Show that $R$ is an equivalence relation.

Show, by giving a counter-example, that the statement ${r_1}R{r_2} \Rightarrow r_1^2Rr_2^2$ is false.

Determine

(i)     the equivalence class $E$ containing $1 + \sqrt 2$ ;

(ii)     the equivalence class $F$ containing $1 - \sqrt 2$ .

Show that

(i)     ${(1 + \sqrt 2 )^3} \in F$ ;

(ii)     ${(1 + \sqrt 2 )^6} \in E$ .

Determine whether the set $E$ forms a group under

  (i)     the operation of addition;

  (ii)     the operation of multiplication.

reflexive: if ${r_{}} = {a_{}} + {b_{}}\sqrt 2  \in S$ then $a \equiv a(\bmod 2)$ and $b \equiv b(\bmod 3)$

$( \Rightarrow rRr)$     A1

symmetric: if ${r_1}R{r_2}$ then ${a_1} \equiv {a_2}(\bmod 2)$ and ${b_1} \equiv {b_2}(\bmod 3)$ , and     M1

${a_2} \equiv {a_1}(\bmod 2)$ and ${b_2} \equiv {b_1}(\bmod 3)$ , (so that ${r_2}R{r_1}$ )     A1

transitive: if ${r_1}R{r_2}$ and ${r_2}R{r_3}$ then

$2|{a_1} - {a_2}$ and $2|{a_2} - {a_3}$     M1

$\Rightarrow 2|{a_1} - {a_2} + {a_2} - {a_3} \Rightarrow 2|{a_1} - {a_3}$     M1A1

$3|{b_1} - {b_2}$ and $3|{b_2} - {b_3}$

$\Rightarrow 3|{b_1} - {b_2} + {b_2} - {b_3} \Rightarrow 3|{b_1} - {b_3}( \Rightarrow {r_1}R{r_3})$     A1AG

[7 marks]

consider, for example, ${r_1} = 1 + \sqrt 2$ , ${r_2} = 3 + \sqrt 2$ $({r_1}R{r_2})$     M1

Note: Only award M1 if the two numbers are related and neither $a$ nor $b = 0$ .

$r_1^2 = 3 + 2\sqrt 2$ , $r_2^2 = 11 + 6\sqrt 2$     A1

the squares are not equivalent because $2 \ne 6(\bmod 3)$     A1

[3 marks]

(i)     $E = \left\{ {2k + 1 + (3m + 1)\sqrt 2 } :k,m \in \mathbb{Z} \right\}$     A1A1

(ii)     $F = \left\{ {2k + 1 + (3m - 1)\sqrt 2 } :k,m \in \mathbb{Z} \right\}$     A1

[3 marks]

(i)     ${(1 + \sqrt 2 )^3} = 7 + 5\sqrt 2$     A1

$= 2 \times 3 + 1 + (3 \times 2 - 1)\sqrt 2  \in F$     R1AG

(ii)     ${(1 + \sqrt 2 )^6} = 99 + 70\sqrt 2$     A1

$= 2 \times 49 + 1 + (3 \times 23 + 1)\sqrt 2  \in E$     R1AG

[4 marks]

(i)     $E$ is not a group under addition     A1

any valid reason eg $0 \notin E$     R1

(ii)     $E$ is not a group under multiplication     A1

any valid reason eg $1 \notin E$     R1

[4 marks]

The majority of candidates earned significant marks on this question. However, many lost marks in part (a) by assuming that equivalence modulo $2$ and $3$ is transitive. This is a non-trivial true result but requires proof.

The majority of candidates earned significant marks on this question.

The majority of candidates earned significant marks on this question.

The majority of candidates earned significant marks on this question.

The majority of candidates earned significant marks on this question.