My Question is how does this proof workout? Unable to understand $m\mid n$ and its relation to $a$ and $b$

Question

Let $m, n \in N$ and $a, b \in Z$ so that $ m\mid n$ and $a \equiv b \mod n.$ Then $a \equiv b \mod m$

Answer 1

To say $m|n$ means there exists some natural number $x$ such that $mx=n$.

To say $a \equiv b$ $mod n$ means that $(a-b)=yn$ for some integer $y$.

We want to show there exists some value $z$ such that $(a-b)=zm$.

Since we know $(a-b)=yn$ and $mx=n$, we can substitute like so: $(a-b)=y(mx)$. Observe that $z=yx$

Answer 2

just do it.

$a \equiv b \pmod n \implies$

$n|a-b$.

And $m|n$ and $n|a-b$ so $m|a-b$.

So $a\equiv b \pmod m$

.....

Alternatively.

$a \equiv b \pmod n\implies$ there is a $k $ so that $a = b + kn$.

And $m|n \implies$ there is a $j$ so that $n= jm$.

So $a=b + k(jm) = b+(kj) m$.

So $a \equiv b\pmod m$.

.........

Consider an example:

$26 \equiv 5 \pmod {21}$.

we know that is true because $26 = 5 + 21$.

But $26 \equiv 5 \pmod 7$ because $26 = 5 + 3*7$.

And $26 \equiv 5 \pmod 3$ because $26 = 5 + 7*3$.

.....

Does that not make sense?