Answer :

Consider the case when n is odd then n–1 should be even

Using the definition

f: W → W is such that

f(f(n))=f(n–1)=n–1+1=n

Consider the case when n is even then n+1 should be odd

Using the definition

fof: W → W is such that

f(f(n))=f(n+1)=n+1–1=n

Therefore fof = I

So f is invertible and f = f^{–1}.

**OR**

To prove R is an equivalence relation we need to prove that R is reflexive, symmetric and transitive

Let (x, y) є N × N

So, x^{2}+y^{2}= y^{2}+ x^{2}

Here (x, y)R(x, y)

Therefore R is reflexive.

Let (x, y), (s, t) є N × N such that (x, y)R(s, t)

So x^{2}+t^{2}= y^{2}+ s^{2}

⇒ s^{2}+y^{2}= t^{2}+ x^{2}

⇒ (s, t)R(x, y)

Therefore R is symmetric

Let (x, y), (s, t), (m, n) є N × N such that (x, y)R(s, t) and (s, t)R(m, n)

⇒ x^{2}+t^{2} = y^{2}+s^{2} and s^{2}+n^{2}= t^{2}+m^{2}

Adding both

x^{2}+t^{2}+s^{2}+n^{2}= y^{2}+ s^{2}+t^{2}+ m^{2}

⇒ x^{2}+n^{2}= y^{2}+ m^{2}

⇒ (x, y)R(m, n)

Therefore R is transitive.

R is reflexive, symmetric and transitive and thus an equivalence relation.

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