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Square Root of 2 Is Irrational — Definition, Formula & Examples

The square root of 2 is irrational means that 2\sqrt{2} cannot be written as a fraction ab\frac{a}{b} where aa and bb are integers. This is typically proved by contradiction and is one of the most famous results in all of mathematics.

There exist no integers aa and bb with b≠0b \neq 0 such that (ab)2=2\left(\frac{a}{b}\right)^2 = 2. Equivalently, 2∉Q\sqrt{2} \notin \mathbb{Q}, where Q\mathbb{Q} denotes the set of rational numbers.

How It Works

The proof uses contradiction: assume 2=ab\sqrt{2} = \frac{a}{b} in lowest terms, meaning aa and bb share no common factor. Squaring both sides gives a2=2b2a^2 = 2b^2, which means a2a^2 is even. Since the square of an odd number is always odd, aa itself must be even. Write a=2ka = 2k, substitute back, and you get 4k2=2b24k^2 = 2b^2, so b2=2k2b^2 = 2k^2, meaning bb is also even. But now both aa and bb are even, contradicting the assumption that the fraction was in lowest terms. Therefore no such fraction exists.

Example

Problem: Prove that 2\sqrt{2} is irrational.
Assume the opposite: Suppose 2\sqrt{2} is rational, so 2=ab\sqrt{2} = \frac{a}{b} where a,ba, b are integers with no common factor and b≠0b \neq 0.
2=ab,gcd⁡(a,b)=1\sqrt{2} = \frac{a}{b}, \quad \gcd(a, b) = 1
Square both sides: Squaring gives 2=a2b22 = \frac{a^2}{b^2}, so a2=2b2a^2 = 2b^2. This means a2a^2 is even, which forces aa to be even. Write a=2ka = 2k.
a2=2b2  ⟹  (2k)2=2b2  ⟹  4k2=2b2a^2 = 2b^2 \implies (2k)^2 = 2b^2 \implies 4k^2 = 2b^2
Derive a contradiction: Simplifying gives b2=2k2b^2 = 2k^2, so b2b^2 is even, meaning bb is also even. Both aa and bb are even, so they share a factor of 2. This contradicts gcd⁡(a,b)=1\gcd(a, b) = 1.
b2=2k2  ⟹  b is even  ⟹  gcd⁡(a,b)≥2  — contradictionb^2 = 2k^2 \implies b \text{ is even} \implies \gcd(a,b) \geq 2 \; \text{— contradiction}
Answer: The assumption leads to a contradiction, so 2\sqrt{2} is irrational.

Why It Matters

This proof is often a student's first encounter with proof by contradiction, a technique used throughout algebra, analysis, and number theory. The same strategy generalizes to show that p\sqrt{p} is irrational for any prime pp, and it connects to deeper ideas about which numbers can be expressed as ratios of integers.

Common Mistakes

Mistake: Claiming that because 2=1.41421…\sqrt{2} = 1.41421\ldots never terminates, it must be irrational.
Correction: A non-terminating decimal can still be rational if it repeats (e.g., 13=0.333…\frac{1}{3} = 0.333\ldots). The proof requires showing no fraction equals 2\sqrt{2}, not just observing the decimal expansion.