Congruences - Rotman - Proposition 1.58

[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 (ii) ...

Proposition 1.58 reads as follows:https://www.physicsforums.com/attachments/4521
View attachment 4522In the above text ... specifically, in the proof of Part (ii) we read:

" ... ... (ii) If \(\displaystyle r = r' \text{ mod } m\), then \(\displaystyle m \mid (r - r')\) and \(\displaystyle m \le r - r'\) . ... ... "


Can someone show me precisely and formally how \(\displaystyle r = r' \text{ mod } m\) implies that \(\displaystyle m \le r - r'\) ...

It seems quite plausible ... but how do we formally and rigorously show this ...

Peter

 

[Evgeny.Makarov]

If $m>0$ and $n\ge0$ are integers, then $m\mid n$ means that $n=km$ for some nonnegative integer $k$. If $k=0$, then $n=0$; if $k\ge1$, then $n\ge m$. Therefore, if $r\equiv r'\pmod{m}$, then $m\mid (r-r')$ and therefore either $r-r'=0$ or $r-r'\ge m$. The first option is impossible because by assumption $r'<r$, and the second option is impossible because $r<m$.

 

[Math Amateur]

If $m>0$ and $n\ge0$ are integers, then $m\mid n$ means that $n=km$ for some nonnegative integer $k$. If $k=0$, then $n=0$; if $k\ge1$, then $n\ge m$. Therefore, if $r\equiv r'\pmod{m}$, then $m\mid (r-r')$ and therefore either $r-r'=0$ or $r-r'\ge m$. The first option is impossible because by assumption $r'<r$, and the second option is impossible because $r<m$.

Thanks Evgeny ... I very much appreciate your help ...

Peter