TITLE: Irrationality proofs not by contradiction
QUESTION [24 upvotes]: 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?

REPLY [3 votes]: Consider the sum of two reduced fractions. The variables a₁ and a₂ are integers. The variables b₁ and b₂ are positive integers.
$$
\displaystyle \frac{a_1}{b_1}+\frac{a_2}{b_2}=\frac{a_1b_2+a_2 b_1}{b_1 b_2}
$$
If the sum is an integer, then $ b_1|b_2 $
and $ b_2|b_1 $, thus $ b_1=b_2 $.
Summarizing, if the sum of two reduced fractions is an integer, then the denominators are equal. Contrapositively, if the denominators are not equal, then the sum of two reduced fractions is not an integer.
Now let us apply this idea to the nth root of a positive integer, m.
$$
\displaystyle m^{1/n}=\frac{a}{b}\\
\displaystyle m=\frac{a^n}{b^n}\\
\displaystyle \frac{a^n}{b^n}+\frac{-m}{1}=0
$$
Because the sum is an integer, the denominators must be equal.
$$
\displaystyle b^n=1\\
\displaystyle b=1
$$
In conclusion, the only rational solution occurs when $ m^{1/n} $
is an integer. Therefore, $ m^{1/n} $ is either an integer or irrational.