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Date May 2014 Marks available 9 Reference code 14M.3srg.hl.TZ0.2
Level HL only Paper Paper 3 Sets, relations and groups Time zone TZ0
Command term Justify, Show that, and Write down Question number 2 Adapted from N/A

Question

Consider the set S defined by S={sQ:2sZ}.

You may assume that + (addition) and × (multiplication) are associative binary operations

on Q.

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

(ii)     Show that {S, +} is a group.

(iii)     Give a reason why {S, ×} is not a group. Justify your answer.

[9]
a.

The relation R is defined on S by s1Rs2 if 3s1+5s2Z.

(i)     Show that R is an equivalence relation.

(ii)     Determine the equivalence classes.

[10]
b.

Markscheme

(i)     0, 12, 1, 32, 2, 52     A2

 

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

 

(ii)     EITHER

closure: if s1, s2S, then s1=m2 and s2=n2 for some m, n¢.     M1

 

Note:     Accept two distinct examples (eg, 12+12=1; 12+1=32) for the M1.

 

s1+s2=m+n2S     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 sS     A1

it is associative (since S§)     A1

the group axioms are satisfied     AG

(iii)     EITHER

the set is not closed under multiplication,     A1

for example, 12×12=14, but 14S     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¢ reflexive     A1

symmetric: if s1Rs2, consider 3s2+5s1     M1

for example, =3s1+5s2+(2s12s2)¢symmetric     A1

transitive: if s1Rs2 and s2Rs3, consider     (M1)

3s1+5s3=(3s1+5s2)+(3s2+5s3)8s2     M1

¢transitive     A1

so R is an equivalence relation     AG

(ii)     C1=¢     A1

C2={±12, ±32, ±52, }     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 » 8.7 » The definition of a group {G,} .

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