claimstandardinferred_dep
Dirichlet pigeonhole gives lower bound exponent 2
Dirichlet gives infinitely often, so exponent ≥ 2
the easy lower bound half of the theorem
Claim·§4 Choice of parameters and proof of Theorem~thm:main·inside thmIrrationality exponent of pi is 2
Statement (paper)
For every integer there exist and with . After reduction to lowest terms the error is at most the inverse square of the reduced denominator, and since these errors tend to zero as and is irrational, the reduced denominators are unbounded. Hence every exponent occurs infinitely often in the defining inequality, so .
In plain English
For every there are integers and with . Since is irrational, the reduced denominators of these approximations are unbounded, so every exponent occurs infinitely often and .
Why it's here
This is the easy half of the theorem: Dirichlet's pigeonhole approximation supplies infinitely many rationals with error at most the inverse square of the denominator, which shows the exponent cannot be below 2. Irrationality of (the previous claim) ensures the errors are nonzero and the reduced denominators grow, and it complements the contradiction argument that gives the upper bound.
Classified standard:
This is Dirichlet’s approximation theorem, proved by the pigeonhole principle; its consequence that every irrational number has irrationality exponent at least 2 is standard (Hardy and Wright, An Introduction to the Theory of Numbers).