Show that $J$ is closed.

State the identity in $J$.

Show that

(i)     $1 - \sqrt 2$ has an inverse in $J$;

(ii)     $2 + 4\sqrt 2$ has no inverse in $J$.

Show that the subset, $G$, of elements of $J$ which have inverses, forms a group of infinite order.

(i)     Find the inverse of $a + b\sqrt 2$.

(ii)     Hence show that ${a^2} - 2{b^2}$ divides exactly into $a$ and $b$.

(iii)     Deduce that ${a^2} - 2{b^2} =  \pm 1$.

$\left( {a + b\sqrt 2 } \right) \times \left( {c + d\sqrt 2 } \right) = ac + bc\sqrt 2  + ad\sqrt 2  + 2bd$    M1

$= ac + 2bd + (bc + ad)\sqrt 2  \in J$    A1

hence $J$ is closed     AG

Note:     Award M0A0 if the general element is squared.

[2 marks]

the identity is $1(a = 1,{\text{ }}b = 0)$     A1

[1 mark]

(i)     $\left( {1 - \sqrt 2 } \right) \times a = 1$

$a = \frac{1}{{1 - \sqrt 2 }}$    M1

$= \frac{{1 + \sqrt 2 }}{{\left( {1 - \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}} = \frac{{1 + \sqrt 2 }}{{ - 1}} =  - 1 - \sqrt 2$    A1

hence $1 - \sqrt 2$ has an inverse in $J$     AG

(ii)     $\left( {2 + 4\sqrt 2 } \right) \times a = 1$

$a = \frac{1}{{2 + 4\sqrt 2 }}$    M1

$= \frac{{2 - 4\sqrt 2 }}{{\left( {2 - 4\sqrt 2 } \right)\left( {2 + 4\sqrt 2 } \right)}} = \frac{{2 - 4\sqrt 2 }}{{ - 28}}$    A1

which does not belong to $J$     R1

hence $2 + 4\sqrt 2$ has no inverse in $J$     AG

[5 marks]

multiplication is associative     A1

let ${g_1}$ and ${g_2}$ belong to $G$, then $g_1^{ - 1},{\text{ }}g_2^{ - 1}$ and $g_2^{ - 1}g_1^{ - 1}$ belong to $J$     M1

then $({g_1}{g_2}) \times (g_2^{ - 1}g_1^{ - 1}) = 1 \times 1 = 1$     A1

so ${g_1}{g_2}$ has inverse $g_2^{ - 1}g_1^{ - 1}$ in $J \Rightarrow G$ is closed     A1

$G$ contains the identity     A1

$G$ possesses inverses     A1

$G$ contains all integral powers of $1 - \sqrt 2$     A1

hence $G$ is an infinite group     AG

[7 marks]

(i)     ${\left( {a + b\sqrt 2 } \right)^{ - 1}} = \frac{1}{{a + b\sqrt 2 }} = \frac{1}{{a + b\sqrt 2 }} \times \frac{{a - b\sqrt 2 }}{{a - b\sqrt 2 }}$     M1

$= \frac{a}{{{a^2} - 2{b^2}}} - \frac{b}{{{a^2} - 2{b^2}}}\sqrt 2$    A1

(ii)     above number belongs to $J$ and ${a^2} - 2{b^2} \in \mathbb{Z}$     R1

implies ${a^2} - 2{b^2}$ divides exactly into $a$ and $b$     AG

(iii)     since $\gcd (a,{\text{ }}b) = 1$     R1

${a^2} - 2{b^2} =  \pm 1$    AG

[4 marks]

Parts (a), (b) and (c) were generally well done. In a few cases, squaring a general element was thought, erroneously, to be sufficient to prove closure in part (a).

Parts (a), (b) and (c) were generally well done. In a few cases, squaring a general element was thought, erroneously, to be sufficient to prove closure in part (a).

Parts (a), (b) and (c) were generally well done. In a few cases, squaring a general element was thought, erroneously, to be sufficient to prove closure in part (a).

In part (d) closure was rarely established satisfactorily.

Part (e) was often tackled well.