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.