Given that A is the interval \(\left\{ {x\,{\text{:}}\,0 \leqslant x \leqslant 3} \right\}\) and B is the interval \(\left\{ {y\,{\text{:}}\,0 \leqslant x \leqslant 4} \right\}\) then describe A × B in geometric form.

Show that the function \(f\) is a bijection.

Hence find the inverse function \({f^{ - 1}}\).

A × B is a rectangle       A1

vertices at (0, 0), (3, 0), (0, 4) and (3, 4) or equivalent description      A1

and its interior      A1

Note: Accept diagrammatic answers.

[3 marks]

need to prove it is injective and surjective        R1

need to show if \(f\left( {x,\,y} \right) = f\left( {u,\,v} \right)\) then \(\left( {x,\,y} \right) = \left( {u,\,v} \right)\)      M1

\( \Rightarrow x + 3y = u + 3v\)

\(2x - y = 2u - v\)      A1

Equation 2 – 2 Equation 1 \( \Rightarrow y = v\)

Equation 1 + 3 Equation 2 \( \Rightarrow x = u\)    A1

thus \(\left( {x,\,y} \right) = \left( {u,\,v} \right) \Rightarrow f\) is injective

let \(\left( {s,\,t} \right)\) be any value in the co-domain \(\mathbb{R} \times \mathbb{R}\)

we must find \(\left( {x,\,y} \right)\) such that \(f\left( {x,\,y} \right) = \left( {s,\,t} \right)\)    M1

\(s = x + 3y\) and \(t = 2x - y\)    M1

\( \Rightarrow y = \frac{{2s - t}}{7}\)      A1

and \(x = \frac{{s + 3t}}{7}\)      A1

hence \(f\left( {x,\,y} \right) = \left( {s,\,t} \right)\) and is therefore surjective

[8 marks]

\({f^{ - 1}}\left( {x,\,y} \right) = \left( {\frac{{x + 3y}}{7},\,\frac{{2x - y}}{7}} \right)\)      A1A1

[2 marks]

Draw the Cayley table for \(\left\{ {G,\left.  *  \right\}} \right.\) .

(i)     Determine the order of each element of \(\left\{ {G,\left.  *  \right\}} \right.\) .

(ii)     Find all the proper subgroups of \(\left\{ {G,\left.  *  \right\}} \right.\) .

Solve the equation \(x * 6 * x = 3\) where \(x \in G\) .

     A3

Note: Award A2 for 1 error, A1 for 2 errors, A0 for 3 or more errors.

[3 marks]

(i)     We first identify \(1\) as the identity     (A1)

Order of \(1 = 1\)

Order of \(2 = 3\)

Order of \(3 = 6\)

Order of \(4 = 3\)

Order of \(5 = 6\)

Order of \(6 = 2\)     A3

Note: Award A2 for 1 error, A1 for 2 errors, A0 for more than 2 errors.

(ii)     \(\left\{ {1,\left. 6 \right\}} \right.\) ; \(\left\{ {1,\left. {2,4} \right\}} \right.\)     A1A1

[6 marks]

The equation is equivalent to

\(6 * x * x = 3\)     M1

\(x * x = 4\)

\(x = 2\) or \(5\)     A1A1

[3 marks]

Show that \(\{ G,{\text{ }} * \} \) is a group.

Determine whether or not \(\{ H,{\text{ }} * \} \)  is a subgroup of \(\{ G,{\text{ }} * \} \).

closure: let A, B \( \in G\)

(because AB is a \(2 \times 2\) matrix)

and det(AB) = det(A)det(B) \( = 1 \times 1 = 1\)     M1A1

identity: the \(2 \times 2\) identity matrix has determinant 1     R1

inverse: let A \( \in G\). Then A has an inverse because it is non-singular     (R1)

since AA\(^{ - 1} = \) I, det(A)det(A\(^{ - 1}\)) = det(I) = 1 therefore A\(^{ - 1} \in G\)     R1

associativity is assumed

the four axioms are satisfied therefore \(\{ G,{\text{ }} * \} \) is a group     AG

[5 marks]

closure: let A, B \( \in H\). Then AB \( \in H\) because the arithmetic involved produces elements that are integers     R1

inverse: A\(^{ - 1} \in H\) because the calculation of the inverse involves interchanging the elements and dividing by the determinant which is 1     R1

the identity (and associativity) follow as above     R1

therefore \(\{ H,{\text{ }} * \} \) is a subgroup of \(\{ G,{\text{ }} * \} \)     A1

Note:     Award the A1 only if the first two R1 marks are awarded but not necessarily the third R1.

Note:     Accept subgroup test.

[4 marks]

For \({\rho _1}\) and \({\rho _2}\) determine whether or not each is reflexive, symmetric and transitive.

For each of \({\rho _1}\) and \({\rho _2}\) which is an equivalence relation, describe the equivalence classes.

\({\rho _1}\)

\(({x_1},{\text{ }}{y_1}){\rho _1}({x_1},{\text{ }}{y_1}) \Rightarrow 0 = 0\;\;\;\)hence reflexive.     R1

\(({x_1},{\text{ }}{y_1}){\rho _1}({x_2},{\text{ }}{y_2}) \Rightarrow x_1^2 - x_2^2 = y_1^2 - y_2^2\)

\( \Rightarrow (x_1^2 - x_2^2) =  - (y_1^2 - y_2^2)\)

\( \Rightarrow x_2^2 - x_1^2 = y_2^2 - y_1^2 \Rightarrow ({x_2},{\text{ }}{y_2}){\rho _1}({x_1},{\text{ }}{y_1})\;\;\;\)hence symmetric     M1A1

\(({x_1},{\text{ }}{y_1}){\rho _1}({x_2},{\text{ }}{y_2}) \Rightarrow x_1^2 - x_2^2 = y_1^2 - y_2^2{\text{ - i}}\)

\(({x_2},{\text{ }}{y_2}){\rho _1}({x_3},{\text{ }}{y_3}) \Rightarrow x_2^2 - x_3^2 = y_2^2 - y_3^2{\text{ - ii}}\)     M1

\({\text{i}} + {\text{ii}} \Rightarrow x_1^2 - x_3^2 = y_1^2 - y_3^2 \Rightarrow ({x_1},{\text{ }}{y_1}){\rho _1}({x_3},{\text{ }}{y_3})\;\;\;\)hence transitive     A1

\({\rho _2}\)

\(({x_1},{\text{ }}{y_1}){\rho _2}({x_1},{\text{ }}{y_1}) \Rightarrow \sqrt {2x_1^2}  \leqslant \sqrt {2y_1^2} \;\;\;\)This is not true in the case of (3,1)

hence not reflexive.     R1

\(({x_1},{\text{ }}{y_1}){\rho _2}({x_2},{\text{ }}{y_2}) \Rightarrow \sqrt {x_1^2 + x_2^2}  \leqslant \sqrt {y_1^2 + y_2^2} \)

\( \Rightarrow \sqrt {x_2^2 + x_1^2}  \leqslant \sqrt {y_2^2\_y_1^2}  \Rightarrow ({x_2},{\text{ }}{x_2}){\rho _2}({x_1},{\text{ }}{y_1})\;\;\;\)hence symmetric.     A1

it is not transitive.     A1

attempt to find a counterexample     (M1)

for example \((1,{\text{ }}0){\rho _2}(0,{\text{ 1)}}\) and \((0,{\text{ }}1){\rho _2}(1,{\text{ 0)}}\)     A1

however, it is not true that \((1,{\text{ }}0){\rho _2}(1,{\text{ 0)}}\)     A1

\({\rho _1}\) is an equivalence relation     A1

the equivalence classes for \({\rho _1}\) form a family of curves of the form

\({y^2} - {x^2} = k\)     A1

Most candidates attempted this question with many showing correctly that \({\rho _1}\) is an equivalence relation. Most candidates, however, were unable to find a counterexample to show that \({\rho _2}\) is not transitive although many suspected that was the case. Most candidates were unable to describe the equivalence classes.

Most candidates attempted this question with many showing correctly that \({\rho _1}\) is an equivalence relation. Most candidates, however, were unable to find a counterexample to show that \({\rho _2}\) is not transitive although many suspected that was the case. Most candidates were unable to describe the equivalence classes.

Determine the order of \(P\), justifying your answer.

Find \({P^2}\).

The permutation group \(G\) is generated by \(P\). Determine the element of \(G\) that is of order 2, giving your answer in cycle notation.

the order is 6     A1

tracking 1 through successive powers of \(P\) returns to 1 after 6 transitions (or equivalent)     R1

[2 marks]

\({P^2} = (1{\text{ }}5{\text{ }}4)(2{\text{ }}6{\text{ }}3){\text{ or }}\left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 5&6&2&1&4&3 \end{array}} \right)\)     (M1)A1

[2 marks]

since \(P\) is of order 6, \({P^3}\) will be of order 2     R1

\({P^3} = \left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 2&1&4&3&6&5 \end{array}} \right)\)     (M1)(A1)

\({P^3} = (1{\text{ }}2)(3{\text{ }}4)(5{\text{ }}6)\)     A1

[4 marks]

Explain why only one of the following statements is true

(i)     \(17 \subset P\);

(ii)     \(\{ 7,{\text{ }}17,{\text{ }}37,{\text{ }}47,{\text{ }}57\}  \in Q\);

(iii)     \(\phi  \subset Q\) and \(\phi  \in Q\), where \(\phi \) is the empty set.

(i)     State the value of \(f(1)\), giving a reason for your answer.

(ii)     Find \(n\left( {f(2310)} \right)\).

Determine whether or not \(f\) is

(i)     injective;

(ii)    surjective.

(i)     17 is an element not a subset of \(P\)     R1

(ii)     57 is not a prime number     R1

(iii)     any demonstration that this is the true statement     A1

because every set contains the empty set as a subset     R1

[4 marks]

(i)     \(f(1) = \phi \)     A1

because 1 has no prime factors     R1

(ii)     \(f(2310) = f(2 \times 3 \times 5 \times 7 \times 11){\text{ }}\left( { = \{ 2,{\text{ }}3,{\text{ }}5,{\text{ }}7,{\text{ }}11\} } \right)\)     A1

\(n\left( {f(2310)} \right) = 5\)    A1

[4 marks]

(i)     not injective     A1

because, for example, \(f(2) = f(4) = \{ 2\} \)     R1

(ii)     not surjective     A1

\({f^{ - 1}}(2,{\text{ }}3,{\text{ }}5,{\text{ }}7,{\text{ }}11,{\text{ }}13)\) does not belong to S because

\(2 \times 3 \times 5 \times 7 \times 11 \times 13 > 2500\)    R1

Note:     Accept any appropriate example.

[4 marks]

The question caused a number of problems for candidates. In part (a) a number of candidates thought part (i) was correct as they did not realise it was an element and a number thought part (ii) was correct as they did not recognise 57 as a prime number. In both of these two cases, candidates then suggested part (iii) was false giving a variety of incorrect justifications. Part (b) was more successful for most candidates with many wholly correct answers seen. Part (c) again saw many correct answers, but some candidates tried to argue the opposite, incorrect viewpoint or in other cases gave no reason for their decisions, showing a complete misunderstanding of the command term “determine”.

The question caused a number of problems for candidates. In part (a) a number of candidates thought part (i) was correct as they did not realise it was an element and a number thought part (ii) was correct as they did not recognise 57 as a prime number. In both of these two cases, candidates then suggested part (iii) was false giving a variety of incorrect justifications. Part (b) was more successful for most candidates with many wholly correct answers seen. Part (c) again saw many correct answers, but some candidates tried to argue the opposite, incorrect viewpoint or in other cases gave no reason for their decisions, showing a complete misunderstanding of the command term “determine”.

The question caused a number of problems for candidates. In part (a) a number of candidates thought part (i) was correct as they did not realise it was an element and a number thought part (ii) was correct as they did not recognise 57 as a prime number. In both of these two cases, candidates then suggested part (iii) was false giving a variety of incorrect justifications. Part (b) was more successful for most candidates with many wholly correct answers seen. Part (c) again saw many correct answers, but some candidates tried to argue the opposite, incorrect viewpoint or in other cases gave no reason for their decisions, showing a complete misunderstanding of the command term “determine”.

The group \(\{ G,{\text{ }} * \} \) has a subgroup \(\{ H,{\text{ }} * \} \). The relation \(R\) is defined, for \(x,{\text{ }}y \in G\), by \(xRy\) if and only if \({x^{ - 1}} * y \in H\).

(a)     Show that \(R\) is an equivalence relation.

(b)     Given that \(G = \{ 0,{\text{ }} \pm 1,{\text{ }} \pm 2,{\text{ }} \ldots \} \), \(H = \{ 0,{\text{ }} \pm 4,{\text{ }} \pm 8,{\text{ }} \ldots \} \) and \( * \) denotes addition, find the equivalence class containing the number \(3\).

(a)     \(\underline {{\text{reflexive}}} \)

\({x^{ - 1}}x = e \in H\)     A1

therefore \(xRx\) and \(R\) is reflexive     R1

\(\underline {{\text{symmetric}}} \)

Note: Accept the word commutative.

let \(xRy\) so that \({x^{ - 1}}y \in H\)     M1

the inverse of \({x^{ - 1}}y\) is \({y^{ - 1}}x \in H\)     A1

therefore \(yRx\) and \(R\) is symmetric     R1

\(\underline {{\text{transitive}}} \)

let \(xRy\) and \(yRz\) so \({x^{ - 1}}y \in H\) and \({y^{ - 1}}z \in H\)     M1

therefore \({x^{ - 1}}y\,{y^{ - 1}}z = {x^{ - 1}}z \in H\)     A1

therefore \(xRz\) and \(R\) is transitive     R1

hence \(R\) is an equivalence relation     AG

[8 marks]

(b)     the identity is \(0\) so the inverse of \(3\) is \(-3\)     (R1)

the equivalence class of 3 contains \(x\) where \( - 3 + x \in H\)     (M1)

\( - 3 + x = 4n{\text{ }}(n \in \mathbb{Z})\)     (M1)

\(x = 3 + 4n{\text{ (n}} \in \mathbb{Z})\)     A1

Note: Accept \(\{  \ldots  - 5,{\text{ }} - 1,{\text{ }}3,{\text{ }}7,{\text{ }} \ldots \} \) or \(x \equiv 3(\bmod 4)\).

Note: If no other relevant working seen award A3 for \(\{ 3 + 4n\} \) or \(\{  \ldots  - 5,{\text{ }} - 1,{\text{ }}3,{\text{ }}7,{\text{ }} \ldots \} \) seen anywhere.

[4 marks]

Show that \({(ba)^2} = e\) .

Express \({(bab)^{ - 1}}\) in its simplest form.

Given that \(a \ne e\) ,

  (i)     show that \(b \ne e\) ;

  (ii)     show that \(G\) is not Abelian.

EITHER

\(baba = ba{a^2}b\)     M1

\( = b{a^3}b\)     (A1)

\( = {b^2}\)     A1

\( = e\)     AG

OR

\(baba = {a^2}bba\)     M1

\( = {a^2}{b^2}a\)      (A1)

\( = {a^3}\)     A1

\( = e\)     AG

[3 marks]

\(bab = {a^2}bb\)     (M1)

\( = {a^2}\)     (A1)

\({(bab)^{ - 1}} = a\)     A1

[3 marks]

(i)     assume \(b = e\)     M1

then \(a = {a^2}\)     A1

\( \Rightarrow a = e\) which is a contradiction     R1

(ii)     if \(ab = ba\)     M1

then \(ab = {a^2}b\)     A1

\( \Rightarrow a = e\) which is a contradiction     R1

[6 marks]

This question was started by the majority of candidates, but only successfully completed by a few. Many candidates seemed to be aware of this style of question, but were either unable to make significant progress or manipulated the algebra in a contorted manner and hence lost valuable time. Also a number of candidates made assumptions about commutativity which were not justified. Overall, the level and succinctness of meaningful algebraic manipulation shown by candidates was disappointing.

This question was started by the majority of candidates, but only successfully completed by a few. Many candidates seemed to be aware of this style of question, but were either unable to make significant progress or manipulated the algebra in a contorted manner and hence lost valuable time. Also a number of candidates made assumptions about commutativity which were not justified. Overall, the level and succinctness of meaningful algebraic manipulation shown by candidates was disappointing.

This question was started by the majority of candidates, but only successfully completed by a few. Many candidates seemed to be aware of this style of question, but were either unable to make significant progress or manipulated the algebra in a contorted manner and hence lost valuable time. Also a number of candidates made assumptions about commutativity which were not justified. In part (c) the idea of a proof by contradiction was used by stronger candidates, but weaker candidates were often at a loss as how to start. Overall, the level and succinctness of meaningful algebraic manipulation shown by candidates was disappointing.

The set \({{\rm{S}}_1} = \left\{ {2,4,6,8} \right\}\) and \({ \times _{10}}\) denotes multiplication modulo \(10\).

  (i)     Write down the Cayley table for \(\left\{ {{{\rm{S}}_1},{ \times _{10}}} \right\}\) .

  (ii)     Show that \(\left\{ {{{\rm{S}}_1},{ \times _{10}}} \right\}\) is a group.

  (iii)     Show that this group is cyclic.

Now consider the group \(\left\{ {{{\rm{S}}_1},{ \times _{20}}} \right\}\) where \({{\rm{S}}_2} = \left\{ {1,9,11,19} \right\}\) and \({{ \times _{20}}}\) denotes multiplication modulo \(20\). Giving a reason, state whether or not \(\left\{ {{{\rm{S}}_1},{ \times _{10}}} \right\}\) and \(\left\{ {{{\rm{S}}_1},{ \times _{20}}} \right\}\) are isomorphic.

(i)

     A2

Note: Award A1 for one error.

(ii)     closure: it is closed because no new elements are formed     A1

identity: \(6\) is the identity element    A1

inverses: \(4\) is self-inverse and (\(2\), \(8\)) form an inverse pair     A1

associativity: multiplication is associative     A1

the four group axioms are satisfied

(iii)     any valid reason, e.g.

\(2\) (or \(8\)) has order \(4\), or \(2\) (or \(8\)) is a generator     A2

[8 marks]

the groups are not isomorphic     A1

any valid reason, e.g. \({{\rm{S}}_2}\) is not cyclic or all its elements are self-inverse     R2

[3 marks]

Parts (a) (i) and (a) (iii) were well answered in general. However, in (a) (ii), some candidates lost marks by not showing convincingly that \(\left\{ {{{\rm{S}}_1},{ \times _{10}}} \right\}\) was a group. For example, in verifying the group axioms, some candidates just made bald statements such as "\(\left\{ {{{\rm{S}}_1},{ \times _{10}}} \right\}\) is closed". This was not convincing because the question indicated that it was a group so that closure was implied by the question. It was necessary here to make some reference to the Cayley table which showed that no new elements were formed by the binary operation. To gain full marks on this style of question candidates need to clearly explain the reasoning used for deductions.

In (b), most candidates realised that the quickest way to establish isomorphism (or not) was to determine the order of each element. Candidates who knew that there are essentially only two different groups of order four had a slight advantage in this question.

Show that \(R\) is an equivalence relation.

The relationship between \(a\) , \(b\) , \(c\) and \(d\) is changed to \(ad - bc = n\) . State, with a reason, whether or not there are any non-zero values of \(n\) , other than \(1\), for which \(R\) is an equivalence relation.

since \(\boldsymbol{A} = \boldsymbol{AI}\) where \(\boldsymbol{I}\) is the identity     A1

and \(\det (\boldsymbol{I}) = 1\) ,     A1

\(R\) is reflexive

\(\boldsymbol{ARB} \Rightarrow \boldsymbol{A} = \boldsymbol{BX}\) where \(\det (\boldsymbol{X}) = 1\)     M1

it follows that \(\boldsymbol{B} = \boldsymbol{A}{\boldsymbol{X}^{ - 1}}\)     A1

and \(\det ({\boldsymbol{X}^{ - 1}}) = \det{(\boldsymbol{X})^{ - 1}} = 1\)     A1

\(R\) is symmetric

\(\boldsymbol{ARB}\) and \(\boldsymbol{BRC} \Rightarrow \boldsymbol{A} = \boldsymbol{BX}\) and \(\boldsymbol{B} = \boldsymbol{CY}\) where \(\det (\boldsymbol{X}) = \det (\boldsymbol{Y}) = 1\)     M1

it follows that \(\boldsymbol{A} = \boldsymbol{CYX}\)     A1

\(\det (\boldsymbol{YX}) = \det (\boldsymbol{Y})\det (\boldsymbol{X}) = 1\)     A1

\(R\) is transitive

hence \(R\) is an equivalence relation     AG

[8 marks]

for reflexivity, we require \(\boldsymbol{ARA}\) so that \(\boldsymbol{A} = \boldsymbol{AI}\) (for all \(\boldsymbol{A} \in S\) )     M1

since \(\det (\boldsymbol{I}) = 1\) and we require \(\boldsymbol{I} \in S\) the only possibility is \(n = 1\)     A1

[2 marks]

This question was not well done in general, again illustrating that questions involving both matrices and equivalence relations tend to cause problems for candidates. A common error was to assume, incorrectly, that ARB and BRC \( \Rightarrow A = BX\) and \(B = CX\) , not realizing that a different "\(x\)" is required each time. In proving that \(R\) is an equivalence relation, consideration of the determinant is necessary in this question although many candidates neglected to do this.

In proving that \(R\) is an equivalence relation, consideration of the determinant is necessary in this question although many candidates neglected to do this.

Construct the Cayley table for \(\{ S,{\text{ }}{ + _6}\} \).

Show that \(\{ S,{\text{ }}{ + _6}\} \) forms an Abelian group.

State the order of each element.

Explain whether or not the group is cyclic.

    A2

Note: A1 for one or two errors in the table, A0 otherwise.

closed no new elements     A1

\(0\) is identity (since \(0 + a = a + 0 = a,{\text{ }}a \in S\))     A1

\(0\), \(3\) self inverse, \(1 \Leftrightarrow 5\) inverse pair, \(2 \Leftrightarrow 4\) inverse pair     A1

all elements have an inverse

associativity is assumed over addition     A1

since symmetry on leading diagonal in table or commutativity of addition     A1

\( \Rightarrow \{ S,{\text{ }}{ + _6}\} \) is an Abelian group     AG

    A2

Note: A1 for one or two errors in the table, A0 otherwise.

since there is an element with order \(6\) OR \(1\) or \(5\) are generators     R1

the group is cyclic A1

This question was well answered in general although some candidates showed only commutativity, not realising that they also had to prove that it was a group.

This question was well answered in general although some candidates showed only commutativity, not realising that they also had to prove that it was a group.

This question was well answered in general although some candidates showed only commutativity, not realising that they also had to prove that it was a group.

This question was well answered in general although some candidates showed only commutativity, not realising that they also had to prove that it was a group.

Prove that the function \(f:\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}\) defined by \(f(x,{\text{ }}y) = (2x + y,{\text{ }}x + y)\) is a bijection.

to be a bijection it must be injective and surjective     R1

Note: This R1 may be awarded at any stage

suppose \(f(x,{\text{ }}y) = f(u,{\text{ }}v)\)     M1

\(2x + y = 2u + v\;\;\;({\text{ - i}})\)

\(x + y = u + v\;\;\;({\text{ - ii}})\)

\({\text{i - ii}} \Rightarrow x = u\)

\({\text{i - 2(ii)}} \Rightarrow  - y =  - v\)

\( \Rightarrow x = u,{\text{ }}y = v\)     A1

thus \((x,{\text{ }}y) = (u,{\text{ }}v)\) hence injective     A1

let \(2x + y = s\;\;\;({\text{ - i}})\)

\(x + y = t\;\;\;({\text{ - ii}})\)     M1

\({\text{i - ii }}x = s - t\)

\( \Rightarrow y = 2t - s\)

both \(x\) and \(y\) are integer if \(s\) and \(t\) are integer     R1

hence it is surjective     A1

hence  \(f\) is a bijection     AG

Note: Accept a valid argument based on matrices

Most candidates were able to show that \(f\) was an injection although some candidates appear to believe that it is sufficient to show that \(f(x,{\text{ }}y)\) is unique. A significant minority failed to show that \(f\) is a surjection and most candidates failed to note that it had to be checked that all values were integers. Some candidates introduced a matrix to define the transformation which was often a successful alternative method.

Show that the set \(S\) of numbers of the form \({2^m} \times {3^n}\) , where \(m,n \in \mathbb{Z}\) , forms a group \(\left\{ {S, \times } \right\}\) under multiplication.

Show that \(\left\{ {S, \times } \right\}\) is isomorphic to the group of complex numbers \(m + n{\rm{i}}\) under addition, where \(m\), \(n \in \mathbb{Z}\) .

Closure: Consider the numbers \({2^{{m_1}}} \times {3^{{m_1}}}\) and \({2^{{m_2}}} \times {3^{{n_2}}}\) where     M1

\({m_1},{m_2},{n_1},{n_2}, \in \mathbb{Z}\) . Then,

Product \( = {2^{{m_1} + {m_2}}} \times {3^{{n_1} + {n_2}}}\) which \( \in S\)     A1

Identity: \({2^0} \times {3^0} = 1 \in S\)     A1

Since \(({2^m} \times {3^n}) \times ({2^{ - m}} \times {3^{ - n}}) = 1\) and \({2^{ - m}} \times {3^{ - n}} \in S\)     R1

then \({2^{ - m}} \times {3^{ - n}}\) is the inverse.     A1

Associativity: This follows from the associativity of multiplication.     R1

[6 marks]

Consider the bijection

\(f({2^m} \times {3^n}) = m + n{\rm{i}}\)    (M1)

Then

\(f({2^{{m_1}}} \times {3^{{n_1}}}) \times ({2^{{m_2}}} \times {3^{{n_2}}}) = f({2^{{m_1} + {m_2}}} \times {3^{{n_1} + {n_2}}})\)     M1A1

\( = {m_1} + {m_2} + ({n_1} + {n_2}){\rm{i}}\)     A1

\( = ({m_1} + {n_1}{\rm{i}}) + ({m_2} + {n_2}{\rm{i}})\)     (A1)

\( = f({2^{{m_1}}} \times {3^{{n_1}}}) + f({2^{{m_2}}} \times {3^{{n_2}}})\)     A1

[6 marks]

Show that \(R\) is an equivalence relation.

State the equivalence classes of \(R\) .

\(4a + b = 5n\) for \(a,b,n \in \mathbb{Z}\)

reflexive:

\(4a + a = 5a\) so \(aRa\) , and \(R\) is reflexive     A1

symmetric:

\(4a + b = 5n\)

\(4b + a = 5b - b + 5a - 4a\)     M1

\( = 5b + 5a - (4a + b)\)     A1

\( = 5m\) so \(bRa\) , and \(R\) is symmetric     A1

transitive:

\(4a + b = 5n\)     M1

\(4b + c = 5k\)     M1

\(4a + 5b + c = 5n + 5k\)     A1

\(4a + c = 5(n + k - b)\) so \(aRc\) , and \(R\) is transitive     A1

therefore \(R\) is an equivalence relation     AG

[8 marks]

equivalence classes are

\(\left\{ { \ldots , - 10, - 5,0,5,10,\left.  \ldots  \right\}} \right.\)     (M1)

\(\left\{ { \ldots , - 9, - 4,1,6,11,\left.  \ldots  \right\}} \right.\)

\(\left\{ { \ldots , - 8, - 3,2,7,12,\left.  \ldots  \right\}} \right.\)

\(\left\{ { \ldots , - 7, - 2,3,8,13,\left.  \ldots  \right\}} \right.\)

\(\left\{ { \ldots , - 6, - 1,4,9,14,\left.  \ldots  \right\}} \right.\)

or \(\left\{ {\left\langle 0 \right\rangle ,\left\langle 1 \right\rangle ,\left\langle 2 \right\rangle ,\left\langle 3 \right\rangle ,\left. {\left\langle 4 \right\rangle } \right\}} \right.\)     A2

Note: Award A2 for all classes, A1 for at least 2 correct classes.

[3 marks]

Part (a) was generally well done but not always in the most direct manner.

Too many missed the equivalence classes in part (b).

The function \(f:{\mathbb{R}^ + } \times {\mathbb{R}^ + } \to {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) is defined by \(f(x,{\text{ }}y) = \left( {xy,{\text{ }}\frac{x}{y}} \right)\).

Prove that \(f\) is a bijection.

we need to show that \(f\) is injective and surjective     (R1)

Note: Award R1 if seen anywhere in the solution.

\(\underline {{\text{injective}}} \)

let \((a,{\text{ }}b)\) and \((c,{\text{ }}d) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\), and let \(f(a,{\text{ }}b) = f(c,{\text{ }}d)\)     M1

it follows that

\(ab = cd\) and \(\frac{a}{b} = \frac{c}{d}\)     A1

multiplying these equations,

\({a^2} = {c^2} \Rightarrow a = c\) and therefore \(b = d\)     A1

since \(f(a,{\text{ }}b) = f(c,{\text{ }}d) \Rightarrow (a,{\text{ }}b) = (c,{\text{ }}d),{\text{ }}f\) is injective     R1

Note: Award R1 if stated anywhere as needing to be shown.

\(\underline {{\text{surjective}}} \)

let \((p,{\text{ }}q) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\)

consider \(f(x,{\text{ }}y) = (p,{\text{ }}q)\) so \(xy = p\) and \(\frac{x}{y} = q\)     M1A1

multiplying these equations,

\({x^2} = pq\) so \(x = \sqrt {pq} \) and therefore \(y = \sqrt {\frac{p}{q}} \)     A1

so given \((p,{\text{ }}q) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + },{\text{ }}\exists (x,{\text{ }}y) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) such that \(f(x,{\text{ }}y) = (p,{\text{ }}q)\) which shows that \(f\) is surjective     R1

Note: Award R1 if stated anywhere as needing to be shown.

\(f\) is therefore a bijection

[9 marks]

Copy and complete the following Cayley table for the transformations of T₁, T₂, T₃, T₄, under the operation of composition of transformations.

Show that T₁, T₂, T₃, T₄ under the operation of composition of transformations form a group. Associativity may be assumed.

Show that this group is cyclic.

Write down the 2 × 2 matrices representing T₃, T₄ and T₅.

Find the 2 × 2 matrix representing T.

Give a geometric description of the transformation T.

      A2

[2 marks]

Note: Award A1 for 6, 7 or 8 correct.

the table is closed – no new elements      A1

T₁ is the identity      A1

T₃ (and T₁) are self-inverse; T₂ and T₄ are an inverse pair. Hence every element has an inverse      A1

hence it is a group      AG

[3 marks]

all elements in the group can be generated by T₂ (or T₄)       R1

hence the group is cyclic      AG

[1 mark]

T₃ is represented by \(\left( {\begin{array}{*{20}{c}}
{ - 1}&0 \\
0&{ - 1}
\end{array}} \right)\)      A1

T₄ is represented by \(\left( {\begin{array}{*{20}{c}}
0&{ - 1} \\
1&0
\end{array}} \right)\)      A1

T₅ is represented by \(\left( {\begin{array}{*{20}{c}}
1&0 \\
0&{ - 1}
\end{array}} \right)\)      A1

[3 marks]

\(\left( {\begin{array}{*{20}{c}}
0&{ - 1} \\
1&0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
1&0 \\
0&{ - 1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{ - 1}&0 \\
0&{ - 1}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
0&{ - 1} \\
{ - 1}&0
\end{array}} \right)\)       (M1)A1

Note: Award M1A0 for multiplying the matrices in the wrong order.

[2 marks]

a reflection in the line \(y =  - x\)      A1

[1 mark]

\(\{ G,{\text{ }} * \} \) is a group of order \(N\) and \(\{ H,{\text{ }} * \} \) is a proper subgroup of \(\{ G,{\text{ }} * \} \) of order \(n\).

(a)     Define the right coset of \(\{ H,{\text{ }} * \} \) containing the element \(a \in G\).

(b)     Show that each right coset of \(\{ H,{\text{ }} * \} \) contains \(n\) elements.

(c)     Show that the union of the right cosets of \(\{ H,{\text{ }} * \} \) is equal to \(G\).

(d)     Show that any two right cosets of \(\{ H,{\text{ }} * \} \) are either equal or disjoint.

(e)     Give a reason why the above results can be used to prove that \(N\) is a multiple of \(n\).

(a)     the right coset containing \(a\) has the form \(\{ ha|h \in H\} \)     A1

[1 mark]

Note: From here on condone the use of left cosets.

(b)     let \(b\), \(c\) be distinct elements of \(H\). Then, given \(a \in G\), by the Latin square property of the Cayley table, \(ba\) and \(ca\) are distinct     A1

therefore each element of \(H\) corresponds to a unique element in the coset which must therefore contain \(n\) elements     R1

[2 marks]

(c)     let \(d\) be any element of \(G\). Then since \(H\) contains the identity \(e\), \(ed = d\) will be in a coset     R1

therefore every element of \(G\) will be contained in a coset which proves that the union of all the cosets is \(G\)     R1

[2 marks]

(d)     let the cosets of \(b\) and \(c{\text{ }}(b,{\text{ }}c \in G)\) contain a common element so that

\(pb = qc\) where \(p,{\text{ }}q \in H\). Let \(r\) denote any other element \( \in H\)     M1

then

\(rb = r{p^{ - 1}}qc\)     A1

since \(r{p^{ - 1}}q \in H\), this shows that all the other elements are common and the cosets are equal     R1

since not all cosets can be equal, there must be other cosets which are disjoint     R1

[4 marks]

(e)     the above results show that \(G\) is partitioned into a number of disjoint subsets containing \(n\) elements so that \(N\) must be a multiple of \(n\)     R1

[1 mark]

Show that any matrix of this form is its own inverse.

Show that \(S\) forms an Abelian group under matrix multiplication.

Giving a reason, state whether or not this group is cyclic.

\(\left( {\begin{array}{*{20}{c}}
a&0&0\\
0&b&0\\
0&0&c
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
a&0&0\\
0&b&0\\
0&0&c
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{{a^2}}&0&0\\
0&{{b^2}}&0\\
0&0&{{c^2}}
\end{array}} \right)\)     A1M1

\( = \left( {\begin{array}{*{20}{c}}
1&0&0\\
0&1&0\\
0&0&1
\end{array}} \right)\)     A1

this shows that each matrix is self-inverse

[3 marks]

closure:

\(\left( {\begin{array}{*{20}{c}}
{{a_1}}&0&0\\
0&{{b_1}}&0\\
0&0&{{c_1}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{{a_2}}&0&0\\
0&{{b_2}}&0\\
0&0&{{c_2}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{{a_1}{a_2}}&0&0\\
0&{{b_1}{b_2}}&0\\
0&0&{{c_1}{c_2}}
\end{array}} \right)\)     M1A1

\( = \left( {\begin{array}{*{20}{c}}
{{a_3}}&0&0\\
0&{{b_3}}&0\\
0&0&{{c_3}}
\end{array}} \right)\)

where each of \({a_3}\), \({b_3}\), \({c_3}\) can only be \( \pm 1\)     A1

this proves closure

identity: the identity matrix is the group identity     A1

inverse: as shown above, every element is self-inverse     A1

associativity: this follows because matrix multiplication is associative     A1

\(S\) is therefore a group     AG

Abelian:

\(\left( {\begin{array}{*{20}{c}}
{{a_2}}&0&0\\
0&{{b_2}}&0\\
0&0&{{c_2}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{{a_1}}&0&0\\
0&{{b_1}}&0\\
0&0&{{c_1}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{{a_2}{a_1}}&0&0\\
0&{{b_2}{b_1}}&0\\
0&0&{{c_2}{c_1}}
\end{array}} \right)\)     A1

\(\left( {\begin{array}{*{20}{c}}
{{a_1}}&0&0\\
0&{{b_1}}&0\\
0&0&{{c_1}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{{a_2}}&0&0\\
0&{{b_2}}&0\\
0&0&{{c_2}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{{a_1}{a_2}}&0&0\\
0&{{b_1}{b_2}}&0\\
0&0&{{c_1}{c_2}}
\end{array}} \right)\)     A1

Note: Second line may have been shown whilst proving closure, however a reference to it must be made here.

we see that the same result is obtained either way which proves commutativity so that the group is Abelian      R1

[9 marks]

since all elements (except the identity) are of order \(2\), the group is not cyclic (since \(S\) contains \(8\) elements)     R1

[1 mark]

Prove that the number \(14 641\) is the fourth power of an integer in any base greater than \(6\).

For \(a,b \in \mathbb{Z}\) the relation \(aRb\) is defined if and only if \(\frac{a}{b} = {2^k}\) , \(k \in \mathbb{Z}\) .

  (i)     Prove that \(R\) is an equivalence relation.

  (ii)     List the equivalence classes of \(R\) on the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

\(14641\) (base \(a > 6\) ) \( = {a^4} + 4{a^3} + 6{a^2} + 4a + 1\) ,     M1A1

\( = {(a + 1)^4}\)     A1

this is the fourth power of an integer     AG

[3 marks]

(i)     \(aRa\) since \(\frac{a}{a} = 1 = {2^0}\) , hence \(R\) is reflexive     A1

\(aRb \Rightarrow \frac{a}{b} = {2^k} \Rightarrow \frac{b}{a} = {2^{ - k}} \Rightarrow bRa\)

so R is symmetric     A1

\(aRb\) and \(bRc \Rightarrow \frac{a}{b} = {2^m}\), \(m \in \mathbb{Z}\) and \(bRc \Rightarrow \frac{b}{c} = {2^n}\) , \(n \in \mathbb{Z}\)     M1

\( \Rightarrow \frac{a}{b} \times \frac{b}{c} = \frac{a}{c} = {2^{m + n}}\) , \(m + n \in \mathbb{Z}\)    A1

\( \Rightarrow aRc\) so transitive     R1

hence \(R\) is an equivalence relation     AG

(ii)     equivalence classes are {1, 2, 4, 8} , {3, 6} , {5, 10} , {7} , {9}     A3

Note: Award A2 if one class missing, A1 if two classes missing, A0 if three or more classes missing.

[8 marks]

This was not difficult but a surprising number of candidates were unable to do it. Care with notation and logic were lacking.

The question was at first straightforward but some candidates mixed up the properties of an equivalence relation with those of a group. The idea of an equivalence class is still not clearly understood by many candidates so that some were missing.

The group \(\left\{ {G, + } \right\}\) is defined by the operation of addition on the set \(G = \left\{ {2n|n \in \mathbb{Z}} \right\}\) .

The group \(\left\{ {H, + } \right\}\) is defined by the operation of addition on the set \(H = \left\{ {4n|n \in \mathbb{Z}} \right\}\) 

Prove that \(\left\{ {G, + } \right\}\) and \(\left\{ {H, + } \right\}\) are isomorphic.

consider the function \(f:G \to H\) defined by \(f(g) = 2g\) where \(g \in G\)     A1

given \({g_1}\), \({g_2} \in G,f({g_1}) = f({g_2}) \Rightarrow 2{g_1} = 2{g_2} \Rightarrow {g_1} = {g_2}\) (injective)     M1

given \(h \in H\) then \(h = 4n\) , so \(f(2n) = h\) and \(2n \in G\) (surjective)     M1

hence f is a bijection     A1

then, for \({g_1}\), \({g_2} \in G\)

\(f({g_1} + {g_2}) = 2({g_1} + {g_2})\)      A1

\(f({g_1}) + f({g_2}) = 2{g_1} + 2{g_2}\)     A1

it follows that \(f({g_1} + {g_2}) = f({g_1}) + f({g_2})\)     R1

which completes the proof that \(\left\{ {G, + } \right\}\) and \(\left\{ {H, + } \right\}\) are isomorphic     AG

[7 marks]

Use the Euclidean algorithm to find \(\gcd (162,{\text{ }}5982)\).

The relation \(R\) is defined on \({\mathbb{Z}^ + }\) by \(nRm\) if and only if \(\gcd (n,{\text{ }}m) = 2\).

(i)     By finding counterexamples show that \(R\) is neither reflexive nor transitive.

(ii)     Write down the set of solutions of \(nR6\).

\(5982 = 162 \times 36 + 150\)    M1A1

\(162 = 150 \times 1 + 12\)    A1

\(150 = 12 \times 12 + 6\)

\(12 = 6 \times 2 + 0 \Rightarrow \gcd \) is 6     A1

[4 marks]

(i)     for example, \(\gcd (4,{\text{ }}4) = 4\)     A1

\(4 \ne 2\)    R1

so \(R\) is not reflexive     AG

for example

\(\gcd (4,{\text{ }}2) = 2\) and \(\gcd (2,{\text{ }}8) = 2\)     M1A1

but \(\gcd (4,{\text{ }}8) = 4{\text{ }}( \ne 2)\)     R1

so \(R\) is not transitive     AG

(ii)     EITHER

even numbers     A1

not divisible by 6     A1

OR

\(\{ 2 + 6n:n \in \mathbb{N}\} {\text{ }} \cup \{ 4 + 6n:n \in \mathbb{N}\} \)   A1A1

OR

\(2,{\text{ }}4,{\text{ }}8,{\text{ }}10,{\text{ }} \ldots \)    A2

[7 marks]

This was a successful question for many students with many wholly correct answers seen. Part (a) was successfully answered by most candidates and those candidates usually had a reasonable understanding of how to complete part (b). A number were not fully successful in knowing how to explain their results.

This was a successful question for many students with many wholly correct answers seen. Part (a) was successfully answered by most candidates and those candidates usually had a reasonable understanding of how to complete part (b). A number were not fully successful in knowing how to explain their results.

An investigator wishes to test the hypotheses H₀ : μ = 65, H₁ : μ > 65.

He decides on the following acceptance criteria:

Accept H₀ if the sample mean \(\bar x\) ≤ 66.5

Accept H₁ if \(\bar x\) > 66.5

Find the probability of a Type I error.

She decides to use the same acceptance criteria as the previous investigator. Find the probability of a Type II error.

Find the critical value for \({\bar x}\) if she wants the probabilities of a Type I error and a Type II error to be equal.

\(\bar X \sim {\text{N}}\left( {\mu ,\,\frac{{{\sigma ^2}}}{n}} \right)\)

\(\bar X \sim {\text{N}}\left( {65,\,\frac{{36}}{{100}}} \right)\)     (A1)

P(Type I Error) \( = {\text{P}}\left( {\bar X > 66.5} \right)\)      (M1)

= 0.00621       A1

[3 marks]

P(Type II Error) = P(accept H₀| H₁ is true)

\( = {\text{P}}\left( {\bar X \leqslant 66.5\left| {\mu  = 67.9} \right.} \right)\)        (M1)

\( = {\text{P}}\left( {\bar X \leqslant 66.5} \right)\) when \(\bar X \sim {\text{N}}\left( {67.9,\,\frac{{36}}{{100}}} \right)\)        (M1)

= 0.00982      A1

[3 marks]

the variances of the distributions given by H₀ and H₁ are equal,       (R1)

by symmetry the value of \({\bar x}\) lies midway between 65 and 67.9      (M1)

\( \Rightarrow \bar x = \frac{1}{2}\left( {65 + 67.9} \right) = 66.45\)       A1

[3 marks]