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About LinkBackscan you help me prove this one
Given m > 0. there are exactly m distinct residue classes modulo m, namely,
[0],[1],[2],...,[m-1].
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Not sure exactly what you mean. But I try anyway.Originally Posted by earlkaize
Given m > 0. there are exactly m distinct residue classes modulo m, namely,
[0],[1],[2],...,[m-1].
Consider the integral domain,, regocnize that it is a commutative ring with unity (by definition) therefore you can speak of ideals. Note that for any
, that the coset
is an ideal in
. And you end with,
,
....
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, in fact, it is a famous result that you should memorize that,
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!)
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