Q. 75.0( 1 Vote )

Let f: W <span la

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, 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


Adding both


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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