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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

Bill Dubuque
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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

2 Answers2

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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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so, you're totally right about what $\mathbb{Z}/m\mathbb{Z}$ is. it's like you've got these buckets, or classes, for all the integers based on their remainder when you divide by $m$. for $m=7$, you have the seven buckets: $$ [0]_7, [1]_7, [2]_7, [3]_7, [4]_7, [5]_7, [6]_7 $$ each bucket $[a]_7$ holds all the numbers that are congruent to $a$ modulo 7.

now, a complete set of residues modulo $m$, let's call it $R = \{r_1, r_2, \dots, r_m\}$, is just a collection of $m$ regular integers. the only rule is that no two integers in that set should be congruent to each other modulo $m$. so, if you pick any two different guys $r_i$ and $r_j$ from your set $R$, then $r_i \not\equiv r_j \pmod{m}$.

your question was, does this set $R$ have to be $\{0, 1, 2, \dots, m-1\}$?

and the answer is nope, it doesn't have to be. that set $\{0, 1, \dots, m-1\}$ definitely is a complete set of residues. it's got $m$ things in it, and they're all different modulo $m$ (like, 3 isn't congruent to 5 mod 7). it's usually the first one everyone thinks of, and sometimes it's called the least non-negative complete set.

but why can other sets work too? well, the main idea is that a complete set of residues just needs to have one number from each of those buckets (the congruence classes). the set $\{0, 1, \dots, m-1\}$ picks the simplest number from each bucket (the actual remainder). but you could pick any number you want from each bucket!

let's look at your example $R = \{0, 2, 4, 6, 8, 10, 12\}$ for modulo 7. it has 7 numbers, which is good. are they all different modulo 7? let's check their remainders: $\begin{align*} 0 &\pmod{7} \text{ gives } 0 \\ 2 &\pmod{7} \text{ gives } 2 \\ 4 &\pmod{7} \text{ gives } 4 \\ 6 &\pmod{7} \text{ gives } 6 \\ 8 &\pmod{7} \text{ gives } 1 \quad (\text{since } 8 = 1 \times 7 + 1) \\ 10 &\pmod{7} \text{ gives } 3 \quad (\text{since } 10 = 1 \times 7 + 3) \\ 12 &\pmod{7} \text{ gives } 5 \quad (\text{since } 12 = 1 \times 7 + 5) \end{align*}$ the remainders we got are $\{0, 1, 2, 3, 4, 5, 6\}$. see? we hit every single possible remainder from 0 to 6 exactly once. that means the original numbers $\{0, 2, 4, 6, 8, 10, 12\}$ must be pairwise incongruent modulo 7. each one belongs to a different bucket.

so, $\{0, 2, 4, 6, 8, 10, 12\}$ totally works as a complete set of residues modulo 7.

think of it like this: you have 7 bins, labeled $[0]_7$ to $[6]_7$. a complete set of residues is just a set you make by taking exactly one integer out of each bin.

  • $\{0, 1, 2, 3, 4, 5, 6\}$ takes the smallest non-negative integer from each bin.
  • $\{0, 2, 4, 6, 8, 10, 12\}$ takes $0$ from bin $[0]_7$, $8$ from bin $[1]_7$, $2$ from bin $[2]_7$, $10$ from bin $[3]_7$, $4$ from bin $[4]_7$, $12$ from bin $[5]_7$, and $6$ from bin $[6]_7$. it still grabbed one from each bin.

here are a couple more examples for $m=7$:

  • $\{7, 1, 9, 3, 11, 5, 13\}$. if you check their remainders mod 7, you get $\{0, 1, 2, 3, 4, 5, 6\}$.
  • $\{-3, -2, -1, 0, 1, 2, 3\}$. their remainders mod 7 are $\{4, 5, 6, 0, 1, 2, 3\}$.

both of those have 7 numbers, and they're all different modulo 7.

so yeah, you weren't wrong about the basics at all. the only mix-up was thinking that $\{0, 1, \dots, m-1\}$ was the only option for a complete set of residues. it's just the most standard one. any set of $m$ numbers that covers all the different remainders modulo $m$ works.

heather milkem
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