OCR MEI Further Extra Pure 2019 June — Question 6 13 marks

Exam BoardOCR MEI
ModuleFurther Extra Pure (Further Extra Pure)
Year2019
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGroups
TypeVerify group axioms
DifficultyChallenging +1.8 This is an abstract algebra question requiring proof of group axioms, understanding of subgroups, and counterexamples. While the algebraic manipulations are manageable (conjugate multiplication, checking closure/inverses), it demands mathematical maturity beyond standard A-level: formal group theory proofs, recognizing why H fails closure, and constructing a finite-order counterexample. The 13-mark allocation and Further Extra Pure context reflect its position as challenging Further Maths content requiring both technical precision and abstract reasoning.
Spec4.01a Mathematical induction: construct proofs8.03c Group definition: recall and use, show structure is/isn't a group8.03f Subgroups: definition and tests for proper subgroups
  1. Given that \(\sqrt{7}\) is an irrational number, prove that \(a^2 - 7b^2 \neq 0\) for all \(a, b \in \mathbb{Q}\) where \(a\) and \(b\) are not both 0. [2]
  2. A set \(G\) is defined by \(G = \{a + b\sqrt{7} : a, b \in \mathbb{Q}, a\) and \(b\) not both 0\(\}\). Prove that \(G\) is a group under multiplication. (You may assume that multiplication is associative.) [7]
  3. A subset \(H\) of \(G\) is defined by \(H = \{1 + c\sqrt{7} : c \in \mathbb{Q}\}\). Determine whether or not \(H\) is a subgroup of \((G, \times)\). [2]
  4. Using \((G, \times)\), prove by counter-example that the statement 'An infinite group cannot have a non-trivial subgroup of finite order' is false. [2]
Question 6:
AnswerMarks Guidance
6(a) p r
Assume a2-7b2 =0 with a= and b= where
q s
p, q, r, s are (non-zero) integers
a ps
± 7 = = which contradicts the given
b qr
AnswerMarks
irrationalityM1
A1
AnswerMarks
[2]2.1
2.1Initial assumption, to set up proof
by contradiction
AnswerMarks
Condone missing oe 𝑎,𝑏 𝜖 ℚ assumed
1 a-b 7
Inverse: =
a+b 7 a2-7b2
and this is an element of G since
 a2-7b2 ¹0
a b
 and (-) are in
a2 -7b2 a2-7b2
a b
 and (-) are not both zero
a2 -7b2 a2-7b2
AnswerMarks
Associativity: given, so (G, ) is a groupM1
A1
B1
AnswerMarks
[7]1.1
2.4
AnswerMarks
2.2aAll three elements are required
Do not award if commutativity is
AnswerMarks
included as one of the axiomsThis mark is
independent but
depends on all 4
axioms stated or
implied as being
necessary
AnswerMarks Guidance
6(c) ( )( )
e.g. 1+ 7 1+ 7 =8+2 7
AnswerMarks
which is not in H so H is not a subgroupM1
A1
AnswerMarks Guidance
[2]2.1
2.2aEvaluation of suitable product
Conclusion correctly justifiedoe, eg finding inverse
6(d) {1, 1}
 (G, ) has infinite order
 {1, 1} is a non-trivial subgroup (of finite order)
 ‘Statement is false’ or ‘{1, 1} is a counter-
AnswerMarks
example’M1
A1
AnswerMarks
[2]3.1a
2.1This set identified
soi
soi
Clear statement
PPMMTT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
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Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
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Registered Office; The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA
Registered Company Number: 3484466
OCR is an exempt Charity
OCR (Oxford Cambridge and RSA Examinations)
Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2019
Question 6:
6 | (a) | p r
Assume a2-7b2 =0 with a= and b= where
q s
p, q, r, s are (non-zero) integers
a ps
± 7 = = which contradicts the given
b qr
irrationality | M1
A1
[2] | 2.1
2.1 | Initial assumption, to set up proof
by contradiction
Condone missing  | oe 𝑎,𝑏 𝜖 ℚ assumed
1 a-b 7
Inverse: =
a+b 7 a2-7b2
and this is an element of G since
 a2-7b2 ¹0
a b
 and (-) are in
a2 -7b2 a2-7b2
a b
 and (-) are not both zero
a2 -7b2 a2-7b2
Associativity: given, so (G, ) is a group | M1
A1
B1
[7] | 1.1
2.4
2.2a | All three elements are required
Do not award if commutativity is
included as one of the axioms | This mark is
independent but
depends on all 4
axioms stated or
implied as being
necessary
6 | (c) | ( )( )
e.g. 1+ 7 1+ 7 =8+2 7
which is not in H so H is not a subgroup | M1
A1
[2] | 2.1
2.2a | Evaluation of suitable product
Conclusion correctly justified | oe, eg finding inverse
6 | (d) | {1, 1}
 (G, ) has infinite order
 {1, 1} is a non-trivial subgroup (of finite order)
 ‘Statement is false’ or ‘{1, 1} is a counter-
example’ | M1
A1
[2] | 3.1a
2.1 | This set identified
soi
soi
Clear statement
PPMMTT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance
programme your call may be recorded or monitored
Oxford Cambridge and RSA Examinations
is a Company Limited by Guarantee
Registered in England
Registered Office; The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA
Registered Company Number: 3484466
OCR is an exempt Charity
OCR (Oxford Cambridge and RSA Examinations)
Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2019
\begin{enumerate}[label=(\alph*)]
\item Given that $\sqrt{7}$ is an irrational number, prove that $a^2 - 7b^2 \neq 0$ for all $a, b \in \mathbb{Q}$ where $a$ and $b$ are not both 0. [2]
\item A set $G$ is defined by $G = \{a + b\sqrt{7} : a, b \in \mathbb{Q}, a$ and $b$ not both 0$\}$.
Prove that $G$ is a group under multiplication. (You may assume that multiplication is associative.) [7]
\item A subset $H$ of $G$ is defined by $H = \{1 + c\sqrt{7} : c \in \mathbb{Q}\}$.
Determine whether or not $H$ is a subgroup of $(G, \times)$. [2]
\item Using $(G, \times)$, prove by counter-example that the statement 'An infinite group cannot have a non-trivial subgroup of finite order' is false. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Extra Pure 2019 Q6 [13]}}
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