Maths in a minute: The square root of 2 is irrational


The proof of the irrationality of root 2 is often attributed to Hippasus of Metapontum, a member of the Pythagorean cult. He is said to have been murdered for his discovery (though historical evidence is rather murky) as the Pythagoreans didn't like the idea of irrational numbers.

Here’s one of the most elegant proofs in the history of maths. It shows that $\sqrt {2}$ is an irrational number, in other words, that it cannot be written as a fraction $a/b$ where $a$ and $b$ are whole numbers.

We start by assuming that $\sqrt {2}$ can be written as a fraction $a/b$ and that $a$ and $b$ have no common factor — if they did, we could simply cancel it out. In symbols,

$\frac{a}{b} = \sqrt {2}.$

Squaring both sides gives

$\frac{a^2}{b^2} = 2.$

and multiplying by $b^2$ gives

$a^2 = 2b^2.$

This means that $a^2$ is an even number: it’s a multiple of $2$. Now if $a^2$ is an even number, then so is $a$ (you can check for yourself that the square of an odd number is odd). This means that $a$ can be written as $2c$ for some other whole number $c$. Therefore,

  \[ 2b^2 = a^2 = (2c)^2 = 4c^2. \]    

Dividing through by $2$ gives

  \[ b^2 = 2c^2. \]    

This means that $b^2$ is even, which again means that $b$ is even. But then, both $a$ and $b$ are even, which contradicts the assumption that they contain no common factor: if they are both even, then they have a common factor of $2$. This contradiction implies that our original assumption, that $\sqrt {2}$ can be written as a fraction $a/b$ must be false. Therefore, $\sqrt {2}$ is irrational.


There's a shorter proof which requires unique factorization of integers, while ignoring the assumption that a and b have no common factors.

Given a^2 = 2b^2, neither have the same number of 2s as a factor, therefore they can't be equal.

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