To prove irrational numbers exist, we will prove √2 is irrational.

We will do this by assuming √2 is rational and proving that this assumption is false.

By definition, a rational number can be represented as a ratio of two integers. Any such ratio can be reduced to the lowest common denominator, i.e.; having no common factors.

If √2 is rational, the following equation can be proven either true or false.

**a/b = √2**

Where **a** and **b** are integers **with no common factors.**

(a/b)^{2} = (√)^{2}

a^{2}/b^{2} = 2

a^{2} = 2b^{2}

ergo a^{2} is even, since it is twice b^{2}

ergo a is even, since only even numbers have even squares

Since a is even, we can replace it with 2c where c is half of a

**Restate the equation:**

2c/b = √2

(2c/b)^{2} = (√2)^{2}

4c^{2}/b^{2} = 2

4c^{2} = 2b^{2}

2c^{2} = b^{2}

ergo b^{2} is even since it is twice c^{2}

ergo b is even, since only even numbers have even squares

We had postulated that a and b had no common factors. However, we proved both are even integers and thus had the common factor of 2. This demonstrates the postulated equation is false.

**Ergo, postulating √2 is rational is false,** and it must therefore be irrational, **Q.E.D.**