The function \(f:{\mathbb{R}^ + } \times {\mathbb{R}^ + } \to {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) is defined by \(f(x,{\text{ }}y) = \left( {xy,{\text{ }}\frac{x}{y}} \right)\).

Prove that \(f\) is a bijection.

we need to show that \(f\) is injective and surjective     (R1)

Note: Award R1 if seen anywhere in the solution.

\(\underline {{\text{injective}}} \)

let \((a,{\text{ }}b)\) and \((c,{\text{ }}d) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\), and let \(f(a,{\text{ }}b) = f(c,{\text{ }}d)\)     M1

it follows that

\(ab = cd\) and \(\frac{a}{b} = \frac{c}{d}\)     A1

multiplying these equations,

\({a^2} = {c^2} \Rightarrow a = c\) and therefore \(b = d\)     A1

since \(f(a,{\text{ }}b) = f(c,{\text{ }}d) \Rightarrow (a,{\text{ }}b) = (c,{\text{ }}d),{\text{ }}f\) is injective     R1

Note: Award R1 if stated anywhere as needing to be shown.

\(\underline {{\text{surjective}}} \)

let \((p,{\text{ }}q) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\)

consider \(f(x,{\text{ }}y) = (p,{\text{ }}q)\) so \(xy = p\) and \(\frac{x}{y} = q\)     M1A1

multiplying these equations,

\({x^2} = pq\) so \(x = \sqrt {pq} \) and therefore \(y = \sqrt {\frac{p}{q}} \)     A1

so given \((p,{\text{ }}q) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + },{\text{ }}\exists (x,{\text{ }}y) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) such that \(f(x,{\text{ }}y) = (p,{\text{ }}q)\) which shows that \(f\) is surjective     R1

Note: Award R1 if stated anywhere as needing to be shown.

\(f\) is therefore a bijection

[9 marks]