# Proof by Contradiction (Edexcel IAL Maths P4)

> Edexcel International A-Level Mathematics · IAL Maths P4
> Source: https://www.owlsprep.com/study/edexcel-ial-math-p4-proof/

This guide covers the only proof method required for Edexcel IAL Mathematics Pure 4 (P4): proof by contradiction. It includes mandatory standard proofs, worked unfamiliar examples, and exam-focused tips to avoid common errors.

**Prerequisites:** [Basic number theory (even/odd, primes, rational/irrational numbers)](https://www.owlsprep.com/study/edexcel-ial-math-p1-number-theory/); [P2 foundational logical reasoning for proofs](https://www.owlsprep.com/study/edexcel-ial-math-p2-proof-basics/)

## Learning objectives

- Recall the formal 4-step structure of proof by contradiction
- Reproduce the mandatory proofs of √2 irrationality and infinite primes
- Apply proof by contradiction to unfamiliar A2-level problems
- Avoid common marking pitfalls to secure full marks on proof questions

## Structure of Proof by Contradiction

**Proof by Contradiction** — A deductive proof method where you assume the negation (opposite) of the statement you want to prove is true. You follow logical steps to derive an impossible result (contradiction), which proves your initial assumption was false, so the original statement is true.

*Example:* Proving there is no largest integer by assuming a largest integer N exists, then showing N+1 is larger, creating a contradiction.

All Edexcel P4 proof by contradiction questions follow a standard 4-step structure you should apply consistently to secure full method marks:

1. State the original statement you intend to prove.
2. Assume the **negation** (opposite) of the statement is true.
3. Use logical algebraic or numerical reasoning to derive a contradiction.
4. Conclude your initial assumption was false, so the original statement is true.

**Worked example:** Prove that there is no smallest positive rational number.

1. Step 1: State the original statement: There is no smallest positive rational number.
2. Step 2: Assume the negation is true: Let $a$ be the smallest positive rational number.
3. $$b = \frac{a}{2}$$
4. Since $a$ is non-zero and rational, $b$ is also positive and rational.
5. Step 3: Derive contradiction: $b < a$, but we assumed $a$ was the smallest positive rational. This is impossible.
6. Step 4: Conclusion: Our initial assumption is false, so there is no smallest positive rational number.

> **Exam tip:** Always explicitly state the contradiction and your final conclusion, as these are explicitly allocated marks in Edexcel P4 mark schemes.

*Calculator:* allowed

## Mandatory Standard Proofs

The Edexcel P4 specification explicitly requires you to memorize and reproduce two standard proofs: the irrationality of √2, and the infinity of primes. These are often tested directly, so learn them word-for-word.

**Worked example:** Prove that √2 is irrational.

1. Step 1: Original statement: √2 is an irrational number.
2. Step 2: Assume negation: √2 is rational, so it can be written as $\sqrt{2} = \frac{a}{b}$ where $a$ and $b$ are positive integers with no common factors (coprime, lowest terms).
3. $$2 = \frac{a^2}{b^2} \implies a^2 = 2b^2$$
4. $a^2$ is even, so $a$ must be even. Let $a = 2k$ where $k$ is an integer.
5. $$(2k)^2 = 2b^2 \implies 4k^2 = 2b^2 \implies b^2 = 2k^2$$
6. $b^2$ is even, so $b$ is also even.
7. Step 3: Contradiction: We assumed $a$ and $b$ are coprime, but we have shown both are even (so they share a common factor of 2). This is impossible.
8. Step 4: Conclusion: Our initial assumption is false, so √2 is irrational.

**Worked example:** Prove that there are infinitely many prime numbers.

1. Step 1: Original statement: There are infinitely many prime numbers.
2. Step 2: Assume negation: There is a finite list of all primes: $p_1, p_2, p_3, ..., p_n$.
3. $$N = p_1 \times p_2 \times p_3 \times ... \times p_n + 1$$
4. Consider the number $N$ constructed as above.
5. Step 3: Contradiction: When $N$ is divided by any prime $p_i$ on our list, it leaves a remainder of 1. So $N$ is either prime itself, or has a prime factor not on our original list. This contradicts the assumption our list contained all primes.
6. Step 4: Conclusion: Our initial assumption is false, so there are infinitely many prime numbers.

> **Exam tip:** For the √2 proof, you must explicitly state that $a$ and $b$ are coprime (in lowest terms) — this is required for the contradiction step to be valid, and omitting it will lose marks.

*Calculator:* allowed

## Unfamiliar Proof by Contradiction Problems

Most P4 exams include an unfamiliar proof by contradiction question, usually relating to number theory, algebra, or inequalities. The same 4-step structure applies, and you should look for contradictions involving parity (even/odd), divisibility, or impossible integer values.

**Worked example:** Prove that if $n$ is an integer and $n^2$ is odd, then $n$ is odd.

1. Step 1: Original statement: For all integers $n$, if $n^2$ is odd then $n$ is odd.
2. Step 2: Assume negation: There exists an integer $n$ where $n^2$ is odd, but $n$ is even.
3. If $n$ is even, we can write $n = 2k$ for integer $k$.
4. $$n^2 = (2k)^2 = 4k^2 = 2(2k^2)$$
5. This means $n^2$ is even, which contradicts our assumption that $n^2$ is odd.
6. Step 3: Contradiction confirmed, so our initial assumption is false. Therefore the original statement is true.

> **Exam tip:** When tackling unfamiliar proofs, start by writing down the exact negation of the statement first. For 'if A then B' statements, the negation is 'A is true and B is false' — this is a common source of error.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Assuming the wrong negation of the original statement, e.g. negating 'all primes are odd' as 'all primes are even' instead of 'there exists at least one even prime'.
  - Why it fails: Incorrect negations lead to invalid proofs, as you are not contradicting the right statement, and you will lose all method marks.
  - Correct: For universal statements ('for all x, P(x) is true'), the negation is 'there exists at least one x where P(x) is false'. For conditional statements ('if A then B'), the negation is 'A is true and B is false'.
- **Wrong:** Omitting the 'coprime' condition when proving √2 is irrational.
  - Why it fails: Without stating that a and b are in lowest terms, you cannot derive a contradiction when you find both are even, so the proof is incomplete.
  - Correct: Explicitly write 'where a and b are positive integers with no common factors (coprime)' when stating your assumption for the √2 proof.
- **Wrong:** Failing to explicitly identify the contradiction and restate the conclusion at the end of the proof.
  - Why it fails: Edexcel mark schemes allocate 1-2 marks specifically for stating the contradiction and final conclusion, so omitting these costs easy marks.
  - Correct: End every proof with two clear sentences: first, state exactly what the contradiction is, second, state that the assumption is false so the original statement is true.
- **Wrong:** Using circular reasoning in the proof, e.g. using the original statement you are trying to prove as part of your reasoning.
  - Why it fails: Circular reasoning is logically invalid, and will result in zero marks for the proof.
  - Correct: Only use known facts (e.g. even numbers are divisible by 2, primes have only 1 and themselves as factors) and your initial assumption in your reasoning steps.
- **Wrong:** For the infinite primes proof, stating that N must be prime, instead of that N has a prime factor not on the original list.
  - Why it fails: N is not always prime (e.g. 2×3×5×7×11×13 + 1 = 30031 = 59×509, which is composite), so this is a factual error that invalidates the proof.
  - Correct: State that N is either prime itself, or has a prime factor not present on your original finite list of primes, both of which contradict the assumption that your list was complete.

## Cheatsheet

| Proof Type | Key Steps | Mandatory Marking Details |
| --- | --- | --- |
| General Proof by Contradiction | 1. State original statement 2. Assume negation is true 3. Derive contradiction 4. Conclude assumption is false | Explicit contradiction statement, final conclusion matching the question |
| √2 is Irrational | 1. Assume √2 = a/b (coprime integers) 2. Show a² even → a even 3. Substitute a=2k, show b even 4. Contradiction (a,b share factor 2) | Explicitly state a and b are coprime (lowest terms) |
| Infinite Primes | 1. Assume finite list of primes p₁ to pₙ 2. Construct N = product of all primes + 1 3. Show N has prime factor not on list 4. Contradiction | Note N may be prime *or* have a new prime factor, do not claim N is always prime |

## What's next

Now that you have mastered proof by contradiction for Edexcel IAL P4, you are ready to move on to the other core topics in the unit. Proof by contradiction is often combined with number theory and algebra concepts from earlier units, so revise those foundations if you struggled with any worked examples. Practice writing full proofs under timed conditions to ensure you do not omit mandatory marking points, and attempt all past P4 proof questions to familiarize yourself with the range of problems that can be asked. This skill will also be useful if you go on to study Further Mathematics, where more advanced proof methods are covered.

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