## Theorem 2.4.1: No Square Roots in Q

### Proof:

Suppose there was such an*x*. Being a rational number, we can write it as

*x = a / b*(with no common divisors)

*x*we have

^{2}= x * x = 2*a*^{2}= 2 b^{2}

*a*is even, and therefore

^{2}*a*must be even as well. (Can you prove this ?). Hence,

*a = 2 c*for some integer*c*.

*4 c*, or^{2}= 2 b^{2}*2 c*^{2}= b^{2}

*b*is even.. But then both

*a*and

*b*are divisible by 2. That's a contradiction, because

*a*and

*b*were supposed to have no common divisors.