TITLE: Why are naturals < modulus equal if congruent? [uniqueness of remainder]
QUESTION [5 upvotes]: Let $a$, $b$, and $n$ be integers with $0 \le a \lt n$ and $0 \le b \lt n$. Suppose $a \equiv b \pmod n$. Then $a-b=0$ and $a=b$. Why does the congruence of two non negative integers imply that they must be equal?

REPLY [5 votes]: The key trick here is the interplay between magnitude and divisibility — every nonzero multiple of $n$ must have a magnitude at least as large as $n$.
Symbolically, if $k \neq 0$, then $|kn| \geq |n|$.
Since integers in the same congruence class are separated by multiples of $n$, this puts a limit on how close they can be. We have
If $a \equiv b \bmod n$ and $a \neq b$, then $|a-b| \geq |n|$
The premise of the problem constrains both $a$ and $b$ to a narrow interval: they are both contained in the set $\{ 0, 1, \ldots, n-1 \}$. The largest possible difference between any two elements of this set has magnitude $n-1$. That is, for any two elements $x,y$ of this set, we have $|x-y| \leq n-1$.
So by the previous observation, the fact that $a \equiv b \bmod n$ implies that one of the following is true:

$a = b$
$|a-b| \geq n$

and we can rule out the second possibility because $|a-b| \leq n-1$.