Suppose that we are given a real number $r\in \mathbb R$, then certainly it is either rational or irrational. Let $S\subset \mathbb R$ be a set of real numbers for which it is possible to prove that $s\in S$ is an irrational number. Now I claim the following:
Claim: $S$ is at most countable.
Proof: Let $P$ be the set of all proofs in this universe. It's clear that $P$ is at most countable. Let $P'\subset P$ be a set of proofs that are used to prove that $S$ consists of real numbers whose irrationality can be proved. By Cantor's theorem and hypothesis, it follows that card$(P')=$ card $(S)\lt$ card$(\mathbb R\setminus\mathbb Q)$ and this proves the claim.
In other words, it's impossible to prove irrationality of all irrational numbers.
Is the above idea correct? Please let me know if something is wrong in this. And if my claim is true then it is also possible in similar lines to show that Euler Mascheroni constant may never be proven to be transcendental as there are not enough proofs (at most countable). Thanks.
I am sorry if this question sounds very silly or trivial.
Best Answer
This claim is fallacious. Just replace the adjective “irrational” with “positive” and you would conclude that any set of reals that is provably positive must be countable. If you phrase it just right you could make this a true statement, but it clearly defies most reasonable interpretations of “proving positivity of all (positive) real numbers”.
It is not necessary for every individual element of $S$ to have a distinct certificate of irrationality in order for the entire set to be provably irrational. The same proof can cover all elements of $S$ without explicitly naming each and every one (which is clearly impossible).
For example, define $f: \mathcal P(\mathbb N) \to \mathbb R$ by $$f(A) := e + \sum_{k\in A} \frac1{2^{k!}}.$$ This is irrational — in fact, transcendental — for every value of $A\subseteq \mathbb N$, and there are uncountably many such values ($f$ is injective so the range of $f$ is indeed uncountable).