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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

Aashiq Raza
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    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, – Bill Dubuque Aug 23 '16 at 16:04
  • @BillDubuque:Yoy mean to say, every integer $t$ is congruent modulo $m$ to some integer $r_i \in R$? – Aashiq Raza Aug 23 '16 at 16:08
  • @BillDubuque:i think,it is the point where i got confused. – Aashiq Raza Aug 23 '16 at 16:10
  • Yes, but congruent to one and only one $,r_\in R,$ – Bill Dubuque Aug 23 '16 at 16:15
  • @BillDubuque:What is the difference b/w remainder & Residue? – Aashiq Raza Aug 23 '16 at 16:16
  • @BillDubuque:As the set {$0,1,2,3,...,(m-1)$} is called set of residue modulo $m$,which is also the set of all possible remainders when integers are divided by $m$. – Aashiq Raza Aug 23 '16 at 16:19
  • Usually (but not always) the remainder refers to a canonical representative chosen from each residue class, which is usually chosen to be the least nonnegative residue (just like we usually choose the least terms rep for each fraction in its equivalence class of equal fractions). – Bill Dubuque Aug 23 '16 at 16:31
  • @BillDubuque:You mean remainder is least non-negative element of each residue class? – Aashiq Raza Aug 23 '16 at 16:42
  • Yes, that is by far the most common canonical choice or representative. – Bill Dubuque Aug 23 '16 at 17:03
  • @BillDubuque:If we consider 15,then $15\equiv 1(mod 7)$,but $1$ is not within the complete set of residue modulo 7,{$0,2,4,6,8,10,12$}? – Aashiq Raza Aug 23 '16 at 17:06
  • But $\ 15\equiv 8\pmod 7$ and $,8\in R.,$ Recall that, by definition $,a\equiv b\pmod n\iff n\mid a-b\iff a-b = kn,$ for some integer $,n.\ $ – Bill Dubuque Aug 23 '16 at 17:28
  • @BillDubuque:But 1=8 in (modulo 7). – Aashiq Raza Aug 23 '16 at 18:54
  • @BillDubuque:Is the set $R$ unique? – Aashiq Raza Aug 23 '16 at 19:07
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    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, – Bill Dubuque Aug 23 '16 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 $7$
J.F. Hernández
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