Q:

Prove that if $a\equiv b \pmod m $ , then $a \bmod m = b \bmod m$

Question

Prove that if $a\equiv b \pmod  m$ , then
$a \bmod  m = b \bmod m$

Approach
Given,

$a\equiv b \pmod m$

$\implies m\mid (a-b)$
$\implies (a-b)=m\cdot k$
$\implies a=b+m\cdot k$
Now,

$a \bmod  m$   can be written as
$\implies b+m\cdot k  \pmod m$

No idea how to move forward to get
$$b \pmod m \impliedby b +m\cdot k  \pmod m$$
Please help me out!!!
Thanks!

A:

Hint $\, $ Show  $\ c\equiv  d\pmod{\!m}\,$  and $\,|c-d|<m\,\Rightarrow\,  c= d$
Apply that to $\,\ \underbrace{a\bmod m}_{\Large c}\!\equiv a\equiv b\equiv \underbrace{b\bmod m}_{\Large d}\,\ \pmod{\!m}$
Remark $ $ Any sequence of $m$ consecutive integers forms a complete system $S$ of representatives (normal forms) of all equivalence classes mod $m,\,$ so the above implies that congruence reduces to equality on these normal forms, i.e.  $\ a\equiv b\iff \bar a = \bar b,\,$ where $\bar n =$ unique normal-form rep in $S$ congruent to $n\,$ (which is analogous to: $ $ fractions are equivalent iff their least form rep's are equal, i.e.  least terms fractions are normal forms for equivalence of fractions).
An alternative convenient choice of normal forms for modular arithmetic arises by choosing the least magnitude reps (signed or balanced residue system), e.g. $\,\{-2,\color{#c00}{-1},0,1,2\}\pmod{\! 5}.\,$ In these signed systems the  normal form of $\,m\!-\!1\,$ is $\,-1,\,$ e.g. above $\,4\equiv\color{#c00}{-1}\pmod{\!5},\,$ simplifying common calculations such as $\,(m\!-\!1)^k\equiv (-1)^k,\,$  e.g. when casting out elevens.

A:

$$a\equiv b \pmod m$$
$$\implies (a-b) = m k $$
$$\implies a = m k + b $$
$$\implies a \bmod m = (mk\bmod m + b\bmod m)\bmod m = b\bmod m \bmod m = b \bmod m$$
$$\implies a \bmod m = b \bmod m$$
Here is an important point raised by @(Bill Dubuque) in this proof. In line $4$, I used $(mk+b)\bmod m = (mk\bmod m + b\bmod m)\bmod m$ which must be proved for the completion of the proof of the original result. There are multiple ways to prove this equality, however if one uses the approach used in this answer, then this leads to a circular dependency i.e. usage of this equality to prove the original result and the usage of the original result in proving this equality. So, to complete the proof of original result one must prove this equality with some other approach. Based on the suggestion by Bill, here is a revised proof,
Let $b = mq + r \implies b \bmod m = r$, then,
$$a\equiv b \pmod m$$
$$\implies (a-b) = m k $$
$$\implies a = m k + b = mk + mq + r = m(k+q) + r $$
$$\implies a \bmod m = r = b \bmod m$$
$$\implies a \bmod m = b \bmod m$$
Note:
Here is an issues found by @(Bill Dubuque) in my first proof.
$$a\equiv b \pmod m$$
$$\implies (a-b) = m k $$
$$\implies (a-b)\bmod m = m k \bmod m $$
$$\implies a \bmod m - b \bmod m = 0$$
$$\implies a \bmod m = b \bmod m$$
But the $4$th equation above cannot follow from the $3$rd equation in general (counterexample $a=m$ and $b=1$).

