For positive integers prove that $a\Big|bc \implies a\Big|b \lor a\Big|c$

$$a\Big | b,\; b = ak.$$ $$a\Big|c, c = al,$$

So do I multiply $$b$$ and $$c$$ to get $$a(kl)$$ to prove that $$bc = a$$ multiplied by some integer $$kl$$ closed under multiplication?

• Welcome to stackexchange. This is not true. $6$ divides $3 \times 4$ but divides neither factor. Make sure you have correctly stated what you have been asked to prove. edit the question to clarify, don't use comments. And use mathjax: math.meta.stackexchange.com/questions/5020/… (as some editors have done). – Ethan Bolker Mar 13 at 1:56
• Need to assume $a$ is prime to show this, then it's a fairly well known property of primes. – coffeemath Mar 13 at 2:39

$$6 \mid 2\cdot 3, 6 \not \mid 2, 6 \not \mid 3$$.