Elementary number theory -Divisibility 1

Question

For arbitrary integers $a, b$ and $c$. Prove or disprove:

If $a|c$ and $b|c$ then $(ab)|c $

My solution was

$c=ma$ & $c=nb$ for some $n,m\in \mathbb{Z}$

∴ we can say that $c=kab$

means $ab|c$

Is my solution correct?

Answer 1

You are correct that $c=ma$ and $c=nb$ for some $n,m \in \mathbb{Z}.$ However, that only imples that $ma=nb$, not the $c=kab$ that you suggested. We need an expression with $ab$ and it doesn't look like we can get there from $ma=nb$ without making things more complex, so that gives a hint that maybe we should try proving it false.

It doesn't take long. For a small example $a=b=c=2$ works. For a less trivial one try mixing combinations of primes. $2\cdot3$ and $2\cdot5$ both divide $2\cdot3\cdot5$, but does their product?

Answer 2

Your argument is not valid.

As a counter example, note that $4$ divides $12$ and $6$ divides $12$ but $6\times 4 = 24$ does not divide $12$.