# IB DP Maths Topic 8.2 Ordered pairs: the Cartesian product of two sets HL Paper 3

## Question

Sets X and Y are defined by $${\text{ }}X = \left] {0,{\text{ }}1} \right[;{\text{ }}Y = \{ 0,{\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5\}$$.

(i)     Sketch the set $$X \times Y$$ in the Cartesian plane.

(ii)     Sketch the set $$Y \times X$$ in the Cartesian plane.

(iii)     State $$(X \times Y) \cap (Y \times X)$$.


a.

Consider the function $$f:X \times Y \to \mathbb{R}$$ defined by $$f(x,{\text{ }}y) = x + y$$ and the function $$g:X \times Y \to \mathbb{R}$$ defined by $$g(x,{\text{ }}y) = xy$$.

(i)     Find the range of the function f.

(ii)     Find the range of the function g.

(iii)     Show that $$f$$ is an injection.

(iv)     Find $${f^{ – 1}}(\pi )$$, expressing your answer in exact form.

(v)     Find all solutions to $$g(x,{\text{ }}y) = \frac{1}{2}$$.


b.

## Markscheme

(i) correct horizontal lines     A1

correctly labelled axes     A1

clear indication that the endpoints are not included     A1

(ii) fully correct diagram     A1

Note:     Do not penalize the inclusion of endpoints twice.

(iii)     the intersection is empty     A1

[5 marks]

a.

(i)     range $$(f) = \left] {0,{\text{ 1}}} \right[ \cup \left] {1,{\text{ 2}}} \right[ \cup \rm{L} \cup \left] {5,{\text{ 6}}} \right[$$     A1A1

Note:     A1 for six intervals and A1 for fully correct notation.

Accept $$0 < x < 6,{\text{ }}x \ne 0{\text{, 1, 2, 3, 4, 5, 6}}$$.

(ii)     range $$(g) = \left[ {0,{\text{ 5}}} \right[$$     A1

(iii)     Attempt at solving

$$f({x_1},{\text{ }}{y_1}) = f({x_2},{\text{ }}{y_2})$$     M1

$$f(x,{\text{ }}y) \in \left] {y,{\text{ }}y + 1} \right[ \Rightarrow {y_1} = {y_2}$$     M1

and then $${x_1} = {x_2}$$     A1

so $$f$$ is injective     AG

(iv)     $${f^{ – 1}}(\pi ) = (\pi – 3,{\text{ }}3)$$     A1A1

(v)     solutions: (0.5, 1), (0.25, 2), $$\left( {\frac{1}{6},{\text{ 3}}} \right)$$, (0.125, 4), (0.1, 5)     A2

Note:     A2 for all correct, A1 for 2 correct.

[10 marks]

b.

[N/A]

a.

[N/A]

b.

## 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.


a.

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


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

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.


a.

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


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.


a.

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


b.

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


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.


a.

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


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.


a.i.

Show that $$R$$ is symmetric.


a.ii.

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


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.