TITLE: Can we remove the absolute value when doing Diophantine Approximation?
QUESTION [5 upvotes]: This is a very general question. For Diophantine Approximation propositions, the statements always include the absolute value sign, but I think if we can remove the absolute value sign, the approximation will become more useful.
For example, we have Hurwitz's theorem:

For any irrational $\zeta$ there are infinitely many pairs of integers $p,q$ such that $|\zeta - p/q|<\frac{1}{\sqrt{5}q^2}$

However, when we remove the absolute value, is the following statement true or not?

For any irrational $\zeta$ there are infinitely many pairs of integers $p,q$ such that $0<\zeta - p/q<\frac{1}{\sqrt{5}q^2}$

This is just example. My main question is: Is there any theory about it? Where can I get the reference on this topic?

REPLY [4 votes]: I don't know whether you can drop the absolute value in the approximation.
However, it will be true if you drop both the absolute value and the factor
$\frac{1}{\sqrt{5}}$.
Quoting material from $\S$ 1.3 of Ivan Niven's book Diophantine Approximations, we have following asymmetric approximation of irrational numbers.

Theorem 1.7 - Given any irrational number $\theta$ and any non-negative real number $\tau$,
  there exists infinitely many rational numbers $\frac{h}{k}$ such that
  $$-\frac{1}{\sqrt{1+4\tau}k^2} < \theta - \frac{h}{k} < \frac{\tau}{\sqrt{1+4\tau}k^2}$$
  Furthermore, the statement holds if we replace the central piece by $\frac{h}{k} - \theta$.

As a corollary, 

Corollary 1.9 - Given any irrational $\theta$, there are infinitely many rationals $\frac{h}{k}$ such that 
  $$-\frac{1}{k^2} < \theta - \frac{h}{k} < 0$$
  and also infinitely many rationals $\frac{h}{k}$ such that
  $$ 0 < \theta - \frac{h}{k} < \frac{1}{k^2}$$