(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