abstract algebra, number theory, modular arithmetic

Question

Let $a,b,m \in \mathbb Z$ with $m \geq 1$. Prove that if $a \operatorname{mod} m = b \operatorname{mod} m$, then $a \equiv b \pmod{m}$

Confused on how to prove this... I was guessing you had to use the division of algebra at this point??

Answer 1

You are assuming $a$ mod $(m)$ =$ b$ mod $(m)$

That means $a = km+r$ and $b=lm +r$ for some integers $k$ and $l$.

Note that $ a-b = (k-l)m$ that is $m$ divides $a -b$.

Thus, $a\equiv b$ mod (m).