# Finding the equivalence classes

The relation X on the set $\{1, 2, 3, 4, 5\}$ is defined by the rule $(a, b) ϵ X$ if 3 divides a – b.

• List the elements of X

These are $\{(4,1),(1,4),(5,2),(2,5),(1,1),(2,2),(3,3),(4,4),(5,5)\}$

• List the equivalence class$\color{red}{\text{es}}$

The answer is $\{1,4\},\{2,5\},\{3\}$

This is where I am confused. I thought equivalence class meant that one should only present the elements that don't result in a similar result.

However, $4-1 = 5-2 = 3$ these are all the same

Would anyone care to explain how the equivalence class can be found in this case?

• you're confusing a set of representatives with the set of classes. – Henno Brandsma Jan 17 '17 at 11:31
• your classes are correct. – Henno Brandsma Jan 17 '17 at 11:32
• Yes, i got them from the answer booklet, I couldn't understand how to get them. – Johny Jan 17 '17 at 11:32

Equivalence classes (mean) that one should only present the elements that don't result in a similar result.

I believe you are mixing up two slightly different questions. Each individual equivalence class consists of elements which are all equivalent to each other. That is why one equivalence class is $\{1,4\}$ - because $1$ is equivalent to $4$. We can refer to this set as "the equivalence class of $1$" - or if you prefer, "the equivalence class of $4$".

Note that we have been talking about individual classes. We are now going to talk about all possible equivalence classes. You could list the complete sets, $$\{1,4\}\quad\hbox{and}\quad\{2,5\}\quad\hbox{and}\quad\{3\}\ .$$ Alternatively, you could name each of them as we did in the previous paragraph, $$\hbox{(the equivalence class of 1)}\quad\hbox{and}\quad \hbox{(the equivalence class of 2)}\quad\hbox{and}\quad \hbox{(the equivalence class of 3)}\ .$$ Or if you prefer, $$\hbox{(the equivalence class of 4)}\quad\hbox{and}\quad \hbox{(the equivalence class of 2)}\quad\hbox{and}\quad \hbox{(the equivalence class of 3)}\ .$$ You see that the "names" we use here are three elements with no two equivalent. I think you are confusing this with the previous paragraph.

Hope this helps!

• I assume as a third set I could any of the $(x,x)$ sets? Like $(1,1)$, $(2,2)$, etc... – Johny Jan 17 '17 at 11:34
• No. Note that each equivalence class contains individual elements from $\{1,2,3,4,5\}$, not pairs of elements. In this example, $1$ cannot be in an equivalence class by itself because $4$ has to be in the same class. The fact that $1$ is related to $1$ is reflected in the fact that $1$ is (obviously) in the same set as $1$. – David Jan 17 '17 at 11:38
• Also, $3$ has to be in the same class as other numbers which are equivalent to it. But there are no such numbers, that's why $3$ is by itself. – David Jan 17 '17 at 11:40

By definition, the equivalence classes

(1) partition the Universe, $\{1,2,3,4,5\}$. Your classes do partition the universe.

(2) Pairs of elements within the equivalence classes are in relation $X$.

(3) Pairs of elements from different equivalence classes are not in relation $X$. This holds in your case.

I edited your question. (Plural of class)

How to find the equivalence classes? In such a finite case, it is easy.

(1) Take the first element of your Universe and compare it to the other elements. List the pairs that you found to be in $X$. This list is your first eqivalence class.

(2) There will be elements that you have not yet paired with the first one. Take the first such element and see those elements that remained after operation (1). Do the comparison as in (1). You'll find the second equivalence class.

(3) Repeat (2) until there will be elements not yet paired with another.