| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Verify group axioms |
| Difficulty | Standard +0.3 This is a standard group axioms verification question from Further Maths. While it requires knowledge of abstract algebra (closure, associativity, identity, inverse), the operation is simple and the proofs are routine. Part (i)(c) requires slightly more thought about what order 2 means, and part (ii) is straightforward counterexample finding. This is typical FP3 material but doesn't require deep insight—just systematic application of definitions. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03e Order of elements: and order of groups |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x + y - a \in \mathbb{R}\) | B1 | For stating closure is satisfied |
| \((x*y)*z = (x+y-a)*z = x+y+z-2a\) | M1 | For using 3 distinct elements bracketed both ways |
| \(x*(y*z) = x*(y+z-a) = x+y+z-2a\) | A1 | For obtaining same result twice for associativity. SR 3 distinct elements bracketed once, expanded, and symmetry noted scores M1 A1 |
| \(x + e - a = x \Rightarrow e = a\) | B1 | For stating identity \(= a\) |
| \(x + x^{-1} - a = a \Rightarrow x^{-1} = 2a - x\) | M1, A1 6 | For attempting to obtain inverse of \(x\); For obtaining inverse \(= 2a - x\), OR for showing that inverses exist, where \(x + x^{-1} = 2a\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x + y - a = y + x - a \Rightarrow\) commutative | B1 1 | For stating commutativity is satisfied, with justification |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x\) order \(2 \Rightarrow x*x = e \Rightarrow 2x - a = e\) | M1 | For obtaining equation for an element of order 2 |
| \(\Rightarrow 2x - a = a \Rightarrow x = a\) | A1 2 | For solving and showing the only solution is the identity (which has order 1), OR for proving there are no self-inverse elements (other than the identity) |
| OR \(x = x^{-1} \Rightarrow x = 2a - x \Rightarrow x = a = e \Rightarrow\) no elements of order 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. \(2 + 1 - 5 = -2 \notin \mathbb{R}^+\) \(\Rightarrow\) not closed | M1, A1 | For attempting to disprove closure; For stating closure is not necessarily satisfied (\(0 < x + y\), 5 required) |
| e.g. \(2 \times 5 - 11 = -1 \notin \mathbb{R}^+\) \(\Rightarrow\) no inverse | M1, A1 4 | For attempting to find an element with no inverse; For stating inverse is not necessarily satisfied (\(x\ldots10\) required) |
# Question 7:
## Part (i)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x + y - a \in \mathbb{R}$ | B1 | For stating closure is satisfied |
| $(x*y)*z = (x+y-a)*z = x+y+z-2a$ | M1 | For using 3 distinct elements bracketed both ways |
| $x*(y*z) = x*(y+z-a) = x+y+z-2a$ | A1 | For obtaining same result twice for associativity. **SR** 3 distinct elements bracketed once, expanded, and symmetry noted scores M1 A1 |
| $x + e - a = x \Rightarrow e = a$ | B1 | For stating identity $= a$ |
| $x + x^{-1} - a = a \Rightarrow x^{-1} = 2a - x$ | M1, A1 **6** | For attempting to obtain inverse of $x$; For obtaining inverse $= 2a - x$, OR for showing that inverses exist, where $x + x^{-1} = 2a$ |
## Part (i)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x + y - a = y + x - a \Rightarrow$ commutative | B1 **1** | For stating commutativity is satisfied, with justification |
## Part (i)(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x$ order $2 \Rightarrow x*x = e \Rightarrow 2x - a = e$ | M1 | For obtaining equation for an element of order 2 |
| $\Rightarrow 2x - a = a \Rightarrow x = a$ | A1 **2** | For solving and showing the only solution is the identity (which has order 1), OR for proving there are no self-inverse elements (other than the identity) |
| OR $x = x^{-1} \Rightarrow x = 2a - x \Rightarrow x = a = e \Rightarrow$ no elements of order 2 | | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. $2 + 1 - 5 = -2 \notin \mathbb{R}^+$ $\Rightarrow$ not closed | M1, A1 | For attempting to disprove closure; For stating closure is not necessarily satisfied ($0 < x + y$, 5 required) |
| e.g. $2 \times 5 - 11 = -1 \notin \mathbb{R}^+$ $\Rightarrow$ no inverse | M1, A1 **4** | For attempting to find an element with no inverse; For stating inverse is not necessarily satisfied ($x\ldots10$ required) |
**Total: 13 marks**
---7 (i) The operation $*$ is defined by $x * y = x + y - a$, where $x$ and $y$ are real numbers and $a$ is a real constant.
\begin{enumerate}[label=(\alph*)]
\item Prove that the set of real numbers, together with the operation $*$, forms a group.
\item State, with a reason, whether the group is commutative.
\item Prove that there are no elements of order 2.\\
(ii) The operation $\circ$ is defined by $x \circ y = x + y - 5$, where $x$ and $y$ are positive real numbers. By giving a numerical example in each case, show that two of the basic group properties are not necessarily satisfied.
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2009 Q7 [13]}}