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.
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.
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]
(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]