User interface language: English | Español

Date May 2014 Marks available 10 Reference code 14M.3srg.hl.TZ0.2
Level HL only Paper Paper 3 Sets, relations and groups Time zone TZ0
Command term Determine and Show that Question number 2 Adapted from N/A

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.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 8 - Option: Sets, relations and groups » 8.2 » Relations: equivalence relations; equivalence classes.
Show 36 related questions

View options