# prove that root 2 is irrational • 128

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!!!

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