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</div><h2>SL Paper 2</h2><div class="specification">
<p>The set of all integer s from 0 to 99 inclusive is denoted by <em>S</em>. The binary operations \( * \) and \( \circ \) are defined on <em>S</em> by</p>
<p style="padding-left: 180px;">\(a * b = \left[ {a + b + 20} \right]\)(mod 100)</p>
<p style="padding-left: 180px;">\(a \circ b = \left[ {a + b - 20} \right]\)(mod 100).</p>
</div>
<div class="specification">
<p>The equivalence relation <em>R</em> is defined by \(aRb \Leftrightarrow \left( {{\text{sin}}\frac{{\pi a}}{5} = {\text{sin}}\frac{{\pi b}}{5}} \right)\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the identity element of <em>S</em> with respect to \( * \).</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that every element of <em>S</em> has an inverse with respect to \( * \).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>State which elements of <em>S</em> are self-inverse with respect to \( * \).</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Prove that the operation \( \circ \) is not distributive over \( * \).</p>
<div class="marks">[5]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine the equivalence classes into which <em>R</em> partitions <em>S</em>, giving the first four elements of each class.</p>
<div class="marks">[5]</div>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find two elements in the same equivalence class which are inverses of each other with respect to \( * \).</p>
<div class="marks">[2]</div>
<div class="question_part_label">f.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(a + e + 20 = a\)(mod 100) <em><strong> (M1)</strong></em></p>
<p>\(e = - 20\)(mod 100) <em><strong>(A1)</strong></em></p>
<p>\(e = 80\) <em><strong>A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(a + {a^{ - 1}} + 20 = 80\)(mod 100) <em><strong> (M1)</strong></em></p>
<p>inverse of \(a\) is \(60 - a\) (mod 100) <em><strong>A1</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>30 and 80 <em><strong>A1A1</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(a \circ \left( {b * c} \right) = a \circ \left( {b + c + 20} \right)\)(mod 100)</p>
<p>\( = a + \left( {b + c + 20} \right) - 20\)(mod 100) <em><strong> (M1)</strong></em></p>
<p>\( = a + b + c\)(mod 100) <em><strong>A1</strong></em></p>
<p>\(\left( {a \circ b} \right) * \left( {a \circ c} \right) = \left( {a + b - 20} \right) * \left( {a + c - 20} \right)\)(mod 100) <em><strong>M1</strong></em></p>
<p>\( = a + b - 20 + a + c - 20 + 20\)(mod 100)</p>
<p>\( = 2a + b + c - 20\)(mod 100) <em><strong>A1</strong></em></p>
<p>hence we have shown that \(a \circ \left( {b * c} \right) \ne \left( {a \circ b} \right) * \left( {a \circ c} \right)\) <em><strong>R1</strong></em></p>
<p>hence the operation \( \circ \) is not distributive over \( * \) <em><strong>AG</strong></em></p>
<p><strong>Note:</strong> Accept a counterexample.</p>
<p><em><strong>[5 marks]</strong></em></p>
<p> </p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>{0,5,10,15...} <em><strong>A1</strong></em></p>
<p>{1,4,11,14...} <em><strong>A1</strong></em></p>
<p>{2,3,12,13...} <em><strong>A1</strong></em></p>
<p>{6,9,16,19...} <em><strong>A1</strong></em></p>
<p>{7,8,17,18...} <em><strong>A1</strong></em></p>
<p><em><strong>[5 marks]</strong></em></p>
<p> </p>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>for example 10 and 50, 20 and 40, 0 and 60… <em><strong>A2</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<p> </p>
<div class="question_part_label">f.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">f.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Consider the set \(J = \left\{ {a + b\sqrt 2 :a,{\text{ }}b \in \mathbb{Z}} \right\}\) under the binary operation multiplication.</p>
</div>
<div class="specification">
<p class="p1">Consider \(a + b\sqrt 2 \in G\)<span class="s1">, where \(\gcd (a,{\text{ }}b) = 1\),</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that \(J\) <span class="s1">is closed.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">State the identity in \(J\).</p>
<div class="marks">[1]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that</p>
<p class="p1">(i) <span class="Apple-converted-space"> \(1 - \sqrt 2 \)</span> has an inverse in \(J\)<span class="s1">;</span></p>
<p class="p1">(ii) <span class="Apple-converted-space"> \(2 + 4\sqrt 2 \)</span> has no inverse in \(J\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">Show that the subset, \(G\)</span>, of elements of <span class="s1">\(J\) </span>which have inverses, forms a group of infinite order.</p>
<div class="marks">[7]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Find the inverse of \(a + b\sqrt 2 \)<span class="s1">.</span></p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Hence show that \({a^2} - 2{b^2}\) divides exactly into \(a\) and \(b\).</p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>Deduce that \({a^2} - 2{b^2} = \pm 1\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\(\left( {a + b\sqrt 2 } \right) \times \left( {c + d\sqrt 2 } \right) = ac + bc\sqrt 2 + ad\sqrt 2 + 2bd\) </span><strong><em>M1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\( = ac + 2bd + (bc + ad)\sqrt 2 \in J\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1">hence \(J\) is closed <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p3"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Award <strong><em>M0A0 </em></strong>if the general element is squared.</p>
<p class="p3"> </p>
<p class="p1"><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">the identity is \(1(a = 1,{\text{ }}b = 0)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[1 mark]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> \(\left( {1 - \sqrt 2 } \right) \times a = 1\)</span></p>
<p class="p1"><span class="Apple-converted-space">\(a = \frac{1}{{1 - \sqrt 2 }}\) </span><strong><em>M1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\( = \frac{{1 + \sqrt 2 }}{{\left( {1 - \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}} = \frac{{1 + \sqrt 2 }}{{ - 1}} = - 1 - \sqrt 2 \) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1">hence \(1 - \sqrt 2 \) has an inverse in \(J\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"> </p>
<p class="p1">(ii) <span class="Apple-converted-space"> \(\left( {2 + 4\sqrt 2 } \right) \times a = 1\)</span></p>
<p class="p1"><span class="Apple-converted-space">\(a = \frac{1}{{2 + 4\sqrt 2 }}\) </span><strong><em>M1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\( = \frac{{2 - 4\sqrt 2 }}{{\left( {2 - 4\sqrt 2 } \right)\left( {2 + 4\sqrt 2 } \right)}} = \frac{{2 - 4\sqrt 2 }}{{ - 28}}\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1">which does not belong to \(J\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"><span class="s2">hence \(2 + 4\sqrt 2 \) </span>has no inverse in \(J\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">multiplication is associative <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">let \({g_1}\) and \({g_2}\) belong to \(G\), then \(g_1^{ - 1},{\text{ }}g_2^{ - 1}\) <span class="s1">and \(g_2^{ - 1}g_1^{ - 1}\) </span>belong to \(J\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">then \(({g_1}{g_2}) \times (g_2^{ - 1}g_1^{ - 1}) = 1 \times 1 = 1\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">so \({g_1}{g_2}\) has inverse \(g_2^{ - 1}g_1^{ - 1}\) in \(J \Rightarrow G\) is closed <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\(G\) contains the identity <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\(G\) possesses inverses <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\(G\) contains all integral powers of \(1 - \sqrt 2 \) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">hence \(G\) is an infinite group <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"><strong><em>[7 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> \({\left( {a + b\sqrt 2 } \right)^{ - 1}} = \frac{1}{{a + b\sqrt 2 }} = \frac{1}{{a + b\sqrt 2 }} \times \frac{{a - b\sqrt 2 }}{{a - b\sqrt 2 }}\)</span> <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\( = \frac{a}{{{a^2} - 2{b^2}}} - \frac{b}{{{a^2} - 2{b^2}}}\sqrt 2 \) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"> </p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>above number belongs to \(J\) and \({a^2} - 2{b^2} \in \mathbb{Z}\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">implies \({a^2} - 2{b^2}\) divides exactly into \(a\) and \(b\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p2"> </p>
<p class="p2">(iii) <span class="Apple-converted-space"> </span>since \(\gcd (a,{\text{ }}b) = 1\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>R1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\({a^2} - 2{b^2} = \pm 1\) </span><span class="s1"><strong><em>AG</em></strong></span></p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">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).</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">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).</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">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).</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">In part (d) closure was rarely established satisfactorily.</p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Part (e) was often tackled well.</p>
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Draw the Cayley table for the set \(S = \left\{ {0,1,2,3,4,\left. 5 \right\}} \right.\) under addition </span><span style="font-family: times new roman,times; font-size: medium;">modulo six \(({ + _6})\) and hence show that \(\left\{ {S, + \left. {_6} \right\}} \right.\) is a group.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Show that the group is cyclic and write down its generators.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) Find the subgroup of \(\left\{ {S, + \left. {_6} \right\}} \right.\) that contains exactly three elements.</span></p>
<div class="marks">[11]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Prove that a cyclic group with exactly one generator cannot have more than </span><span style="font-family: times new roman,times; font-size: medium;">two elements.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(H\) is a group and the function \(\Phi :H \to H\) is defined by \(\Phi (a) = {a^{ - 1}}\) , where \({a^{ - 1}}\) </span><span style="font-family: times new roman,times; font-size: medium;">is the inverse of <strong><em>a</em></strong> under the group operation. Show that \(\Phi \) is an isomorphism </span><span style="font-family: times new roman,times; font-size: medium;"><strong>if and only if</strong> <em><strong>H</strong></em> is Abelian.</span></p>
<div class="marks">[9]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i)<br><img src="images/feet.png" alt></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the table is closed <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the identity is \(0\) <em><strong> A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(0\) is in every row and column once so each element has a unique inverse <em><strong> A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">addition is associative <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">therefore \(\left\{ {S, + \left. {_6} \right\}} \right.\) is a group <em><strong>R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) \(1 + 1 + 1 + 1 + 1 + 1 = 0\) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(1 + 1 + 1 + 1 + 1 = 5\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(1 + 1 + 1 + 1 = 4\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(1 + 1 + 1 = 3\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(1 + 1 = 2\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">so \(1\) is a generator of \(\left\{ {S, + \left. {_6} \right\}} \right.\) and the group is cyclic <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(since \(5\) is the additive inverse of \(1\)) \(5\) is also a generator <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(iii) \(\left\{ {0,2,\left. 4 \right\}} \right.\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong><em> </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[11 marks]</span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">if \(a\) is a generator of group \((G, * )\) then so is \({a^{ - 1}}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">if \((G, * )\) has exactly one generator \(a\) then \(a = {a^{ - 1}}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">so \({a^2} = e\) and \(G = \left\{ {e,\left. a \right\}} \right.\) \(\left\{ {\left. e \right\}} \right.\) <strong><em>A1R1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">so cyclic group with exactly one generator cannot have more than two elements <strong><em>AG</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[4 marks]</span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">every element of a group has a unique inverse so \(\Phi \) is a bijection <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\Phi (ab) = {(ab)^{ - 1}} = {b^{ - 1}}{a^{ - 1}}\) <strong><em>M1A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">if \(H\) is Abelian then it follows that</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({b^{ - 1}}{a^{ - 1}} = {a^{ - 1}}{b^{ - 1}} = \Phi (a)\Phi (b)\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">so \(\Phi \) is an isomorphism <em><strong>R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">if \(\Phi \) is an isomorphism, then <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">for all \(a,b \in H\) , \(\Phi (ab) = \Phi (a)\Phi (b)\) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({(ab)^{ - 1}} = {a^{ - 1}}{b^{ - 1}}\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow {b^{ - 1}}{a^{ - 1}} = {a^{ - 1}}{b^{ - 1}}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">so \(H\) is Abelian <em><strong>R1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[9 marks]</span></strong></em></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(a)(i) This was routine start to the question, but some candidates thought that commutativity was necessary as a group property. </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Showing why 1 and 5 were generators would have been appropriate since this is needed for the cyclic property of the group. </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) This did not prove difficult for most candidates. </span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">There were some long, confused arguments that did not lead anywhere. Candidates often do not appreciate the significance of "if" and "only". </span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">There were some long, confused arguments that did not lead anywhere. Candidates often do not appreciate the significance of "if" and "only". </span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The function \(f:\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\)</span><span style="font-family: times new roman,times; font-size: medium;"> is defined by \(\boldsymbol{X} \mapsto \boldsymbol{AX}\) , where \(\boldsymbol{X} = \left[ \begin{array}{l}<br>x\\<br>y<br>\end{array} \right]\) and \(\boldsymbol{A} = \left[ \begin{array}{l}<br>a\\<br>c<br>\end{array} \right.\left. \begin{array}{l}<br>b\\<br>d<br>\end{array} \right]\) </span><span style="font-family: times new roman,times; font-size: medium;">where \(a\) , \(b\) , \(c\) , \(d\) are all non-zero.</span></p>
</div>
<div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Consider the group \(\left\{ {S,{ + _m}} \right\}\) where \(S = \left\{ {0,1,2 \ldots m - 1} \right\}\) , \(m \in \mathbb{N}\) , \(m \ge 3\) and \({ + _m}\) </span><span style="font-family: times new roman,times; font-size: medium;">denotes addition modulo \(m\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(f\) is a bijection if \(\boldsymbol{A}\) is non-singular.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Suppose now that \(\boldsymbol{A}\) is singular.</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) Write down the relationship between \(a\) , \(b\) , \(c\) , \(d\) .</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Deduce that the second row of \(\boldsymbol{A}\) is a multiple of the first row of \(\boldsymbol{A}\) .</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (iii) Hence show that \(f\) is not a bijection.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(\left\{ {S,{ + _m}} \right\}\) is cyclic for all <strong><em>m</em></strong> .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Given that \(m\) is prime,</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) explain why all elements except the identity are generators of \(\left\{ {S,{ + _m}} \right\}\) ;</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (ii) find the inverse of \(x\) , where <strong><em>x</em></strong> is any element of \(\left\{ {S,{ + _m}} \right\}\) apart from the </span><span style="font-family: times new roman,times; font-size: medium;">identity;</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (iii) determine the number of sets of two distinct elements where each element </span><span style="font-family: times new roman,times; font-size: medium;">is the inverse of the other.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">B.b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Suppose now that \(m = ab\) where \(a\) , \(b\) are unequal prime numbers. Show that </span><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ {S,{ + _m}} \right\}\) has two proper subgroups and identify them.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">B.c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">recognizing that the function needs to be injective and surjective <em><strong>R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>R1</strong></em> if this is seen anywhere in the solution.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">injective:</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">let \(\boldsymbol{U}, \boldsymbol{V} \in ^\circ \times ^\circ \) be 2-D column vectors such that \(\boldsymbol{AU} = \boldsymbol{AV}\) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\boldsymbol{A}^{ - 1}}\boldsymbol{AU} = {\boldsymbol{A}^{ - 1}}\boldsymbol{AV}\) <strong><em>M1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\boldsymbol{U} = \boldsymbol{V}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">this shows that \(f\) is injective</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">surjective:</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">let \(W \in ^\circ \times ^\circ \) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">then there exists \(\boldsymbol{Z} = {\boldsymbol{A}^{ - 1}}\boldsymbol{W} \in ^\circ \times ^\circ \) such that \(\boldsymbol{AZ} = \boldsymbol{W}\) <strong><em>M1A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">this shows that \(f\) is surjective</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">therefore \(f\) is a bijection <em><strong>AG</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[7 marks]</span></strong></em></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) the relationship is \(ad = bc\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) it follows that \(\frac{c}{a} = \frac{d}{b} = \lambda \) </span><span style="font-family: times new roman,times; font-size: medium;">so that \((c,d) = \lambda (a,b)\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(iii) <strong>EITHER</strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">let </span><span style="font-family: times new roman,times; font-size: medium;">\(\boldsymbol{W} = \left[ \begin{array}{l}<br>p\\<br>q<br>\end{array} \right]\) be a 2-D vector</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">then </span><span style="font-family: times new roman,times; font-size: medium;">\(\boldsymbol{AW} = \left[ \begin{array}{l}<br>a\\<br>\lambda a<br>\end{array} \right.\left. \begin{array}{l}<br>b\\<br>\lambda b<br>\end{array} \right]\left[ \begin{array}{l}<br>p\\<br>q<br>\end{array} \right]\) <strong><em>M1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = \left[ \begin{array}{l}<br>ap + bq\\<br>\lambda (ap + bq)<br>\end{array} \right]\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the image always satisfies \(y = \lambda x\) so \(f\) is not surjective and therefore </span><span style="font-family: times new roman,times; font-size: medium;">not a bijection <strong><em>R1</em></strong></span></p>
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">OR</span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">consider</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left[ {\begin{array}{*{20}{c}}<br> a&b \\ <br> {\lambda a}&{\lambda b} <br>\end{array}} \right]\left[ {\begin{array}{*{20}{c}}<br> b \\ <br> 0 <br>\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}<br> {ab} \\ <br> {\lambda ab} <br>\end{array}} \right]\)<br></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left[ {\begin{array}{*{20}{c}}<br> a&b \\ <br> {\lambda a}&{\lambda b} <br>\end{array}} \right]\left[ {\begin{array}{*{20}{c}}<br> 0 \\ <br> a <br>\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}<br> {ab} \\ <br> {\lambda ab} <br>\end{array}} \right]\)<br></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">this shows that \(f\) is not injective and therefore not a bijection <strong><em>R1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[5 marks]</span></em></strong></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the identity element is \(0\) <strong><em> R1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">consider, for \(1 \le r \le m\) ,</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">using \(1\) as a generator <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(1\) combined with itself \(r\) times gives \(r\) and as \(r\) increases from \(1\) to m, the </span><span style="font-family: times new roman,times; font-size: medium;">group is generated ending with \(0\) when \(r = m\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">it is therefore cyclic <strong><em>AG</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></em></strong></p>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) by Lagrange the order of each element must be a factor of \(m\) and if \(m\) </span><span style="font-family: times new roman,times; font-size: medium;">is prime, its only factors are \(1\) and \(m\) <strong><em>R1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">since 0 is the only element of order \(1\), all other elements are of order \(m\) </span><span style="font-family: times new roman,times; font-size: medium;">and are therefore generators <strong><em>R1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) since \(x{ + _m}(m - x) = 0\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">(M1)</span></em></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the inverse of <strong><em>x</em></strong> is \((m - x)\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) consider</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/america.png" alt> </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">there are \(\frac{1}{2}(m - 1)\) </span><span style="font-family: times new roman,times; font-size: medium;">inverse pairs <strong><em>A1 N1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <strong><em>M1</em></strong> for an attempt to list the inverse pairs, <strong><em>A1</em></strong> for completing it </span><span style="font-family: times new roman,times; font-size: medium;">correctly and <strong><em>A1</em></strong> for the final answer.</span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[7 marks]</span></strong></em></p>
<div class="question_part_label">B.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">since \(a\), \(b\) are unequal primes the only factors of \(m\) are \(a\) and \(b\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">there are therefore only subgroups of order \(a\) and \(b\) <em><strong> R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">they are</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ {0,a,2a, \ldots ,(b - 1)a} \right\}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ {0,b,2b, \ldots ,(a - 1)b} \right\}\) <strong><em>A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></em></strong></p>
<div class="question_part_label">B.c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This proved to be a difficult question for some candidates. Most candidates realised that they had to show that the function was both injective and surjective but many failed to give convincing proofs. Some candidates stated, incorrectly, that <strong><em>f</em></strong> was injective because \(\boldsymbol{AX}\) is uniquely defined, not realising that they had to show that \(\boldsymbol{AX} = \boldsymbol{AY} \Rightarrow \boldsymbol{X} = \boldsymbol{Y}\) . </span></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Solutions to (b) were disappointing with many candidates failing to realise that they had either to show that \(\boldsymbol{AX}\) was confined to a subset of \(\mathbb{R} \times \mathbb{R}\)</span><span style="font-family: times new roman,times; font-size: medium;"> or that two distinct vectors had the same image under \(f\). </span></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was well answered in general with solutions to (c) being the least successful. </span></p>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was well answered in general with solutions to (c) being the least successful. </span></p>
<div class="question_part_label">B.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was well answered in general with solutions to (c) being the least successful. </span></p>
<div class="question_part_label">B.c.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The binary operator \( * \) is defined for <strong><em>a</em></strong> , \(b \in \mathbb{R}\) by \(a * b = a + b - ab\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Show that \( * \) is associative.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Find the identity element.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(iii) Find the inverse of \(a \in \mathbb{R}\) , showing that the inverse exists for all values </span><span style="font-family: times new roman,times; font-size: medium;">of \(a\) except one value which should be identified.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iv) Solve the equation \(x * x = 1\) .</span></p>
<div class="marks">[15]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The domain of \( * \) is now reduced to \(S = \left\{ {0,2,3,4,5,\left. 6 \right\}} \right.\) and the arithmetic is </span><span style="font-family: times new roman,times; font-size: medium;">carried out modulo \(7\).</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (i) Copy and complete the following Cayley table for \(\left\{ {S,\left. * \right\}} \right.\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img style="display: block; margin-left: auto; margin-right: auto;" src="images/disco.png" alt></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Show that \(\left\{ {S,\left. * \right\}} \right.\) is a group.</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (iii) Determine the order of each element in <strong><em>S</em></strong> and state, with a reason, </span><span style="font-family: times new roman,times; font-size: medium;">whether or not \(\left\{ {S,\left. * \right\}} \right.\) is cyclic.</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (iv) Determine all the proper subgroups of \(\left\{ {S,\left. * \right\}} \right.\) and explain how your results </span><span style="font-family: times new roman,times; font-size: medium;">illustrate Lagrange’s theorem.</span></p>
<p style="margin-left: 30px;"><span style="font-size: medium;"><span style="font-family: times new roman,times;"> (v) Solve the equation</span> <span style="font-family: times new roman,times;">\(2 * x * x = 5\) .</span></span></p>
<div class="marks">[17]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) \(a * (b * c) = a * (b + c - bc)\) <strong><em> M1</em></strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = a + b + c - bc - a(b + c - bc)\) <strong><em> A1</em> </strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = a + b + c - bc - ca - ab + abc\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\((a * b) * c = (a + b - ab) * c\) <strong><em>M1</em> </strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = a + b - ab + c - (a + b - ab)c\) <em><strong>A1</strong> </em></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = a + b + c - bc - ca - ab + abc\) , hence associative <em><strong>AG</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) let \(e\) be the identity element, so that \(a * e = a\) <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">then, </span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\(a + e - ae = a\) <em><strong>A1 </strong></em></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\(e(1 - a) = 0\)</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\(e = 0\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) let \({a^{ - 1}}\) be the inverse of \(a\), so that \(a * {a^{ - 1}} = 0\) <strong><em>(M1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">then, </span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\(a + {a^{ - 1}} - a{a^{ - 1}} = 0\) <strong><em>A1</em> </strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\({a^{ - 1}} = \frac{a}{{a - 1}}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">this gives an inverse for all elements except 1 which has no inverse <em><strong>R1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iv) \(2x - {x^2} = 1\) <em><strong>M1</strong></em> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({(x - 1)^2} = 0\) <strong><em>(A1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(x = 1\) <strong><em>A1 </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong><em> </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[15 marks] </span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><img src="images/mcds.png" alt></span><strong> <span style="font-family: times new roman,times; font-size: medium;"><em>A3</em> </span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <strong><em>A3</em></strong> for correct table, <strong><em>A2</em></strong> for one error, <strong><em>A1</em></strong> for two errors and <strong><em>A0</em></strong> for more than two errors. </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) there are no new elements in the table so it is closed <strong><em>A1 </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">there is an identity element, \(0\) <strong><em>A1 </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">every row (column) has a \(0\) so every element has an inverse <strong><em>A1 </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">associativity has been proved earlier <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">therefore \(\left\{ {S,\left. * \right\}} \right.\) is a group <strong><em>AG</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><img src="images/mid.png" alt></span><em><strong> <span style="font-family: times new roman,times; font-size: medium;">A3</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <strong><em>A3</em></strong> for correct table, <strong><em>A2</em></strong> for one error, <strong><em>A1</em></strong> for two errors and <strong><em>A0</em></strong> for more than two errors. </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">it is cyclic because there are elements of order \(6\) <strong><em> R1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iv) the proper subgroups are \(\left\{ {0,\left. 2 \right\}} \right.\) , \(\left\{ {0,\left. {4,6} \right\}} \right.\) <strong><em>A1A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">the orders of the subgroups (\(2\), \(3\)) are factors of the order of the group (6) <em><strong> A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(v) recognizing \(x * x = 4\) <strong><em>(M1)</em> </strong></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(x = 3\) , \(6\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1A1 </span></em></strong></p>
<p> </p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[17 marks]</span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was well answered by many candidates. The most common error in (a) was confusing associativity with commutativity. </span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Many wholly correct or almost wholly correct answers to part (b) were seen. Those who did make errors in part (b) were usually unable to fully justify the properties of a group, could not explain why the group was cyclic or could not relate subgroups to Lagrange’s theorem. Some candidates made errors in calculating the orders of the elements. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The set \(S\) consists of real numbers <strong><em>r</em></strong> of the form \(r = a + b\sqrt 2 \) , where \(a,b \in \mathbb{Z}\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The relation \(R\) is defined on \(S\) by \({r_1}R{r_2}\) if and only if \({a_1} \equiv {a_2}\) (mod2) and </span><span style="font-family: times new roman,times; font-size: medium;">\({b_1} \equiv {b_2}\) (mod3), where \({r_1} = {a_1} + {b_1}\sqrt 2 \) and \({r_2} = {a_2} + {b_2}\sqrt 2 \) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(R\) is an equivalence relation.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show, by giving a counter-example, that the statement \({r_1}R{r_2} \Rightarrow r_1^2Rr_2^2\) is false.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Determine</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) the equivalence class \(E\) containing \(1 + \sqrt 2 \) ;</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) the equivalence class \(F\) containing \(1 - \sqrt 2 \) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Show that</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \({(1 + \sqrt 2 )^3} \in F\)</span><span style="font-family: times new roman,times; font-size: medium;"> ;</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) \({(1 + \sqrt 2 )^6} \in E\)</span><span style="font-family: times new roman,times; font-size: medium;"> .</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Determine whether the set \(E\) forms a group under</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) the operation of addition;</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) the operation of multiplication.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">reflexive: if \({r_{}} = {a_{}} + {b_{}}\sqrt 2 \in S\) then \(a \equiv a(\bmod 2)\) and \(b \equiv b(\bmod 3)\) </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(( \Rightarrow rRr)\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">symmetric: if \({r_1}R{r_2}\) then \({a_1} \equiv {a_2}(\bmod 2)\) and \({b_1} \equiv {b_2}(\bmod 3)\) , and <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({a_2} \equiv {a_1}(\bmod 2)\) and \({b_2} \equiv {b_1}(\bmod 3)\) , (so that \({r_2}R{r_1}\) ) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">transitive: if \({r_1}R{r_2}\) and \({r_2}R{r_3}\) then </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(2|{a_1} - {a_2}\) and \(2|{a_2} - {a_3}\) <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow 2|{a_1} - {a_2} + {a_2} - {a_3} \Rightarrow 2|{a_1} - {a_3}\) <strong><em>M1A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(3|{b_1} - {b_2}\) and \(3|{b_2} - {b_3}\)<br></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow 3|{b_1} - {b_2} + {b_2} - {b_3} \Rightarrow 3|{b_1} - {b_3}( \Rightarrow {r_1}R{r_3})\) <strong><em>A1AG </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[7 marks]</span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">consider, for example, \({r_1} = 1 + \sqrt 2 \) , \({r_2} = 3 + \sqrt 2 \) \(({r_1}R{r_2})\) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Only award <strong><em>M1</em></strong> if the two numbers are related and neither \(a\) nor \(b = 0\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(r_1^2 = 3 + 2\sqrt 2 \) , \(r_2^2 = 11 + 6\sqrt 2 \) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the squares are not equivalent because \(2 \ne 6(\bmod 3)\) <em><strong>A1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \(E = \left\{ {2k + 1 + (3m + 1)\sqrt 2 } :k,m \in \mathbb{Z} \right\}\) <em><strong>A1A1</strong> </em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) \(F = \left\{ {2k + 1 + (3m - 1)\sqrt 2 } :k,m \in \mathbb{Z} \right\}\) <strong><em>A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \({(1 + \sqrt 2 )^3} = 7 + 5\sqrt 2 \) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = 2 \times 3 + 1 + (3 \times 2 - 1)\sqrt 2 \in F\) <strong><em>R1AG</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) \({(1 + \sqrt 2 )^6} = 99 + 70\sqrt 2 \) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = 2 \times 49 + 1 + (3 \times 23 + 1)\sqrt 2 \in E\) <strong><em>R1AG</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[4 marks]</span></em></strong></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \(E\) is not a group under addition <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">any valid reason <em>eg</em> \(0 \notin E\) <strong><em>R1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) \(E\) is not a group under multiplication <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">any valid reason <em>eg</em> \(1 \notin E\) <strong><em>R1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[4 marks]</span></em></strong></p>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">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.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The majority of candidates earned significant marks on this question.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The majority of candidates earned significant marks on this question.</span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The majority of candidates earned significant marks on this question.</span></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The majority of candidates earned significant marks on this question.</span></p>
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="specification">
<p>The set \({S_n} = \{ 1,{\text{ }}2,{\text{ }}3,{\text{ }} \ldots ,{\text{ }}n - 2,{\text{ }}n - 1\} \), where \(n\) is a prime number greater than 2, and \({ \times _n}\) denotes multiplication modulo \(n\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that there are no elements \(a,{\text{ }}b \in {S_n}\) such that \(a{ \times _n}b = 0\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that, for \(a,{\text{ }}b,{\text{ }}c \in {S_n},{\text{ }}a{ \times _n}b = a{ \times _n}c \Rightarrow b = c\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \({G_n} = \{ {S_n},{\text{ }}{ \times _n}\} \) is a group. You may assume that \({ \times _n}\) is associative.</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that the order of the element \((n - 1)\) is 2.</p>
<div class="marks">[1]</div>
<div class="question_part_label">c.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that the inverse of the element 2 is \(\frac{1}{2}(n + 1)\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Explain why the inverse of the element 3 is \(\frac{1}{3}(n + 1)\) for some values of \(n\) but not for other values of \(n\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.iii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine the inverse of the element 3 in \({G_{11}}\).</p>
<div class="marks">[1]</div>
<div class="question_part_label">c.iv.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine the inverse of the element 3 in \({G_{31}}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.v.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(a{ \times _n}b = 0 \Rightarrow ab = \) a multiple of \(n\) (or vice versa) <strong><em>R1</em></strong></p>
<p>since \(n\) is prime, this can only occur if \(a = 1\) and \(b = \) multiple of \(n\) which is impossible because the multiple of \(n\) would not belong to \({S_n}\) <strong><em>R1</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(a{ \times _n}b = a{ \times _n}c \Rightarrow a{ \times _n}(b - c) = 0\) <strong><em>M1</em></strong></p>
<p>suppose \(b \ne c\) and let \(b > c\) (without loss of generality)</p>
<p>\((b - c) \in {S_n}\) and from (i), \(a{ \times _n}(b - c) = 0\) is a contradiction <strong><em>R1</em></strong></p>
<p>therefore \(b = c\) <strong><em>AG</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\({G_n}\) is associative because modular multiplication is associative <strong><em>A1</em></strong></p>
<p>\({G_n}\) is closed because the value of \(a{ \times _n}b\) always lies between 1 and \(n - 1\) <strong><em>A1</em></strong></p>
<p>the identity is 1 <strong><em>A1</em></strong></p>
<p>consider \(a{ \times _n}b\) where \(b\) can take \(n - 1\) possible values. Using the result from (a)(ii), this will result in \(n - 1\) different values, one of which will be 1, which will give the inverse of \(a\) <strong><em>R1</em></strong></p>
<p>\({G_n}\) is therefore a group <strong><em>AG</em></strong></p>
<p><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\({(n - 1)^2} = {n^2} - 2n + 1 \equiv 1(\bmod n)\) <strong><em>M1</em></strong></p>
<p>so that \((n - 1){ \times _n}(n - 1) = 1\) and \(n - 1\) has order 2 <strong><em>R1AG</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">c.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>consider \(2 \times \frac{1}{2}(n + 1) = n + 1 = 1(\bmod n)\) <strong><em>A1</em></strong></p>
<p>since \(\frac{1}{2}(n + 1)\) is an integer for al \(n\), it is the inverse of 2 <strong><em>R1AG</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">c.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>consider \(3 \times \frac{1}{3}(n + 1) = n + 1 = 1(\bmod n)\) <strong><em>M1</em></strong></p>
<p>therefore \(\frac{1}{3}(n + 1)\) is the inverse of 3 if it is an integer but not otherwise <strong><em>R1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">c.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the inverse of 3 in \({G_{11}}\) is 4 <strong><em>A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">c.iv.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the inverse of 3 in \({G_{31}}\) is 21 <strong><em>(M1)A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">c.v.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.iv.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.v.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">The set of all permutations of the list of the integers \(1,{\text{ }}2,{\text{ }}3{\text{ }} \ldots {\text{ }}n\) is a group, \({S_n}\), under the operation of composition of permutations.</p>
</div>
<div class="specification">
<p class="p1"><span class="s1">Each element of \({S_4}\) </span>can be represented by a \(4 \times 4\) matrix. For example, the cycle \({\text{(1 2 3 4)}}\) is represented by the matrix</p>
<p class="p1">\(\left( {\begin{array}{*{20}{c}} 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \\ 1&0&0&0 \end{array}} \right)\) acting on the column vector \(\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 3 \\ 4 \end{array}} \right)\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">(i) <span class="Apple-converted-space"> </span></span>Show that the order of \({S_n}\) is \(n!\);</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>List the <span class="s2">6 </span>elements of \({S_3}\) <span class="s1">in cycle form;</span></p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>Show that \({S_3}\) <span class="s1">is not Abelian;</span></p>
<p class="p1">(iv) <span class="Apple-converted-space"> </span>Deduce that \({S_n}\) is not Abelian for \(n \geqslant 3\).</p>
<div class="marks">[9]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Write down the matrices <span class="s1"><strong><em>M</em></strong>\(_1\), <strong><em>M</em></strong>\(_2\)</span> representing the permutations \((1{\text{ }}2),{\text{ }}(2{\text{ }}3)\)<span class="s1">, </span><span class="s3">respectively;</span></p>
<p class="p2"><span class="s4">(ii) <span class="Apple-converted-space"> </span>Find </span><span class="s1"><strong><em>M</em></strong>\(_1\)<strong><em>M</em></strong>\(_2\)</span> and state the permutation represented by this matrix;</p>
<p class="p1"><span class="s3">(</span>iii) <span class="Apple-converted-space"> </span>Find \(\det (\)<span class="s1"><strong><em>M</em></strong></span><span class="s5">\(_1)\)</span><span class="s1">, \(\det (\)<strong><em>M</em></strong></span><span class="s5">\(_2)\) </span>and deduce the value of \(\det (\)<span class="s1"><strong><em>M</em></strong>\(_1\)<strong><em>M</em></strong></span>\(_2)\)<span class="s5">.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">(i) <span class="Apple-converted-space"> </span></span>Use mathematical induction to prove that</p>
<p class="p1"><span class="s1">\((1{\text{ }}n)(1{\text{ }}n{\text{ }} - 1)(1{\text{ }}n - 2) \ldots (1{\text{ }}2) = (1{\text{ }}2{\text{ }}3 \ldots n){\text{ }}n \in {\mathbb{Z}^ + },{\text{ }}n > 1\).</span></p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Deduce that every permutation can be written as a product of cycles of length 2.</p>
<div class="marks">[8]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="s1">(i) <span class="Apple-converted-space"> </span>1 </span>has \(n\) <span class="s1">possible new positions; 2 </span>then has \(n - 1\) possible new positions…</p>
<p class="p1">\(n\) has only one possible new position <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">the number of possible permutations is \(n \times (n - 1) \times \ldots \times 2 \times 1\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( = n!\) </span><strong><em>AG</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Give no credit for simply stating that the number of permutations is \(n!\)</p>
<p class="p2"> </p>
<p class="p3">(ii) <span class="Apple-converted-space"> \((1)(2)(3);{\text{ }}(1{\text{ }}2)(3);{\text{ }}(1{\text{ }}3)(2);{\text{ }}(2{\text{ }}3)(1);{\text{ }}(1{\text{ }}2{\text{ }}3);{\text{ }}(1{\text{ }}3{\text{ }}2)\)</span> <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A2</em></strong></span></p>
<p class="p2"> </p>
<p class="p1"><strong>Notes: <em>A1 </em></strong>for 4 or 5 correct.</p>
<p class="p1">If single bracket terms are missing, do not penalize.</p>
<p class="p1">Accept \(e\) in place of the identity.</p>
<p class="p2"> </p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>attempt to compare \({\pi _1} \circ {\pi _2}\) <span class="s1">with \({\pi _2} \circ {\pi _1}\) </span>for two permutations <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p3">for example \((1{\text{ }}2)(1{\text{ }}3) = (1{\text{ }}3{\text{ }}2)\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p3">but \((1{\text{ }}3)(1{\text{ }}2) = (1{\text{ }}2{\text{ }}3)\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p1">hence \({S_3}\) is not Abelian <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"> </p>
<p class="p1">(iv) <span class="Apple-converted-space"> \({S_3}\)</span> is a subgroup of \({S_n}\), <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">so \({S_n}\) contains non-commuting elements <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">\( \Rightarrow {S_n}\) is not Abelian for \(n \geqslant 3\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"><strong><em>[9 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>M</em></strong>\(_1 = \left( {\begin{array}{*{20}{c}} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{array}} \right)\)</span>, <span class="s1"><strong><em>M</em></strong>\(_2 = \left( {\begin{array}{*{20}{c}} 1&0&0&0 \\ 0&0&1&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{array}} \right)\) <span class="Apple-converted-space"> </span></span><strong><em>A1A1</em></strong></p>
<p class="p2"> </p>
<p class="p2"><span class="s2">(ii) <span class="Apple-converted-space"> </span></span><strong><em>M</em></strong>\(_1\)<strong><em>M</em></strong>\(_2 = \left( {\begin{array}{*{20}{c}} 0&0&1&0 \\ 1&0&0&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{array}} \right)\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p2">this represents \((1{\text{ }}3{\text{ }}2)\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p1"> </p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>by, for example, interchanging a pair of rows <span class="Apple-converted-space"> </span><span class="s3">(</span><strong><em>M1</em></strong><span class="s3">)</span></p>
<p class="p2">\(\det (\)<strong><em>M</em></strong>\(_1) = \det (\)<strong><em>M</em></strong>\(_2) = - 1\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p1">then \(\det (\)<span class="s1"><strong><em>M</em></strong></span>\(_1\)<span class="s1"><strong><em>M</em></strong></span>\(_2) = ( - 1) \times ( - 1) = 1\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[7 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>let \({\text{P}}(n)\) be the proposition that</p>
<p class="p1">\((1{\text{ }}n)(1{\text{ }}n - 1)(1{\text{ }}n - 2) \ldots (1{\text{ }}2) = (1{\text{ }}2{\text{ }}3 \ldots n){\text{ }}n \in {\mathbb{Z}^ + }\)</p>
<p class="p1">the statement that \({\text{P}}(2)\) <span class="s1">is true <em>eg</em> \((1{\text{ }}2) = (1{\text{ }}2)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="s2">assume \({\text{P}}(k)\) </span>is true for some \(k\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2">consider \((1{\text{ }}k + 1)(1{\text{ }}k)(1{\text{ }}k - 1)(1{\text{ }}k - 2) \ldots (1{\text{ }}2)\)</p>
<p class="p2"><span class="Apple-converted-space">\( = (1{\text{ }}k + 1)(1{\text{ }}2{\text{ }}3 \ldots k)\) </span><strong><em>M1</em></strong></p>
<p class="p2">then the composite permutation has the following effect on the first \(k + 1\) <span class="s2">integers: \(1 \to 2,{\text{ }}2 \to 3 \ldots k - 1 \to k,{\text{ }}k \to 1 \to k + 1,{\text{ }}k + 1 \to 1\) <span class="Apple-converted-space"> </span></span><strong><em>A1</em></strong></p>
<p class="p1">this is \((1{\text{ }}2{\text{ }}3 \ldots k{\text{ }}k + 1)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2">hence the assertion is true by induction <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p2"> </p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>every permutation is a product of cycles <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p2">generalizing the result in (i) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p2">every cycle is a product of cycles of length 2 <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p2">hence every permutation can be written as a product of cycles of length 2 <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p2"><strong><em>[8 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">In part (a)(i), many just wrote down \(n!\) without showing how this arises by a sequential choice process. Part (ii) was usually correctly answered, although some gave their answers in the unwanted 2-dimensional form. Part (iii) was often well answered, though some candidates failed to realise that they need to explicitly evaluate the product of two elements in both orders.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Part (b) was often well answered. A number of candidates found \(2 \times 2\) matrices – this gained no marks.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Nearly all candidates knew how to approach part (c)(i), but failed to be completely convincing. Few candidates seemed to know that every permutation can be written as a product of non-overlapping cycles, as the first step in part (ii).</p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Let \(f\) be a homomorphism of a group \(G\) onto a group \(H\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that if \(e\) is the identity in \(G\), then \(f(e)\) is the identity in \(H\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that if \(x\) is an element of \(G\), then \(f({x^{ - 1}}) = {\left( {f(x)} \right)^{ - 1}}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that if \(G\) is Abelian, then \(H\) must also be Abelian.</p>
<div class="marks">[5]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that if \(S\) is a subgroup of \(G\), then \(f(S)\) is a subgroup of \(H\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">\(f(a) = f(ae) = f(a)f(e)\) <span class="Apple-converted-space"> </span><strong><em>M1A1</em></strong></p>
<p class="p1">hence \(f(e)\) is the identity in \(H\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\(e' = f(e)\)</p>
<p class="p1">\( = f(x{x^{ - 1}})\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\( = f(x)f({x^{ - 1}})\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">hence \(f({x^{ - 1}}) = {\left( {f(x)} \right)^{ - 1}}\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">let \(a',{\text{ }}b' \in H\), we need to show that \(a'b' = b'a'\) <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1">since \(f\) is onto \(H\) there exists \(a,{\text{ }}b \in G\) such that \(f(a) = a'\)</p>
<p class="p1">and \(f(b) = b'\) <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1">now \(a'b' = f(a)f(b) = f(ab)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">since \(f(ab) = f(ba)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\(f(ba) = f(b)f(a) = b'a'\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">hence Abelian <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>METHOD 1</strong></p>
<p class="p1">\(e' = f(e)\) and \(f({x^{ - 1}}) = {\left( {f(x)} \right)^{ - 1}}\) from above <span class="Apple-converted-space"> </span><strong><em>A1A1</em></strong></p>
<p class="p1">let \(f(a)\) and \(f(b)\) be two elements in \(f(S)\)</p>
<p class="p1">then \(f(a)f(b) = f(ab)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\( \Rightarrow f(a)f(b) \in f(S)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">hence closed under the operation of \(H\)</p>
<p class="p1">\(f(S)\) is a subgroup of \(H\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"> </p>
<p class="p1"><strong>METHOD 2</strong></p>
<p class="p1">\(f(S)\) contains the identity, so is non empty <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">Suppose \(f(a),{\text{ }}f(b) \in f(S)\)</p>
<p class="p1">Consider \(f(a)f{(b)^{ - 1}}\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\( = f(a)f({b^{ - 1}})\) <span class="Apple-converted-space"> </span>(from (b)) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\( = f(a{b^{ - 1}})\) <span class="Apple-converted-space"> </span>(homomorphism) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\( \in f(S)\) as \(a{b^{ - 1}} \in H\)</p>
<p class="p1">So \(f(S)\) is a subgroup of \(H\) (by a subgroup theorem) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">It was pleasing to see a small number of wholly correct responses on this final question. Although the majority of candidates gained some marks, the majority failed to gain full marks because they failed to show full formal understanding of the situation.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">It was pleasing to see a small number of wholly correct responses on this final question. Although the majority of candidates gained some marks, the majority failed to gain full marks because they failed to show full formal understanding of the situation.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">It was pleasing to see a small number of wholly correct responses on this final question. Although the majority of candidates gained some marks, the majority failed to gain full marks because they failed to show full formal understanding of the situation.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">It was pleasing to see a small number of wholly correct responses on this final question. Although the majority of candidates gained some marks, the majority failed to gain full marks because they failed to show full formal understanding of the situation.</p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p>Consider the special case in which \(G = \{ 1,{\text{ }}3,{\text{ }}4,{\text{ }}9,{\text{ }}10,{\text{ }}12\} ,{\text{ }}H = \{ 1,{\text{ }}12\} \) and \( * \) denotes multiplication modulo 13.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>The group \(\{ G,{\text{ }} * \} \) has a subgroup \(\{ H,{\text{ }} * \} \). The relation \(R\) is defined such that for \(x\), \(y \in G\), \(xRy\) if and only if \({x^{ - 1}} * y \in H\). Show that \(R\) is an equivalence relation.</p>
<div class="marks">[8]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that 3\(R\)10.</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine the three equivalence classes.</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.ii.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>Reflexive: \(xRx\) <strong>(<em>M1) </em></strong></p>
<p>because \({x^{ - 1}}x = {\text{e}} \in H\) <strong><em>R1</em></strong></p>
<p>therefore reflexive <strong><em>AG</em></strong></p>
<p>Symmetric: Let \(xRy\) so that \({x^{ - 1}}y \in H\) <strong><em>M1</em></strong></p>
<p>it follows that \({({x^{ - 1}}y)^{ - 1}} = {y^{ - 1}}x \in H \Rightarrow yRx\) <strong><em>M1A1</em></strong></p>
<p>therefore symmetric <strong><em>AG</em></strong></p>
<p>Transitive: Let \(xRy\) and \(yRz\) so that \({x^{ - 1}}y \in H\) and \({y^{ - 1}}z \in H\) <strong><em>M1</em></strong></p>
<p>it follows that \({x^{ - 1}}y{\text{ }}{y^{ - 1}}z = {x^{ - 1}}z \in H \Rightarrow xRz\) <strong><em>M1A1</em></strong></p>
<p>therefore transitive (therefore \(R\) is an equivalence relation on the set \(G\)) <strong><em>AG</em></strong></p>
<p><strong><em>[8 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>attempt at inverse of 3: since \(3 \times 9 = 27 = 1(\bmod 13)\) <strong><em>(M1)</em></strong></p>
<p>it follows that \({3^{ - 1}} = 9\) <strong><em>A1</em></strong></p>
<p>since \(9 \times 10 = 90 = 12(\bmod 13) \in H\) <strong><em>M1A1</em></strong></p>
<p>it follows that 3\(R\)10 <strong><em>AG</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the three equivalence classes are \(\{ 3,{\text{ }}10\} ,{\text{ }}\{ 1,{\text{ }}12\} \) and \(\{ 4,{\text{ }}9\} \) <strong><em>A1A1A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.ii.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">\(S\) is defined as the set of all \(2 \times 2\) <span class="s1">non-singular matrices. </span>\(A\) <span class="s1">and </span>\(B\) <span class="s1">are two elements of the set </span>\(S\)<span class="s1">.</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Show that \({({A^T})^{ - 1}} = {({A^{ - 1}})^T}\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Show that \({(AB)^T} = {B^T}{A^T}\).</p>
<div class="marks">[8]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">A relation \(R\) is defined on \(S\) such that \(A\) is related to \(B\) if and only if there exists an element \(X\) of \(S\) such that \(XA{X^T} = B\). Show that \(R\) is an equivalence relation.</p>
<div class="marks">[8]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>\(A = \left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right)\)</p>
<p class="p1">\({A^T} = \left( {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\({({A^T})^{ - 1}} = \frac{1}{{ad - bc}}\left( {\begin{array}{*{20}{c}} d&{ - c} \\ { - b}&a \end{array}} \right)\;\;\;\)(which exists because \(ad - bc \ne 0\)) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\({A^{ - 1}} = \frac{1}{{ad - bc}}\left( {\begin{array}{*{20}{c}} d&{ - b} \\ { - c}&a \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\({({A^{ - 1}})^T} = \frac{1}{{ad - bc}}\left( {\begin{array}{*{20}{c}} d&{ - c} \\ { - b}&a \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">hence \({({A^T})^{ - 1}} = {({A^{ - 1}})^T}\) as required <strong><em>AG</em></strong></p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>\(A = \left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right)\;\;\;B = \left( {\begin{array}{*{20}{c}} e&f \\ g&h \end{array}} \right)\)</p>
<p class="p1">\(AB = \left( {\begin{array}{*{20}{c}} {ae + bg}&{af + bh} \\ {ce + dg}&{cf + dh} \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\({(AB)^T} = \left( {\begin{array}{*{20}{c}} {ae + bg}&{ce + dg} \\ {af + bh}&{cf + dh} \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\({B^T} = \left( {\begin{array}{*{20}{c}} e&g \\ f&h \end{array}} \right)\;\;\;{A^T} = \left( {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\({B^T}{A^T} = \left( {\begin{array}{*{20}{c}} e&g \\ f&h \end{array}} \right)\left( {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {ae + bg}&{ce + dg} \\ {af + bh}&{cf + dh} \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">hence \({(AB)^T} = {B^T}{A^T}\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\(R\) is reflexive since \(I \in S\) and \(IA{I^T} = A\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\(XA{X^T} = B \Rightarrow A = {X^{ - 1}}B{({X^T})^{ - 1}}\) <span class="Apple-converted-space"> </span><strong><em>M1A1</em></strong></p>
<p class="p1">\( \Rightarrow A = {X^{ - 1}}B{({X^{ - 1}})^T}\) from a (i) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">which is of the correct form, hence symmetric <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1">\(ARB \Rightarrow XA{X^T} = B\) and \(BRC = YB{Y^T} = C\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><strong>Note:<span class="Apple-converted-space"> </span></strong>Allow use of \(X\) rather than \(Y\) in this line.</p>
<p class="p2"> </p>
<p class="p1">\( \Rightarrow YXA{X^T}{Y^T} = YB{Y^T} = C\) <span class="Apple-converted-space"> </span><strong><em>M1A1</em></strong></p>
<p class="p1">\( \Rightarrow (YX)A{(YX)^T} = C\) from a (ii) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">this is of the correct form, hence transitive</p>
<p class="p1">hence \(R\) is an equivalence relation <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">Part a) was successfully answered by the majority of candidates..</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">There were some wholly correct answers seen to part b) but a number of candidates struggled with the need to formally explain what was required.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">A group has exactly three elements, the identity element \(e\) , \(h\) and \(k\) . Given the </span><span style="font-family: times new roman,times; font-size: medium;">operation is denoted by \( \otimes \) , show that</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Show that \({\mathbb{Z}_4}\) (the set of integers modulo 4) together with the operation </span><span style="font-family: times new roman,times; font-size: medium;">\({ + _4}\) (addition modulo 4) form a group \(G\) . You may assume associativity.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Show that \(G\) is cyclic.</span></p>
<div class="marks">[9]</div>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Using Cayley tables or otherwise, show that \(G\) and \(H = \left( {\left\{ {1,2,3,\left. 4 \right\},{ \times _5}} \right.} \right)\) are </span><span style="font-family: times new roman,times; font-size: medium;">isomorphic where \({{ \times _5}}\) is multiplication modulo 5. State clearly all the possible </span><span style="font-family: times new roman,times; font-size: medium;">bijections.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">the group is cyclic.<br></span></p>
<div class="marks">[3]</div>
<div class="question_part_label">B.b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">the group is cyclic.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i)<br></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><img src="images/judy.png" alt></span><em><strong><span style="font-family: times new roman,times; font-size: medium;"> A2</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <strong><em>A1</em></strong> for table if exactly one error and <strong><em>A0</em></strong> if more than one error.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">all elements belong to \({\mathbb{Z}_4}\) so it is closed <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(0\) is the identity element <strong><em> A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(2\) is self inverse <strong><em>A1 </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(1\) and \(3\) are an inverse pair <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">hence every element has an inverse </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">hence \(\left\{ {{\mathbb{Z}_4},{ + _4}} \right\}\) form a group \(G\) <strong><em>AG</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(1{ + _4}1 \equiv 2(\bmod 4)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(1{ + _4}1{ + _4}1 \equiv 3(\bmod 4)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(1{ + _4}1{ + _4}1{ + _4}1 \equiv 0(\bmod 4)\) <strong><em>M1A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">hence \(1\) is a generator <em><strong>R1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">therefore \(G\) is cyclic <strong><em>AG</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(\(3\) is also a generator) </span></p>
<p> </p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;"><em>[9 marks]</em> </span></strong></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;"><img src="images/two.png" alt> </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong> A1A1</strong> </span></em></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;"><strong>EITHER</strong> </span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">for the group \(\left( {\left\{ {1,2,3,\left. 4 \right\},{ \times _5}} \right.} \right)\)</span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(1\) is the identity and \(4\) is self inverse <strong><em>A1 </em></strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(2\) and \(3\) are an inverse pair <strong><em>A1 </em></strong></span></p>
<p align="JUSTIFY"><strong><span style="font-family: times new roman,times; font-size: medium;">OR </span></strong></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">for \(G\), for \(H\), </span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">0 has order 1 1 has order 1 </span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">1 has order 4 2 has order 4 </span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">2 has order 2 3 has order 4 </span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">3 has order 4 4 has order 2 <em><strong>A1A1 </strong></em></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;"><strong>THEN</strong> </span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">hence there is a bijection <strong><em> R1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(h(1) \to 0\) , \(h(2) \to 1\) , \(h(3) \to 3\) , \(h(4) \to 2\) <strong><em>A1</em></strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">the groups are isomorphic <strong><em>AG</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(k(1) \to 0\) , \(k(2) \to 3\) , \(k(3) \to 1\) , \(k(4) \to 2\) <strong><em>A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">is also a bijection </span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[7 marks] </span></em></strong></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">if cyclic then the group is {\(e\), \(h\), \({h^2}\)} <em><strong>R1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({h^2} = e\) or \(h\) or \(k\) <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({h^2} = e \Rightarrow h \otimes h = h \otimes k\)<br></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow h = k\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">but \(h \ne k\) so \({h^2} \ne e\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({h^2} = h \Rightarrow h \otimes h = h \otimes e \Rightarrow h = e\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">but \(h \ne e\) so \({h^2} \ne h\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">so \({h^2} = k\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">also \({h^3} = h \otimes k = e\)<br></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">hence the group is cyclic <em><strong>AG</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note:</strong> An alternative proof is possible based on order of elements and Lagrange.</span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[5 marks]</span></strong></em></p>
<div class="question_part_label">B.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">if cyclic then the group is \(\left\{ {e,h,\left. {{h^2}} \right\}} \right.\) <strong><em>R1 </em></strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\({h^2} = e\) or \(h\) or \(k\) <strong><em>M1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\({h^2} = e \Rightarrow h \otimes h = h \otimes k\)</span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow h = k\)</span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">but \(h \ne k\) so \({h^2} \ne e\) <strong><em>A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\({h^2} = h \Rightarrow h \otimes h = h \otimes e \Rightarrow h = e\)</span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">but \(h \ne e\) so \({h^2} \ne h\)</span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">so \({h^2} = k\) <em><strong> A1</strong> </em></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">also \({h^3} = h \otimes k = e\) <strong><em>A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">hence the group is cyclic <strong><em>AG </em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: An alternative proof is possible based on order of elements and Lagrange. </span></p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;"><em>[5 marks]</em> </span></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Most candidates drew a table for this part and generally achieved success in both (i) and (ii). </span></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">In (b) most did use Cayley tables and managed to match element order but could not clearly state the two possible bijections. Sometimes showing that the two groups were isomorphic was missed. </span></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Part B was not well done and the properties of a three element group were often quoted without any proof.<br></span></p>
<div class="question_part_label">B.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Part B was not well done and the properties of a three element group were often quoted without any proof. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The relation \({R_1}\) is defined for \(a,b \in {\mathbb{Z}^ + }\) by \(a{R_1}b\) if and only if \(n\left| {({a^2} - {b^2})} \right.\) where \(n\) </span><span style="font-family: times new roman,times; font-size: medium;">is a fixed positive integer.</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) Show that \({R_1}\) is an equivalence relation.</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Determine the equivalence classes when \(n = 8\) .</span></p>
<div class="marks">[11]</div>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Consider the group \(\left\{ {G, * } \right\}\) and let \(H\) be a subset of \(G\) defined by</span></p>
<p style="text-align: center;"><span style="font-family: times new roman,times; font-size: medium;">\(H = \left\{ {x \in G} \right.\) such that \(x * a = a * x\) for all \(a \in \left. G \right\}\) .<br></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(\left\{ {H, * } \right\}\) is a subgroup of \(\left\{ {G, * } \right\}\) .</span></p>
<div class="marks">[12]</div>
<div class="question_part_label">B.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The relation \({R_2}\) is defined for \(a,b \in {\mathbb{Z}^ + }\) by \(a{R_2}b\) if and only if \((4 + \left| {a - b} \right|)\) is the </span><span style="font-family: times new roman,times; font-size: medium;">square of a positive integer. Show that \({R_2}\) is not transitive.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">B.b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Since \({a^2} - {a^2} = 0\) is divisible by <strong><em>n</em></strong>, it follows that \(a{R_1}a\) so \({R_1}\) is reflexive. <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(a{R_1}b \Rightarrow {a^2} - {b^2}\) </span><span style="font-family: times new roman,times; font-size: medium;">divisible by \(n \Rightarrow {b^2} - {a^2}\) divisible by \(n \Rightarrow b{R_1}a\) so</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">symmetric. <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(a{R_1}b\) and \(b{R_1}c \Rightarrow {a^2} - {b^2} = pn\)</span><span style="font-family: times new roman,times; font-size: medium;"> and \({b^2} - {c^2} = qn\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(({a^2} - {b^2}) + ({b^2} - {c^2}) = pn + qn\) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">so \({a^2} - {c^2} = (p + q)n \Rightarrow a{R_1}c\) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Therefore \({R_1}\) is transitive.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">It follows that \({R_1}\) is an equivalence relation. <strong><em> AG</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) When \(n = 8\) , the equivalence classes are</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ {1,3,5,7,9, \ldots } \right\}\) , i.e. the odd integers <strong><em>A2</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ {2,6,10,14, \ldots } \right\}\) <em><strong>A2</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">and \(\left\{ {4,8,12,16, \ldots } \right\}\) <strong><em>A2</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: If finite sets are shown award <em><strong>A1A1A1</strong></em>.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[11 marks]</span></strong></em></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Associativity follows since <strong><em>G</em></strong> is associative. <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Closure: Let \(x,y \in H\) so \(ax = xa\) , \(ay = ya\) for \(a \in G\) <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Consider \(axy = xay = xya \Rightarrow xy \in H\) <strong><em>M1A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The identity \(e \in H\) since \(ae = ea\) for \(a \in G\) <em><strong>A2</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Inverse: Let \(x \in H\) so \(ax = xa\) for \(a \in G\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Then</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({x^{ - 1}}a = {x^{ - 1}}ax{x^{ - 1}}\) <em><strong>M1A1</strong></em></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = {x^{ - 1}}xa{x^{ - 1}}\) <strong><em>M1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = a{x^{ - 1}}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">so \( \Rightarrow {x^{ - 1}} \in H\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The four group axioms are satisfied so \(H\) is a subgroup. <strong><em>R1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[12 marks]</span></em></strong></p>
<div class="question_part_label">B.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Attempt to find a counter example. <strong><em> (M1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">We note that \(1{R_2}6\) and \(6{R_2}11\) but 1 not \({R_2}11\) . <strong><em>A2</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Accept any valid counter example.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The relation is not transitive. <strong><em>AG</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></em></strong></p>
<div class="question_part_label">B.b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">B.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">B.b.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The relation \(R\) is defined on \({\mathbb{R}^ + } \times {\mathbb{R}^ + }\) such that \(({x_1},{y_1})R({x_2},{y_2})\) if and only if \(\frac{{{x_1}}}{{{x_2}}} = \frac{{{y_2}}}{{{y_1}}}\) </span><span style="font-family: times new roman,times; font-size: medium;">.</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(R\) is an equivalence relation.</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Determine the equivalence class containing \(({x_1},{y_1})\) and interpret it geometrically.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{x_1}}}{{{x_1}}} = \frac{{{y_1}}}{{{y_1}}} \Rightarrow ({x_1},{y_1})R({x_1},{y_1})\) so \(R\) is reflexive <strong><em>R1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(({x_1},{y_1})R({x_2},{y_2}) \Rightarrow \frac{{{x_1}}}{{{x_2}}} = \frac{{{y_2}}}{{{y_1}}} \Rightarrow \frac{{{x_2}}}{{{x_1}}} = \frac{{{y_1}}}{{{y_2}}} \Rightarrow ({x_2},{y_2})R({x_1},{y_1})\) <strong><em>M1A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">so \(R\) is symmetric</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(({x_1},{y_1})R({x_2},{y_2})\) and \(({x_2},{y_2})R({x_3},{y_3}) \Rightarrow \frac{{{x_1}}}{{{x_2}}} = \frac{{{y_2}}}{{{y_1}}}\) and \(\frac{{{x_2}}}{{{x_3}}} = \frac{{{y_3}}}{{{y_2}}}\) <strong><em>M1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">multiplying the two equations, <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow \frac{{{x_1}}}{{{x_3}}} = \frac{{{y_3}}}{{{y_1}}} \Rightarrow ({x_1},{y_1})R({x_3},{y_3})\) so \(R\) is transitive <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">thus \(R\) is an equivalence relation <em><strong>AG</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [6 marks]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\((x,y)R({x_1},{y_1}) \Rightarrow \frac{{{x_{}}}}{{{x_1}}} = \frac{{{y_1}}}{{{y_{}}}} \Rightarrow xy = {x_1}{y_1}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">(M1)</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the equivalence class is therefore \(\left\{ {(x,y)|xy = {x_1}{y_1}} \right\}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">geometrically, the equivalence class is (one branch of) a (rectangular) hyperbola <strong><em>A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The set \(S\) contains the eighth roots of unity given by \(\left\{ {{\text{cis}}\left( {\frac{{n\pi }}{4}} \right),{\text{ }}n \in \mathbb{N},{\text{ }}0 \leqslant n \leqslant 7} \right\}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Show that \(\{ S,{\text{ }} \times \} \) is a group where \( \times \) denotes multiplication of complex numbers.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Giving a reason, state whether or not \(\{ S,{\text{ }} \times \} \) is cyclic.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) closure: let \({a_1} = {\text{cis}}\left( {\frac{{{n_1}\pi }}{4}} \right)\) and \({a_2} = {\text{cis}}\left( {\frac{{{n_2}\pi }}{4}} \right) \in S\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">then \({a_1} \times {a_2} = {\text{cis}}\left( {\frac{{({n_1} + {n_2})\pi }}{4}} \right)\) (which \( \in S\) because the addition is carried out modulo 8) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">identity: the identity is 1 (and corresponds to \(n = 0\)) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">inverse: the inverse of \({\text{cis}}\left( {\frac{{n\pi }}{4}} \right)\) is \({\text{cis}}\left( {\frac{{(8 - n)\pi }}{4}} \right) \in S\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">associatively: multiplication of complex numbers is associative <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">the four group axioms are satisfied so \(S\) is a group <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \(S\) is cyclic <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">because \({\text{cis}}\left( {\frac{\pi }{4}} \right)\), for example, is a generator <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The binary operation multiplication modulo \(9\), denoted by \({ \times _9}\) , is defined on the set </span><span style="font-family: times new roman,times; font-size: medium;">\(S = \left\{ {1,2,3,4,5,6,7,8} \right\}\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Copy and complete the following Cayley table.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/cal.png" alt></span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(\left\{ {S,{ \times _9}} \right\}\) is not a group.</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Prove that a group \(\left\{ {G,{ \times _9}} \right\}\) can be formed by removing two elements from the </span><span style="font-family: times new roman,times; font-size: medium;">set \(S\) .</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Find the order of all the elements of \(G\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Write down all the proper subgroups of \(\left\{ {G,{ \times _9}} \right\}\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(iii) Determine the coset containing the element \(5\) for each of the subgroups in </span><span style="font-family: times new roman,times; font-size: medium;">part (ii).</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Solve the equation \(4{ \times _9}x{ \times _9}x = 1\) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><em><strong><span style="font-family: times new roman,times; font-size: medium;"><img 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" alt> A3</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A2</strong></em> if one error, <em><strong>A1</strong></em> if two errors and <em><strong>A0</strong></em> if three or more errors.</span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">any valid reason, <em><strong>R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><em>e.g.</em> not closed</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(3\) or \(6\) has no inverse,</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">it is not a Latin square</span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[1 mark]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">remove \(3\) and \(6\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">for the remaining elements,</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the table is closed <em><strong>R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">associative because multiplication is associative <strong><em>R1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the identity is \(1\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">every element has an inverse, (\(2\), \(5\)) and (\(4\), \(7\)) are </span><span style="font-family: times new roman,times; font-size: medium;">inverse pairs and \(8\) (and \(1\)) are self-inverse <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">thus it is a group <strong><em>AG</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[5 marks]</span></em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) the orders are</span></p>
<p align="LEFT"> <em><strong><span style="font-family: times new roman,times; font-size: medium;">A3</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A2</strong></em> if one error, <em><strong>A1</strong></em> if two errors and <em><strong>A0</strong></em> if three or more errors.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) the proper subgroups are</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ {1,8} \right\}\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ {1,4,7} \right\}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Do not penalize inclusion of \(\left\{ 1 \right\}\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(iii) the cosets are \(\left\{ {5,4} \right\}\) <em><strong>(M1)A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ {5,2,8} \right\}\) <em><strong>A1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[8 marks]</span></strong></em></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(x{ \times _9}x = 7\) <em><strong>(A1)</strong></em></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(x = 4,5\) <strong><em>A1A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></em></strong></p>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The relation \(R\) is defined for \(x,y \in {\mathbb{Z}^ + }\) such that \(xRy\) if and only if </span><span style="font-family: times new roman,times; font-size: medium;">\({3^x} \equiv {3^y}(\bmod 10)\) .</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) Show that \(R\) is an equivalence relation.</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Identify all the equivalence classes.</span></p>
<div class="marks">[11]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Let \(S\) denote the set \(\left\{ {x\left| {x = a + b\sqrt 3 ,a,b \in \mathbb{Q},{a^2} + {b^2} \ne 0} \right.} \right\}\) .</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) Prove that \(S\) is a group under multiplication.</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Give a reason why \(S\) would not be a group if the conditions on \({a,b}\) were </span><span style="font-family: times new roman,times; font-size: medium;">changed to \({a,b \in \mathbb{R},{a^2} + {b^2} \ne 0}\) .</span></p>
<div class="marks">[15]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) \({3^x} \equiv {3^x}(\bmod 10) \Rightarrow xRx\) so <strong><em>R</em></strong> is reflexive. <strong><em>R1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(xRy \Rightarrow {3^x} \equiv {3^y}(\bmod 10) \Rightarrow {3^y} \equiv {3^x}(\bmod 10) \Rightarrow yRx\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">so \(R\) is symmetric. <em><strong>R2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(xRy\) and \(yRz \Rightarrow {3^x} - {3^y} = 10M\) and \({3^y} - {3^z} = 10N\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Adding \({3^x} - {3^z} = 10(M + N) \Rightarrow {3^x} \equiv {3^z}(\bmod 10)\) hence transitive <em><strong>R2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Consider \({3^1} = 3,{3^2} = 9,{3^3} = 27,{3^4} = 81,{3^5} = 243\) , etc. <em><strong> (M2)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">It is evident from this sequence that there are 4 equivalence classes, </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(1\), \(5\), \(9\), … <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(2\), \(6\), \(10\), … <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(3\), \(7\), \(11\), … <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(4\), \(8\), \(12\), … <strong><em>A1 </em></strong></span></p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;"><em> </em></span></strong></p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;"><em>[11 marks]</em> </span></strong></p>
<div class="question_part_label">a.</div>
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<p><span style="font-family: times new roman,times; font-size: medium;">(i) Consider \(a + b\sqrt 3 \) \(c + d\sqrt 3 \) \( = (ac + 3bd) + (bc + ad)\sqrt 3 \) <strong><em>M1A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">This establishes closure since products of rational numbers are rational. <em><strong>R1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Since if \(a\) and \(b\) are not both zero and \(c\) and \(d\) are not both zero, it follows that \(ac + 3bd\) and \(bc + ad\) are not both zero. <em><strong>R1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">The identity is \(1( \in S)\) . <em><strong>R1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Consider \(a + b{\sqrt 3 ^{ - 1}} = \frac{1}{{a + b\sqrt 3 }}\) <strong><em>M1A1</em> </strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{1}{{a + b\sqrt 3 }} \times \frac{{a - b\sqrt 3 }}{{a - b\sqrt 3 }}\) <strong><em>A1 </em></strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{a}{{({a^2} - 3{b^2})}} \times \frac{b}{{({a^2} - 3{b^2})}}\sqrt 3 \) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">This inverse \( \in S\) because \({({a^2} - 3{b^2})}\) cannot equal zero since \(a\) and \(b\) cannot both be zero <strong><em>R1 </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">and \(({a^2} - 3{b^2}) = 0\) would require \(\frac{a}{b} = \pm \sqrt 3 \) which is impossible because a rational number cannot equal \(\sqrt 3 \) . <em><strong>R2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Finally, multiplication of numbers is associative. <em><strong>R1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) If \(a\) and \(b\) are both real numbers, \(a + b\sqrt 3 \) would have no inverse if \({a^2} = 3{b^2}\) . <em><strong>R2 </strong></em></span></p>
<p><em><span style="font-family: times new roman,times; font-size: medium;"><strong> </strong></span></em></p>
<p><em><span style="font-family: times new roman,times; font-size: medium;"><strong>[15 marks]</strong> </span></em></p>
<div class="question_part_label">b.</div>
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<h2 style="margin-top: 1em">Examiners report</h2>
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[N/A]
<div class="question_part_label">a.</div>
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<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
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