| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Non-group structures |
| Difficulty | Challenging +1.3 This is an FP3 abstract algebra question on binary operations. While it requires understanding of group axioms (associativity, identity, inverses, commutativity), the actual manipulations are straightforward algebraic substitutions. Part (i) is routine verification, parts (ii)-(iv) involve solving simple equations, and part (v) requires identifying why not all elements have inverses (m≠0 required). The topic is conceptually more advanced than core A-level, but the execution is mechanical rather than requiring deep insight. |
| Spec | 8.03a Binary operations: and their properties on given sets8.03b Cayley tables: construct for finite sets under binary operation8.03c Group definition: recall and use, show structure is/isn't a group |
| Answer | Marks | Guidance |
|---|---|---|
| \(((a, b)*(c, d))*(e, f) = (ac, ad + b)*(e, f)\) | M1 | For 3 distinct elements bracketed and attempt to expand |
| \(= (ace, acf + ad + b)\) | A1 | For correct expression |
| \((a, b)*((c, d)*(e, f)) = (a, b)*(ce, cf + d) = (ace, acf + ad + b)\) | A1 3 | For correct expression again |
| Answer | Marks | Guidance |
|---|---|---|
| \((a, b)*(1, 0) = (a, a + b), \quad (1, 1)*(a, b) = (a, b + 1)\) | M1 | For combining both ways round |
| \(a + b = b + 1 \Rightarrow a = 1\) | M1 | For equating components (allow from incorrect pairs) |
| \(\Rightarrow (1, b) \forall b\) | A1 3 | For correct elements AEF |
| Answer | Marks | Guidance |
|---|---|---|
| \((mp, mq + n)\) OR \((pm, pn + q) = (1, 0)\) | M1 | For either element on LHS |
| \(\Rightarrow (p, q) = (\frac{1}{m}, -\frac{n}{m})\) | A1 2 | For correct inverse |
| Answer | Marks | Guidance |
|---|---|---|
| \((a, b)*(a, b) = (a^2, ab + b) = (1, 0)\) | M1 | For attempt to find self-inverses |
| OR \((a, b) = (\frac{1}{a}, -\frac{b}{a}) \Rightarrow a^2 = 1, \quad ab = -b\) | ||
| \(\Rightarrow\) self-inverse elements \((1, 0)\) and \((-1, b) \forall b\) | B1 A1 | For \((1, 0)\). For \((-1, b)\) AEF |
| Answer | Marks | Guidance |
|---|---|---|
| \((0, y)\) has no inverse for any \(y \Rightarrow\) not a group | B1 1 | For stating any one element with no inverse. Allow \(x \neq 0\) required, provided reference to inverse is made "Some elements have no inverse" B0 |
## (i)
$((a, b)*(c, d))*(e, f) = (ac, ad + b)*(e, f)$ | M1 | For 3 distinct elements bracketed and attempt to expand
$= (ace, acf + ad + b)$ | A1 | For correct expression
$(a, b)*((c, d)*(e, f)) = (a, b)*(ce, cf + d) = (ace, acf + ad + b)$ | A1 3 | For correct expression again
## (ii)
$(a, b)*(1, 0) = (a, a + b), \quad (1, 1)*(a, b) = (a, b + 1)$ | M1 | For combining both ways round
$a + b = b + 1 \Rightarrow a = 1$ | M1 | For equating components (allow from incorrect pairs)
$\Rightarrow (1, b) \forall b$ | A1 3 | For correct elements **AEF**
## (iii)
$(mp, mq + n)$ OR $(pm, pn + q) = (1, 0)$ | M1 | For either element on LHS
$\Rightarrow (p, q) = (\frac{1}{m}, -\frac{n}{m})$ | A1 2 | For correct inverse
## (iv)
$(a, b)*(a, b) = (a^2, ab + b) = (1, 0)$ | M1 | For attempt to find self-inverses
OR $(a, b) = (\frac{1}{a}, -\frac{b}{a}) \Rightarrow a^2 = 1, \quad ab = -b$ |
$\Rightarrow$ self-inverse elements $(1, 0)$ and $(-1, b) \forall b$ | B1 A1 | For $(1, 0)$. For $(-1, b)$ **AEF**
## (v)
$(0, y)$ has no inverse for any $y \Rightarrow$ not a group | B1 1 | For stating any one element with no inverse. Allow $x \neq 0$ required, provided reference to inverse is made "Some elements have no inverse" B0
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# Question 9 – Not included in provided pagesThe operation $*$ is defined on the elements $(x, y)$, where $x, y \in \mathbb{R}$, by
$$(a, b) * (c, d) = (ac, ad + b).$$
It is given that the identity element is $(1, 0)$.
\begin{enumerate}[label=(\roman*)]
\item Prove that $*$ is associative. [3]
\item Find all the elements which commute with $(1, 1)$. [3]
\item It is given that the particular element $(m, n)$ has an inverse denoted by $(p, q)$, where
$$(m, n) * (p, q) = (p, q) * (m, n) = (1, 0).$$
Find $(p, q)$ in terms of $m$ and $n$. [2]
\item Find all self-inverse elements. [3]
\item Give a reason why the elements $(x, y)$, under the operation $*$, do not form a group. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2011 Q8 [12]}}