| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Prove group-theoretic identities |
| Difficulty | Challenging +1.2 This is a Further Pure Mathematics 3 group theory question requiring algebraic manipulation and proof by induction. Part (i) is straightforward manipulation using the given relation ar = r²a. Part (ii) is a standard induction proof in group theory. While this requires abstract reasoning beyond standard A-level, it's a relatively routine FP3 question with clear structure and only 6 marks total—moderately above average difficulty but not exceptionally challenging for further maths students. |
| Spec | 4.01a Mathematical induction: construct proofs8.03c Group definition: recall and use, show structure is/isn't a group |
| Answer | Marks | Guidance |
|---|---|---|
| \(ar = r^5a \Rightarrow rar = r^6a\) | M1 | Pre-multiply \(ar = r^5a\) by \(r\) |
| \(r^6 = e \Rightarrow rar = a\) | A1 | Use \(r^6 = e\) and obtain answer AG |
| Answer | Marks | Guidance |
|---|---|---|
| For \(n = 1\), \(rar = a\) OR For \(n = 0\), \(r^0ar^0 = a\) | B1 | For stating true for \(n = 1\) OR for \(n = 0\) |
| Assume \(r^ka r^k = a\) | ||
| EITHER Assumption \(\Rightarrow r^{k+1}ar^{k+1} = rar = a\) | M1 | For attempt to prove true for \(k + 1\) |
| OR \(r^{k+1}ar^{k+1} = r.r^kar.r = rar = a\) | A1 | For obtaining correct form |
| OR \(r^{k+1}ar^{k+1} = r^k.rar.r^k = r^kar^k = a\) | ||
| Hence true for all \(n \in \mathbb{Z}^+\) | A1 4 | For statement of induction conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| \(r^2ar^2 = r.rar.r = rar = a\), similarly for \(r^3ar^3 = a\) | M1 | For attempt to prove for \(n = 2, 3\) |
| \(r^4ar^4 = r.r^3ar^3.r = r.ar = rar = a\), similarly for \(r^5ar^5 = a\) | A1 | For proving true for \(n = 2, 3, 4, 5\) |
| \(r^6ar^6 = eae = a\) | B1 | For showing true for \(n = 6\) |
| For \(n > 6\), \(r^n = r^{n\bmod 6}\), hence true for all \(n \in \mathbb{Z}^+\) | A1 | For using \(n \bmod 6\) and correct conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| \(r^na r^n = r^{n-1}.rar.r^{n-1}\) | M1 | Starting from \(n\), for attempt to prove true for \(n-1\) |
| OR \(r^na r^n = r^n.r^5a r^{n-1} = r^{n+5}a r^{n-1}\) | A1 | For proving true for \(n-1\) |
| \(= r^{n-1}ar^{n-1}\) | ||
| \(= r^{n-2}ar^{n-2} = \ldots\) | A1 | For continuation from \(n-2\) downwards |
| \(= rar = a\) | B1 | For final use of \(rar = a\) |
| SR can be done in reverse |
| Answer | Marks | Guidance |
|---|---|---|
| \(ar = r^5a \Rightarrow ar^2 = r^5ar = r^{10}a\) etc. | M1 | For attempt to derive \(ar^n = r^{5n}a\) |
| \(\Rightarrow ar^n = r^{5n}a\) | A1 | For correct equation |
| SR may be stated without proof | ||
| \(\Rightarrow r^na r^n = r^{6n}a\) | B1 | For pre-multiplication by \(r^n\) |
| \(= ea = a\) | A1 | For obtaining \(a\) (\(r^6 = e\) may be implied) |
## (i)
$ar = r^5a \Rightarrow rar = r^6a$ | M1 | Pre-multiply $ar = r^5a$ by $r$
$r^6 = e \Rightarrow rar = a$ | A1 | Use $r^6 = e$ and obtain answer AG
## (ii) METHOD 1
For $n = 1$, $rar = a$ OR For $n = 0$, $r^0ar^0 = a$ | B1 | For stating true for $n = 1$ OR for $n = 0$
Assume $r^ka r^k = a$ | |
EITHER Assumption $\Rightarrow r^{k+1}ar^{k+1} = rar = a$ | M1 | For attempt to prove true for $k + 1$
OR $r^{k+1}ar^{k+1} = r.r^kar.r = rar = a$ | A1 | For obtaining correct form
OR $r^{k+1}ar^{k+1} = r^k.rar.r^k = r^kar^k = a$ | |
Hence true for all $n \in \mathbb{Z}^+$ | A1 4 | For statement of induction conclusion
## (ii) METHOD 2
$r^2ar^2 = r.rar.r = rar = a$, similarly for $r^3ar^3 = a$ | M1 | For attempt to prove for $n = 2, 3$
$r^4ar^4 = r.r^3ar^3.r = r.ar = rar = a$, similarly for $r^5ar^5 = a$ | A1 | For proving true for $n = 2, 3, 4, 5$
$r^6ar^6 = eae = a$ | B1 | For showing true for $n = 6$
For $n > 6$, $r^n = r^{n\bmod 6}$, hence true for all $n \in \mathbb{Z}^+$ | A1 | For using $n \bmod 6$ and correct conclusion
## (ii) METHOD 3
$r^na r^n = r^{n-1}.rar.r^{n-1}$ | M1 | Starting from $n$, for attempt to prove true for $n-1$
OR $r^na r^n = r^n.r^5a r^{n-1} = r^{n+5}a r^{n-1}$ | A1 | For proving true for $n-1$
$= r^{n-1}ar^{n-1}$ | |
$= r^{n-2}ar^{n-2} = \ldots$ | A1 | For continuation from $n-2$ downwards
$= rar = a$ | B1 | For final use of $rar = a$
| | SR can be done in reverse
## (ii) METHOD 4
$ar = r^5a \Rightarrow ar^2 = r^5ar = r^{10}a$ etc. | M1 | For attempt to derive $ar^n = r^{5n}a$
$\Rightarrow ar^n = r^{5n}a$ | A1 | For correct equation
SR may be stated without proof | |
$\Rightarrow r^na r^n = r^{6n}a$ | B1 | For pre-multiplication by $r^n$
$= ea = a$ | A1 | For obtaining $a$ ($r^6 = e$ may be implied)
---A multiplicative group with identity $e$ contains distinct elements $a$ and $r$, with the properties $r^6 = e$ and $ar = r^2a$.
\begin{enumerate}[label=(\roman*)]
\item Prove that $rar = a$. [2]
\item Prove, by induction or otherwise, that $r^n ar^n = a$ for all positive integers $n$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2010 Q2 [6]}}