# Famous Theorems of Mathematics/√2 is irrational

The square root of 2 is irrational, $\sqrt{2} \notin \mathbb{Q}$

## Proof

Assume for the sake of contradiction that $\sqrt{2} \in \mathbb{Q}$. Hence $\sqrt{2} = \frac{a}{b}$ holds for some a and b that are coprime.

This implies that $2 = \frac{a^2}{b^2}$. Rewriting this gives $2b^2 = a^2 \!\,$.

Since the left-hand side of the equation is divisible by 2, then so must the right-hand side, i.e., $2 | a^2$. Since 2 is prime, we must have that $2 | a$.

So we may substitute a with $2A$, and we have that $2b^2 = 4A^2 \!\,$.

Dividing both sides with 2 yields $b^2 = 2A^2 \!\,$, and using similar arguments as above, we conclude that $2 | b$. However, we assumed that $\sqrt{2} = \frac{a}{b}$ such that that a and b were coprime, and have now found that $2 | a$ and $2 | b$; a contradiction.

Therefore, the assumption was false, and $\sqrt{2}$ cannot be written as a rational number. Hence, it is irrational.

## Another Proof

The following reductio ad absurdum argument is less well-known. It uses the additional information √2 > 1.

1. Assume that √2 is a rational number. This would mean that there exist integers m and n with n ≠ 0 such that m/n = √2.
2. Then √2 can also be written as an irreducible fraction m/n with positive integers, because √2 > 0.
3. Then $\sqrt{2} = \frac{\sqrt{2}\cdot n(\sqrt{2}-1)}{n(\sqrt{2}-1)} = \frac{2n-\sqrt{2}n}{\sqrt{2}n-n} = \frac{2n-m}{m-n}$, because $\sqrt{2}n=m$.
4. Since √2 > 1, it follows that m > n, which in turn implies that m > 2nm.
5. So the fraction m/n for √2, which according to (2) is already in lowest terms, is represented by (3) in strictly lower terms. This is a contradiction, so the assumption that √2 is rational must be false.

Similarly, assume an isosceles right triangle whose leg and hypotenuse have respective integer lengths n and m. By the Pythagorean theorem, the ratio m/n equals √2. It is possible to construct by a classic compass and straightedge construction a smaller isosceles right triangle whose leg and hypotenuse have respective lengths m − n and 2n − m. That construction proves the irrationality of √2 by the kind of method that was employed by ancient Greek geometers.

## Notes

• As a generalization one can show that the square root of every prime number is irrational.
• Another way to prove the same result is to show that $x^2-2$ is an irreducible polynomial in the field of rationals using Eisenstein's criterion.