Proof Without contradiction $x^2 = 2$ has no rational solutions

Question

I am in the process of learning real analysis and I was wondering if there was a way to prove no $x \in \mathbb{Q}$ satisfies $x^2 = 2$ without a proof by contradiction.

Answer 1

It depends on what you mean by "without contradiction" - when we reduce proofs of irrationality to their most basic form, they look like:

Let $p/q$ be any rational number. We will show that $(p/q)^2$ is not $2$.

The trouble is "it's not $2$" is often defined to mean "we could derive a contradiction if it were equal to $2$." More or less, one talks about proof by contradiction when a contradiction is used to prove a statement that is not the negation of some other statement (e.g. "all trees have one less edge than vertex") - where essentially we try to prove this statement by saying "it's not true that not all trees have one less edge than vertex" and then cancelling the two "nots" (which is not valid in intuitionistic logic).

The usual proof of this is a direct proof of a negated statement, not a proof by contradiction. In particular, consider the argument:

Let $p/q$ be a rational number. We claim that assuming $(p/q)^2=2$ leads to a contradiction. To see this, first reduce $p/q$ to a fraction $p'/q'$ with coprime numerator and denominator. Note $(p')^2 = 2(q')^2$. Therefore $p'$ must be even so $p'=2r$. Then note $2r^2 = (q')^2$. So $q'$ must be even. This creates a contradiction since $2$ divides both $p'$ and $q'$.

While this formalizes in different ways depending on your formal system, you can think of it as a recipe that says "Give me any rational number and allow me to assume it is $\sqrt{2}$. I will give back a contradiction" - which is exactly what it means to show that $\sqrt{2}$ is irrational, with no fancy steps like double negation elimination involved.

In other words: our goal was to define a sort of "function" that takes two inputs:

1. A rational number $x$

2. The assumption that $x=\sqrt{2}$.

and gives as output a contradiction. That is what our proof describes. (For reference: this particular interpretation as a function is in line with a branch of logic known as type theory)

Answer 2

Only irrational numbers have infinite continued fraction expansions. See https://en.wikipedia.org/wiki/Continued_fraction for more information about continued fractions.

We can write $\sqrt{2}$ as $[1,2,2,2,2,\ldots]$ and therefore it is irrational.

Alternative method using some ring theory:

By Eisenstein's Criterion, the polynomial $x^2-2$ is irreducible over $\mathbb{Z}$. Therefore it is irreducible over $\mathbb{Q}$ by Gauss' Lemma. Hence $x^2-2$ has no rational solutions.