## Question

The binary operation \( * \) is defined on the set *S* = {0, 1, 2, 3} by

\[a * b = a + 2b + ab(\bmod 4){\text{ .}}\]

(a) (i) Construct the Cayley table.

(ii) Write down, with a reason, whether or not your table is a Latin square.

(b) (i) Write down, with a reason, whether or not \( * \) is commutative.

(ii) Determine whether or not \( * \) is associative, justifying your answer.

(c) Find all solutions to the equation \(x * 1 = 2 * x\) , for \(x \in S\) .

**Answer/Explanation**

## Markscheme

(a) (i)

*A3*

**Note:** Award ** A3** for no errors,

**for one error,**

*A2***for two errors and**

*A1***for three or more errors.**

*A0*

(ii) it is not a Latin square because some rows/columns contain the same digit more than once *A1*

*[4 marks]*

* *

(b) (i) **EITHER**

it is not commutative because the table is not symmetric about the leading diagonal *R2*

**OR**

it is not commutative because \(a + 2b + ab \ne 2a + b + ab\) in general *R2*

**Note:** Accept a counter example *e.g.* \(1 * 2 = 3\) whereas \(2 * 1 = 2\) .

(ii) **EITHER**

for example \((0 * 1) * 1 = 2 * 1 = 2\) *M1*

and \(0 * (1 * 1) = 0 * 0 = 0\) *A1*

so \( * \) is not associative *A1*

**OR**

associative if and only if \(a * (b * c) = (a * b) * c\) *M1*

which gives

\(a + 2b + 4c + 2bc + ab + 2ac + abc = a + 2b + ab + 2c + ac + 2bc + abc\) *A1*

so \( * \) is not associative as \(2ac \ne 2c + ac\) , in general *A1*

*[5 marks]*

* *

(c) *x* = 0 is a solution *A2*

*x* = 2 is a solution *A2*

*[4 marks]*

*Total [13 marks]*

## Examiners report

This question was generally well answered.

## Question

Let *c *be a positive, real constant. Let *G *be the set \(\{ \left. {x \in \mathbb{R}} \right| – c < x < c\} \) . The binary operation \( * \) is defined on the set *G *by \(x * y = \frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}}\).

Simplify \(\frac{c}{2} * \frac{{3c}}{4}\) .

State the identity element for *G *under \( * \).

For \(x \in G\) find an expression for \({x^{ – 1}}\) (the inverse of *x *under \( * \)).

Show that the binary operation \( * \) is commutative on *G *.

Show that the binary operation \( * \) is associative on *G *.

(i) If \(x,{\text{ }}y \in G\) explain why \((c – x)(c – y) > 0\) .

(ii) Hence show that \(x + y < c + \frac{{xy}}{c}\) .

Show that *G *is closed under \( * \).

Explain why \(\{ G, * \} \) is an Abelian group.

**Answer/Explanation**

## Markscheme

\(\frac{c}{2} * \frac{{3c}}{4} = \frac{{\frac{c}{2} + \frac{{3c}}{4}}}{{1 + \frac{1}{2} \cdot \frac{3}{4}}}\) *M1*

\( = \frac{{\frac{{5c}}{4}}}{{\frac{{11}}{8}}} = \frac{{10c}}{{11}}\) *A1*

*[2 marks]*

identity is 0 *A1*

*[1 mark]*

inverse is –*x* *A1*

*[1 mark]*

\(x * y = \frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}},{\text{ }}y * x = \frac{{y + x}}{{1 + \frac{{yx}}{{{c^2}}}}}\) *M1*

(since ordinary addition and multiplication are commutative)

\(x * y = y * x{\text{ so }} * \) is commutative *R1*

**Note: **Accept arguments using symmetry.

* *

*[2 marks]*

\((x * y) * z = \frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}} * z = \frac{{\left( {\frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}}} \right) + z}}{{1 + \left( {\frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}}} \right)\frac{z}{{{c^2}}}}}\) *M1*

\( = \frac{{\frac{{\left( {x + y + z + \frac{{xyz}}{{{c^2}}}} \right)}}{{\left( {1 + \frac{{xy}}{{{c^2}}}} \right)}}}}{{\frac{{\left( {1 + \frac{{xy}}{{{c^2}}} + \frac{{xz}}{{{c^2}}} + \frac{{yz}}{{{c^2}}}} \right)}}{{\left( {1 + \frac{{xy}}{{{c^2}}}} \right)}}}} = \frac{{\left( {x + y + z + \frac{{xyz}}{{{c^2}}}} \right)}}{{\left( {1 + \left( {\frac{{xy + xz + yz}}{{{c^2}}}} \right)} \right)}}\) *A1*

\(x * (y * z) = x * \left( {\frac{{y + z}}{{1 + \frac{{yz}}{{{c^2}}}}}} \right) = \frac{{x + \left( {\frac{{y + z}}{{1 + \frac{{yz}}{{{c^2}}}}}} \right)}}{{1 + \frac{x}{{{c^2}}}\left( {\frac{{y + z}}{{1 + \frac{{yz}}{{{c^2}}}}}} \right)}}\)

\( = \frac{{\frac{{\left( {x + \frac{{xyz}}{{{c^2}}} + y + z} \right)}}{{\left( {1 + \frac{{yz}}{{{c^2}}}} \right)}}}}{{\frac{{\left( {1 + \frac{{yz}}{{{c^2}}} + \frac{{xy}}{{{c^2}}} + \frac{{xz}}{{{c^2}}}} \right)}}{{\left( {1 + \frac{{yz}}{{{c^2}}}} \right)}}}} = \frac{{\left( {x + y + z + \frac{{xyz}}{{{c^2}}}} \right)}}{{\left( {1 + \left( {\frac{{xy + xz + yz}}{{{c^2}}}} \right)} \right)}}\) **A1**

since both expressions are the same \( * \) is associative *R1*

**Note**: After the initial ** M1A1**, correct arguments using symmetry also gain full marks.

* *

*[4 marks]*

(i) \(c > x{\text{ and }}c > y \Rightarrow c – x > 0{\text{ and }}c – y > 0 \Rightarrow (c – x)(c – y) > 0\) *R1AG*

* *

(ii) \({c^2} – cx – cy + xy > 0 \Rightarrow {c^2} + xy > cx + cy \Rightarrow c + \frac{{xy}}{c} > x + y{\text{ (as }}c > 0)\)

so \(x + y < c + \frac{{xy}}{c}\) *M1AG*

*[2 marks]*

if \(x,{\text{ }}y \in G{\text{ then }} – c – \frac{{xy}}{c} < x + y < c + \frac{{xy}}{c}\)

thus \( – c\left( {1 + \frac{{xy}}{{{c^2}}}} \right) < x + y < c\left( {1 + \frac{{xy}}{{{c^2}}}} \right){\text{ and }} – c < \frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}} < c\) *M1*

\(({\text{as }}1 + \frac{{xy}}{{{c^2}}} > 0){\text{ so }} – c < x * y < c\) *A1*

proving that *G *is closed under \( * \) *AG*

*[2 marks]*

as \(\{ G, * \} \) is closed, is associative, has an identity and all elements have an inverse *R1*

it is a group *AG*

as \( * \) is commutative *R1*

it is an Abelian group *AG*

*[2 marks]*

## Examiners report

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully – with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully – with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully – with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

## Question

The binary operation \( * \) is defined on \(\mathbb{N}\) by \(a * b = 1 + ab\).

Determine whether or not \( * \)

is closed;

is commutative;

is associative;

has an identity element.

**Answer/Explanation**

## Markscheme

\( * \) is closed *A1*

because \(1 + ab \in \mathbb{N}\) (when \(a,b \in \mathbb{N}\)) *R1*

*[2 marks]*

consider

\(a * b = 1 + ab = 1 + ba = b * a\) *M1A1*

therefore \( * \) is commutative

*[2 marks]*

**EITHER**

\(a * (b * c) = a * (1 + bc) = 1 + a(1 + bc){\text{ }}( = 1 + a + abc)\) *A1*

\((a * b) * c = (1 + ab) * c = 1 + c(1 + ab){\text{ }}( = 1 + c + abc)\) *A1*

(these two expressions are unequal when \(a \ne c\)) so \( * \) is not associative *R1*

**OR**

proof by counter example, for example

\(1 * (2 * 3) = 1 * 7 = 8\) *A1*

\((1 * 2) * 3 = 3 * 3 = 10\) *A1*

(these two numbers are unequal) so \( * \) is not associative *R1*

*[3 marks]*

let *e* denote the identity element; so that

\(a * e = 1 + ae = a\) gives \(e = \frac{{a – 1}}{a}\) (where \(a \ne 0\)) *M1*

then any valid statement such as: \(\frac{{a – 1}}{a} \notin \mathbb{N}\) or *e* is not unique *R1*

there is therefore no identity element *A1*

**Note:** Award the final ** A1** only if the previous

**is awarded.**

*R1** *

*[3 marks]*

## Examiners report

For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

## Question

The binary operation \(\Delta\) is defined on the set \(S =\) {1, 2, 3, 4, 5} by the following Cayley table.

(a) State whether *S *is closed under the operation Δ and justify your answer.

(b) State whether Δ is commutative and justify your answer.

(c) State whether there is an identity element and justify your answer.

(d) Determine whether Δ is associative and justify your answer.

(e) Find the solutions of the equation \(a\Delta b = 4\Delta b\), for \(a \ne 4\).

**Answer/Explanation**

## Markscheme

(a) yes *A1*

because the Cayley table only contains elements of *S **R1*

*[2 marks]*

* *

(b) yes *A1*

because the Cayley table is symmetric *R1*

*[2 marks]*

* *

(c) no *A1*

because there is no row (and column) with 1, 2, 3, 4, 5 *R1*

*[2 marks]*

* *

(d) attempt to calculate \((a\Delta b)\Delta c\) and \(a\Delta (b\Delta c)\) for some \(a,{\text{ }}b,{\text{ }}c \in S\) *M1*

counterexample: for example, \((1\Delta 2)\Delta 3 = 2\)

\(1\Delta (2\Delta 3) = 1\) *A1*

Δ is not associative *A1*

**Note: **Accept a correct evaluation of \((a\Delta b)\Delta c\) and \(a\Delta (b\Delta c)\) for some \(a,{\text{ }}b,{\text{ }}c \in S\) for the ** M1**.

**[3 marks]**

** **

(e) for example, attempt to enumerate \(4\Delta b\) for *b* = 1, 2, 3, 4, 5 and obtain (3, 2, 1, 4, 1) *(M1)*

find \((a,{\text{ }}b) \in \left\{ {{\text{(2, 2), (2, 3)}}} \right\}\) for \(a \ne 4\) (or equivalent) *A1A1*

**Note:** Award ** M1A1A0 **if extra ‘solutions’ are listed.

*[3 marks]*

* *

*Total [12 marks]*

## Examiners report

## Question

The binary operations \( \odot \) and \( * \) are defined on \({\mathbb{R}^ + }\) by

\[a \odot b = \sqrt {ab} {\text{ and }}a * b = {a^2}{b^2}.\]

Determine whether or not

\( \odot \) is commutative;

\( * \) is associative;

\( * \) is distributive over \( \odot \) ;

\( \odot \) has an identity element.

**Answer/Explanation**

## Markscheme

\(a \odot b = \sqrt {ab} = \sqrt {ba} = b \odot a\) *A1*

since \(a \odot b = b \odot a\) it follows that \( \odot \) is commutative *R1*

*[2 marks]*

\(a * (b * c) = a * {b^2}{c^2} = {a^2}{b^4}{c^4}\) *M1A1*

\((a * b) * c = {a^2}{b^2} * c = {a^4}{b^4}{c^2}\) *A1*

these are different, therefore \( * \) is not associative *R1*

**Note:** Accept numerical counter-example.

* *

*[4 marks]*

\(a * (b \odot c) = a * \sqrt {bc} = {a^2}bc\) *M1A1*

\((a * b) \odot (a * c) = {a^2}{b^2} \odot {a^2}{c^2} = {a^2}bc\) *A1*

these are equal so \( * \) is distributive over \( \odot \) *R1*

*[4 marks]*

the identity e would have to satisfy

\(a \odot e = a\) for all *a* *M1*

now \(a \odot e = \sqrt {ae} = a \Rightarrow e = a\) *A1*

therefore there is no identity element *A1*

*[3 marks]*

## Examiners report

[N/A]

[N/A]

[N/A]

[N/A]