TITLE: Are some irrational numbers closer to rational numbers than others?
QUESTION [6 upvotes]: I'm not even sure this question is meaningful because any irrational number can be approximated by a rational as close as you want.  But given that there are infinitely many irrationals between any two rationals, surely some must be closer to a rational number than others?
Is there a metric by which irrationals can be ranked, or can we compare two irrationals, based on how far they are from a rational?
I'm having trouble wrapping my head around this, can anyone assist?

REPLY [2 votes]: Aside from a set of measure zero there is no much difference.

On the one hand, classical result due to Dirichlet asserts that for
any irrational x (say x between 0 and 1), there are infinite many integers p,q such that  $${|x - p/q|}  < 1/q^{2}$$
On the other hand, for fixed $\epsilon$ > 0 the set of those numbers which can be approximated by order 2 + $\epsilon$ (denoted below by A) has measure zero.
To see this define
$$A_{k,N}^{\epsilon} = \{x; x\in [0,1], |x-k/N| < 1/N^{2 + \epsilon}, 1 \leqq k\leqq N\}$$
One has
$$A = \bigcap_{n=1}^{^\infty}\bigcup_{k,N \geq n}^{\infty}A_{k,N}^{\epsilon}$$
Since
$$|\bigcup_{k,N \geq n}^{\infty}A_{k,N}^{\epsilon}| \leqq \sum_{k,N\geq n}^{}|A_{k,N}^{\epsilon}|$$
and
$$|A_{k,N}^{\epsilon}|\leqq 1/N^{1+\epsilon}$$
we get
$$|\bigcup_{k,N \geq n}^{\infty}A_{k,N}^{\epsilon}| \leqq \sum_{k,N\geq n}^{}|A_{k,N}^{\epsilon}| \leqq \sum_{N=n}^{\infty}1/N^{1+\epsilon}$$
In words, A has really measure zero as advertised.