Question

(iv) f : N → Y such that f(n) = 2n + 1,
where Y = {y ∈ N : y = 4n + 3 for some n ∈ N}

Answer 1

f(n)=2n+1

Put n =y

f(y)=2y+1

and y = 4n+3

F(y)= 2(4n+3)+1=8n+7

Ans: 8n+7