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

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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

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Question ID: mtid-5-stid-8-sqid-3792-qpid-2491