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Lemma. Let $y \in [0,1] \setminus \mathbb{Q}$, $\epsilon > 0$ and $N \in \mathbb{N}$ such that $\frac{1}{N} < \epsilon$.

Then one can choose $\delta > 0$ small enough such that

$$ \frac{p}{q} \in (y - \delta, y + \delta) \Rightarrow q > N, $$

where $\frac{p}{q} \in \mathbb{Q}$ and $gcd(p, q) = 1$.

I can see that any $\frac{p}{q}$ that satisfies the condition equals $y + t$, where $\delta \le |t| \in [0,1] \setminus \mathbb{Q}$. I just don't see how we can make $\frac{1}{q}$ arbitrarily small.

Adam Rubinson
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IsaacR24
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1 Answers1

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I'll outline how to prove the larger theorem:

If $y \in [0,1] \setminus \mathbb{Q},\quad p,q\in\mathbb{N},\ q\neq 0\ $ and $\ N\in\mathbb{N}\ $ then one can find $\ \delta>0\ $ small enough so that

$$ \frac{p}{q} \in (y - \delta, y + \delta) \Rightarrow q > N.$$

This encompasses your theorem, as the $\ gcd\ $ condition is satisfied also.

  • Use the fact that $\ y\ $ is irrational to show:

$\exists \delta_1>0\ $ such that $\ \frac{1}{1},\ \frac{2}{1},\ \frac{3}{1},\ \ldots\ $ are all outside $\ (y-\delta_1, y+\delta_1)$;

$\exists \delta_2>0\ $ such that $\ \frac{1}{2},\ \frac{2}{2},\ \frac{3}{2},\ \ldots\ $ are all outside $\ (y-\delta_2, y+\delta_2)$;

$\exists \delta_3>0\ $ such that $\ \frac{1}{3},\ \frac{2}{3},\ \frac{3}{3},\ \ldots\ $ are all outside $\ (y-\delta_3, y+\delta_3)$;

$\ldots$

$\exists \delta_{N}>0\ $ such that $\ \frac{1}{N},\ \frac{2}{N},\ \frac{3}{N},\ \ldots\ $ are all outside $\ (y-\delta_{N}, y+\delta_{N}).$

  • Let $\ \delta = \min\{\delta_1,\delta_2,\ldots,\delta_{N+1}\}.$

  • If $\ q\in\mathbb{N}\ $ and $\ q<N+1,\ $ then $\ \frac{p}{q}\ \notin (y-\delta, y+\delta).$

  • The contrapositive statement, that is, $\ \frac{p}{q}\ \in (y-\delta, y+\delta) \implies q \geq N+1\ $ must also be true.

Adam Rubinson
  • 20,052