# Prove that if a|b, c|d, then ac|bd [duplicate]

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

• We can also formulate the claim in terms of mods. Specifically if $b \equiv 0 \pmod a$ and $d \equiv 0 \pmod c$ then $bd \equiv 0 \pmod {ac}$ – john Oct 20 '19 at 19:40

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 $ac$.

• 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 – JorgeeFG Sep 27 '15 at 21:12
• See my latest edit – Omnomnomnom Sep 27 '15 at 21:17
• Thanks that was what I was needing. I've tried for a time, it's not that easy for me. – JorgeeFG Sep 27 '15 at 21:20

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 $bd=klac$.