Question

(a) Prove that the following relation on the set R of real numbers is an equivalence relation:$\mathit{a}\mathbf{~}\mathit{b}$ if and only if$\mathit{c}\mathit{o}\mathit{s}\mathit{a}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{b}$ .

(b) Describe the equivalence class of 0 and the equivalence class of$\mathit{\pi }\mathbf{/}\mathbf{2}$ .

Short answer

a) It is proved that $~$is an equivalence relation.

b) The class 0 of is$\left\{0\right\}$ , and the equivalence class of$\frac{\mathrm{\pi }}{2}$ is$\left\{-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right\}$ .

Step-by-step solution

Definitions of Equivalence relation

A relation R defined on a set is called an equivalence relation if it is reflexive, symmetric and transitive.

Prove that ~ is an equivalence relation(a)

Check$~$is reflexive:

Let$a\in R$then$\cos a=\cos a$that is$a~a$. So,$~$is reflexive.

Check$~$is symmetric:

Let$a,b\in R$and$a~b$that is$\cos a=\cos b$which is same as$\cos b=\cos a$implies$b~a$. That ,is$a~b$implies$b~a$. So,$~$is symmetric.

Check$~$is transitive:

Let $a,b,c\in R$and $a~b$implies $\cos a=\cos b$and$b~c$implies$\cos b=\cos c$on equating $\cos a=\cos c$implies $a~c$.

Thus,$a~b$, and$b~c$implies$a~c$. So,$~$is transitive.

Since all the three conditions are satisfies.

Therefore, the $~$is an equivalence relation.

Describe the equivalence class of 0 and π/2 .(b)

If$x\ne 0$ , then the equivalence class of x isrole="math" localid="1659160938470" $\overline{x}=\left\{-x,x\right\}$ .

The equivalence class of 0 is$\overline{0}=\left\{0\right\}$ .

The equivalence class of$\frac{\mathrm{\pi }}{2}$ is$\overline{\frac{\mathrm{\pi }}{2}}=\left\{-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right\}$ .