Functions-21

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Prove that the function f: Q → Q given by f(x) = 2x – 3 for all x ∈ Q is a bijection.

Solution: Injectivity: Let x, y be two arbitrary elements in Q. Then,

f(x) = f(y) => 2x – 3 = 2y – 3 => 2x = 2y => x = y

Thus, f(x) = f(y) => x = y for all x, y ∈ Q.

So, f is an injective map.

Surjectivity: Let y be an arbitrary element of Q. Then,

f(x) = y => 2x – 3 = y => x = (y + 3) / 2

Clearly, for all y ∈ Q, x = (y + 3) / 2 ∈ Q. Thus, for all y ∈ Q (co-domain) there exists x ∈ Q (domain) given by

x = (y + 3) / 2 such that f(x) = f[(y + 3) / 2] = 2[(y + 3) / 2] – 3 = y

Thus , every element in the co-domain has its pre-image in x. So, f is a surjection.

Hence, f : Q → Q is a bijection.

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