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if $a\mid x$ and $b\mid x$ and $\gcd(a,b) = 1$, prove that $a\cdot b\mid x$

Well, I've started by saying that $x = q_1a$ and $x = q_2b$, and I know that $a,b$ are prime numbers. And, I'm not sure how to proceed from here.

Edit: The answer here is better than in the posts listed here

Ilan Aizelman WS
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    a and b need not be prime numbers. They are coprime. E.g. 6 and 7 are coprime, but not both primes. – Bram28 Nov 17 '16 at 15:31
  • $\quad\uparrow\quad$ Or $14$ and $15$ if you want an example where neither is prime. $\qquad$ – Michael Hardy Nov 17 '16 at 15:34
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    For me the first thing that comes to mind is that if $\gcd(a,b)=1$ then $a$ and $b$ have no prime factors in common, so in the prime factorization of $x$ the parts corresponding to $a$ and to $b$ are disjoint from each other. – Michael Hardy Nov 17 '16 at 15:35

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Suppose $x=ak=bl$ where $k$ and $l$ are integers. Since $\text{gcd}(a,b)=1$, there are also integers $r,s$ such that $$ 1=ra+sb. $$ Then, $$ x=rax+sbx=ra(bl)+sb(ak)=ab(rl+sk) $$ proving that $ab|x$.

yurnero
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  • I knew that $1 = ra + sb$, but didn't know how to continue from there. but multiplying by $x$ solves this so easy :) – Ilan Aizelman WS Nov 17 '16 at 15:32
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    @IlanAizelmanWS The proof is more simple, more conceptual (and more general) if you use Euclid;s Lemma (or unique factorization) as in the other answers. For example those proofs work over certain polynomial rings whereas the above proof fails. – Bill Dubuque Nov 17 '16 at 15:51
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If $\gcd(a,b)=1$ then $a$ and $b$ have no prime factors in common, so in the prime factorization of $x$ the parts corresponding to $a$ and to $b$ are disjoint from each other.

This is probably the simplest way to infer the desired conclusion from well known results, but as Bill Dubuque points out in comments, it may not be the simplest was to do it from scratch because it does not include a proof of the uniqueness of prime factorizations. That uniqueness itself takes some work to prove.

Michael Hardy
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  • @OP Beware that it in some approaches this way might be circular since it is used to prove the Fundamental Theorem of Arithmetic, i.e. the existance and uniqueness of prime factorizations. – Bill Dubuque Nov 17 '16 at 15:45
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By Euclid's Lemma $\, (\color{#c00}{a,b})=1,\,\ \color{#c00}a\mid \color{#c00}b\,(x/b)\,\Rightarrow\, a\mid x/b\,\Rightarrow\, ab\mid x\ $

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Bill Dubuque
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