lemma to prove the Fundamental Theorem of Arithmetic, not the other way around. – user448810 Sep 18 '12 at 13:44
• Euclid's proof is based on proportions of numbers. Reinterpreted in modern language, using fractions, it is essentially equivalent to unique fractionization, e.g. see here and here. If you wish to understand these matters conceptually, then Euclid is not the place to look. If, instead, you wish to understand the history, then the required background is much too long for an MSE answer - there are many subtleties. See some of Franz Lemmermeyer's expositions for an introduction. – Bill Dubuque Sep 18 '12 at 14:08

I think this is the proof of Euclid's Lemma, and then I go on to prove the Fundamental Theorem of Arithmetic, which is what I first set out to do. The first paragraph proves Euclid's Lemma, the second paragraph proves that all positive integers greater than 1 can be factored into primes, and the third paragraph proves that the factorization is unique, which is the Fundamental Theorem of Arithmetic:

Suppose that a ratio a:b reduces to c:d in lowest terms, assume that c does not divide a, and assume that c(m/n) = a. Since a:b is the same ration as c:d, then d(m/n) = b, which implies that c/m = (1/n)a and d/m = (1/n)b. Therefore c/m:d/m is the same ratio as a:b, which shows that c:d is not in lowest terms. But that contradicts the earlier assumption. Therefore c does divide a, and d divides b the same number of times.

Suppose that a prime p divides the product ab but that p does not divide a, so that p is relatively prime to a. Let n = ab/p; this must be an integer because p|ab. Then p/a = b/n, and p/a must be in lowest terms, because p is relatively prime to a. Thus, by the previous paragraph, there must be some x for which px=b and ax=n, because the two ratios are the same, and therefore p|b. Likewise, if we suppose that p does not divide b, then p|a. Thus, if p is prime and p|ab, then either p|a or p|b.

Suppose that n is the smallest positive integer greater than 1 that cannot be writen as the product of primes. Now n cannot be prime because such a number is the product of a single prime, itself. Thus the composite is n = ab, where both a and b are positive integers less than n. Since n is the smallest number that cannot be written as the product of primes, both a and b must be able to be written as the product of primes. But then n = ab can be written as the product of primes simply by combining the factorizations of a and b. But that contradicts our supposition. Therefore, all positive integers greater than 1 can be written as the product of primes.

Furthermore, that factorization is unique, ignoring the order in which the primes are written. Now suppose that s is the smallest positive integer greater than 1 that can be written as two different products of prime numbers, so that s = p1 * p2 * ... * pm = q1 * q2 * ... * qn. By Euclid's Lemma either p1 divides q1 or p1 divides q2 * ... * qn. Therefore p1 = qk for some k. But removing p1 and qk from the initial equivalence leaves a smaller integer that can be factored in two ways, contradicting the initial supposition. Thus there can be no such s, and all integers greater than 1 have a unique factorization.

Please let me know if I got anything wrong.

EDIT: I added a first paragraph, and modified the second paragraph at the point "there must be some x". The new first paragraph is VII.20, the now-second paragraph is VII.30 (Euclid's Lemma), and the third and fourth paragraphs are the Fundamental Theorem of Arithmetic.

Now is it right?

• The proof is circular unless you supply a proof of "thus there must be some x...". That's what I refer to as unique fractionization. It is equivalent to Euclid's Lemma and Uniqueness of Prime Factorizations. For proofs and further discussion see the links in my comment to your question. – Bill Dubuque Sep 18 '12 at 16:21