Irrationality proofs not by contradiction

Per now, I have basically come upon proofs of the irrationality of $\sqrt{2}$ (and so on) and the proof of the irrationality of $e$. However, both proofs were by contradiction.

When thinking about it, it seems like the definition of irrationality itself demands proofs by contradiction. An irrational number is a number that is not a rational number. It seems then that if we were to find direct irrationality proofs, this would rely on some equivalent definition of irrational numbers, not involving rational numbers themselves.

Are there any irrationality proofs not using contradiction?

Solution 1:

Cantor's diagonal argument shows that for any countable list of numbers, one can construct a number not on that list. Cantor used this argument to show, for example, that there are transcendental numbers, since one may describe a way to list all the polynomials with integer coefficients and their roots and hence to list all the algebraic numbers.

One sometimes hears it asserted that Cantor's proof of the existence of transcendental numbers is a "pure-existence" proof, showing merely that transcendental numbers exist, but not that any particular number is transcendental. But that view is not correct, for the argument is completely constructive: one may explicitly describe an enumeration of the algebraic numbers and the diagonal procedure produces a definite real number that is not algebraic. (I once saw an article, I think in one of the MAA volumes, with the title something like "0.24543... is transcendental", where they implemented Cantor's actual algorithm.)

The relevance to this question is that Cantor diagonalization also can be used to prove that specific real numbers are not rational, by producing real numbers that are explicitly different from every rational number. Specifically, we may enumerate the rational numbers as $q_k$ in any of the usual effective manners, and define a real number $z$ so that the $k$-th digit of $z$ is $4$, say, if $r_k$ does not have $k$-th digit $4$, and otherwise the $k$-th digit of $z$ is $5$. It now follows by construction that $z\neq r_k$ for each $k$, which is what it means for $z$ to be irrational.

Solution 2:

An irrational number can be defined as having an infinite continued fraction expansion.

The continued fraction of $\sqrt{2}$ is [1, 2, 2, 2, ...] so it's irrational.

Solution 3:

A very simple proof follows from Gauss's lemma (polynomial)

If $p$ is prime, then clearly, $\sqrt{p}$ is a root of $f(x)=x^2-p$. Gauss's lemma shows that $f(x)=x^2-p$ has no rational roots (since it clearly has no integer roots). Thus $\sqrt{p}$ is irrational.

Also note the proof of Gauss' lemma is not a proof by contradiction so this entire result can be proven directly.