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I was reading about modular arithmetic. The congruencies modulo $m$ are equivalence relations. The textbook says that for the relation $a\equiv b\pmod m$ the quotient set of the equivalence classes are $Z/m={0,1,...,m-1}$ where each number represents the modulo $m$. So my question is: Why are there $m$ equivalence classes and why they end at $m-1$? If I have $a\equiv b\pmod 1$ then the equivalence classes are $$\begin{split} R(0)&= [a \in \mathbb{Z}]\\ R(1)&=[a \in \mathbb{Z}, b \in \mathbb{Z}: a=b-1]\\ \end{split}$$

which are two, not 1 .

Bill Dubuque
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Mr. Nicolas
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    You start counting at $0$. – Randall Mar 19 '23 at 21:38
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    Mod $1$ there is only ONE class since $1$ divides every integer. – Randall Mar 19 '23 at 21:39
  • I know that it starts counting at 0. But why it ends at m-1? – Mr. Nicolas Mar 19 '23 at 21:40
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    Because $m \equiv 0 \bmod m$ so you don't need a "new" class for $m$. It's just the class of $0$ again. Think of it this way: if I divide by $3$ there are only $3$ possible remainders: $0$, $1$, and $2$. If you divide by $3$ and get remainder $3$ you did it wrong – Randall Mar 19 '23 at 21:41
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    You have the right idea looking at the trivial case of mod $1$. But you've made a mistake in thinking it has two equivalence classes. Every number is $0$ mod $1$, so every number is equivalent. – Zoe Allen Mar 19 '23 at 21:41
  • I see. Thank you. – Mr. Nicolas Mar 19 '23 at 21:41
  • The point is that $0,1,\ldots,m-1,$ is a complete residue system $\bmod m,,$ i.e. every integer $,k,$ is congruent to exacly one of them, as follows from the existence and uniqueness of the remainder in the division algorithm $,k\div n.\ $ Said equivalently we can choose the least nonnegative element of $,k+m\Bbb Z,$ as a normal form (reduced) representative of each coset. These ideas are explained at length in the linked dupes. – Bill Dubuque Mar 20 '23 at 00:32
  • They stop at $m-1$ because going further we have $,m\equiv 0,\ m+1\equiv 1,\ldots$ so including these further elements would destroy uniqueness. But we could instead use $,1,2,\ldots m,$ (or any consecutive sequence of $m$ integers). $, n\equiv 0\pmod{!1},$ by $,1\mid (n-0),$ so $,0,$ is a complete residue system (special case of above). – Bill Dubuque Mar 20 '23 at 00:39

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