Proof that √2 is Irrational: GCSE Higher Maths Step by Step

What does it mean for a number to be rational or irrational?

A rational number can be written as p/q where p and q are integers and q ≠ 0. Examples: ½, −3, 0.75 (= 3/4), and all terminating or recurring decimals.

An irrational number cannot be written in that form. Its decimal expansion continues forever without repeating. Examples: √2, √3, π, and e.

What is proof by contradiction?

Proof by contradiction starts by assuming the opposite of what you want to prove. You then do careful algebra until you arrive at a statement that is clearly false (a contradiction). Since the assumption led to a false statement, the assumption itself must be wrong — so the original claim is true.

At GCSE Higher, this is the only context where you are expected to use this technique.

How do you prove that √2 is irrational?

Method at a glance:

1. Assume the opposite. Suppose √2 is rational, so √2 = p/q where p and q are integers with no common factor (the fraction is in its lowest terms).
2. Square both sides. 2 = p²/q², so p² = 2q².
3. Deduce p is even. Since p² is even (it equals 2q²), p itself must be even — write p = 2k for some integer k.
4. Substitute back. (2k)² = 2q² → 4k² = 2q² → q² = 2k².
5. Deduce q is even. Since q² = 2k² is even, q must also be even.
6. State the contradiction. Both p and q are even, so 2 is a common factor — contradicting our assumption that p/q is in its lowest terms.
7. Conclude. The assumption is false. Therefore √2 is irrational. □

Why does "p² is even → p is even" hold?

Suppose p were odd, so p = 2m + 1. Then p² = (2m + 1)² = 4m² + 4m + 1, which is odd. So if p² is even, p cannot be odd — it must be even.

What does the full written proof look like?

Here is the proof written out as you would in an exam:

Assume, for contradiction, that √2 is rational. Then √2 = p/q for integers p and q with no common factor.

Squaring: 2 = p²/q², so p² = 2q².

Since p² is a multiple of 2, p must be even. Write p = 2k.

Substituting: (2k)² = 2q², so 4k² = 2q², giving q² = 2k².

Since q² is a multiple of 2, q must also be even.

But then p and q are both even, so they share the factor 2. This contradicts the assumption that p/q has no common factor.

Therefore, √2 is irrational. □

Every line earns marks in an exam. Do not skip steps.

Which parts do GCSE mark schemes reward?

A single-sentence conclusion — "This contradicts the fact that p/q is in its lowest terms, so √2 is irrational" — is essential for full marks.

Can you extend this method to other square roots?

Yes. The same argument works for any √n where n is not a perfect square: assume √n = p/q, square, deduce that p² is a multiple of n, argue about shared factors, and reach a contradiction. However, the argument must be adapted for primes: when n is prime, "p² is a multiple of n → p is a multiple of n" follows directly from the Fundamental Theorem of Arithmetic. At GCSE, you are only expected to know the √2 case.

Frequently asked questions

Does this proof appear at Foundation or Higher tier?

Proof that √2 is irrational is a Higher tier only topic. It appears on both AQA and OCR Higher specifications as an example of proof by contradiction. Foundation students are expected to classify numbers as rational or irrational, but do not need to prove it algebraically.

Why does "p² even → p even" not follow just from an example?

You cannot prove a statement for all integers by checking a few examples — that is exactly the point of algebraic proof. The argument works by ruling out the alternative: if p were odd, p² would be odd (as shown), so p² cannot be even if p is odd. Contrapositive reasoning is valid and complete.

Is π also irrational and can you prove it the same way?

π is irrational, but its proof is far beyond GCSE — it requires tools from university-level mathematics. The proof that √2 is irrational is special because it uses only integer arithmetic and the idea of lowest terms, both accessible at GCSE Higher.

What if I assumed √2 = p/q without saying "no common factor"?

The proof would not work — you would not reach a contradiction, because any fraction can be written with a common factor. The condition "no common factor" (i.e. p/q in its lowest terms) is what makes the final step — "both even, so they share 2" — a genuine contradiction.

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