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If you build a simple square with sides of length 11, its diagonal is 2\sqrt{2} by the Pythagorean theorem. Around 500 BCE, the Greek philosopher Hippasus discovered that no fraction on Earth can ever measure this exact length, shattering the Pythagorean belief that whole numbers ruled everything.
A rational number is any value that can be written as a ratio of two integers ab\frac{a}{b}, where b≠0b \neq 0. If a number cannot be written this way, it is irrational.
To prove 2\sqrt{2} is irrational, we use proof by contradictionβ€”we assume the opposite of what we want to prove, and show that this assumption logically destroys itself.
We also need one foundational lemma: if a2a^2 is even, then aa must be even. An odd number times an odd number is always odd ((2k+1)2=4k2+4k+1=2(2k2+2k)+1(2k+1)^2 = 4k^2+4k+1 = 2(2k^2+2k)+1), so an even square can only come from an even integer.
How does this simple parity rule unravel the nature of 2\sqrt{2}?

The Classic Proof: 2\sqrt{2}

Assume 2\sqrt{2} is rational. That means we can write 2=ab\sqrt{2} = \frac{a}{b}, where aa and bb are positive integers with no common factors other than 11 (meaning the fraction is in lowest terms, or gcd⁑(a,b)=1\gcd(a, b) = 1).
Squaring both sides gives 2=a2b22 = \frac{a^2}{b^2}, which rearranges to a2=2b2a^2 = 2b^2. Because a2a^2 is a multiple of 22, a2a^2 is even, meaning aa itself must be even.
Since aa is even, we can rewrite it as a=2ka = 2k for some integer kk. Substituting this back gives (2k)2=2b2(2k)^2 = 2b^2, which simplifies to 4k2=2b24k^2 = 2b^2, or b2=2k2b^2 = 2k^2.
Now b2b^2 is also a multiple of 22, which forces bb to be even. But if both aa and bb are even, they share a common factor of 22, directly contradicting our starting condition that gcd⁑(a,b)=1\gcd(a, b) = 1.
A common mistake is forgetting to state that ab\frac{a}{b} is in lowest terms at the very beginning. Without assuming gcd⁑(a,b)=1\gcd(a, b) = 1, finding that both aa and bb are even wouldn't be a contradictionβ€”it would just mean the fraction could be reduced.