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For $i, j, n \in \mathbb{N}$, if $i \equiv j \pmod{n}$ and $i, j \leq n$, then $i = j$.

Might be a trivial question, but I don't see why this holds. Can someone explain to me why this is true?

1 Answers1

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If $i\equiv j\pmod{n}$, then by definition $n\mid i-j$, i.e., $i-j=kn$ for some integer $k$. If, however, $i$ and $j$ are positive integers, and $i,j\le n$, then

$$-(n-1)=1-n\le i-j\le n-1\;,$$

and the only multiple of $n$ between $-(n-1)$ and $n-1$ is $0=0\cdot n$. Thus, $k=0$, and $i=j.$

Brian M. Scott
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  • I would have said this as $1>\frac {n-1}{n}\ge \frac {\max(i,j)-1}{n}\ge \frac {\max (i,j)-\min (i,j)}{n}=\frac {|i-j|}{n}\in {0}\cup \Bbb Z^+.$ – DanielWainfleet Jun 13 '20 at 08:49