Proof That the Square Root of 2 is Irrational — Definition, Formula & Examples

The proof that √2 is irrational is a classic argument by contradiction showing that no fraction

\frac{a}{b}

with integer numerator and denominator can ever equal

\sqrt{2}

. It works by assuming such a fraction exists in lowest terms, then showing both

a

and

b

must be even — contradicting the lowest-terms assumption.

Assume for contradiction that $\sqrt{2} = \frac{a}{b}$ where $a, b \in \mathbb{Z}$ , $b \neq 0$ , and $\gcd(a, b) = 1$ . Squaring both sides gives $2b^2 = a^2$ , so $a^2$ is even, which implies $a$ is even. Writing $a = 2k$ , we get $2b^2 = 4k^2$ , hence $b^2 = 2k^2$ , so $b$ is also even. This contradicts $\gcd(a, b) = 1$ , so no such fraction exists and $\sqrt{2}$ is irrational.

How It Works

The proof uses a technique called proof by contradiction (also known as reductio ad absurdum). You start by assuming the opposite of what you want to prove — that

\sqrt{2}

is rational. You then follow logical steps until you reach an impossible situation. The key insight is that if

a^2

is even, then

a

itself must be even, because the square of any odd number is always odd. Once both

a

and

b

turn out to be even, the fraction

\frac{a}{b}

could not have been in lowest terms, which is the contradiction that completes the proof.

Example

Problem: Prove that √2 is irrational.

Step 1: Assume the opposite: Suppose √2 is rational, so we can write it as a fraction a/b in lowest terms, meaning gcd(a, b) = 1.

\sqrt{2} = \frac{a}{b}, \quad \gcd(a, b) = 1

Step 2: Square both sides: Squaring gives 2 = a²/b², which rearranges to a² = 2b². This means a² is even, so a must be even. Write a = 2k.

a^2 = 2b^2 \implies a = 2k

Step 3: Substitute and derive contradiction: Substituting a = 2k gives (2k)² = 2b², so 4k² = 2b², which simplifies to b² = 2k². Now b² is even, so b is also even.

b^2 = 2k^2 \implies b \text{ is even}

Step 4: Contradiction: Both a and b are even, so they share a factor of 2. This contradicts gcd(a, b) = 1. Therefore our assumption was false.

Answer: Since the assumption that √2 is rational leads to a contradiction, √2 must be irrational.

Why It Matters

This proof is often the first example of proof by contradiction that students encounter, making it a gateway to formal mathematical reasoning. The same technique generalizes to prove that

\sqrt{p}

is irrational for any prime

p

, and it appears in courses from discrete mathematics to real analysis.

Common Mistakes

Mistake: Claiming that because a² is even, a is even without justification.

Correction: You need to verify this fact. If a were odd, then a = 2m + 1 and a² = 4m² + 4m + 1, which is odd — contradicting a² being even. This step is essential to the proof's validity.