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.
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
⇒ a2+ c2= 2-2b2 ≠ 1
⇒ (a, c) ∉ R
Thus, R is not transitive.
Thus, R is symmetric but neither reflexive nor transitive.