Edexcel FP2 2021 June — Question 2 8 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2021
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGroups
TypeNon-group structures
DifficultyStandard +0.3 This is a straightforward group axiom verification question requiring students to check closure, identity, and inverses for a simple absolute difference operation. While it involves Further Maths content (group theory), the operation is elementary and each part follows directly from the definition with minimal algebraic manipulation. Part (d) requires recognizing that associativity fails, but this is a standard check that students at this level should routinely perform.
Spec8.03a Binary operations: and their properties on given sets8.03c Group definition: recall and use, show structure is/isn't a group
  1. A binary operation ★ on the set of non-negative integers, \(\mathbb { Z } _ { 0 } ^ { + }\), is defined by
$$m \star n = | m - n | \quad m , n \in \mathbb { Z } _ { 0 } ^ { + }$$
  1. Explain why \(\mathbb { Z } _ { 0 } ^ { + }\)is closed under the operation
  2. Show that 0 is an identity for \(\left( \mathbb { Z } _ { 0 } ^ { + } , \star \right)\)
  3. Show that all elements of \(\mathbb { Z } _ { 0 } ^ { + }\)have an inverse under ★
  4. Determine if \(\mathbb { Z } _ { 0 } ^ { + }\)forms a group under ★, giving clear justification for your answer.
Question 2:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
For \(m,n \in \mathbb{Z}_0^+\) we have \(m-n \in \mathbb{Z}\) (difference of integers is an integer) and so \(\m-n\ \in \mathbb{Z}_0^+\), hence closed under \(\star\)
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
For \(m \in \mathbb{Z}_0^+\), \(0\star m = \0-m\ = \
and \(m\star 0 = \m-0\ = \
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
For \(m \in \mathbb{Z}_0^+\), we need \(\m-n\ =0 \Rightarrow n=...\) or shows \(\
As \(\m-m\ =0\) for all \(m \in \mathbb{Z}_0^+\), each \(m\) is self-inverse
Part (d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Checks associativity – evaluates \(m\star(n\star p)\) and \((m\star n)\star p\) with letter or numbersM1 May be by counter example or attempting to evaluate both sides of associativity axiom
E.g. \(1\star(2\star3) = 1\star\2-3\ = 1\star1 = 0\) but \((1\star2)\star3 = \
\(1\star(2\star3) \neq (1\star2)\star3\) hence not associative so not a groupA1 Must deduce associativity does not hold and conclude \(\mathbb{Z}_0^+\) is not a group under \(\star\) 
## Question 2:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| For $m,n \in \mathbb{Z}_0^+$ we have $m-n \in \mathbb{Z}$ (difference of integers is an integer) and so $\|m-n\| \in \mathbb{Z}_0^+$, hence closed under $\star$ | **B1** | "Always positive" as a conclusion is B0 without consideration of the equal zero case |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| For $m \in \mathbb{Z}_0^+$, $0\star m = \|0-m\| = \|-m\| = m$ | **M1** | Checks that 0 is a left or a right identity |
| and $m\star 0 = \|m-0\| = \|m\| = m$, hence 0 is an identity | **A1\*** | Checks 0 works both sides and makes conclusion it is an identity |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| For $m \in \mathbb{Z}_0^+$, we need $\|m-n\|=0 \Rightarrow n=...$ or shows $\|m-m\|=\|0\|=0$ | **M1** | Realises $m$ must be its own inverse; accept if just stated $m$ is self-inverse with no proof, or attempt to solve $\|m-n\|=0$ |
| As $\|m-m\|=0$ for all $m \in \mathbb{Z}_0^+$, each $m$ is self-inverse | **A1** | Each element is self-inverse with a **full** proof given |

### Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Checks associativity – evaluates $m\star(n\star p)$ and $(m\star n)\star p$ with letter or numbers | **M1** | May be by counter example or attempting to evaluate both sides of associativity axiom |
| E.g. $1\star(2\star3) = 1\star\|2-3\| = 1\star1 = 0$ but $(1\star2)\star3 = \|1-2\|\star3 = 1\star3 = \|1-3\| = 2$ | **M1** | Produces a suitable counter example and evaluates both sides |
| $1\star(2\star3) \neq (1\star2)\star3$ hence not associative so not a group | **A1** | Must deduce associativity does not hold and conclude $\mathbb{Z}_0^+$ is not a group under $\star$ |
\begin{enumerate}
  \item A binary operation ★ on the set of non-negative integers, $\mathbb { Z } _ { 0 } ^ { + }$, is defined by
\end{enumerate}

$$m \star n = | m - n | \quad m , n \in \mathbb { Z } _ { 0 } ^ { + }$$

(a) Explain why $\mathbb { Z } _ { 0 } ^ { + }$is closed under the operation\\
(b) Show that 0 is an identity for $\left( \mathbb { Z } _ { 0 } ^ { + } , \star \right)$\\
(c) Show that all elements of $\mathbb { Z } _ { 0 } ^ { + }$have an inverse under ★\\
(d) Determine if $\mathbb { Z } _ { 0 } ^ { + }$forms a group under ★, giving clear justification for your answer.

\hfill \mbox{\textit{Edexcel FP2 2021 Q2 [8]}}
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