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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.

[2]
a.

State the identity in \(J\).

[1]
b.

Show that

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

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

[5]
c.

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

[7]
d.

(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 \({a^2} - 2{b^2} =  \pm 1\).

[4]
e.

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]

a.

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

[1 mark]

b.

(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]

c.

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]

d.

(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]

e.

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).

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).

b.

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).

c.

In part (d) closure was rarely established satisfactorily.

d.

Part (e) was often tackled well.

e.

Syllabus sections

Topic 4 - Sets, relations and groups » 4.6 » The identity element \(e\).

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