Let's suppose √2 were a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero. We additionally make it so that this a/b is simplified to the lowest terms, since that can obviously be done with any fraction.

It follows that 2 = a^{2}/b^{2}, or a^{2} = 2 * b^{2}. So the square of *a* is an even number since it is two times something. From this we can know that *a* itself is also an even number. Why? Because it can't be odd; if a itself was odd, then *a* * *a* would be odd too. Odd number times odd number is always odd. Check if you don't believe that!

Okay, if *a* itself is an even number, then *a* is 2 times some other whole number, or *a* = 2k where k is this other number. We don't need to know exactly what k is; it won't matter. Soon is coming the contradiction:

If we substitute a = 2k into the original equation 2 = a^{2}/b^{2}, this is what we get:

2 | = | (2k)^{2}/b^{2} |

2 | = | 4k^{2}/b^{2} |

2*b^{2} | = | 4k^{2} |

b^{2} | = | 2k^{2}. |

This means b^{2} is even, from which follows again that b itself is an even number!!!