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 = a2/b2, or a2 = 2 * b2. 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 = a2/b2, this is what we get:
| 2 | = | (2k)2/b2 |
| 2 | = | 4k2/b2 |
| 2*b2 | = | 4k2 |
| b2 | = | 2k2. |
This means b2 is even, from which follows again that b itself is an even number!!!
