# $\mid\theta-\frac{a}{b}\mid< \frac{1}{b^{1.0000001}}$, question related to the dirichlet theorem

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?

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no, it's positive –  Henry W Jun 6 '12 at 19:47
With the exponent positive, it isn't true, since there are always infinitely many rational numbers in $(\theta-1,\theta+1)$, and for all such rationals, you have $\frac{a}{b}$, $b^{1.0000001}\geq 1>\left | \theta -\frac{a}{b}\right|$. –  Thomas Andrews Jun 6 '12 at 19:56

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.

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yea, I think there's a typo –  Henry W Jun 6 '12 at 20:05
Unless it is a joke "prove or disprove," what is intended is the inequality with $b^{1.0000001}$ in the denominator. Then it is a real exercise in diophantine approximation ideas. –  André Nicolas Jun 6 '12 at 20:07