Date | May 2008 | Marks available | 10 | Reference code | 08M.3srg.hl.TZ2.4 |
Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ2 |
Command term | Find and Show that | Question number | 4 | Adapted from | N/A |
Question
The relation aRb is defined on {1, 2, 3, 4, 5, 6, 7, 8, 9} if and only if ab is the square of a positive integer.
(i) Show that R is an equivalence relation.
(ii) Find the equivalence classes of R that contain more than one element.
Given the group \((G,{\text{ }} * )\), a subgroup \((H,{\text{ }} * )\) and \(a,{\text{ }}b \in G\), we define \(a \sim b\) if and only if \(a{b^{ - 1}} \in H\). Show that \( \sim \) is an equivalence relation.
Markscheme
(i) \(aRa \Rightarrow a \cdot a = {a^2}\) so R is reflexive A1
\(aRb = {m^2} \Rightarrow bRa\) so R is symmetric A1
\(aRb = ab = {m^2}{\text{ and }}bRc = bc = {n^2}\) M1A1
so \(a = \frac{{{m^2}}}{b}{\text{ and }}c = \frac{{{n^2}}}{b}\)
\(ac = \frac{{{m^2}{n^2}}}{{{b^2}}} = {\left( {\frac{{mn}}{b}} \right)^2},\) A1
ac is an integer hence \({\left( {\frac{{mn}}{b}} \right)^2}\) is an integer R1
so aRc, hence R is transitive R1
R is therefore an equivalence relation AG
(ii) 1R4 and 4R9 or 2R8 M1
so {1, 4, 9} is an equivalence class A1
and {2, 8} is an equivalence class A1
[10 marks]
\(a \sim a{\text{ since }}a{a^{ - 1}} = e \in H\), the identity must be in H since it is a subgroup. M1
Hence reflexivity. R1
\(a \sim b \Leftrightarrow a{b^{ - 1}} \in H\) but H is a subgroup so it must contain \({(a{b^{ - 1}})^{ - 1}} = b{a^{ - 1}}\) M1R1
i.e. \(b{a^{ - 1}} \in H{\text{ so }} \sim \) is symmetric A1
\(a \sim b{\text{ and }}b \sim c \Rightarrow a{b^{ - 1}} \in H{\text{ and }}b{c^{ - 1}} \in H\) M1
But H is closed, so
\((a{b^{ - 1}})(b{c^{ - 1}}) \in H{\text{ or }}a({b^{ - 1}}b){c^{ - 1}} \in H\) R1
\(a{c^{ - 1}} \in H \Rightarrow a \sim c\) A1
Hence \( \sim \) is transitive and is thus an equivalence relation R1AG
[9 marks]
Examiners report
Not a difficult question although using the relation definition to fully show transitivity was not well done. It was good to see some students use an operation binary matrix to show transitivity. This was a nice way given that the set was finite. The proof in (b) proved difficult.
Not a difficult question although using the relation definition to fully show transitivity was not well done. It was good to see some students use an operation binary matrix to show transitivity. This was a nice way given that the set was finite. The proof in (b) proved difficult.