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I'm trying to prove it, but I can't find how.

If a divides b, and c divides d, then a*c divides b*d

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  • $\begingroup$ We can also formulate the claim in terms of mods. Specifically if $b \equiv 0 \pmod a $ and $d \equiv 0 \pmod c $ then $ bd \equiv 0 \pmod {ac} $ $\endgroup$ – john Oct 20 '19 at 19:40
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Hint: $a\mid b$ means that there exists an integer $k$ such that $b = ka$.


You seem to have written the essential step as a comment. Here's how it would fit into a complete proof:

Suppose that $a\mid b$ and $c \mid d$. It follows that we have $b = k_1a$ and $d = k_2d$ for integers $k_1,k_2$. It follows that $$ bd = (k_1a)(k_2c) = (k_1k_2)(ac) $$ Let $k$ be equal to the integer $k_1k_2$. We see that $(bd) = k(ac)$. Thus, $bd$ is divisible by $ac$.

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  • $\begingroup$ I've reached k1* a * k2 * c = k * a * c but then I don't know how to procede, because it would be k1 * k2 = k $\endgroup$ – JorgeeFG Sep 27 '15 at 21:12
  • $\begingroup$ See my latest edit $\endgroup$ – Omnomnomnom Sep 27 '15 at 21:17
  • $\begingroup$ Thanks that was what I was needing. I've tried for a time, it's not that easy for me. $\endgroup$ – JorgeeFG Sep 27 '15 at 21:20
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It goes directly from the definition of divisibility. If $a|b$ then exists $k$ such that $b=ka$. If $c|d$ then exists $l$ such that $d=lc$. Hence $bd=klac$.

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