Finding the no of equivalent classes

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Homework Statement

please consider the relation over the set of integers Z , given by m=n mod (p) where p is a positive integer . Prove that it is an equivalent relation.
find the elements in the equivalent class of m.
find the no of such equivalent classes.

Homework Equations





3.equivalent relation-proved by showing reflexivity, symmetry and transitivity.
the equivalent class of m consists of elements of the type
m +k p where k = 0,+/-1,+/-2...
but i am not able to think how to find the no of such classes. any help will be highly appreciated.
 
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hi commutator! :smile:

(have a ± :wink:)
commutator said:
… the equivalent class of m consists of elements of the type
m +k p where k = 0,+/-1,+/-2...

that's right! :smile:

ok, when you're stuck it's often useful to try an easy example …

if p = 10, what are the equivalence classes? :wink:
 
thanks . that was really encouraging.

for p=10, i got something like [-1]=(...,-11,-21-1,9,19,29...)
[0]=(...-10,-20...0,10,20,...)
[1]=(...-29,-19...1,11,21,31...]
interestingly , [11] is coming same as [1]. [-1] is coming the same as [9].so i think there are 10 distinct classes, which are repeating over and over. then the answer has got to be 10. ie. p. but how do i sketch a formal proof ? by induction?
 
hi commutator! :smile:
commutator said:
… so i think there are 10 distinct classes, which are repeating over and over. then the answer has got to be 10. ie. p. but how do i sketch a formal proof ? by induction?

that's right, there's exactly p classes

the formal proof isn't the problem …

the problem is the formal definition

once you have that, the proof is obvious

eg your definition {...-29,-19...1,11,21,31...} is only a list, so that's not helpful, but your earlier definition (I'm expanding it a little) [m] = {m +kp : k in Z} should do it :smile:

(btw, always use squirly brackets for listing the elements of sets :wink:)
 
thanks a lot !
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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