 # Mark the tick against the correct answer in the following: Question:

Mark the tick against the correct answer in the following:

Let $S$ be the set of all real numbers and let $R$ be a relation on $S$ defined by $a R b \Leftrightarrow a^{2}+b^{2}=1 .$ Then, $R$ is

A. symmetric but neither reflexive nor transitive

B. reflexive but neither symmetric nor transitive

C. transitive but neither reflexive nor symmetric

D. none of these

Solution:

According to the question,

Given set $S=\{\ldots \ldots,-2,-1,0,1,2 \ldots \ldots\}$

And $R=\left\{(a, b): a, b \in S\right.$ and $\left.a^{2}+b^{2}=1\right\}$

Formula

For a relation $R$ in set $A$

Reflexive

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

Symmetric

The relation is Symmetric if $(a, b) \in R$, then $(b, a) \in R$

Transitive

Relation is Transitive if $(a, b) \in R \&(b, c) \in R$, then $(a, c) \in R$

Equivalence

If the relation is reflexive, symmetric and transitive, it is an equivalence relation.

Check for reflexive

Consider, $(a, a)$

$\therefore a^{2}+a^{2}=1$ which is not always true

Ex_if $a=2$

$\therefore 2^{2}+2^{2}=1 \Rightarrow 4+4=1$ which is false.

Therefore , R is not reflexive ……. (1)

Check for symmetric

$a R b \Rightarrow a^{2}+b^{2}=1$

b $R a \Rightarrow b^{2}+a^{2}=1$

Both the equation are the same and therefore will always be true.

Therefore , R is symmetric ……. (2)

Check for transitive

$a R b \Rightarrow a^{2}+b^{2}=1$

b $\mathrm{R} \mathrm{c} \Rightarrow \mathrm{b}^{2}+\mathrm{c}^{2}=1$

$\therefore a^{2}+c^{2}=1$ will not always be true

Ex $_{-} \mathrm{a}=-1, \mathrm{~b}=0$ and $\mathrm{c}=1$

$\therefore(-1)^{2}+0^{2}=1,0^{2}+1^{2}=1$ are true

But $(-1)^{2}+1^{2}=1$ is false.

Therefore , R is not transitive ……. (3)

Now, according to the equations (1), (2), (3)

Correct option will be (A)