Quotient set of an equivalence relation

[Luna=Luna]

On the set of Z of integers define a relation by writing m [itex]\triangleright[/itex] n for m, n [itex]\in[/itex] Z.

m[itex]\triangleright[/itex]n if m-n is divisble by k, where k is a fixed integer.

Show that the quotient set under this equivalence relation is:

Z/[itex]\triangleright[/itex] = {[0], [1], ... [k-1]}

I'm a bit new the subject of Set Theory so I'm a bit unsure as to how to go about solving this.

 

[mathman]

On the set of Z of integers define a relation by writing m [itex]\triangleright[/itex] n for m, n [itex]\in[/itex] Z.

If m-n is divisble by k, where k is a fixed integer then show that the quotient set under this equivalence relation is:

Z/[itex]\triangleright[/itex] = {[0], [1], ... [k-1]}

I'm a bit new the subject of Set Theory so I'm a bit unsure as to how to go about solving this.

[itex]\triangleright[/itex] seems to be undefined. You say it is a relationship without saying what it is.

 

[Luna=Luna]

sorry it wasn't clear from my post, I've rewritten the post to be a bit more clear.

The relationship is:
m[itex]\triangleright[/itex]n, if m-n is divisble by k, where k is a fixed integer.

 

[HallsofIvy]

But you also haven't said what "[1]", "[2]", ... "[k]" are. Of course, they are the equivalence classes but unless your question is "why are there k classes" rather than "why are the classes what they are", you need to specify "what they are"! Essentially what you wrote was that you want to prove that the "quotient set" is the "set of equivalence classes" which is basically the definition of "quotient set"!
(Similar to writing "Luna= Luna"!)

Start with k= 1. All numbers in the same equivalence class with it ([1]) are numbers n such that n- 1 is divisible by k. That is the same as saying "n- 1= mk" for some m or n= mk+ 1. That is, all numbers that are 1 more than a multiple of k: [1]= {..., -2k+ 1, -k+ 1, 1, k+ 1, 2k+ 1, ...}. Similarly, [2]= {..., -2k+ 2, -k+ 2, 2, k+ 2, 2k+ 2, ...}. You can do that until you get to [k]= {..., -2k, -k, 0, k, 2k, ...} which is the same as [0]. If we try to do the same thing with k+ 1, we get [k+1]= {..., -2k+ (k+1), -k+ (k+1), k+ 1, k+ (k+ 1), 2k+ (k+ 1), ...}= {..., -k+ 1, 1, k+ 1, 2k+ 1, 3k+ 1, ...} which is exactly the same as [1].

 

[blue_raver22]

However, from my understanding, a quotient set is a set that contains all possible equivalence classes of a given equivalence relation. In this case, the relation \triangleright is defined as m\trianglerightn if m-n is divisible by k. This means that for any two integers m and n, if their difference is divisible by k, then they are in the same equivalence class.

To find the quotient set, we need to determine all the possible equivalence classes. Let's start with the equivalence class [0]. This means that for any integer m, m-0 must be divisible by k. This is only possible if k divides m, so the equivalence class [0] contains all integers that are divisible by k.

Next, let's consider the equivalence class [1]. This means that for any integer m, m-1 must be divisible by k. In other words, m must be 1 more than a multiple of k. So, the equivalence class [1] contains all integers that are 1 more than a multiple of k.

Following this pattern, we can see that each equivalence class will contain integers that are a certain number more than a multiple of k, where that number ranges from 0 to k-1. Therefore, the quotient set under this equivalence relation will be {[0], [1], ..., [k-1]}.

In summary, the quotient set of the given equivalence relation on the set of integers Z is a set containing all possible equivalence classes, which are determined by the remainder when dividing by k. This can be represented as {[0], [1], ..., [k-1]}.