Let's suppose √2is a rational number. Then we can write it √2 =a/bwherea,bare whole numbers,bnot zero.

We additionally assume that thisa/bis simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order fora/bto be in simplest terms, both ofaandbcannot be even. One or both must be odd. Otherwise, we could simplifya/bfurther.

From the equality √2 =a/bit follows that 2 =a2/b2, or a2= 2 ·b2. So the square ofais an even number since it is two times something.

From this we know thataitself isalsoan even number. Why? Because it can't be odd; ifaitself was odd, thena·awould be odd too. Odd number times odd number is always odd. Check it if you don't believe me!

Okay, ifaitself is an even number, thenais 2 times some other whole number. In symbols,a= 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction.

If we substitutea= 2k into the original equation 2 =a2/b2, this is what we get:

2=(2k)2/b22=4k2/b22*b2=4k2b2=2k2

This means thatb2is even, from which follows again thatbitself is even. And that is a contradiction!!!

WHY is that a contradiction? Because we started the whole process assuming thata/bwas simplified to lowest terms, and now it turns out thataandbboth would be even. We ended at a contradiction; thus our original assumption (that √2is rational) is not correct. Therefore √2cannot be rational