Pearson Edexcel IAL Mathematics A2 P4.1 Proof

[Maximum number: 4]

Prove by contradiction that for all positive numbers k

k+\frac{9}{k} \geqslant 6

Assume that there exists a positive number k such that $k+\frac9k<6$ .
$k+\frac9k<6\Rightarrow k^2+9<6k\Rightarrow k^2-6k+9<0$ $(k-3)^2<0$
But numbers squared are $\ge 0$ , hence $k+\frac9k\ge 6$ .
A1*

[Maximum number: 4]

\mathrm{f}(x)=(8-3 x)^{\frac{4}{3}} \quad 0<x<\frac{8}{3}

Use proof by contradiction to prove that the curve with equation

y=8+8 x-\frac{15}{2} x^{2}

does not intersect the curve with equation

y=A-8 x+\frac{x^{2}}{2}+B x^{3} \quad 0<x<\frac{8}{3}

where A and B are the constants found in part (a).
(Solutions relying on calculator technology are not acceptable.)

Assume the curves meet:
$8x-\frac{15x^2}{2}+8=16-8x+\frac{x^2}{2}+\frac{x^3}{24}$ $0=8-16x+8x^2+\frac{x^3}{24} =8(x-1)^2+\frac{x^3}{24}$
For $0<x<\frac83$ ,
$8(x-1)^2\ge0,\qquad \frac{x^3}{24}>0$
so
$8(x-1)^2+\frac{x^3}{24}>0$
This contradicts the equation $0=8(x-1)^2+\frac{x^3}{24}$ , so the curves do not meet.
(4)

[Maximum number: 8]

A student was asked to prove, for $p \in \mathbb{N}$ , that
"if $p^{3}$ is a multiple of 3, then p must be a multiple of 3 "

The start of the student's proof by contradiction is shown in the box below.

Assumption:

There exists a number $p, p \in \mathbb{N}$ , such that $p^{3}$ is a multiple of 3, and p is NOT a multiple of 3

Let $p=3 k+1, k \in \mathbb{N}$ .

\text { Consider } \begin{aligned} p^{3}=(3 k+1)^{3} & =27 k^{3}+27 k^{2}+9 k+1 \\ & =3\left(9 k^{3}+9 k^{2}+3 k\right)+1 \quad \text { which is not a multiple of } 3 \end{aligned}

(a)

Show the calculations and statements that are required to complete the proof.

[ 3 ]

Let p=3 k+2 then $(3 k+2)^{3}=27 k^{3}+54 k^{2}+36 k+8$ $=3 \times\left(9 k^{3}+18 k^{2}+12 k+3\right)-1$ not a multiple of 3
So p cannot be of form 3 k+1 or 3 k+2, since $p^{3}$ is a multiple of 3. Hence
p must be a multiple of 3, a contradiction of our assumption, hence for all
integers p, when $p^{3}$ is a multiple of 3, then p is a multiple of 3
(3)

(b)

Hence prove, by contradiction, that $\sqrt[3]{3}$ is an irrational number.

[ 5 ]

Assumption: there exist (integers) p and q such that $\sqrt[3]{3}=\frac{p}{q}$
(where p and q have no (non-trivial) common factors.)
Then $\quad \sqrt[3]{3}=\frac{p}{q} \Rightarrow p^{3}=3 q^{3}$
So $p^{3}$ is a multiple of 3 and (so) p is a multiple of 3
But $p=3 k \Rightarrow 27 k^{3}=3 q^{3} \Rightarrow q^{3}=9 k^{3}$
Hence $q^{3}$ is a multiple of 3 so q is a multiple of 3, but as p and q have no
(non-trivial) common factors, this is a contradiction.
Hence $\sqrt[3]{3}$ is an irrational number.*

A1*

(5)

(8 marks)

Notes:

(a)

M1: Attempts to expand $(3 k+2)^{3}$ or $(3 k-1)^{3}$ .

Look for a cubic expression with 4 terms with at least two correct (allowing for incorrect signs).

A1: Achieves a correct $3 \times(\ldots)+r,|r|<10$ form for the expansion and states not a multiple of 3.

Suitable forms include $(3 k+2)^{3}=3\left(9 k^{3}+18 k^{2}+12 k+2\right)+2=3\left(9 k^{3}+18 k^{2}+12 k+3\right)-1$ or $3\left(9 k^{3}+18 k^{2}+12 k\right)+8$ or $(3 k-1)^{3}=3\left(9 k^{3}-9 k^{2}+3 k\right)-1$ etc.

Alternatively, achieves correct $(3 k+2)^{3}=27 k^{3}+54 k^{2}+36 k+8$ or $(3 k-1)^{3}=27 k^{3}-27 k^{2}+9 k-1$ with a reason why it is not a multiple of 3 e.g 3 divides 27, 54 and 36, but not 8 hence not divisible by 3.

A1: Completes the proof. Must have scored both previous marks and a reference to both cases (in some form) leading to a "contradiction" and some indication that proof is complete. It is unlikely to be as complete as that shown in the scheme, but all three bold points must be conveyed. E.g. as a minimum after satisfying the first A "both cases give a contradiction hence the original statement is true".

(b)

May use different letters throughout.

B1: Sets up algebraically the initial statement to be contradicted. Essentially for showing they know what a rational number is algebraically. There is no requirement for this mark to state that p and q are integers with no (non-trivial) common factors (this may be implied for this mark).
M1: Cubes correctly and multiplies through by $q^{3}$
A1: Deduces that both $p^{3}$ is a multiple of 3 and hence p is a multiple of 3. Jumping directly to p is a multiple of 3 is A0.
dM 1: Sets p=3 k and proceeds to find $q^{3}$ in terms of k. (May use a different letter.)
A1: Completes the proof. This requires
- Correct algebraic statements
- Correct deductions in correct order. E.g. $p^{3}$ is a multiple of 3 so p is a multiple of 3
- initial statement must have included that p and q are integers (or accept natural numbers or a $\frac{p}{q}$ is a fraction) and have no common factors (or are co-prime, or in simplest form)
- correct reason for contradiction and acceptable conclusion
- There must have been no contrary statements during the proof (e.g. that p and q are prime)

Edexcel A-Level Mathematics A2 P4.1 Proof Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question 2

Question 3

Question (a)

Question (b)