Date | May 2016 | Marks available | 5 | Reference code | 16M.2.hl.TZ0.6 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Show that | Question number | 6 | Adapted from | N/A |
Question
Consider the set \(J = \left\{ {a + b\sqrt 2 :a,{\text{ }}b \in \mathbb{Z}} \right\}\) under the binary operation multiplication.
Consider \(a + b\sqrt 2 \in G\), where \(\gcd (a,{\text{ }}b) = 1\),
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 \({a^2} - 2{b^2} = \pm 1\).
Markscheme
\(\left( {a + b\sqrt 2 } \right) \times \left( {c + d\sqrt 2 } \right) = ac + bc\sqrt 2 + ad\sqrt 2 + 2bd\) 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]
Examiners report
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.