Let S be the set of all real numbers. Show that the relation R = {(a,

1.	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» Given that, ∀ a, b ∈ S, R = {(a, b) : a² + b² = 1 } Now, R is Reflexive if (a,a) ∈ R ∀ a ∈ S For any a ∈ S, we have a²+a² = 2 a² ≠ 1 ⇒ (a,a) ∉ R Thus, R is not reflexive. R is Symmetric if (a,b) ∈ R ⇒ (b,a) ∈ R ∀ a,b ∈ S (a,b) ∈ R ⇒ a² + b²= 1 ⇒ b² + a² = 1 ⇒ (b,a) ∈ R Thus, R is symmetric . R is Transitive if (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) ∈ R ∀ a,b,c ∈ S Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ S ⇒ a² + b² = 1 and b²+ c² = 1 Adding both, we get a²+ c²+2b² = 2 ⇒ (a, c) ∉ R Thus, R is not transitive. Thus, R is symmetric but neither reflexive nor transitive.

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»

Given that, ∀ a, b ∈ S, R = {(a, b) : a² + b² = 1 }

Now,

R is Reflexive if (a,a) ∈ R ∀ a ∈ S

For any a ∈ S, we have

a²+a² = 2 a² ≠ 1

⇒ (a,a) ∉ R

Thus, R is not reflexive.

R is Symmetric if (a,b) ∈ R ⇒ (b,a) ∈ R ∀ a,b ∈ S

(a,b) ∈ R

⇒ a² + b²= 1

⇒ b² + a² = 1

⇒ (b,a) ∈ R

Thus, R is symmetric .

R is Transitive if (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) ∈ R ∀ a,b,c ∈ S

Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ S

⇒ a² + b² = 1 and b²+ c² = 1

Adding both, we get

a²+ c²+2b² = 2

⇒ (a, c) ∉ R

Thus, R is not transitive.

Thus, R is symmetric but neither reflexive nor transitive.