Question

Let S be the set of all real numbers and let R be a relation on S defined by $a$ R $b\iff a^2+b^2=1$. 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

Answer 1

The correct option is A Symmetric but neither reflexive nor transitive

According to the question,

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

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

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 $(a,b)\in R$ and $(b,c)\in R$, 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.

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

$aRb\Rightarrow a^2+b^2=1$

$bRa\Rightarrow b^2+a^2=1$

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

$\therefore$ $R$ is symmetric ---- ( 2 )

Check for transitive

$aRb\Rightarrow a^2+b^2=1$

$bRc\Rightarrow b^2+c^2=1$

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

Let $a=-1,b=0$ and $c=1$

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

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

$\therefore$ $R$ is not transitive ---- ( 3 )

Now, according to ( 1 ), ( 2 ) and ( 3 )

Correct answer is option $A.$