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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!

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    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|.2012-06-06

1 Answers 1

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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.

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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    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.2012-06-06