Question

Let S be the set of all real numbers. Show that the relation R = {(a, b): a² + b² = 1} is symmetric but neither reflexive nor transitive.

Answer 1

We know that

R = {(a, b): a² + b² = 1}

Reflexive-

R is the set of real numbers

If a ∈ R then a² + a² ≠ 1 where a = 2, 3 ….

Therefore, R is not reflexive.

Symmetric-

Consider (a, b) ∈ R where a² + b² = 1

We get

(b, a) ∈ R and (a, b) ∈ R

Hence, R is symmetric.

Transitive-

Consider (a, b) ∈ R and (b, c) ∈ R

We know that

(cos 30^o, sin 30^o) ∈ R and (sin 30^o, cos 30^o) ∈ R

So (cos 30^o, cos 30^o) ∉ R

Therefore, R is not transitive.

Here, R is symmetric but neither reflexive nor transitive.