I was working on this proof for school and my professor told me that i proved the "if" part twice and never proved the "only if" part, but I really don't follow his logic. To me it all makes sense. Can someone point out to me what I did wrong.
Let $m, n \in \mathbb{N}$, let $a,b \in \mathbb{Z}$ . $a \equiv b $(mod $n$) if and only if $ma \equiv mb$(mod $mn$).
$\Rightarrow$. Suppose that $a \equiv b $(mod $n$), this means that there is some integer $k$ such that $a-b=kn$. For $ma \equiv mb$(mod $mn$) there must exist an integer $j$ such that $ma-mb=mnj$. We can divide each term by $m$ to obtain $a-b=nj$, and by our assumption there does exist and integer $j$ such that $a-b=nj$, and here $j=k$, proving the forward implication.
$\Leftarrow$. Suppose $ma \equiv mb$(mod $mn$). Then there exists some integer $k$ such that $ma-mb=mnk$. We can factor an $m$ out of this equation to get $m(a-b)=mnk$ and if we divide both sides by $m$ we are left with $a-b=nk$. This is precisely what it means for $a \equiv b $(mod $n$), thus proving the backwards implication and completing the proof.
Best Answer
You cannot say: "We can divide each term by $m$ to obtain $a - b = nj$." This assumes that $j$ already exists, which is what you are trying to prove.
Instead, just write:
"Multiply by $m$ so that $m(a-b) = ma - mb = kmn$. Then it follows that there is an integer, namely $j = k$, such that $ma - mb = jmn$."