| Exam Board | OCR MEI |
|---|---|
| Module | Further Extra Pure (Further Extra Pure) |
| Year | 2024 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Contradiction proof of irrationality |
| Difficulty | Standard +0.8 This is a Further Maths proof requiring students to construct a contradiction argument using prime factorization and the given lemma. While the technique is standard for irrationality proofs, it requires careful logical reasoning and the insight to raise both sides to power b, then apply unique prime factorization. The deduction in part (b) is straightforward once (a) is proven. Moderately challenging for Further Maths students. |
| Spec | 1.01d Proof by contradiction4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | a |
| Answer | Marks |
|---|---|
| assumption must be wrong. | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 3.2a | Making initial assumption. |
| Answer | Marks |
|---|---|
| is even for a > 0 while RHS is not). | a, b ∈ must be seen before a = 0 & |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (b) | a |
| Answer | Marks |
|---|---|
| 2 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 3.2a | Making initial assumption and using |
| definition of rationality | If not by contradiction, argument |
Question 5:
5 | (a) | a
Assume that 3=2b for some a, b∈ (b ≠ 0).
So 2a =3bwhich gives a = b = 0 from the given
statement which is not a valid solution to the
original equation (or which cannot be so as b ≠ 0)
and hence is a contradiction. Hence our original
assumption must be wrong. | M1
A1
[2] | 3.1a
3.2a | Making initial assumption.
Raising both sides to the power of b
and expressing in a form from which
the given statement can be used to
show the contradiction. Argument
must be complete.
Could use different argument (eg LHS
is even for a > 0 while RHS is not). | a, b ∈ must be seen before a = 0 &
b = 0 claimed from given statement.
ℤ
If M0 then SCB1 if correct argument
made except a, b ∈ missed or late.
Argument must not ℤbe based on eg
log 3 being irrational.
2
5 | (b) | a
If log 3∈then log 3= for some integers a
2 2 b
and b.
a
⇒3=2b for a, b∈which, (from (a)), is a
contradiction and so log 3∉.
2 | M1
A1
[2] | 3.1a
3.2a | Making initial assumption and using
definition of rationality | If not by contradiction, argument
could run: (From (a)) It is not
possible to find a pair a, b∈such
a a
that 3=2b and 3=2b is equivalent
a
to log 3= (M1).
2
b
Completing argument A1.
But it must be clear that no such
integers a and b can exist, not simply
a
that 3≠2b for some a and b.
Argument must not depend on
a
a, b∉⇒ ∉. Argument must be
b
rigorous and contain proper
conclusion.
If M0 then SCB1 if correct argument
made except a, b ∈ missed.
PMT
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Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.5 In this question you may assume that if $p$ and $q$ are distinct prime numbers and $\mathbf { p } ^ { \alpha } = \mathbf { q } ^ { \beta }$ where $\alpha , \beta \in \mathbb { Z }$, then $\alpha = 0$ and $\beta = 0$.
\begin{enumerate}[label=(\alph*)]
\item Prove that it is not possible to find $a$ and $b$ for which $\mathrm { a } , \mathrm { b } \in \mathbb { Z }$ and $3 = 2 ^ { \frac { \mathrm { a } } { \mathrm { b } } }$.
\item Deduce that $\log _ { 2 } 3 \notin \mathbb { Q }$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Extra Pure 2024 Q5 [4]}}