TITLE: $\mid\theta-\frac{a}{b}\mid< \frac{1}{b^{1.0000001}}$, question related to the dirichlet theorem
QUESTION [6 upvotes]: The question is: 
A certain real number $\theta$ has the following property: There exist inﬁnitely many rational numbers $\frac{a}{b}$(in reduced form) such that:
$$\mid\theta-\frac{a}{b}\mid< \frac{1}{b^{1.0000001}}$$
Prove that $\theta$ is irrational.
I just don't know how I could somehow relate $b^{1.0000001}$ to $b^2$ or $2b^2$ so that the dirichlet theorem can be applied. Or is there other ways to approach the problem? 
Thank you in advance for your help!

REPLY [5 votes]: Hint: Let $\theta=\frac{p}{q}$, where $p$ and $q$ are relatively prime.  Look at 
$$\left|\frac{p}{q}-\frac{a}{b}\right|.\tag{$1$}$$
Bring to the common denominator $bq$. Then if the top is non-zero, it is $\ge 1$, and therefore Expression $(1)$ is $\ge \frac{1}{bq}$. 
But if $b$ is large enough, then $bq<b^{1.0000001}$.  
Edit: The above shows that if $\theta$ is rational, there cannot be arbitrarily large $b$ such that
$$\left|\theta-\frac{a}{b}\right|<\frac{1}{b^{1.0000001}}.\tag{$2$}$$
Of course, if we replace the right-hand side by $b^{1.0000001}$, then there are arbitrarily large such $b$. Indeed if we replace it by any fixed $\epsilon\gt 0$, there are arbitrarily large such $b$, since any real number can be approximated arbitrarily closely by rationals.  Thus if in the original problem one has $b^{1.0000001}$, and not its reciprocal, it must be a typo.