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Let $f$ the function defined on $\mathbb Q$ by : $ f(n/m) = n $.

I would like to know whether it is true that : $ \forall q\in \mathbb Q - \{ 0 \} \quad \forall R>0 \quad \exists \delta >0\quad \forall x\in \mathbb Q \quad :\quad 0<|x-q|<\delta \Rightarrow |f(x)|>R $

(That is : $ \lim_{x \to q, x \in \mathbb Q} f(x) = \infty$, for all rational $q ≠ 0$).

Thanks in advance !

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I'd clarify your definition of $f$; it sounds like $f\left(\frac{1}{1}\right) = 1$ and $f\left(\frac{2}{2}\right) = 2$... –  MartianInvader Apr 18 at 19:26

1 Answer 1

You need to clarify that $x \neq q$ when you are quantifying over all $x$, and that fractions are always given in lowest terms. If you do that, then it is true.

Lemma: If $x = m/n \neq q$, where $q = r/s$, then $|x - q| \geq 1/ns$.

Corollary: If $\delta < 1/ns$ for some positive integer $n$, and $|x - q| < \delta$ for $x$ rational, then $x = m'/n'$ in lowest terms where $|n'| > n$.

Lemma: If $|x - q| < \delta$ where $q \neq 0$ and $x = m'/n'$, then $|m'| \geq (|q| - \delta)|n'|$.

Corollary: $|f(x)| \to \infty$ as $|x - q| \to 0$ for $x \neq q$.

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