If $b\mid x-y$, $b\in \mathbb{N}$, $b\geq 2$, in this inequation,$-(b-1)\leq x-y\leq b-1$, why the only integer divisible by $b$ is zero?

Question

I didn't understand how they reached at this conclusion:

If $b\mid x-y$, $b\in \mathbb{N}$, $b\geq 2$

In this inequation:

$$-(b-1)\leq x-y\leq b-1$$

The only integer divisible by $b$ is zero. (Why only zero??)

Then $x=y$

Answer 1

Notice that if $0 < a < b$, then $b$ doesn't divide $a$. So $b$ doesn't divide $-a$ either. This covers all integral values between $-(b-1)$ and $b-1$ except $0$.

Answer 2

What integer divisible by $b$ lies between $-(b-1)$ and $(b-1)$?

For example, when $b=4$, what number divisible by $4$ lies between $-3$ and $3$?

Further example, when $b=11$, what number divisible by $11$ lies between $-10$ and $10$?

The answer is always only $0$, for all integer values of $b$.