# Proof: $\sqrt{2}$ is irrational

## by Subhomoy Haldar

# Theorem

$\sqrt{2}$ is irrational.

# Proof by Contradiction

We start by assuming the opposite. Suppose $\sqrt{2}$ is rational and can be represented as a reduced fraction: $\frac{p}{q}$ where, $p, q \in \mathbb{Z}$, $\gcd(p, q) = 1$ and $q \neq 0$. Therefore, if we write down the equality and square both sides, we must have

$$ 2 = (\sqrt{2})^2 = \frac{p^2}{q^2} $$

Multiplying both sides by $q^2$, we have

$$ 2q^2 = p^2 $$

Since $p^2$ is of the form $2k$ where $k = q^2 \in \mathbb{Z}$, it must be true that $2 | p^2$. Since $2$ is a prime and $p^2$ is a square number, $2$ must also divide $p$ or $4 | p^2$. This is because the prime factorisation of a square integer has even multiples of all the constituent primes.

In other words, $p$ is **even**.

Therefore, we can write $p^2 = 4l$ where $l=\frac{p^2}{4}, l \in \mathbb{Z}$. The equation eventually becomes

$$ 2q^2 = 4l $$ $$ \text{or, }q^2 = 2l $$

which implies $q^2$ is **even** and therefore, $q$ is also **even**. This conclusion is derived by applying a chain of reasoning similar to that exercised earlier for $p^2$ and $p$.

The contradiction that arises is that *both* $p$ and $q$ appear to be even (that is, have a common factor of $2$), even though we had assumed that $\gcd(p, q) = 1$. Therefore, our assumption is false, and $\sqrt{2}$ is indeed, irrational.

## Q.E.D.

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