Q. 75.0( 1 Vote )

# Let f: W <span la

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, x2+y2= y2+ x2

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 x2+t2= y2+ s2

s2+y2= t2+ x2

(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)

x2+t2 = y2+s2 and s2+n2= t2+m2

x2+t2+s2+n2= y2+ s2+t2+ m2

x2+n2= y2+ m2

(x, y)R(m, n)

Therefore R is transitive.

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

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