Mathematical Proof – A Level Maths Revision

Mathematical proof at A Level is a specialised area of study that requires patience, practice, and the odd bit of guidance here and there doesn’t hurt! Let’s get started…

Mathematical proof tests your ability to use the maths you’ve learnt to prove or disprove mathematical statements.

This is quite a broad topic – it can involve using a wide range of other techniques and can vary a lot in difficulty (often within the same question – it might be easy if you spot the right way to approach it but seemingly impossible if you don’t). For this reason, it’s important to practise questions to get in the right mindset.

If you’d like to see this blog in PDF or PowerPoint form, they’re available here.

To practice some pre-requisite skills, try these multiple-choice questions.

If you want more practice, try these shorter problems or these exam-style questions.

Mathematical Proof for A Level Maths

A mathematical proof is a structured argument that shows that a statement (conjecture) is always true. Once the conjecture has been proved, the statement can be called a theorem.

A number of methods can be used to prove or disprove a statement:

• Proof by deduction
• Proof by exhaustion
• Disproving with a counter-example

Proof by Deduction

In proof by deduction, you use a known mathematical fact (or several of them) to reach the proof. The process is usually algebraic but can also be geometric.

Example Question 1

Prove that π2 β 4π + 9 is positive for all real values of π.

The expression in this question is quadratic – there are several ways to manipulate these. One way is to complete the square.

\begin{aligned} n^2 - 4n + 9 &=(n-2)^2-(-2)^2+9\\&=(n-2)^2+5 \end{aligned}

(π – 2)2 is non-negative for all real values of π and adding a positive number (5) to a non-negative number will result in a positive number.

Therefore, (π β 2)2 + 5 is positive for all real values of π.

Sometimes, you will need to recall a definition for your starting point.

Example Question 2

Prove that the sum of two rational numbers is always a rational number.

A rational number is one that can be written in the form $\frac{a}{b}$, where π and π are integers such that π β  0.

Let the two rational numbers be $\frac{a}{b}$ and $\frac{c}{d}$ where π, π, π and π are integers where π and π β  0.

\begin{aligned} \frac{a}{b} +\frac{c}{d} &= \frac{ad}{bd} + \frac{bc}{bd} \\[0.5em] &= \frac{ad+bc}{bd} \end{aligned}

Since π, π, π and π are integers, their products ππ, ππ and ππ are also integers. The sum of two integers is an integer, so ππ + ππ is an integer. π and π β  0, so ππ β  0.

By definition, $\frac{ad\,+\,bc}{bd}$ is a rational number since its numerator and denominator are integers and the denominator is not equal to zero.

Proof by Exhaustion

Proof by exhaustion involves listing all possible outcomes and checking that they match the statement you’re trying to prove. This proof works better in instances where there are a small number of possible options.

Example Question 3

Prove that (π β 1)3 is not divisible by 5 for integer values of π such that 2 β€ π β€ 5.

Substitute π = 2, π = 3, π = 4 and π = 5 into the expression (π β 1)3:

If π = 2, (2 β 1)3 = 1
If π = 3, (3 β 1)3 = 8
If π = 4, (4 β 1)3 = 27
If π = 5, (5 β 1)3 = 64

1, 8, 27 and 64 are not divisible by 5, so (π β 1)3 is not divisible by 5 for 2 β€ π β€ 5.

Disproving with a Counter Example

To disprove a conjecture, you only need to find one example which doesn’t fit. Often, fractional and negative values are a good place to start.

Example Question 4

Prove that the following statement is not true for all values of π and π.

$\frac{1}{a} > \frac{1}{b} \text{ if } a < b$

Start by setting π = -1 and π = 2:

\begin{aligned} &\frac{1}{a}=-1\\&\frac{1}{b}=\frac{1}{2}\end{aligned}

-1 < $\frac{1}{2}$ and -1 < 2, so the statement is not true.

Practice Questions

1. Prove that (π + 1)2 β (π β 1)2 = 4π for all values of π.

Deduction:
(π + 1)2 = (π2 + 2π + 1)
(π β 1)2 = (π2 β 2π + 1)
(π2 + 2π + 1) β (π2 β 2π + 1) = 4π as required.

2. Prove that the statement (π₯ + π¦)2 = π₯2 + π¦2 is not always true for non-zero real numbers π₯ and π¦.

Method 1: Counter example
Let π₯ = 1, π¦ = 2
(1 + 2)2 = 9
12 + 22 = 5
5 β  9 as required

Method 2: Deduction
(π₯ + π¦)2 = (π₯ + π¦)(π₯ + π¦)= π₯2 + 2π₯π¦ + π¦2
π₯2 + 2π₯π¦ + π¦2 β  π₯2 + π¦2 as required

3. Prove that every even number between 6 and 20 inclusive can be expressed as the sum of two prime numbers. The prime numbers can be repeated.

Exhaustion:
6 = 3 + 3
8 = 3 + 5
10 = 3 + 7
12 = 5 + 7
14 = 7 + 7
16 = 3 + 13
18 = 5 + 13
20 = 7 + 13
By exhaustion, every even number between 6 and 20 inclusive can be expressed as the sum of two prime numbers

4. Prove that (π β 2)2 > 2π β 11 for all real values of π.

Deduction:

$\boldsymbol{(n-2)^2=n^2-4n+4}$
\boldsymbol{\begin{aligned}(n-2)^2-(2n-11)&=n^2-6n+15\\&=(n-3)^2+6\\&>0\end{aligned}}

Therefore (π β 2)2 > 2π β 11

5. Three points A(2, 1), B(6, 5) and C(-1, 4) are joined to form a triangle. Prove that the triangle ABC is right-angled.

Method 1: Find the gradients of the line segments AB, BC and AC.

\boldsymbol{\begin{aligned}&m_{AB} = \frac{5\,-\,1}{6\,-\,2}=1\\&m_{BC} = \frac{4\,-\,5}{(-1)\,-\,6}=\frac{1}{7}\\&m_{AC}=\frac{4\,-\,1}{(-1)\,-\,2}=-1\end{aligned}}

πAB x πAC = -1, so the line segments AB and AC are perpendicular. Triangle ABC is right-angled.

Method 2: Use Pythagorasβ theorem.

\boldsymbol{\begin{aligned}&\textbf{AB} = \sqrt{(6-2)^2+(5-1)^2}=4\sqrt{2} \textbf{ units}\\&\textbf{BC} = \sqrt{(-1-6)^2+(4-5)^2}=5\sqrt{2} \textbf{ units}\\&\textbf{AC} = \sqrt{(-1-2)^2+(4-1)^2}=3\sqrt{2} \textbf{ units}\end{aligned}}

$\boldsymbol{(4\sqrt{2})^2+(3\sqrt{2})^2=(5\sqrt{2})^2}$

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