Let S be the set of all real numbers and let R = {(a, b) : a, b ∈ S

1.	Let S be the set of all real numbers and let R = {(a, b) : a, b ∈ S and a = ± b}. Show that R is an equivalence relation on S.
Answer» In order to show R is an equivalence relation we need to show R is Reflexive, Symmetric and Transitive. Given that, ∀ a, b ∈ S, R = {(a, b) : a = ± b } Now, R is Reflexive if (a,a) ∈ R ∀ a ∈ S For any a ∈ S, we have a = ±a ⇒ (a,a) ∈ R Thus, R is reflexive. R is Symmetric if (a,b) ∈ R ⇒ (b,a) ∈ R ∀ a,b ∈ S (a,b) ∈ R ⇒ a = ± b ⇒ b = ± a ⇒ (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 ⇒ a = ± b and b = ± c ⇒ a = ± c ⇒ (a, c) ∈ R Thus, R is transitive. Hence, R is an equivalence relation.

Let S be the set of all real numbers and let R = {(a, b) : a, b ∈ S and a = ± b}. Show that R is an equivalence relation on S.

Answer»

In order to show R is an equivalence relation we need to show R is Reflexive, Symmetric and Transitive.

Given that, ∀ a, b ∈ S, R = {(a, b) : a = ± b }

Now,

R is Reflexive if (a,a) ∈ R ∀ a ∈ S

For any a ∈ S, we have

a = ±a

⇒ (a,a) ∈ R

Thus, R is reflexive.

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

(a,b) ∈ R

⇒ a = ± b

⇒ b = ± a

⇒ (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

⇒ a = ± b and b = ± c

⇒ a = ± c

⇒ (a, c) ∈ R

Thus, R is transitive.

Hence, R is an equivalence relation.