The fact that there exists irrational number $a,b$ such that $a^b$ is rational is proved by the law of excluded middle, but I read somewhere that irrationality of $\sqrt{2}^{\sqrt{2}}$ is proved constructively. Do you know the proof?
[Math] irrationality of $\sqrt{2}^{\sqrt{2}}$.
constructive-mathematicsirrational-numbersrationality-testing
Related Solutions
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.
The distinction you are missing is that there are numbers, which have certain properties, and then there are numerals, which are sequences of symbols that we use to represent numbers.
The base is a choice about how to represent a number as a numeral. For example, in base 10, the number 100 is represented with the symbol 100. In base 7, it is represented as 202; in base 13 it is represented as 79. In all cases, it is a perfect square; it is an even number; it is a multiple of 5, and so forth, because it is the same number. It is still equal to $36+64$, regardless of whether we write that equation as 36 + 64 = 100 or as 51 + 121 = 202 or as 2A + 4C = 79.
The only questions that change with the base have to do with the representation, with the particular symbols that are used to write down the numeral. The base-10 representation of the number 100 ends with a zero, and the base-7 representation of the same number does not. A number is divisible by 10 if and only if its base-10 numeral ends with a zero, so you can conclude that 100 is divisible by 10; it is divisible by 7 if and only if its base-7 numeral ends in a 0, and so you can conclude that 100 is not divisible by 7.
3 is a rational number and an integer, regardless of how it is written or in what base. It is true that a rational number will have a repeating or terminating representation in an integer base, and an irrational number will not, but that is separate from the question of whether it is a rational number.
Writing 3 in "base $2.5$" means observing:
$$ 3 = \color{red}1\cdot 2.5 + \color{red}0\cdot 1 + \frac{\color{red}1}{2.5} + \frac{\color{red}0}{(2.5)^2} + \cdot\frac{\color{red}1}{(2.5)^3} + \cdots \\ = \color{red}{10.101\ldots}_{(2.5)} $$
but the fact that this representation does or doesn't repeat has nothing to do with whether the number 3 is an integer.
Best Answer
Since this is a well-established result, this is a community wiki post.
Relevant question: Deciding whether $2^{\sqrt2}$ is irrational/transcendental
Kuzmin proved the following claim in 1930:
Unfortunately the paper is in Russian and I failed to find an English translation. A corollary of this is that $2^{\sqrt{2}}$ is transcendental, and so is its square root $\sqrt{2}^{\sqrt{2}}$.
The outlines of both Gelfond and Kuzmin's constructive proof can be found here.
As David Mitra pointed out the comments, Niven's book had a section dedicated to this. I love Niven's book so much. The technique is similar to the adapted proof I posted here, proof by contradiction.
Rough idea about the construction: First assuming $\alpha^{\sqrt{\beta}}$ is algebraic. Then using sufficient large degree Lagrange interpolation polynomial to approximate $e^{(\ln \alpha)x}$ at points $\{a+ b\sqrt{2}\}$ for $a,b\in \mathbb{Z}$. Let the number of points go to infinity the error will go to zero, this shows a transcendental function $\alpha^x$ can be interpolate using countably many algebraic points. Contradiction.