Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In the book of Richard Hammack, I come accross with the following question:

There are two different equivalence relations on the set $A = \{a,b\}$. Describe them.

OK, I found that the solution is,

$$R_1 = \{(a,a),(b,b),(a,b),(b,a)$$ and $$R_2 = \{(a,a),(b,b)\}$$

Then I thought two more equivalence classes $R_3 = \{(a,a)\}$, $R_4 = \{(b,b)\}$. But when I looked the answer, I saw that, $R_1$ and $R_2$ are true but others are false. Why is that?

share|improve this question
add comment

4 Answers 4

up vote 4 down vote accepted

Because they're not reflexive. An equivalence relation is reflexive, i.e. it contains all pairs of the form $(x,x)$.

share|improve this answer
add comment

A relation is an equivalence relation if and only if it is reflexive, symmetric, and transitive. Your first two relations are indeed equivalence relations.

A relation $R$ is reflexive on a set $A$ if and only if for all $x \in A, (x, x) \in R$.

In $R_3$, we do not have that for $b \in A$, $(b, b) \in R_3$.

And in $R_4$, we do not have that for $a \in A$, $(a, a) \in R_4$.

So neither $R_3$ nor $R_4$ are reflexive, hence neither can be an equivalence relation.

share|improve this answer
add comment

Read the reflexive property again.

Reflexive property is not conditional.

share|improve this answer
add comment

$R_3 = \{(a, a)\}$ and $R_4 = \{(b, b)\}$ are not equivalence relations because they are not reflexive; $b$ must be related to itself.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.