MHB Congruences - Rotman - Proposition 1.58 - Second Question

[Math Amateur]

I am reading Joseph J.Rotman's book, A First Course in Abstract Algebra.

I am currently focused on Section 1.5 Congruences.

I need help with the proof of Proposition 1.58 part (iii) ...

Proposition 1.58 reads as follows:View attachment 4523
View attachment 4524

In the above text we read the following:" ... ... Therefore, if $$a \equiv b \text{ mod } m$$, then $$a - b = 0 \text{ mod } m$$, hence$$ r - r' \equiv 0 \text{ mod } m$$, hence $$r - r' \equiv 0 \text{ mod } m$$, and $$r \equiv r' \text{ mod } m$$; by part (ii), $$r = r'$$. ... ... "


My question is ... how exactly does it follow from part (ii) of Proposition 1.58 that $$r = r'$$ ...Note: I suspect Rotman is asking us to use the contrapositive of (ii) ... in other words the negative of $$r \nequiv r' \text{ mod m }$$ ( which is presumably $$r \equiv r'$$ ) implies the negative of $$0 \le r' \lt r \lt m$$ ... but what exactly is the negative of $$0 \le r' \lt r \lt m$$?

 

[caffeinemachine]

I am reading Joseph J.Rotman's book, A First Course in Abstract Algebra.

I am currently focused on Section 1.5 Congruences.

I need help with the proof of Proposition 1.58 part (iii) ...

Proposition 1.58 reads as follows:In the above text we read the following:" ... ... Therefore, if $$a \equiv b \text{ mod } m$$, then $$a - b = 0 \text{ mod } m$$, hence$$ r - r' \equiv 0 \text{ mod } m$$, hence $$r - r' \equiv 0 \text{ mod } m$$, and $$r \equiv r' \text{ mod } m$$; by part (ii), $$r = r'$$. ... ... "


My question is ... how exactly does it follow from part (ii) of Proposition 1.58 that $$r = r'$$ ...Note: I suspect Rotman is asking us to use the contrapositive of (ii) ... in other words the negative of $$r \nequiv r' \text{ mod m }$$ ( which is presumably $$r \equiv r'$$ ) implies the negative of $$0 \le r' \lt r \lt m$$ ... but what exactly is the negative of $$0 \le r' \lt r \lt m$$?

Since both $r$ and $r'$ are non-negative integers which are strictly less than $m$, we can have $m|(r-r')$ only if $r=r'$.Try this with small values of $m$.