Bijective Functions

Question

Write down, with proof, bijective functions with the indicated domains and codomains:

(a) f : {2 ^−k : k ∈ N} → {2 ^ (-k-1): k ∈ N}

(b) g : (0, 1) → (0, 1]—you may use the function f from the previous part in your work.

Answer 1

⁽¹⁾

^{f(x)=x/2 ∀x=2⁻ᵏ: k∈ℕ}

^{Let y=f(x)=x/2, so x=2y.}

^{Let g(y)=x=2y, then g(x)=2x ∀x=2¹⁻ᵏ: k∈ℕ, and g(x)=f⁻¹(x).}

^{Therefore f(x) is bijective because it has an inverse.}

^{Range of f(x) is (0,½]. Range of g(x) is (0,1] which is the range of f⁻¹(x).}

⁽²⁾ 

^{g:(0,1)→(0,1]}

^{Consider (2¹⁻²ᵏ+2⁻ᵏ-1)/(2¹⁻ᵏ-1) for k∈ℕ.}

^{When k=1, the expression is undefined.}

^{As k→∞, the expression → 1.}

^{For x=2⁻ᵏ, the expression becomes (2x²+x-1)/(2x-1)=(2x-1)(x+1)/(2x-1).}

^{This expression reduces to x+1 for x≠½. When k=1, x=½.}

^{Let g(x)=(2x²+x-1)/(2x-1) which has the range (0,1) and domain (0,½), since x=2⁻ᵏ: k∈ℕ.}

^{If y=g(x)=x+1, then x=y-1, so if h(y)=x, h(y)=y-1, h(x)=x-1. And h(x)=g⁻¹(x), so g(x) has an inverse which makes g bijective.}