Question

If $r$ is a relation from $R$ (set of real numbers) to $R$ defined by $r=\left\{(a,b)|a,b\in R;a-b+\sqrt{3}\text{is an irrational number}\right\}$. Then, the relation r is

• A. an equivalence relation
• B. only reflexive
• C. only symmetric
• D. only transitive

Answer 1

The correct option is A

an equivalence relation

Explanation for the correct options:

A relation is reflexive if $(a,a)\in R$ for every $a\in A$.

A relation is symmetric if $(a,b)\in R$ then $(b,a)\in R$.

A relation is transitive if $(a,b)\in R$ and $(b,c)\in R$ then $(a,c)\in R$.

Check for reflexive relation:

The relation $r$ is defined as $r=\left\{(a,b)|a,b\in R;a-b+\sqrt{3}\text{is an irrational number}\right\}$

if $a\in R$ then $a-a+\sqrt{3}=\sqrt{3}$ is an irrational number

Therefore, $(a,a)\in r$ which implies that relation $r$ is reflexive.

Check for symmetric relation:

Assume $(a,b)\in r$ that is $a-b+\sqrt{3}$ is an irrational number

Also $b-a+\sqrt{3}$ is an irrational then $(b,a)\in r$ also exists.

Therefore, $r$ is symmetric.

Check for transitive relation:

Assume $(a,b)\in r$ and $(b,c)\in r$

Therefore, both $a-b+\sqrt{3}$ and $a-c+\sqrt{3}$ are irrational

Adding the above two relation

$\left(a-b+\sqrt{3}\right)+\left(b-c+\sqrt{3}\right)=a-c+2\sqrt{3}$ an irrational number.

Therefore, $(a,c)\in r$ exists, which implies that $r$ is transitive.

Conclude that $r$ being reflexive, symmetric and transitive, then it is an equivalence relation.

Hence, the correct option is (A).