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Set $\mathbb Z/m \Bbb Z$ is called the set of Congruence Classes modulo $m$(also called Residue class modulo $m$).

Now,$\mathbb Z/m \Bbb Z=${$0+\mathbb Z/m \Bbb Z$,$1+\mathbb Z/m \Bbb Z$,$2+\mathbb Z/m \Bbb Z$,...,$(m-1)+\mathbb Z/m \Bbb Z$}.

The set $R=${$r_1,r_2,r_3,...,r_m$} is called a complete set of Residue modulo $m$ if $r_1,r_2,r_3,...,r_m$ are pairwise incongrent modulo $m$.

Does here $R$ is the set of all possible remainders when integers are divided by $m$?If $Yes$,then the complete set of residues modulo $m$ should be {$0,1,2,3,...,(m-1)$}.

Considering it as a true result ,the complete set of residue modulo $7$ will be {$0,1,2,3,4,5,6$},but it is not so.(It is {$0,2,4,6,8,10,12$}).

I think i've understood this concept in a wrong way.I need to know where i'm wrong.

If anyone have different viewpoint for this,plese share it with me.

NOTE:I do not have any any background for number theory.

Thank you

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    $\begingroup$ By your definition both are complete sets of residues. An equivalent definition is that every integer is congruent to exactly one element of the set. Don't confuse "residue' with "remainder". Remainders are the least nonegative elements in the residue class, $\endgroup$ Aug 23, 2016 at 16:04
  • $\begingroup$ @BillDubuque:Yoy mean to say, every integer $t$ is congruent modulo $m$ to some integer $r_i \in R$? $\endgroup$ Aug 23, 2016 at 16:08
  • $\begingroup$ @BillDubuque:i think,it is the point where i got confused. $\endgroup$ Aug 23, 2016 at 16:10
  • $\begingroup$ Yes, but congruent to one and only one $\,r_\in R\,$ $\endgroup$ Aug 23, 2016 at 16:15
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    $\begingroup$ There are infinitely many complete residue systems since there are infinitely many possibilities to choose as rep for each residue class, e.g. we can choose any element of $\,\{1+km\,:\, k\in\Bbb Z\}\,$ for the rep in $R$ that is $\equiv 1\pmod m$. You might be confusing mod the relation with mod the operator. There are many answers explaining the difference, e.g. here, $\endgroup$ Aug 23, 2016 at 19:24

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  1. $\mathbb{Z}/m\mathbb{Z}$ is $$\{0+m\mathbb{Z},1+m\mathbb{Z},\dots,m-1+m\mathbb{Z}\}$$
  2. $R=\{0,1,2,3,4,5,6\}=\{0,2,4,6,8,10,12\}$. Because in this context $1$ means $$1+7\mathbb{Z}=\{\dots,-13,-6,1,8,15,\dots\}$$ and equally for all the rest numbers: $0=0+7\mathbb{Z},\;1=1+7\mathbb{Z},\;2=2+7\mathbb{Z},\dots$. Therefore, $8=1$, because $1+7\mathbb{Z}=8+7\mathbb{Z}$; $10=3$, because $10+7\mathbb{Z}=3+7\mathbb{Z}$ and $12=5$ because $12+7\mathbb{Z}=5+7\mathbb{Z}$. Therefore, both sets $R=\{0,1,2,3,4,5,6\}$ and $R=\{0,2,4,6,8,10,12\}$ are complete set of residue modulo $7$
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