# IB DP Maths Topic 8.2 Relations: equivalence relations; equivalence classes. HL Paper 3

## Question

The relation R is defined on ordered pairs by

$(a,{\text{ }}b)R(c,{\text{ }}d){\text{ if and only if }}ad = bc{\text{ where }}a,{\text{ }}b,{\text{ }}c,{\text{ }}d \in {\mathbb{R}^ + }.$

(a)     Show that R is an equivalence relation.

(b)     Describe, geometrically, the equivalence classes.

## Markscheme

(a)     Reflexive: $$(a,{\text{ }}b)R(a,{\text{ }}b)$$ because $$ab = ba$$     R1

Symmetric: $$(a,{\text{ }}b)R(c,{\text{ }}d) \Rightarrow ad = bc \Rightarrow cb = da \Rightarrow (c,{\text{ }}d)R(a,{\text{ }}b)$$     M1A1

Transitive: $$(a,{\text{ }}b)R(c,{\text{ }}d) \Rightarrow ad = bc$$     M1

$$(c,{\text{ }}d)R(e,{\text{ }}f) \Rightarrow cf = de$$

Therefore

$$\frac{{ad}}{{de}} = \frac{{bc}}{{cf}}{\text{ so }}af = be$$     A1

It follows that $$(a,{\text{ }}b)R(e,{\text{ }}f)$$     R1

[6 marks]

(b)     $$(a,{\text{ }}b)R(c,{\text{ }}d) \Rightarrow \frac{a}{b} = \frac{c}{d}$$     (M1)

Equivalence classes are therefore points lying, in the first quadrant, on straight lines through the origin.     A2

Notes: Accept a correct sketch.

Award A1 if “in the first quadrant” is omitted.

Do not penalise candidates who fail to exclude the origin.

[3 marks]

Total [9 marks]

## Examiners report

Part (a) was well answered by many candidates although some misunderstandings of the terminology were seen. Some candidates appeared to believe, incorrectly, that reflexivity was something to do with $$(a,{\text{ }}a)R(a,{\text{ }}a)$$ and some candidates confuse the terms ‘reflexive’ and ‘symmetric’. Many candidates were unable to describe the equivalence classes geometrically.

## Question

The relation aRb is defined on {1, 2, 3, 4, 5, 6, 7, 8, 9} if and only if ab is the square of a positive integer.

(i)     Show that R is an equivalence relation.

(ii)     Find the equivalence classes of R that contain more than one element.

[10]
a.

Given the group $$(G,{\text{ }} * )$$, a subgroup $$(H,{\text{ }} * )$$ and $$a,{\text{ }}b \in G$$, we define $$a \sim b$$ if and only if $$a{b^{ – 1}} \in H$$. Show that $$\sim$$ is an equivalence relation.

[9]
b.

## Markscheme

(i)     $$aRa \Rightarrow a \cdot a = {a^2}$$ so R is reflexive     A1

$$aRb = {m^2} \Rightarrow bRa$$ so R is symmetric     A1

$$aRb = ab = {m^2}{\text{ and }}bRc = bc = {n^2}$$     M1A1

so $$a = \frac{{{m^2}}}{b}{\text{ and }}c = \frac{{{n^2}}}{b}$$

$$ac = \frac{{{m^2}{n^2}}}{{{b^2}}} = {\left( {\frac{{mn}}{b}} \right)^2},$$     A1

ac is an integer hence $${\left( {\frac{{mn}}{b}} \right)^2}$$ is an integer     R1

so aRc, hence R is transitive     R1

R is therefore an equivalence relation     AG

(ii)     1R4 and 4R9 or 2R8     M1

so {1, 4, 9} is an equivalence class     A1

and {2, 8} is an equivalence class     A1

[10 marks]

a.

$$a \sim a{\text{ since }}a{a^{ – 1}} = e \in H$$, the identity must be in H since it is a subgroup.     M1

Hence reflexivity.     R1

$$a \sim b \Leftrightarrow a{b^{ – 1}} \in H$$ but H is a subgroup so it must contain $${(a{b^{ – 1}})^{ – 1}} = b{a^{ – 1}}$$     M1R1

i.e. $$b{a^{ – 1}} \in H{\text{ so }} \sim$$ is symmetric     A1

$$a \sim b{\text{ and }}b \sim c \Rightarrow a{b^{ – 1}} \in H{\text{ and }}b{c^{ – 1}} \in H$$     M1

But H is closed, so

$$(a{b^{ – 1}})(b{c^{ – 1}}) \in H{\text{ or }}a({b^{ – 1}}b){c^{ – 1}} \in H$$     R1

$$a{c^{ – 1}} \in H \Rightarrow a \sim c$$     A1

Hence $$\sim$$ is transitive and is thus an equivalence relation     R1AG

[9 marks]

b.

## Examiners report

Not a difficult question although using the relation definition to fully show transitivity was not well done. It was good to see some students use an operation binary matrix to show transitivity. This was a nice way given that the set was finite. The proof in (b) proved difficult.

a.

Not a difficult question although using the relation definition to fully show transitivity was not well done. It was good to see some students use an operation binary matrix to show transitivity. This was a nice way given that the set was finite. The proof in (b) proved difficult.

b.

## Question

The relation R is defined on $$\mathbb{Z} \times \mathbb{Z}$$ such that $$(a,{\text{ }}b)R(c,{\text{ }}d)$$ if and only if ac is divisible by 3 and bd is divisible by 2.

(a)     Prove that R is an equivalence relation.

(b)     Find the equivalence class for (2, 1) .

(c)     Write down the five remaining equivalence classes.

## Markscheme

(a)     consider $$(x,{\text{ }}y)R(x,{\text{ }}y)$$

since xx = 0 and yy = 0 , R is reflexive     A1

assume $$(x,{\text{ }}y)R(a,{\text{ }}b)$$

$$\Rightarrow x – a = 3M$$ and $$y – b = 2N$$     M1

$$\Rightarrow a – x = – 3M$$ and $$b – y = – 2N$$     A1

$$\Rightarrow (a,{\text{ }}b)R(x,{\text{ }}y)$$

hence R is symmetric

assume $$(x,{\text{ }}y)R(a,{\text{ }}b)$$

$$\Rightarrow x – a = 3M$$ and $$y – b = 2N$$

assume $$(a,{\text{ }}b)R(c,{\text{ }}d)$$

$$\Rightarrow a – c = 3P$$ and $$b – d = 2Q$$     M1

$$\Rightarrow x – c = 3(M + P)$$ and $$y – d = 2(N + Q)$$     A1

hence $$(x,{\text{ }}y)R(c,{\text{ }}d)$$     A1

hence R is transitive

therefore R is an equivalence relation     AG

[7 marks]

(b)     $$\left\{ {(x,{\text{ }}y):x = 3m + 2,{\text{ }}y = 2n + 1,{\text{ }}m,{\text{ }}n \in \mathbb{Z}} \right\}$$     A1A1

[2 marks]

(c)     $$\{ 3m,{\text{ }}2n\} {\text{ \{ }}3m + 1,{\text{ }}2n\} {\text{ }}\{ 3m + 2,{\text{ }}2n\}$$

$$\{ 3m,{\text{ }}2n + 1\} {\text{ \{ }}3m + 1,{\text{ }}2n + 1\} {\text{ }}m,{\text{ }}n \in \mathbb{Z}$$     A1A1A1A1A1

[5 marks]

Total [14 marks]

## Examiners report

Stronger candidates had little problem with part (a) of this question, but proving an equivalence relation is still difficult for many. Equivalence classes still cause major problems and few fully correct answers were seen to this question.

## Question

The relations R and S are defined on quadratic polynomials P of the form

$P(z) = {z^2} + az + b{\text{ , where }}a{\text{ , }}b \in \mathbb{R}{\text{ , }}z \in \mathbb{C}{\text{ .}}$

(a)     The relation R is defined by $${P_1}R{P_2}$$ if and only if the sum of the two zeros of $${P_1}$$ is equal to the sum of the two zeros of $${P_2}$$ .

(i)     Show that R is an equivalence relation.

(ii)     Determine the equivalence class containing $${z^2} – 4z + 5$$ .

(b)     The relation S is defined by $${P_1}S{P_2}$$ if and only if $${P_1}$$ and $${P_2}$$ have at least one zero in common. Determine whether or not S is transitive.

## Markscheme

(a)     (i)     R is reflexive, i.e. PRP because the sum of the zeroes of P is equal to the sum of the zeros of P     R1

R is symmetric, i.e. $${P_1}R{P_2} \Rightarrow {P_2}R{P_1}$$ because the sums of the zeros of $${P_1}$$ and $${P_2}$$ are equal implies that the sums of the zeros of $${P_2}$$ and $${P_1}$$ are equal     R1

suppose that $${P_1}R{P_2}$$ and $${P_2}R{P_3}$$     M1

it follows that $${P_1}R{P_3}$$ so R is transitive, because the sum of the zeros of $${P_1}$$ is equal to the sum of the zeros of $${P_2}$$ which in turn is equal to the sum of the zeros of $${P_3}$$ , which implies that the sum of the zeros of $${P_1}$$ is equal to the sum of the zeros of $${P_3}$$     R1

the three requirements for an equivalence relation are therefore satisfied     AG

(ii)     the zeros of $${z^2} – 4z + 5$$ are $$2 \pm {\text{i}}$$ , for which the sum is 4     M1A1

$${{\text{z}}^2} + az + b$$ has zeros of $$\frac{{ – a \pm \sqrt {{a^2} – 4b} }}{2}$$ , so the sum is –a     (M1)

Note: Accept use of the result (although not in the syllabus) that the sum of roots is minus the coefficient of z.

hence – a = 4 and so a = – 4     A1

the equivalence class is $${z^2} – 4z + k{\text{ , }}(k \in \mathbb{R})$$     A1

[9 marks]

(b)     for example, $$(z – 1)(z – 2)S(z – 1)(z – 3)$$ and

$$(z – 1)(z – 3)S(z – 3)(z – 4)$$ but $$(z – 1)(z – 2)S(z – 3)(z – 4)$$ is not true     M1A1

so S is not transitive     A1

[3 marks]

Total [12 marks]

## Examiners report

Most candidates were able to show, in (a), that R is an equivalence relation although few were able to identify the required equivalence class. In (b), the explanation that S is not transitive was often unconvincing.

## Question

Let R be a relation on the set $$\mathbb{Z}$$ such that $$aRb \Leftrightarrow ab \geqslant 0$$, for a, b $$\in \mathbb{Z}$$.

(a)     Determine whether R is

(i)     reflexive;

(ii)     symmetric;

(iii)     transitive.

(b)     Write down with a reason whether or not R is an equivalence relation.

## Markscheme

(a)     (i)     $${a^2} \geqslant 0$$ for all $$a \in \mathbb{Z}$$, hence R is reflexive     R1

(ii)     $$aRb \Rightarrow ab \geqslant 0$$     M1

$$\Rightarrow ba \geqslant 0$$     R1

$$\Rightarrow bRa$$, hence R is symmetric     A1

(iii)     $$aRb{\text{ and }}bRc \Rightarrow ab \geqslant 0{\text{ and }}bc \geqslant 0,{\text{ is }}aRc?$$     M1

no, for example, $$– 3R0$$ and $$0R5$$, but $$– 3R5$$ is not true     A1

aRc is not generally true, hence R is not transitive     A1

[7 marks]

(b)     R does not satisfy all three properties, hence R is not an equivalence relation     R1

[1 mark]

Total [8 marks]

## Examiners report

Although the properties of an equivalence relation were well known, few candidates provided a counter-example to show that the relation is not transitive. Some candidates interchanged the definitions of the reflexive and symmetric properties.

## Question

The relation R is defined for a , $$b \in {\mathbb{Z}^ + }$$ such that aRb if and only if $${a^2} – {b^2}$$ is divisible by 5.

Show that R is an equivalence relation.

[6]
a.

Identify the three equivalence classes.

[4]
b.

## Markscheme

reflexive: aRa because $${a^2} – {a^2} = 0$$ (which is divisible by 5)     A1

symmetric: let aRb so that $${a^2} – {b^2} = 5M$$     M1

it follows that $${a^2} – {b^2} = – 5M$$ which is divisible by 5 so bRa     A1

transitive: let aRb and bRc so that $${a^2} – {b^2} = 5M$$ and $${b^2} – {c^2} = 5N$$     M1

$${a^2} – {b^2} + {b^2} – {c^2} = 5M + 5N$$     A1

$${a^2} – {c^2} = 5M + 5N$$ which is divisible by 5 so aRc     A1

$$\Rightarrow$$ R is an equivalence relation     AG

[6 marks]

a.

the equivalence classes are

{1, 4, 6, 9, …}     A2

{2, 3, 7, 8, …}     A1

{5, 10, …}     A1

Note: Do not award any marks for classes containing fewer elements than shown above.

[4 marks]

b.

## Examiners report

Many candidates solved (a) correctly but solutions to (b) were generally poor. Most candidates seemed to have a weak understanding of the concept of equivalence classes and were unaware of any systematic method for finding the equivalence classes. If all else fails, a trial and error approach can be used. Here, starting with 1, it is easily seen that 4, 6,… belong to the same class and the pattern can be established.

a.

Many candidates solved (a) correctly but solutions to (b) were generally poor. Most candidates seemed to have a weak understanding of the concept of equivalence classes and were unaware of any systematic method for finding the equivalence classes. If all else fails, a trial and error approach can be used. Here, starting with 1, it is easily seen that 4, 6, … belong to the same class and the pattern can be established.

b.

## Question

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

Show that R is an equivalence relation.

[8]
a.

The Cayley table for G is shown below.

The subgroup H is given as $$H = \{ e,{\text{ }}{a^2}b\}$$.

(i)     Find the equivalence class with respect to R which contains ab.

(ii)     Another equivalence relation $$\rho$$ is defined on G by $$x\rho y$$ if and only if $${x^{ – 1}}y \in H$$, for $$x,{\text{ }}y \in G$$. Find the equivalence class with respect to $$\rho$$ which contains ab.

[6]
b.

## Markscheme

$$x{x^{ – 1}} = e \in H$$     M1

$$\Rightarrow xRx$$

hence R is reflexive     A1

if xRy then $$x{y^{ – 1}} \in H$$

$$\Rightarrow {(x{y^{ – 1}})^{ – 1}} \in H$$     M1

now $$(x{y^{ – 1}}){(x{y^{ – 1}})^{ – 1}} = e$$ and $$x{y^{ – 1}}y{x^{ – 1}} = e$$

$$\Rightarrow {(x{y^{ – 1}})^{ – 1}} = y{x^{ – 1}}$$     A1

hence $$y{x^{ – 1}} \in H \Rightarrow yRx$$

hence R is symmetric     A1

if xRy, yRz then $$x{y^{ – 1}} \in H,{\text{ }}y{z^{ – 1}} \in H$$     M1

$$\Rightarrow (x{y^{ – 1}})(y{z^{ – 1}}) \in H$$     M1

$$\Rightarrow x({y^{ – 1}}y){z^{ – 1}} \in H$$

$$\Rightarrow {x^{ – 1}}z \in H$$

hence R is transitive     A1

hence R is an equivalence relation     AG

[8 marks]

a.

(i)     for the equivalence class, solving:

EITHER

$$x{(ab)^{ – 1}} = e{\text{ or }}x{(ab)^{ – 1}} = {a^2}b$$     (M1)

$$\{ ab,{\text{ }}a\}$$     A2

OR

$$ab{(x)^{ – 1}} = e{\text{ or }}ab{(x)^{ – 1}} = {a^2}b$$     (M1)

$$\{ ab,{\text{ }}a\}$$     A2

(ii)     for the equivalence class, solving:

EITHER

$${x^{ – 1}}(ab) = e{\text{ or }}{x^{ – 1}}(ab) = {a^2}b$$     (M1)

$$\{ ab,{\text{ }}{a^2}\}$$     A2

OR

$${(ab)^{ – 1}}x = e{\text{ or }}{(ab)^{ – 1}}x = {a^2}b$$     (M1)

$$\{ ab,{\text{ }}{a^2}\}$$     A2

[6 marks]

b.

## Examiners report

Stronger candidates made a reasonable start to (a), and many were able to demonstrate that the relation was reflexive and transitive. However, the majority of candidates struggled to make a meaningful attempt to show the relation was symmetric, with many making unfounded assumptions. Equivalence classes still cause major problems and few fully correct answers were seen to (b).

a.

Stronger candidates made a reasonable start to (a), and many were able to demonstrate that the relation was reflexive and transitive. However, the majority of candidates struggled to make a meaningful attempt to show the relation was symmetric, with many making unfounded assumptions. Equivalence classes still cause major problems and few fully correct answers were seen to (b).

b.

## Question

The relation R is defined on the set $$\mathbb{N}$$ such that for $$a{\text{ }},{\text{ }}b \in \mathbb{N}{\text{ }},{\text{ }}aRb$$ if and only if $${a^3} \equiv {b^3}(\bmod 7)$$.

Show that R is an equivalence relation.

[6]
a.

Find the equivalence class containing 0.

[2]
b.

Denote the equivalence class containing n by Cn .

List the first six elements of $${C_1}$$.

[3]
c.

Denote the equivalence class containing n by Cn .

Prove that $${C_n} = {C_{n + 7}}$$ for all $$n \in \mathbb{N}$$.

[3]
d.

## Markscheme

reflexive: $${a^3} – {a^3} = 0{\text{ }},{\text{ }} \Rightarrow R$$ is reflexive     R1

symmetric: if $${a^3} \equiv {b^3}(\bmod 7)$$ , then $${b^3} \equiv {a^3}(\bmod 7)$$     M1

$$\Rightarrow R$$ is symmetric     R1

transitive: $${a^3} = {b^3} + 7n$$ and $${b^3} = {c^3} + 7m$$     M1

then $${a^3} = {c^3} + 7(n + m)$$

$$\Rightarrow {a^3} \equiv {c^3}(\bmod 7)$$     R1

$$\Rightarrow R$$ is transitive     A1

and is an equivalence relation     AG

Note: Allow arguments that use $${a^3} – {b^3} \equiv 0(\bmod 7)$$ etc.

[6 marks]

a.

$$\{ 0,{\text{ }}7,{\text{ }}14,{\text{ }}21,{\text{ }}…\}$$     A2

[2 marks]

b.

$$\{ 1,{\text{ }}2,{\text{ }}4,{\text{ }}8,{\text{ }}9,{\text{ }}11\}$$     A3

Note: Deduct 1 mark for each error or omission.

[3 marks]

c.

consider $${(n + 7)^3} = {n^3} + 21{n^2} + 147n + 343 = {n^3} + 7N$$     M1A1

$$\Rightarrow {n^3} \equiv {(n + 7)^3}(\bmod 7) \Rightarrow n$$ and $$n + 7$$ are in the same equivalence class     R1

[3 marks]

d.

## Examiners report

Candidates were mostly aware of the conditions required to show an equivalence relation although many seemed unsure as to the degree of detail required to show that the different conditions are met for the example in this question. In part (b) many candidates found the correct set although a number were unable to write down the set correctly, including or excluding elements that were not part of the equivalence class. Part (c) saw candidate being less successful than (b) and relatively few candidates were able to prove the equivalence class in part (d) although there were a number of very good solutions.

a.

Candidates were mostly aware of the conditions required to show an equivalence relation although many seemed unsure as to the degree of detail required to show that the different conditions are met for the example in this question. In part (b) many candidates found the correct set although a number were unable to write down the set correctly, including or excluding elements that were not part of the equivalence class. Part (c) saw candidate being less successful than (b) and relatively few candidates were able to prove the equivalence class in part (d) although there were a number of very good solutions.

b.

Candidates were mostly aware of the conditions required to show an equivalence relation although many seemed unsure as to the degree of detail required to show that the different conditions are met for the example in this question. In part (b) many candidates found the correct set although a number were unable to write down the set correctly, including or excluding elements that were not part of the equivalence class. Part (c) saw candidate being less successful than (b) and relatively few candidates were able to prove the equivalence class in part (d) although there were a number of very good solutions.

c.

Candidates were mostly aware of the conditions required to show an equivalence relation although many seemed unsure as to the degree of detail required to show that the different conditions are met for the example in this question. In part (b) many candidates found the correct set although a number were unable to write down the set correctly, including or excluding elements that were not part of the equivalence class. Part (c) saw candidate being less successful than (b) and relatively few candidates were able to prove the equivalence class in part (d) although there were a number of very good solutions.

d.

## Question

All of the relations in this question are defined on $$\mathbb{Z}\backslash \{ 0\}$$.

Decide, giving a proof or a counter-example, whether $$xRy \Leftrightarrow x + y > 7$$ is

(i)     reflexive;

(ii)     symmetric;

(iii)     transitive.

[4]
a.

Decide, giving a proof or a counter-example, whether $$xRy \Leftrightarrow – 2 < x – y < 2$$ is

(i)     reflexive;

(ii)     symmetric;

(iii)     transitive.

[4]
b.

Decide, giving a proof or a counter-example, whether $$xRy \Leftrightarrow xy > 0$$ is

(i)     reflexive;

(ii)     symmetric;

(iii)     transitive.

[4]
c.

Decide, giving a proof or a counter-example, whether $$xRy \Leftrightarrow \frac{x}{y} \in \mathbb{Z}$$ is

(i)     reflexive;

(ii)     symmetric;

(iii)     transitive.

[4]
d.

One of the relations from parts (a), (b), (c) and (d) is an equivalence relation.

For this relation, state what the equivalence classes are.

[3]
e.

## Markscheme

(i)     not reflexive e.g. 1 + 1 = 2     R1

(ii)     symmetric since x + y = y + x     R1

(iii)     e.g. 1 + 11 > 7, 11 + 2 > 7 but 1 + 2 = 3, so not transitive     M1A1

Note: For each R1 mark the correct decision and a valid reason must be given.

[4 marks]

a.

(i)     reflexive since $$x – x = 0$$     R1

(ii)     symmetric since $$\left| {x – y} \right| = \left| {y – x} \right|$$     R1

(iii)     e.g. 1R2, 2R3 but 1 − 3 = −2 , so not transitive     M1A1

Note: For each R1 mark the correct decision and a valid reason must be given.

[4 marks]

b.

(i)     reflexive since $${x^2} > 0$$     R1

(ii)     symmetric since $$xy = yx$$     R1

(iii)     $$xy > 0{\text{ and }}yz > 0 \Rightarrow x{y^2}z > 0 \Rightarrow xz > 0{\text{ since }}{y^2} > 0$$, so transitive     M1A1

Note: For each R1 mark the correct decision and a valid reason must be given.

[4 marks]

c.

(i)     reflexive since $$\frac{x}{x} = 1$$     R1

(ii)     not symmetric e.g. $$\frac{2}{1} = 2{\text{ but }}\frac{1}{2} = 0.5$$     R1

(iii)     $$\frac{x}{y} \in \mathbb{Z}{\text{ and }}\frac{y}{z} \in \mathbb{Z} \Rightarrow \frac{{xy}}{{yz}} = \frac{x}{z} \in \mathbb{Z}$$, so transitive     M1A1

Note: For each R1 mark the correct decision and a valid reason must be given.

[4 marks]

d.

only (c) is an equivalence relation     (A1)

the equivalence classes are

{1, 2, 3,…} and {−1,−2,−3,…}     A1A1

[3 marks]

e.

## Examiners report

Generally this question was well answered, with students showing a sound knowledge of relations. There were a few candidates who mixed reflexive and symmetric qualities and marks were also lost because reasoning was either unclear or absent. Most students were able to offer counterexamples for transitivity in parts (a) and (b) but a number lost marks in failing to give adequate working to show transitivity in parts (c) and (d). That said, there were a pleasing number of good solutions here showing all the required rigour. Whilst most students were able to identify part (c) as an equivalence relation, surprisingly few gave the correct equivalence classes.

a.

Generally this question was well answered, with students showing a sound knowledge of relations. There were a few candidates who mixed reflexive and symmetric qualities and marks were also lost because reasoning was either unclear or absent. Most students were able to offer counterexamples for transitivity in parts (a) and (b) but a number lost marks in failing to give adequate working to show transitivity in parts (c) and (d). That said, there were a pleasing number of good solutions here showing all the required rigour. Whilst most students were able to identify part (c) as an equivalence relation, surprisingly few gave the correct equivalence classes.

b.

Generally this question was well answered, with students showing a sound knowledge of relations. There were a few candidates who mixed reflexive and symmetric qualities and marks were also lost because reasoning was either unclear or absent. Most students were able to offer counterexamples for transitivity in parts (a) and (b) but a number lost marks in failing to give adequate working to show transitivity in parts (c) and (d). That said, there were a pleasing number of good solutions here showing all the required rigour. Whilst most students were able to identify part (c) as an equivalence relation, surprisingly few gave the correct equivalence classes.

c.

Generally this question was well answered, with students showing a sound knowledge of relations. There were a few candidates who mixed reflexive and symmetric qualities and marks were also lost because reasoning was either unclear or absent. Most students were able to offer counterexamples for transitivity in parts (a) and (b) but a number lost marks in failing to give adequate working to show transitivity in parts (c) and (d). That said, there were a pleasing number of good solutions here showing all the required rigour. Whilst most students were able to identify part (c) as an equivalence relation, surprisingly few gave the correct equivalence classes.

d.

Generally this question was well answered, with students showing a sound knowledge of relations. There were a few candidates who mixed reflexive and symmetric qualities and marks were also lost because reasoning was either unclear or absent. Most students were able to offer counterexamples for transitivity in parts (a) and (b) but a number lost marks in failing to give adequate working to show transitivity in parts (c) and (d). That said, there were a pleasing number of good solutions here showing all the required rigour. Whilst most students were able to identify part (c) as an equivalence relation, surprisingly few gave the correct equivalence classes.

e.

## Question

The relation R is defined on {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} by aRb if and only if $$a(a + 1) \equiv b(b + 1)(\bmod 5)$$.

Show that R is an equivalence relation.

[6]
a.

Show that the equivalence defining R can be written in the form

$(a – b)(a + b + 1) \equiv 0(\bmod 5).$

[3]
b.

Hence, or otherwise, determine the equivalence classes.

[4]
c.

## Markscheme

reflexive: $$a(a + 1) \equiv a(a + 1)(\bmod 5)$$, therefore aRa     R1

symmetric: $$aRb \Rightarrow a(a + 1) = b(b + 1) + 5N$$     M1

$$\Rightarrow b(b + 1) = a(a + 1) – 5N \Rightarrow bRa$$     A1

transitive:

EITHER

$$aRb{\text{ and }}bRc \Rightarrow a(a + 1) = b(b + 1) + 5M{\text{ and }}b(b + 1) = c(c + 1) + 5N$$     M1

it follows that $$a(a + 1) = c(c + 1) + 5(M + N) \Rightarrow aRc$$     M1A1

OR

$$aRb{\text{ and }}bRc \Rightarrow a(a + 1) \equiv b(b + 1)(\bmod 5){\text{ and}}$$

$$b(b + 1) \equiv c(c + 1)(\bmod 5)$$     M1

$$a(a + 1) – b(b + 1) \equiv 0(\bmod 5);{\text{ }}b(b + 1) – c(c + 1) \equiv 0(\bmod 5)$$     M1

$$a(a + 1) – c(c + 1) \equiv 0\bmod 5 \Rightarrow a(a + 1) \equiv c(c + 1)\bmod 5 \Rightarrow aRc$$     A1

[6 marks]

a.

the equivalence can be written as

$${a^2} + a – {b^2} – b \equiv 0(\bmod 5)$$     M1

$$(a – b)(a + b) + a – b \equiv 0(\bmod 5)$$     M1A1

$$(a – b)(a + b + 1) \equiv 0(\bmod 5)$$     AG

[3 marks]

b.

the equivalence classes are

{1, 3, 6, 8, 11}

{2, 7, 12}

{4, 5, 9, 10}     A4

Note: Award A3 for 2 correct classes, A2 for 1 correct class.

[4 marks]

c.

## Examiners report

Candidates knew the properties of equivalence relations but did not show sufficient working out in the transitive case. Others did not do the modular arithmetic correctly, still others omitted the $$\bmod(5)$$ in part or throughout.

a.

Candidates knew the properties of equivalence relations but did not show sufficient working out in the transitive case. Others did not do the modular arithmetic correctly, still others omitted the $$\bmod (5)$$ in part or throughout.

b.

Candidates knew the properties of equivalence relations but did not show sufficient working out in the transitive case. Others did not do the modular arithmetic correctly, still others omitted the $$\bmod(5)$$ in part or throughout.

c.

## Question

Let $$(H,{\text{ }} * {\text{)}}$$ be a subgroup of the group $$(G,{\text{ }} * {\text{)}}$$.

Consider the relation $$R$$ defined in $$G$$ by $$xRy$$ if and only if $${y^{ – 1}} * x \in H$$.

(a)     Show that $$R$$ is an equivalence relation on $$G$$.

(b)     Determine the equivalence class containing the identity element.

## Markscheme

(a)     $$R$$ is reflexive as $${x^{ – 1}} * x = e \in H \Rightarrow xRx$$ for any $$x \in G$$     A1

if $$xRy$$ then $${y^{ – 1}} * x = h \in H$$

but $$h \in H \Rightarrow {h^{ – 1}} \in H$$, ie$$\underbrace {{{({y^{ – 1}} * x)}^{ – 1}}}_{{x^{ – 1}}{\text{*}}y} \in H$$     M1

therefore $$yRx$$

$$R$$ is symmetric     A1

if $$xRy$$ then $${y^{ – 1}} * x = h \in H$$ and if $$yRz$$ then $${z^{ – 1}} * y = k \in H$$     M1

$$k * h \in H$$, ie, $$\underbrace {({z^{ – 1}} * y) * ({y^{ – 1}} * x)}_{{z^{ – 1}} * x} \in H$$     A1

therefore $$xRz$$

$$R$$ is transitive     A1

so $$R$$ is an equivalence relation on $$G$$     AG

[6 marks]

(b)     $$xRe \Leftrightarrow {e^{ – 1}} * x \in H$$     M1

$$x \in H$$     A1

$$[e] = H$$     A1     N0

[3 marks]

## Examiners report

Part (a) was fairly well answered by many candidates. They knew how to apply the equivalence relations axioms in this particular example. Part (b) however proved to be very challenging and hardly any correct answers were seen.

## Question

Consider the set S defined by $$S = \{ s \in \mathbb{Q}:2s \in \mathbb{Z}\}$$.

You may assume that $$+$$ (addition) and $$\times$$ (multiplication) are associative binary operations

on $$\mathbb{Q}$$.

(i)     Write down the six smallest non-negative elements of $$S$$.

(ii)     Show that $$\{ S,{\text{ }} + \}$$ is a group.

(iii)     Give a reason why $$\{ S,{\text{ }} \times \}$$ is not a group. Justify your answer.

[9]
a.

The relation $$R$$ is defined on $$S$$ by $${s_1}R{s_2}$$ if $$3{s_1} + 5{s_2} \in \mathbb{Z}$$.

(i)     Show that $$R$$ is an equivalence relation.

(ii)     Determine the equivalence classes.

[10]
b.

## Markscheme

(i)     $${\text{0, }}\frac{{\text{1}}}{{\text{2}}}{\text{, 1, }}\frac{{\text{3}}}{{\text{2}}}{\text{, 2, }}\frac{{\text{5}}}{{\text{2}}}$$     A2

Notes:     A2 for all correct, A1 for three to five correct.

(ii)     EITHER

closure: if $${s_1},{\text{ }}{s_2} \in S$$, then $${s_1} = \frac{m}{2}$$ and $${s_2} = \frac{n}{2}$$ for some $$m,{\text{ }}n \in {\text{¢}}$$.     M1

Note:     Accept two distinct examples (eg, $$\frac{1}{2} + \frac{1}{2} = 1;{\text{ }}\frac{1}{2} + 1 = \frac{3}{2}$$) for the M1.

$${s_1} + {s_2} = \frac{{m + n}}{2} \in S$$     A1

OR

the sum of two half-integers     A1

is a half-integer     R1

THEN

identity: 0 is the (additive) identity     A1

inverse: $$s + ( – s) = 0$$, where $$– s \in S$$     A1

it is associative (since $$S \subset \S$$)     A1

the group axioms are satisfied     AG

(iii)     EITHER

the set is not closed under multiplication,     A1

for example, $$\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$, but $$\frac{1}{4} \notin S$$     R1

OR

not every element has an inverse,     A1

for example, 3 does not have an inverse     R1

[9 marks]

a.

(i)     reflexive: consider $$3s + 5s$$     M1

$$= 8s \in {\text{¢}} \Rightarrow$$ reflexive     A1

symmetric: if $${s_1}R{s_2}$$, consider $$3{s_2} + 5{s_1}$$     M1

for example, $$= 3{s_1} + 5{s_2} + (2{s_1} – 2{s_2}) \in {\text{¢}} \Rightarrow$$symmetric     A1

transitive: if $${s_1}R{s_2}$$ and $${s_2}R{s_3}$$, consider     (M1)

$$3{s_1} + 5{s_3} = (3{s_1} + 5{s_2}) + (3{s_2} + 5{s_3}) – 8{s_2}$$     M1

$$\in {\text{¢}} \Rightarrow$$transitive     A1

so R is an equivalence relation     AG

(ii)     $${C_1} = {\text{¢}}$$     A1

$${C_2} = \left\{ { \pm \frac{1}{2},{\text{ }} \pm \frac{3}{2},{\text{ }} \pm \frac{5}{2},{\text{ }} \ldots } \right\}$$     A1A1

Note: A1 for half odd integers and A1 for ±.

[10 marks]

b.

[N/A]

a.

[N/A]

b.

## Question

$$\{ G,{\text{ }} * \}$$ is a group with identity element $$e$$. Let $$a,{\text{ }}b \in G$$.

State Lagrange’s theorem.

[2]
a.

Verify that the inverse of $$a * {b^{ – 1}}$$ is equal to $$b * {a^{ – 1}}$$.

[3]
b.

Let $$\{ H,{\rm{ }} * {\rm{\} }}$$ be a subgroup of $$\{ G,{\rm{ }} * {\rm{\} }}$$. Let $$R$$ be a relation defined on $$G$$ by

$aRb \Leftrightarrow a * {b^{ – 1}} \in H.$

Prove that $$R$$ is an equivalence relation, indicating clearly whenever you are using one of the four properties required of a group.

[8]
c.

Let $$\{ H,{\rm{ }} * {\rm{\} }}$$ be a subgroup of $$\{ G,{\rm{ }} * {\rm{\} }}$$ .Let $$R$$ be a relation defined on $$G$$ by

$aRb \Leftrightarrow a * {b^{ – 1}} \in H.$

Show that $$aRb \Leftrightarrow a \in Hb$$, where $$Hb$$ is the right coset of $$H$$ containing $$b$$.

[3]
d.

Let $$\{ H,{\rm{ }} * {\rm{\} }}$$ be a subgroup of $$\{ G,{\rm{ }} * {\rm{\} }}$$ .Let $$R$$ be a relation defined on $$G$$ by

$aRb \Leftrightarrow a * {b^{ – 1}} \in H.$

It is given that the number of elements in any right coset of $$H$$ is equal to the order of $$H$$.

Explain how this fact together with parts (c) and (d) prove Lagrange’s theorem.

[3]
e.

## Markscheme

in a finite group the order of any subgroup (exactly) divides the order of the group     A1A1

[2 marks]

a.

METHOD 1

$$(a * {b^{ – 1}}) * (b * {a^{ – 1}}) = a * {b^{ – 1}} * b * {a^{ – 1}} = a * e * {a^{ – 1}} = a * {a^{ – 1}} = e$$     M1A1A1

Note:     M1 for multiplying, A1 for at least one of the next 3 expressions,

A1 for $$e$$.

Allow $$(b * {a^{ – 1}}) * (a * {b^{ – 1}}) = b * {a^{ – 1}} * a * {b^{ – 1}} = b * e * {b^{ – 1}} = b * {b^{ – 1}} = e$$.

METHOD 2

$${(a * {b^{ – 1}})^{ – 1}} = {({b^{ – 1}})^{ – 1}} * {a^{ – 1}}$$     M1A1

$$= b * {a^{ – 1}}$$A1

[3 marks]

b.

$$a * {a^{ – 1}} = e \in H\;\;\;$$(as $$H$$ is a subgroup)     M1

so $$aRa$$ and hence $$R$$ is reflexive

$$aRb \Leftrightarrow a * {b^{ – 1}} \in H$$. $$H$$ is a subgroup so every element has an inverse in $$H$$ so

$${(a * {b^{ – 1}})^{ – 1}} \in H$$     R1

$$\Leftrightarrow b * {a^{ – 1}} \in H \Leftrightarrow bRa$$     M1

so $$R$$ is symmetric

$$aRb,{\text{ }}bRc \Leftrightarrow a * {b^{ – 1}} \in H,{\text{ }}b * {c^{ – 1}} \in H$$     M1

as $$H$$ is closed $$(a * {b^{ – 1}}) * {\text{(}}b * {c^{ – 1}}) \in H$$     R1

and using associativity     R1

$$(a * {b^{ – 1}}) * {\text{(}}b * {c^{ – 1}}) = a * ({b^{ – 1}} * b) * {c^{ – 1}} = a * {c^{ – 1}} \in H \Leftrightarrow aRc$$     A1

therefore $$R$$ is transitive

$$R$$ is reflexive, symmetric and transitive

Note:     Can be said separately at the end of each part.

hence it is an equivalence relation     AG

[8 marks]

c.

$$aRb \Leftrightarrow a * {b^{ – 1}} \in H \Leftrightarrow a * {b^{ – 1}} = h \in H$$     A1

$$\Leftrightarrow a = h * b \Leftrightarrow a \in Hb$$     M1R1

[3 marks]

d.

(d) implies that the right cosets of $$H$$ are equal to the equivalence classes of the relation in (c)     R1

hence the cosets partition $$G$$     R1

all the cosets are of the same size as the subgroup $$H$$ so the order of $$G$$ must be a multiple of $$\left| H \right|$$     R1

[3 marks]

Total [19 marks]

e.

## Examiners report

Many students obtained just half marks in (a) for not stating the requirement of the order to be finite.

a.

Part (b) should have been more straightforward than many found.

b.

In part (c) it was evident that most candidates knew what to do, but being a more difficult question fell down on a lack of rigour. Nonetheless, many candidates obtained full or partial marks on this question part.

c.

Part (d) enabled many candidates to obtain, at least partial marks, but there were few students with the insight to be able to answer part (e) satisfactorily.

d.

Part (d) enabled many candidates to obtain, at least partial marks, but there were few students with the insight to be able to answer part (e) satisfactorily.

e.

## Question

The relation $$R$$ is defined on $$\mathbb{Z}$$ by $$xRy$$ if and only if $${x^2}y \equiv y\bmod 6$$.

Show that the product of three consecutive integers is divisible by $$6$$.

[2]
a.

Hence prove that $$R$$ is reflexive.

[3]
b.

Find the set of all $$y$$ for which $$5Ry$$.

[3]
c.

Find the set of all $$y$$ for which $$3Ry$$.

[2]
d.

Using your answers for (c) and (d) show that $$R$$ is not symmetric.

[2]
e.

## Markscheme

in a product of three consecutive integers either one or two are even     R1

and one is a multiple of $$3$$     R1

so the product is divisible by $$6$$     AG

[2 marks]

a.

to test reflexivity, put $$y = x$$     M1

then $${x^2}x – x = (x – 1)x(x + 1) \equiv 0\bmod 6$$     M1A1

so $$xRx$$     AG

[3 marks]

b.

if $$5Ry$$ then $$25y \equiv y\bmod 6$$     (M1)

$$24y \equiv 0\bmod 6$$     (M1)

the set of solutions is $$\mathbb{Z}$$     A1

Note:     Only one of the method marks may be implied.

[3 marks]

c.

if $$3Ry$$ then $$9y \equiv y\bmod 6$$

$$8y \equiv 0\bmod 6 \Rightarrow 4y \equiv 0\bmod 3$$     (M1)

the set of solutions is $$3\mathbb{Z}$$ (ie multiples of $$3$$)     A1

[2 marks]

d.

from part (c) $$5R3$$     A1

from part (d) $$3R5$$ is false     A1

$$R$$ is not symmetric     AG

Note:     Accept other counterexamples.

[2 marks]

Total [12 marks]

e.

## Examiners report

A surprising number of candidates thought that an example was sufficient evidence to answer this part.

a.

Again, a lack of confidence with modular arithmetic undermined many candidates’ attempts at this part.

b.

(c) and (d) Most candidates started these parts, but some found solutions as fractions rather than integers or omitted zero and/or negative integers.

c.

(c) and (d) Most candidates started these parts, but some found solutions as fractions rather than integers or omitted zero and/or negative integers.

d.

Some candidates regarded $$R$$ as an operation, rather than a relation, so returned answers of the form $$aRb \ne bRa$$.

e.

## Question

The function $$f:\mathbb{R} \to \mathbb{R}$$ is defined as $$f:x \to \left\{ {\begin{array}{*{20}{c}} {1,{\text{ }}x \ge 0} \\ { – 1,{\text{ }}x < 0} \end{array}} \right.$$.

Prove that $$f$$ is

(i)     not injective;

(ii)     not surjective.

[4]
a.

The relation $$R$$ is defined for $$a,{\text{ }}b \in \mathbb{R}$$ so that $$aRb$$ if and only if $$f(a) \times f(b) = 1$$.

Show that $$R$$ is an equivalence relation.

[8]
b.

The relation $$R$$ is defined for $$a,{\text{ }}b \in \mathbb{R}$$ so that $$aRb$$ if and only if $$f(a) \times f(b) = 1$$.

State the equivalence classes of $$R$$.

[2]
c.

## Markscheme

(i)     eg$$\;\;\;f(2) = f(3)$$     M1

hence $$f(a) = f(b) \Rightarrow a = b$$     R1

so not injective     AG

(ii)     eg$$\;\;\;$$Codomain is $$\mathbb{R}$$ and range is $$\{ – 1,{\text{ }}1\}$$     M1

these not the same so not surjective     R1AG

Note:     if counter example is given it must be stated it is not in the range to obtain the R1. Eg $$f(x) = 2$$ has no solution as $$f(x) \in \{ – 1,{\text{ }}1\} \forall x$$.

[4 marks]

a.

if $$a \ge 0$$ then $$f(a) \times f(a) = 1 \times 1 = 1$$     A1

if $$a < 0$$ then $$f(a) \times f(a) = – 1 \times – 1 = 1$$     A1

in either case $$aRa$$ so $$R$$ is reflexive     R1

$$aRb \Rightarrow f(a) \times f(b) = 1 \Rightarrow f(b) \times f(a) = 1 \Rightarrow bRa$$     A1

so $$R$$ is symmetric     R1

if $$aRb$$ then either $$a \ge 0$$ and $$b \ge 0$$ or $$a < 0$$ and $$b < 0$$

if $$a \ge 0$$ and $$b \ge 0$$ and $$bRc$$ then $$c \ge 0$$ so $$f(a) \times f(c) = 1 \times 1 = 1$$ and $$aRc$$     A1

if $$a < 0$$ and $$b < 0$$ and $$bRc$$ then $$c < 0$$ so $$f(a) \times f(c) = – 1 \times – 1 = 1$$ and $$aRc$$     A1

in either case $$aRb$$ and $$bRc \Rightarrow aRc$$ so $$R$$ is transitive     R1

Note:     Accept

$$f(a) \times f(b) \times f(b) \times f(c) = 1 \times 1 = 1 \Rightarrow f(a) \times 1 \times {\text{ }}f(c) = 1 \Rightarrow {\text{ }}f(a) \times f(c) = 1$$

Note:     for each property just award R1 if at least one of the A marks is awarded.

as $$R$$ is reflexive, symmetric and transitive it is an equivalence relation     AG

[8 marks]

b.

equivalence classes are $$[0,{\text{ }}\infty [$$ and $$] – \infty ,{\text{ }}0[$$     A1A1

Note:     Award A1A0 for both intervals open.

[2 marks]

Total [14 marks]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.

## Question

The relation $$R=$$ is defined on $${\mathbb{Z}^ + }$$ such that $$aRb$$ if and only if $${b^n} – {a^n} \equiv 0(\bmod p)$$ where $$n,{\text{ }}p$$ are fixed positive integers greater than 1.

Show that $$R$$ is an equivalence relation.

[7]
a.

Given that $$n = 2$$ and $$p = 7$$, determine the first four members of each of the four equivalence classes of $$R$$.

[5]
b.

## Markscheme

since $${a^n} – {a^n} = 0$$,     A1

it follows that $$(aRa)$$ and $$R$$ is reflexive     R1

if $$aRb$$ so that $${b^n} – {a^n} \equiv 0(\bmod p)$$     M1

then, $${a^n} – {b^n} \equiv 0(\bmod p)$$ so that $$bRa$$ and $$R$$ is symmetric     A1

if $$aRb$$ and $$bRc$$ so that $${b^n} – {a^n} \equiv 0(\bmod p)$$ and $${c^n} – {b^n} \equiv 0(\bmod p)$$     M1

adding, (it follows that $${c^n} – {a^n} \equiv 0(\bmod p)$$)     M1

so that $$aRc$$ and $$R$$ is transitive     A1

Note:     Only accept the correct use of the terms “reflexive, symmetric, transitive”.

[7 marks]

a.

we are now given that $$aRb$$ if $${b^2} – {a^2} \equiv 0(\bmod 7)$$

attempt to find at least one equivalence class     (M1)

the equivalence classes are

$$\{ 1,{\text{ }}6,{\text{ }}8,{\text{ }}13,{\text{ }} \ldots \}$$    A1

$$\{ 2,{\text{ }}5,{\text{ }}9,{\text{ }}12,{\text{ }} \ldots \}$$    A1

$$\{ 3,{\text{ }}4,{\text{ }}10,{\text{ }}11,{\text{ }} \ldots \}$$    A1

$$\{ 7,{\text{ }}14,{\text{ }}21,{\text{ }}28,{\text{ }} \ldots \}$$    A1

[5 marks]

b.

## Examiners report

Most candidates were familiar with the terminology of the required conditions to be satisfied for a relation to be an equivalence relation. The execution of the proofs was variable. It was grating to see such statements as $$R$$ is symmetric because $$aRb = bRa$$ or $$aRa = {a^n} – {a^n} = 0$$, often without mention of $$\bmod p$$, and such responses were not fully rewarded.

a.

This was not well answered. Few candidates displayed a strategy to find the equivalence classes.

b.

## Question

A relation $$S$$ is defined on $$\mathbb{R}$$ by $$aSb$$ if and only if $$ab > 0$$.

A relation $$R$$ is defined on a non-empty set $$A$$. $$R$$ is symmetric and transitive but not reflexive.

Show that $$S$$ is

(i)     not reflexive;

(ii)     symmetric;

(iii)     transitive.

[4]
a.

Explain why there exists an element $$a \in A$$ that is not related to itself.

[1]
b.

Hence prove that there is at least one element of $$A$$ that is not related to any other element of $$A$$.

[6]
c.

## Markscheme

(i)     $$0S0$$ is not true so $$S$$ is not reflexive     A1AG

(ii)     $$aSb \Rightarrow ab > 0 \Rightarrow ba > 0 \Rightarrow bSa$$ so $$S$$ is symmetric     R1AG

(iii)     $$aSb$$ and $$bSc \Rightarrow ab > 0$$ and $$bc > 0 \Rightarrow a{b^2}c > 0 \Rightarrow ac > 0$$     M1

since $${b^2} > 0$$ (as $$b$$ could not be 0) $$\Rightarrow aSc$$ so $$S$$ is transitive     R1AG

Note: R1 is for indicating that $${b^2} > 0$$.

[4 marks]

a.

since $$R$$ is not reflexive there is at least one element $$a$$ belonging to $$A$$ such that $$a$$ is not related to $$a$$     R1AG

[1 mark]

b.

argue by contradiction: suppose that $$a$$ is related to some other element $$b$$, ie, $$aRb$$     M1

since $$R$$ is symmetric $$aRb$$ implies $$bRa$$     R1A1

since $$R$$ is transitive $$aRb$$ and $$bRa$$ implies $$aRa$$     R1A1

hence there is at least one element of $$A$$ that is not related to any other member of $$A$$     AG

[6 marks]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.

## Question

The relation R is defined on $$\mathbb{R} \times \mathbb{R}$$ such that $$({x_1},{\text{ }}{y_1})R({x_2},{\text{ }}{y_2})$$ if and only if $${x_1}{y_1} = {x_2}{y_2}$$.

Show that R is an equivalence relation.

[5]
a.

Determine the equivalence class of R containing the element $$(1,{\text{ }}2)$$ and illustrate this graphically.

[4]
b.

## Markscheme

R is an equivalence relation if

R is reflexive, symmetric and transitive     A1

$${x_1}{y_1} = {x_1}{y_1} \Rightarrow ({x_1},{\text{ }}{y_1})R({x_1},{\text{ }}{y_1})$$     A1

so R is reflexive

$$({x_1},{\text{ }}{y_1})R({x_2},{\text{ }}{y_2}) \Rightarrow {x_1}{y_1} = {x_2}{y_2} \Rightarrow {x_2}{y_2} = {x_1}{y_1} \Rightarrow ({x_2},{\text{ }}{y_2})R({x_1},{\text{ }}{y_1})$$     A1

so R is symmetric

$$({x_1},{\text{ }}{y_1})R({x_2},{\text{ }}{y_2})$$ and $$({x_2},{\text{ }}{y_2})R({x_3},{\text{ }}{y_3}) \Rightarrow {x_1}{y_1} = {x_2}{y_2}$$ and $${x_2}{y_2} = {x_3}{y_3}$$     M1

$$\Rightarrow {x_1}{y_1} = {x_3}{y_3} \Rightarrow ({x_1},{\text{ }}{y_1})R({x_3},{\text{ }}{y_3})$$    A1

so R is transitive

R is an equivalence relation     AG

[5 marks]

a.

$$(x,{\text{ }}y)R(1,{\text{ }}2)$$     (M1)

the equivalence class is $$\{ (x,{\text{ }}y)|xy = 2\}$$     A1

correct graph     A1

$$(1,{\text{ }}2)$$ indicated on the graph     A1

Note:     Award last A1 only if plotted on a curve representing the class.

[4 marks]

b.

[N/A]

a.

[N/A]

b.

## Question

The relation $$R$$ is defined such that $$xRy$$ if and only if $$\left| x \right| + \left| y \right| = \left| {x + y} \right|$$ for $$x$$, $$y$$, $$y \in \mathbb{R}$$.

Show that $$R$$ is reflexive.

[2]
a.i.

Show that $$R$$ is symmetric.

[2]
a.ii.

Show, by means of an example, that $$R$$ is not transitive.

[4]
b.

## Markscheme

(for $$x \in \mathbb{R}$$), $$\left| x \right| + \left| x \right| = 2\left| x \right|$$    A1

and $$\left| x \right| + \left| x \right| = \left| {2x} \right| = 2\left| x \right|$$    A1

hence $$xRx$$

so $$R$$ is reflexive    AG

Note: Award A1 for correct verification of identity for $$x$$ > 0; A1 for correct verification for $$x$$ ≤ 0.

[2 marks]

a.i.

if $$xRy \Rightarrow \left| x \right| + \left| y \right| = \left| {x + y} \right|$$

$$\left| x \right| + \left| y \right| = \left| y \right| + \left| x \right|$$    A1

$$\left| {x + y} \right| = \left| {y + x} \right|$$     A1

hence $$yRx$$

so $$R$$ is symmetric      AG

[2 marks]

a.ii.

recognising a condition where transitivity does not hold     (M1)

(eg, $$x$$ > 0, $$y$$ = 0 and $$z$$ < 0)

for example, 1$$R$$0 and 0$$R$$(−1)    A1

however $$\left| 1 \right| + \left| { – 1} \right| \ne \left| {1 + – 1} \right|$$      A1

so 1$$R$$(−1) (for example) is not true      R1

hence $$R$$ is not transitive       AG

[4 marks]

b.

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.

## Question

The relation R is defined on $${\mathbb{Z}^ + }$$ by aRb if and only if ab is even. Show that only one of the conditions for R to be an equivalence relation is satisfied.

[5]
a.

The relation S is defined on $${\mathbb{Z}^ + }$$ by aSb if and only if $${a^2} \equiv {b^2}(\bmod 6)$$ .

(i)     Show that S is an equivalence relation.

(ii)     For each equivalence class, give the four smallest members.

[9]
b.

## Markscheme

reflexive: if a is odd, $$a \times a$$ is odd so R is not reflexive     R1

symmetric: if ab is even then ba is even so R is symmetric     R1

transitive: let aRb and bRc; it is necessary to determine whether or not aRc     (M1)

for example 5R2 and 2R3     A1

since $$5 \times 3$$ is not even, 5 is not related to 3 and R is not transitive     R1

[5 marks]

a.

(i)     reflexive: $${a^2} \equiv {a^2}(\bmod 6)$$ so S is reflexive     R1

symmetric: $${a^2} \equiv {b^2}(\bmod 6) \Rightarrow 6|({a^2} – {b^2}) \Rightarrow 6|({b^2} – {a^2}) \Rightarrow {b^2} \equiv {a^2}(\bmod 6)$$     R1

so S is symmetric

transitive: let aSb and bSc so that $${a^2} = {b^2} + 6M$$ and $${b^2} = {c^2} + 6N$$     M1

it follows that $${a^2} = {c^2} + 6(M + N)$$ so aSc and S is transitive     R1

S is an equivalence relation because it satisfies the three conditions     AG

(ii)     by considering the squares of integers (mod 6), the equivalence     (M1)

classes are

{1, 5, 7, 11, $$\ldots$$}     A1

{2, 4, 8, 10, $$\ldots$$}     A1

{3, 9, 15, 21, $$\ldots$$}     A1

{6, 12, 18, 24, $$\ldots$$}     A1

[9 marks]

b.

[N/A]

a.

[N/A]

b.