Question
The binary operator \( * \) is defined for a , \(b \in \mathbb{R}\) by \(a * b = a + b – ab\) .
(i) Show that \( * \) is associative.
(ii) Find the identity element.
(iii) Find the inverse of \(a \in \mathbb{R}\) , showing that the inverse exists for all values of \(a\) except one value which should be identified.
(iv) Solve the equation \(x * x = 1\) .
The domain of \( * \) is now reduced to \(S = \left\{ {0,2,3,4,5,\left. 6 \right\}} \right.\) and the arithmetic is carried out modulo \(7\).
(i) Copy and complete the following Cayley table for \(\left\{ {S,\left. * \right\}} \right.\) .
(ii) Show that \(\left\{ {S,\left. * \right\}} \right.\) is a group.
(iii) Determine the order of each element in S and state, with a reason, whether or not \(\left\{ {S,\left. * \right\}} \right.\) is cyclic.
(iv) Determine all the proper subgroups of \(\left\{ {S,\left. * \right\}} \right.\) and explain how your results illustrate Lagrange’s theorem.
(v) Solve the equation \(2 * x * x = 5\) .
Answer/Explanation
Markscheme
(i) \(a * (b * c) = a * (b + c – bc)\) M1
\( = a + b + c – bc – a(b + c – bc)\) A1
\( = a + b + c – bc – ca – ab + abc\) A1
\((a * b) * c = (a + b – ab) * c\) M1
\( = a + b – ab + c – (a + b – ab)c\) A1
\( = a + b + c – bc – ca – ab + abc\) , hence associative AG
(ii) let \(e\) be the identity element, so that \(a * e = a\) (M1)
then,
\(a + e – ae = a\) A1
\(e(1 – a) = 0\)
\(e = 0\) A1
(iii) let \({a^{ – 1}}\) be the inverse of \(a\), so that \(a * {a^{ – 1}} = 0\) (M1)
then,
\(a + {a^{ – 1}} – a{a^{ – 1}} = 0\) A1
\({a^{ – 1}} = \frac{a}{{a – 1}}\) A1
this gives an inverse for all elements except 1 which has no inverse R1
(iv) \(2x – {x^2} = 1\) M1
\({(x – 1)^2} = 0\) (A1)
\(x = 1\) A1
[15 marks]
(i)
A3
Note: Award A3 for correct table, A2 for one error, A1 for two errors and A0 for more than two errors.
(ii) there are no new elements in the table so it is closed A1
there is an identity element, \(0\) A1
every row (column) has a \(0\) so every element has an inverse A1
associativity has been proved earlier A1
therefore \(\left\{ {S,\left. * \right\}} \right.\) is a group AG
(iii)
A3
Note: Award A3 for correct table, A2 for one error, A1 for two errors and A0 for more than two errors.
it is cyclic because there are elements of order \(6\) R1
(iv) the proper subgroups are \(\left\{ {0,\left. 2 \right\}} \right.\) , \(\left\{ {0,\left. {4,6} \right\}} \right.\) A1A1
the orders of the subgroups (\(2\), \(3\)) are factors of the order of the group (6) A1
(v) recognizing \(x * x = 4\) (M1)
\(x = 3\) , \(6\) A1A1
[17 marks]
Question
Consider the set \(J = \left\{ {a + b\sqrt 2 :a,{\text{ }}b \in \mathbb{Z}} \right\}\) under the binary operation multiplication.
Consider \(a + b\sqrt 2 \in G\), where \(\gcd (a,{\text{ }}b) = 1\),
Show that \(J\) is closed.
State the identity in \(J\).
Show that
(i) \(1 – \sqrt 2 \) has an inverse in \(J\);
(ii) \(2 + 4\sqrt 2 \) has no inverse in \(J\).
Show that the subset, \(G\), of elements of \(J\) which have inverses, forms a group of infinite order.
(i) Find the inverse of \(a + b\sqrt 2 \).
(ii) Hence show that \({a^2} – 2{b^2}\) divides exactly into \(a\) and \(b\).
(iii) Deduce that \({a^2} – 2{b^2} = \pm 1\).
Answer/Explanation
Markscheme
\(\left( {a + b\sqrt 2 } \right) \times \left( {c + d\sqrt 2 } \right) = ac + bc\sqrt 2 + ad\sqrt 2 + 2bd\) M1
\( = ac + 2bd + (bc + ad)\sqrt 2 \in J\) A1
hence \(J\) is closed AG
Note: Award M0A0 if the general element is squared.
[2 marks]
the identity is \(1(a = 1,{\text{ }}b = 0)\) A1
[1 mark]
(i) \(\left( {1 – \sqrt 2 } \right) \times a = 1\)
\(a = \frac{1}{{1 – \sqrt 2 }}\) M1
\( = \frac{{1 + \sqrt 2 }}{{\left( {1 – \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}} = \frac{{1 + \sqrt 2 }}{{ – 1}} = – 1 – \sqrt 2 \) A1
hence \(1 – \sqrt 2 \) has an inverse in \(J\) AG
(ii) \(\left( {2 + 4\sqrt 2 } \right) \times a = 1\)
\(a = \frac{1}{{2 + 4\sqrt 2 }}\) M1
\( = \frac{{2 – 4\sqrt 2 }}{{\left( {2 – 4\sqrt 2 } \right)\left( {2 + 4\sqrt 2 } \right)}} = \frac{{2 – 4\sqrt 2 }}{{ – 28}}\) A1
which does not belong to \(J\) R1
hence \(2 + 4\sqrt 2 \) has no inverse in \(J\) AG
[5 marks]
multiplication is associative A1
let \({g_1}\) and \({g_2}\) belong to \(G\), then \(g_1^{ – 1},{\text{ }}g_2^{ – 1}\) and \(g_2^{ – 1}g_1^{ – 1}\) belong to \(J\) M1
then \(({g_1}{g_2}) \times (g_2^{ – 1}g_1^{ – 1}) = 1 \times 1 = 1\) A1
so \({g_1}{g_2}\) has inverse \(g_2^{ – 1}g_1^{ – 1}\) in \(J \Rightarrow G\) is closed A1
\(G\) contains the identity A1
\(G\) possesses inverses A1
\(G\) contains all integral powers of \(1 – \sqrt 2 \) A1
hence \(G\) is an infinite group AG
[7 marks]
(i) \({\left( {a + b\sqrt 2 } \right)^{ – 1}} = \frac{1}{{a + b\sqrt 2 }} = \frac{1}{{a + b\sqrt 2 }} \times \frac{{a – b\sqrt 2 }}{{a – b\sqrt 2 }}\) M1
\( = \frac{a}{{{a^2} – 2{b^2}}} – \frac{b}{{{a^2} – 2{b^2}}}\sqrt 2 \) A1
(ii) above number belongs to \(J\) and \({a^2} – 2{b^2} \in \mathbb{Z}\) R1
implies \({a^2} – 2{b^2}\) divides exactly into \(a\) and \(b\) AG
(iii) since \(\gcd (a,{\text{ }}b) = 1\) R1
\({a^2} – 2{b^2} = \pm 1\) AG
[4 marks]
Question
The set of all integer s from 0 to 99 inclusive is denoted by S. The binary operations \( * \) and \( \circ \) are defined on S by
\(a * b = \left[ {a + b + 20} \right]\)(mod 100)
\(a \circ b = \left[ {a + b – 20} \right]\)(mod 100).
The equivalence relation R is defined by \(aRb \Leftrightarrow \left( {{\text{sin}}\frac{{\pi a}}{5} = {\text{sin}}\frac{{\pi b}}{5}} \right)\).
Find the identity element of S with respect to \( * \).
Show that every element of S has an inverse with respect to \( * \).
State which elements of S are self-inverse with respect to \( * \).
Prove that the operation \( \circ \) is not distributive over \( * \).
Determine the equivalence classes into which R partitions S, giving the first four elements of each class.
Find two elements in the same equivalence class which are inverses of each other with respect to \( * \).
Answer/Explanation
Markscheme
\(a + e + 20 = a\)(mod 100) (M1)
\(e = – 20\)(mod 100) (A1)
\(e = 80\) A1
[3 marks]
\(a + {a^{ – 1}} + 20 = 80\)(mod 100) (M1)
inverse of \(a\) is \(60 – a\) (mod 100) A1
[2 marks]
30 and 80 A1A1
[2 marks]
\(a \circ \left( {b * c} \right) = a \circ \left( {b + c + 20} \right)\)(mod 100)
\( = a + \left( {b + c + 20} \right) – 20\)(mod 100) (M1)
\( = a + b + c\)(mod 100) A1
\(\left( {a \circ b} \right) * \left( {a \circ c} \right) = \left( {a + b – 20} \right) * \left( {a + c – 20} \right)\)(mod 100) M1
\( = a + b – 20 + a + c – 20 + 20\)(mod 100)
\( = 2a + b + c – 20\)(mod 100) A1
hence we have shown that \(a \circ \left( {b * c} \right) \ne \left( {a \circ b} \right) * \left( {a \circ c} \right)\) R1
hence the operation \( \circ \) is not distributive over \( * \) AG
Note: Accept a counterexample.
[5 marks]
{0,5,10,15…} A1
{1,4,11,14…} A1
{2,3,12,13…} A1
{6,9,16,19…} A1
{7,8,17,18…} A1
[5 marks]
for example 10 and 50, 20 and 40, 0 and 60… A2
[2 marks]