# Exact division and prime numbers

Will really appreciate some guidance.

Let $p$ be a prime and $n$ be a positive number.

Then $p^a$ exactly divides $n$ if $p^a|n$, but $p^{a+1} \not \! | \; n$. We then write $p^a\|n$ if $a$ is the largest component of $p$ such that $p^a|a$.

Prove that if $p^a\|n$ and $p^b\|m$ then $p^{a+b}\|mn$.

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Write $n=p^a u$ and $m = p^b v$ where neither of $u,v$ is divisible by $p$.
Then $$m n = p^{a+b} u v,$$ and from this your conclusion follows immediately.
EDIT: The OP wanted more details. If $p^{a+b+1}$ were to divide $mn$, then we would have $mn=p^{a+b+1}w$ where $w$ is an integer (perhaps itself divisible by $p$.) In that case, $$mn=p^{a+b}uv=p^{a+b+1}w,$$ so that on cancelling $p^{a+b}$ we would have $$uv=pw,$$ which would imply that $p$ divides one of $u,v$, whereas we assumed neither of $u,v$ were divisible by $p$ above.
Thank you but I thought that from this I need to show that $p^{a+b+1} is not divisible by n and this is where I stuck – Mary Nov 19 '12 at 14:57 I'll put a bit more explanation in the above answer. But you really should say you want to show$p^{a+b+1}$does not divide$mn$, rather than that it is not divisible by$n$. – coffeemath Nov 19 '12 at 15:17 @coffeemath But that's the only nontrivial part of the problem (and also the essence of the matter), so it is reasonable to presume that's where the OP was stuck. – Bill Dubuque Nov 19 '12 at 15:50 In my first version of my answer (before what is now in the "EDIT" part) I thought the "exactly divisible by"$p^{a+b}$was clear since neither of$u,v$are divisible by$p$. By his comment it seems to him it was not clear. – coffeemath Nov 19 '12 at 16:39 Hint$\rm\,\ p^{a+b+1}\mid (jp^a)(kp^b)\:\Rightarrow\:p\mid jk\:\Rightarrow\: p\mid j\ \ or\ \ p\mid k,\ $by$\rm\:p\:\$ prime