Suppose that a sequence of rational fractions p/q converge to an irrational number

Question

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

Answer 1

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.

Answer 2

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.