# Proof that Root Two is Irrational: How to Prove that Root Two is Irrational Using Proof By Contradiction

Proving that the square root of two is irrational is an example of a **proof by contradiction**.

The basic method is to suppose that square root two is rational and then show that this leads to a contradiction (an equation that is obviously not true).

## Prove that root two is irrational

A rational number is one that can be written as a ratio of two integers (a fraction):

**Rational number = p / q** , where p and q are integers (whole numbers).

An irrational number can't be written in this way.

**1. Suppose that square root two is rational**

√2 = p / q

where p and q are integers. Assume that this fraction is expressed in its simplest form: i.e.,p and q have no common factors (this will be important later).

**2. Show that this leads to a contradiction**

First, square both sides:

2 = p^{2} / q^{2}

Rearrange:

q^{2} = 2 p^{2}

Now, we can draw some conclusions about p and q.

q^{2} must be even, because it is equal to 2 times an integer.

Therefore, q must also be even. (Because: even x even = even; odd x odd = odd)

Because q is even, we can write q it as 2 times an integer. We call this integer k.

q = 2 k

Therefore,

q^{2} = (2 k)^{2} = 4 k^{2}

We put this back into the equation q^{2} = 2 p^{2 }to get:

4 k^{2} = 2 p^{2}

Rearrange to get an equation for p^{2}

p^{2} = 2 k^{2}

This means that p^{2} is even (because it is equal to two times an integer). Therefore, p is also even.

This is where the contradiction lies. If p is even, and q is even, then they both have q as a factor. Our original assumption stated that p and q had no common factors.

Our assumption that the square root of two is rational has led to a contradiction. Therefore, it must be untrue.

We have proved that the square root of two is irrational.