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

 

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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 \((X \times Y) \cap (Y \times X)\).

[5]
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 \(g(x,{\text{ }}y) = \frac{1}{2}\).

[10]
b.
Answer/Explanation

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.

Examiners report

[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.

[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.
Answer/Explanation

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.

[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.
Answer/Explanation

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.
Answer/Explanation

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

giving the required contradiction     R1

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

[6 marks]

c.

Examiners report

[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.
Answer/Explanation

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.

Examiners report

[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.
Answer/Explanation

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.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.

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