Solution to Let O be the set of odd numbers and O’ = {1, 5, 9, 13, … - Sikademy
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Archangel Macsika

Let O be the set of odd numbers and O’ = {1, 5, 9, 13, 17, ...} be its subset. Define the bijections, f and g as: f : O \to→ O’, f(d) = 2d - 1, \forall∀ d \in∈ O. g : \NuN \to→ O, g(n) = 2n + 1, \forall∀ n \isin∈ \NuN . Using only the concept of function composition, can there be a bijective map from \NuN to O’? If so, compute it. If not, explain in details why not.

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Given that the functions f \text{~and~} g are bijections, defined as f(d) = 2d - 1, \forall d \in O, g(n) = 2n + 1, \forall n \in \N. Here O' = \{4x + 1 : x\in \N \cup \{0\}\} .

But we see that, g : \N \to O is not surjective since there is no preimage of 1 in \N.

For, 1= g(n)=2n+1 \implies n = 0, but 0\notin \N.

Hence there cannot be a bijective function from \N \to O'.

For g must be bijective, it is must be defined as g: \N \cup \{0\} \to O.

Setting g: \N \cup \{0\} \to O and using the fact "composition of two bijective functions is bijective", we have h = f \circ g : \N \cup\{0\}\to O' is also a bijective function. The composition of functions is

h(x) = (f \circ g)(x) = f(g(x)) = f(2x+1) = 2(2x+1)-1 = 4x+1,~\forall x \in \N \cup \{0\}.

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