AQA Paper 2 2020 June — Question 7 7 marks

Exam BoardAQA
ModulePaper 2 (Paper 2)
Year2020
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof
TypeIdentifying errors in proofs
DifficultyStandard +0.3 This is a proof question testing understanding of rational/irrational numbers and basic algebraic manipulation. Part (a)(i) requires spotting an algebraic error in adding fractions (1/a + 1/b ≠ 2/(a+b)), which is straightforward. Part (a)(ii) needs the correct formula (a+b)/(ab) and recognizing rationals are closed under division. Part (b) is a standard proof by contradiction following a template structure. While it involves proof writing, the concepts are elementary and the steps are routine for A-level students who have practiced proof techniques.
Spec1.01a Proof: structure of mathematical proof and logical steps1.01d Proof by contradiction

\(a\) and \(b\) are two positive irrational numbers. The sum of \(a\) and \(b\) is rational. The product of \(a\) and \(b\) is rational. Caroline is trying to prove \(\frac{1}{a} + \frac{1}{b}\) is rational. Here is her proof: Step 1 \quad \(\frac{1}{a} + \frac{1}{b} = \frac{2}{a + b}\) Step 2 \quad \(2\) is rational and \(a + b\) is non-zero and rational. Step 3 \quad Therefore \(\frac{2}{a + b}\) is rational. Step 4 \quad Hence \(\frac{1}{a} + \frac{1}{b}\) is rational.
    1. Identify Caroline's mistake. [1 mark]
    2. Write down a correct version of the proof. [2 marks]
  1. Prove by contradiction that the difference of any rational number and any irrational number is irrational. [4 marks]

Question 7:

AnswerMarks Guidance
7(a)(i)Identifies the error lies in step 1
without contradiction.2.3 E1
1 1 2
𝑎𝑎+ 𝑏𝑏 = 𝑎𝑎+𝑏𝑏
AnswerMarks
Subtotal1

AnswerMarks
7(a)(ii)Recalls correct addition
b a
Accept +
AnswerMarks Guidance
ab ab1.1b M1
a + b is rational and ab is rational
𝑎𝑎+ 𝑏𝑏 = 𝑎𝑎𝑏𝑏
and therefore
1 1
+ is rational.
a b
Completes rigorous argument
to complete proof. Must state
that abis rational (and non-
a+bis
zero) and rational and
1 1
conclude that + or is
a b
𝑎𝑎+𝑏𝑏
AnswerMarks Guidance
r ational 𝑎𝑎𝑏𝑏2.1 R1
Subtotal2

AnswerMarks
7(b)States assumption to begin
proof by contradiction may
PI by
a c a c
−x= or x− =
AnswerMarks Guidance
b d b d3.1a M1
between a rational and an irrational
number is rational.
a c
−x =
b d
Where a,b,cand d are integers,
b,d ≠0 and x is irrational
a c
x= −
b d
ad cb
= −
bd bd
ad −cb
=
bd
Hence x is rational. This is a
contradiction hence the difference
of any rational number and any
irrational number is irrational.
Uses language and notation
correctly to state initial
assumptions:
States their a,b,cand d are
integers
and
x is irrational do not accept the
irrational written as a fraction
AnswerMarks Guidance
Condone missing b,d ≠02.5 A1
Demonstrates that x can be
expressed as a rational number
ad −cb
by obtaining x= OE
AnswerMarks Guidance
bd1.1b M1
Completes rigorous argument
to prove the required result,
clearly explaining where the
contradiction lies with ALL
assumptions correct at the start
AnswerMarks Guidance
(including b,d ≠0)2.1 R1
Subtotal4
Question Total7
QMarking instructions AO
Question 7:
--- 7(a)(i) ---
7(a)(i) | Identifies the error lies in step 1
without contradiction. | 2.3 | E1 | Mistake is
1 1 2
𝑎𝑎+ 𝑏𝑏 = 𝑎𝑎+𝑏𝑏
Subtotal | 1
--- 7(a)(ii) ---
7(a)(ii) | Recalls correct addition
b a
Accept +
ab ab | 1.1b | M1 | 1 1 𝑎𝑎+𝑏𝑏
a + b is rational and ab is rational
𝑎𝑎+ 𝑏𝑏 = 𝑎𝑎𝑏𝑏
and therefore
1 1
+ is rational.
a b
Completes rigorous argument
to complete proof. Must state
that abis rational (and non-
a+bis
zero) and rational and
1 1
conclude that + or is
a b
𝑎𝑎+𝑏𝑏
r ational 𝑎𝑎𝑏𝑏 | 2.1 | R1
Subtotal | 2
--- 7(b) ---
7(b) | States assumption to begin
proof by contradiction may
PI by
a c a c
−x= or x− =
b d b d | 3.1a | M1 | Assume that the difference
between a rational and an irrational
number is rational.
a c
−x =
b d
Where a,b,cand d are integers,
b,d ≠0 and x is irrational
a c
x= −
b d
ad cb
= −
bd bd
ad −cb
=
bd
Hence x is rational. This is a
contradiction hence the difference
of any rational number and any
irrational number is irrational.
Uses language and notation
correctly to state initial
assumptions:
States their a,b,cand d are
integers
and
x is irrational do not accept the
irrational written as a fraction
Condone missing b,d ≠0 | 2.5 | A1
Demonstrates that x can be
expressed as a rational number
ad −cb
by obtaining x= OE
bd | 1.1b | M1
Completes rigorous argument
to prove the required result,
clearly explaining where the
contradiction lies with ALL
assumptions correct at the start
(including b,d ≠0) | 2.1 | R1
Subtotal | 4
Question Total | 7
Q | Marking instructions | AO | Mark | Typical solution
$a$ and $b$ are two positive irrational numbers.

The sum of $a$ and $b$ is rational.

The product of $a$ and $b$ is rational.

Caroline is trying to prove $\frac{1}{a} + \frac{1}{b}$ is rational.

Here is her proof:

Step 1 \quad $\frac{1}{a} + \frac{1}{b} = \frac{2}{a + b}$

Step 2 \quad $2$ is rational and $a + b$ is non-zero and rational.

Step 3 \quad Therefore $\frac{2}{a + b}$ is rational.

Step 4 \quad Hence $\frac{1}{a} + \frac{1}{b}$ is rational.

\begin{enumerate}[label=(\alph*)]
\item 
\begin{enumerate}[label=(\roman*)]
\item Identify Caroline's mistake.
[1 mark]

\item Write down a correct version of the proof.
[2 marks]
\end{enumerate}

\item Prove by contradiction that the difference of any rational number and any irrational number is irrational.
[4 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 2 2020 Q7 [7]}}