Can we remove the absolute value when doing Diophantine Approximation?

Question

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?

Answer 1

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 $$ 0 < \theta - \frac{h}{k} < \frac{1}{k^2}$$