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Suppose that a sequence of (rational) fractions p/q converge to an irrational number r. Show that q converges to infinity.

Spencer Sullivan
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  • Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you write what your thoughts are on the problem and include your efforts (work in progress) in this and future posts and in what context you have encountered the problem; this will prevent people from telling you things you already know, and help them give their answers at the right level. – JKnecht Feb 29 '16 at 19:38
  • For what it's worth, it seems to me that irrationality of the number $r$ is not really the key issue. Indeed, if we exclude sequences that have eventually constant tails (not much of a restriction), then this property holds for EVERY real number $r.$ – Dave L. Renfro Feb 29 '16 at 20:41

2 Answers2

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What a cute question!

Let $r_n = p_n/q_n$. $r_n \rightarrow x$. $x \not \in \mathbb Q$.

Suppose $q_n \rightarrow y$. Then for $1/2>\epsilon>0$ there exists and $N$ such that $|q_n - q_m| \le |q_n - y| + |y-q_m| <\epsilon + \epsilon <1$ for all $m,n > N$. But $q_n,q_m$ are integers so $|q_n - q_m| < 1 \iff q_n = q_m$. So there exists an $N$ where all $q_{n; n>N}$ are an equal constant integer, $y$.

If $r_n \rightarrow x$ then $\{r_n*y\} \rightarrow xy$. For $n > N$, $r_n*y = p_n*y/q_n = p_n*y/y = p_n \in \mathbb Z$. So, by the exact same argument as above: {$r_n*y$} becomes a sequence of integers for $n > N$. As this converges there is an $M$ where $r_n*y$ is an equal constant integer, $p$, for all n > M.

So for $n > M$, $r_n*y = p \implies r_n = p/y$ and $r_n \rightarrow p/y = x \in \mathbb Q$.

This A contradiction.

So $q_n$ doesn't converge.

fleablood
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  • I like this solution too! It very much makes clear why a subsequence of $n \in \mathbb{Z}$ converges to an integer. +1 – Andres Mejia Feb 29 '16 at 20:39
  • For further reference: http://math.stackexchange.com/questions/311955/let-a-n-n-be-a-convergent-sequence-of-integers-what-can-we-say-about-a-n – Andres Mejia Feb 29 '16 at 20:41
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Let $\{(\frac{p}{q})_k\}$ be a sequence of of rational numbers such that $ \{(\frac{p}{q})_k\}=\frac{\{p_n\}}{\{q_n\}}$ and so that the sequence converges to $r \in \mathbb{R}-\mathbb{Q}$.

Suppose that $\lim_{n \to \infty}q_n=Q$.

If $p_n$ converges, we have a contradiction to the irrationality or $r$, since this would imply that it converges to some $z \in \mathbb{Z}$.

If $p_n$ diverges, $\frac{p_n}{Q}$ also diverges to infinity, contradicting the fact that the sequence is convergent.

Andres Mejia
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