Proof of $a \equiv b$ mod m => $a \equiv b$ mod $m'$ with $d \cdot m' = m$

Question

Let $a,b \in \mathbb{Z}$ and $m, m', d \in \mathbb{N}$ with $d \cdot m' = m$.

How can I prove/disprove that

$a \equiv b$ mod m => $a \equiv b$ mod $m'$?

and

$a \equiv b$ mod $m'$ => $a \equiv b$ mod m

For the first statement, can I for example say

$27 \equiv 7$ mod $10$ => $27 \equiv 7 $ mod $5$ and $27 \equiv 7$ mod $2$?

I don't really understand, because for the first statement, it's the same like $a \equiv b$ mod m => $a \equiv b$ mod $\frac{m}{d}$. But what actually is d?

Answer 1

The first statement is true:

$$a \equiv b \bmod m \implies \\ a=mx+k, \space b=my+k\implies \\ a=m'dx+k, \space b=m'dy+k\implies \\ a=m'(dx)+k, \space b=m'(dy)+k\implies \\ a\equiv b \bmod m' $$

The second statement can be disproved with a single example:

Say $m'=5$, $d=2$, $m=m'd=10$, $a=12$, $b=7$:

Obviously: $a\equiv b \bmod m'$ because $12\equiv7\bmod 5$. But it is not true that $a\equiv b \bmod m$ because $12\not\equiv 7\bmod 10$.

Answer 2

By definition $$a\equiv b \mod m \iff \exists k\in\mathbb{Z}: a = mk+b$$ Using $$m = dm'$$ we get $$a = dm'k+b = (dk)m'+b$$ where $kd\in\mathbb{Z}$ and thus $$ a\equiv b \mod m'$$

For the second statement, try to find a counterexample (actually, there are a lot).