Prove that $f$ is

(i)     not injective;

(ii)     not surjective.

The relation $R$ is defined for $a,{\text{ }}b \in \mathbb{R}$ so that $aRb$ if and only if $f(a) \times f(b) = 1$.

Show that $R$ is an equivalence relation.

The relation $R$ is defined for $a,{\text{ }}b \in \mathbb{R}$ so that $aRb$ if and only if $f(a) \times f(b) = 1$.

State the equivalence classes of $R$.

(i)     eg$\;\;\;f(2) = f(3)$     M1

hence $f(a) = f(b) \Rightarrow a = b$     R1

so not injective     AG

(ii)     eg$\;\;\;$Codomain is $\mathbb{R}$ and range is $\{  - 1,{\text{ }}1\}$     M1

these not the same so not surjective     R1AG

Note:     if counter example is given it must be stated it is not in the range to obtain the R1. Eg $f(x) = 2$ has no solution as $f(x) \in \{  - 1,{\text{ }}1\} \forall x$.

[4 marks]

if $a \ge 0$ then $f(a) \times f(a) = 1 \times 1 = 1$     A1

if $a < 0$ then $f(a) \times f(a) =  - 1 \times  - 1 = 1$     A1

in either case $aRa$ so $R$ is reflexive     R1

$aRb \Rightarrow f(a) \times f(b) = 1 \Rightarrow f(b) \times f(a) = 1 \Rightarrow bRa$     A1

so $R$ is symmetric     R1

if $aRb$ then either $a \ge 0$ and $b \ge 0$ or $a < 0$ and $b < 0$

if $a \ge 0$ and $b \ge 0$ and $bRc$ then $c \ge 0$ so $f(a) \times f(c) = 1 \times 1 = 1$ and $aRc$     A1

if $a < 0$ and $b < 0$ and $bRc$ then $c < 0$ so $f(a) \times f(c) =  - 1 \times  - 1 = 1$ and $aRc$     A1

in either case $aRb$ and $bRc \Rightarrow aRc$ so $R$ is transitive     R1

Note:     Accept

$f(a) \times f(b) \times f(b) \times f(c) = 1 \times 1 = 1 \Rightarrow f(a) \times 1 \times {\text{ }}f(c) = 1 \Rightarrow {\text{ }}f(a) \times f(c) = 1$

Note:     for each property just award R1 if at least one of the A marks is awarded.

as $R$ is reflexive, symmetric and transitive it is an equivalence relation     AG

[8 marks]

equivalence classes are $[0,{\text{ }}\infty [$ and $] - \infty ,{\text{ }}0[$     A1A1

Note:     Award A1A0 for both intervals open.

[2 marks]

Total [14 marks]