Proving that there is no rational number whose square is two

[r0bHadz]

Homework Statement

The question is to prove that no rational number squared is = 2

Homework Equations

The Attempt at a Solution

I want to understand why for (a/b)^2 = 2, we assume one of the numbers is odd.

Is this because, from approximation we know that root 2 is not a whole number, and If they were both even, we would end up with a whole number?

 

[phyzguy]

If both numbers are even, we would cancel the common factors of 2 until one of them is odd.

 

[RPinPA]

As @phyzguy said, the assumption is that (a/b) is a fraction reduced to lowest form, and that means that there are no common factors of 2.

This is an example of something you'll see in a lot of mathematical proofs called "without loss of generality..." (WLOG). We can always assume that a rational number can be represented as (a/b) with one of them odd, because so long as both are even you can reduce the fraction by dividing both numerator and denominator by 2. So assuming any rational number at all means there's a rational number with that property.

So the mathematician would say, "Assume that $\sqrt 2$ is rational and WLOG that $\sqrt 2 = a/b$ where either a or b or both are odd."

 

[mathwonk]

if you know the lowest term form of a fraction is unique, you are almost done. i.e. if a/b is in lowest terms, i.e. a and b have no common prime factors, then so is a^2/b^2 in lowest terms, since the same prime factors occur here. But then a^2/b^2 = 2/1, and both sides are in lowest terms, hence tops and bottoms are equal, so a^2 = 2 and b^2 = 1. But no integer a can have a^2 = 2. done.