Help to prove that: if m ≡ n (mod a) then mb ≡ nb (mod ab)

Question

Question asks to use mathematical language to prove that: if m ≡ n (mod a) then mb ≡ nb (mod ab). Question also says if proving an equivalence, each direction should be clear.

Answer 1

If $m\equiv n(\mod a)$ then $a|m-n$, that is, there is an integrer $k\in \mathbb{Z}$ such that $m-n=ak$. Multiplying both sides of the last equality by $b$, we obtain that, $b(m-n)=abk$, hence, $bm-bn=(ab)k$. So, $ab|bm-bn$, that is, $bm\equiv bn(\mod ab)$.

Answer 2

If $m\equiv n\pmod{a}$, then $a\mid m-n$ by definition of "modulo $a$". Thus $m-n=aq$ for some $q\in\Bbb Z$ by definition of divisibility by $a$. But then $(m-n)b=aqb$, i.e., $mb-nb=(ab)q$ by distributivity. Thus $ab\mid mb-nb$ by definition of divisibility by $ab$. Hence $mb\equiv nb\pmod{ab}$ by definition of "modulo $ab$".

Answer 3

Just do it:

$m \equiv n \pmod a \implies$

$m = n + ka$ for some integer $k\implies$

$mb = nb + kab\implies$

$mb \equiv nb \pmod {ab}$.

.....

Definition of $a|m$ is that there exists an integer $k$ so that $m = ak$ and.

Definition of $m\equiv n \pmod a$ is that $a|m-n$.

Using $a = b \implies ma = mb$ (and if $m \ne 0$ then the converse is true as well) the result is immediate.