can you help me prove this one
Given m > 0. there are exactly m distinct residue classes modulo m, namely,
[0],[1],[2],...,[m-1].
thank you in advance!
God bless you!
Not sure exactly what you mean. But I try anyway.Originally Posted by earlkaize
Consider the integral domain, $\displaystyle \mathbb{Z}$, regocnize that it is a commutative ring with unity (by definition) therefore you can speak of ideals. Note that for any $\displaystyle m>0$, that the coset $\displaystyle m\mathbb{Z}$ is an ideal in $\displaystyle \mathbb{Z}$. Form a factor ring, $\displaystyle \mathbb{Z}/m\mathbb{Z}$. And you end with,
$\displaystyle \mathbb{Z}=\{...,-2m,-m,0,m,2m,...\}$
$\displaystyle 1+\mathbb{Z}=\{....,-2m+1,-m,1,m+1,2m+1,...\}$
$\displaystyle 2+\mathbb{Z}=\{...,-2m+2,-m+2,2,m+2,2m+2,...\}$,
....
$\displaystyle (m-1)+\mathbb{Z}=\{...-2m-1,-m-1,-1,m-1,2m-1,...\}$
Note, all the intergers are divided among these cosets. Further, since cosets are equivalence classes they are all disjoint! Proof complete.
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The ring formed by these cosets behaves just like the number under addition modulo $\displaystyle m$, in fact, it is a famous result that you should memorize that,
$\displaystyle \boxed{\mathbb{Z}/m\mathbb{Z}\cong \mathbb{Z}_m}$
Well, consider an integer m. Euclidean division grants us that, for all integers n, there are only m cases for the residue when n is divided by m:
n=km, or n=km+1, or n=km+2,..., or n=km+m-1.
The numbers n included in every case form the residue classes, m in total.
This is the core of the idea, as PH explained (though his tough algebra eludes us!)