Showing there is no $\frac{m}{n} \in \mathbb{Q}\ s.t\ ( \frac{m}{n})^2 = 2$

Question

Thm: There is no $\frac{m}{n} \in \mathbb{Q}\ s.t\ ( \frac{m}{n})^2 = 2$.

I know the proof for this is by contradiction. Starts off with supposing $\exists\ m, n\ s.t\ ( \frac{m}{n})^2 = 2,$ that is $m^2 = 2n^2$.

By cancelling, we may assume $m, n \in \mathbb{N}$ with no common factors. In particular, they are not both even.

Rest of the proof proceeds showing m is even and therefore n must be even which is a contradiction.

I may just be tired and missing something simple but why can we assume $m, n$ have no common factors. Not sure where cancelling comes into play.

Answer 1

Here’s a proof that doesn’t depend on reducing the fraction to lowest terms, but does work by contradiction. However, you need to know and believe the Fundamental Theorem of Arithmetic, according to which every positive integer may be written as a product of primes, and in essentially only one way.

Take your suspect equation $\frac mn=\sqrt2$, in other words, $\frac{m^2}{n^2}=2$. Now clear of fractions, to get the contradictory statement $m^2=2n^2$. Why contradictory? Count how many times the prime $2$ appears on the right and on the left. On the left, evenly many times, while on the right, oddly many times. Contradicts the uniqueness part of FTA.

Answer 2

You are considering a quadratic form and setting it to zero, namely $$ m^2 - 2 n^2 = 0. $$

Lemma: we may demand $\gcd(m,n) = 1.$ Proof: if there is any integer solution to $x^2 - 2 y^2 = 0,$ we may then find $g = \gcd(x,y)$ and divide through, giving integers $$ \left( \frac{x}{g} \right)^2 - 2 \; \left( \frac{y}{g} \right)^2 = 0 \; , $$ now in coprime integrs. Name $m = x/g$ and $n = y/g$