Help me understand the proof of $a \equiv b \mod m \Rightarrow r_m(a)=r_m(b)$

Question

Let:

$r_m:\mathbb{Z}\rightarrow R_m$

where

$r_m(a)=r\Leftrightarrow a \equiv r \mod m $ and $r\in R_m$

where $R_m$ is the set of residues modulo $m$.

I understand the above proof until $r' \equiv r'' \mod m$. I don't know how to use Proposition 19.2.2 to conclude that $r'=r''$.

Can someone please make this step more explicit?

Answer 1

Tell me if it helps:

$a= Q_{m(a)}m +R_{m(a)} \quad,\quad R_{m(a)}<m$

$b= Q_{m(b)}m +R_{m(b)} \quad,\quad R_{m(b)}< m$

$a ≡ R_{m(a)} \pmod m$

$b ≡ R_{m(a)} \pmod m$

So, $a ≡ b \pmod m \implies R_{m(a)} ≡ R_{m(b)} \pmod m \implies R_{m(a)} = R_{m(b)} +Km$, $K \in \mathbb {Z^+}$

$R_{m(a)} < m \implies K=0 \implies R_{m(a)} = R_{m(b)}$

Answer 2

$a=mq_a+r \land b=mq_b+r' \implies 0 \leq r,r'<m$

$a \equiv b \pmod m \implies r\equiv r' \pmod m$

$0 \leq r<m \quad,\quad -m> -r' \geq 0 \implies 0 \leq \lvert r-r'\rvert <m \implies r=r'$