Given the following theorem:
Let $x\in\mathbb{R}.$ Is there a $\delta>0$ and a rational sequence $\{p_n/q_n\}_n$ with $p_n/q_n\neq x$ and $$\left|x-\frac{p_n}{q_n}\right|<\frac{1}{q_n^{1+\delta}}$$ for all $n$, then x is irrational.
Who did find this statement and when? Is there a proof somewhere?
On
The true theorem is
Theorem (Liouville).-For every real algebraic number $x$ of degree $n≥2$ exists a positive number $c (x)$ such that $| x-p / q |>\frac{ c (α)}{ q ^ n}$ holds for all rational $\frac pq (q> 0)$.
By definition, a Liouville number $x$ is such that for all $n$ large enough there is a rational $\frac pq$ with $q\gt 1$ such that $|x- \frac pq|\le \frac{1}{q^n}$ and this such a number $x$ is necessarily transcendental.
The theorem of Roth says that for every irrational $x$ algebraic and all $\epsilon> 0$ the inequality $$|x-\frac pq | <\frac{1}{ q ^{2 + ε}}$$ has only a finite number of solutions in irreducible rational $\frac pq$ which implies that for all $\epsilon\gt 0$ there exists a positive constant $C (x, ε)$ such that the inequality $| x-\frac pq |> \frac{C (x, ε)}{ q^{2 + ε}}$ is holds for all rational $\frac pq$ with $q>0$.
This is a profound theorem by which Roth won the Fields Medal (International Congress of Mathematicians, Edinburgh, 1958).
On
No, not at all, Piquito. You have answered mentioning Liouville theorem. Your enunciation of this theorem, above, is correct, but the correct answer to the original question is Dirichlet's rarional approximations theorem, which differs from Liouville's one in both the exponent and the order "assumption->conclusion": Dirichlet's theorem says "If there are infinitely many distinct rationals $p/q$ such that, for some $\delta >0$, $|x-p/q| < 1/q^{1+\delta}$, then $x$ is irrational.
This is an irrationality criterion, rather than anything like "if x is irrational, then..."
Fabio M. S. Lima
P.S.: Dirichlet theorem was stated in 1842!
I'd suspect this to be known to Liouville (as it "smells" a lot like his approximation theorem of Comptes Rendue 18 (1844)), but this specific observation may be even older.
The proof is simple: Assume $x=\frac ab$ is rational. Then for any rational $\frac pq\ne x$ of a sequence as given, we have $$\frac1{q^{1+\delta}}>\left|x-\frac pq\right|=\frac{|aq-bp|}{bq}\ge \frac 1{bq}, $$ hence $$ q<b^{1/\delta}.$$ But with bounded denominator there are even only finitely many rationals with the weaker condition $\left|x-\frac pq\right|<1$.