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={s∈Q:2s∈Z}.
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.
The relation R is defined on S by s1Rs2 if 3s1+5s2∈Z.
(i) Show that R is an equivalence relation.
(ii) Determine the equivalence classes.
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, s2∈S, 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+n2∈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∈S 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 14∉S R1
OR
not every element has an inverse, A1
for example, 3 does not have an inverse R1
[9 marks]
(i) reflexive: consider 3s+5s M1
=8s∈¢⇒ reflexive A1
symmetric: if s1Rs2, consider 3s2+5s1 M1
for example, =3s1+5s2+(2s1−2s2)∈¢⇒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]