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]