If a sequence of rational numbers converges to an irrational number, the sequence of their denominators is unbounded

Question

So, the question is as follows:

Let $\alpha$ be irrational and let ${a_{j}}$ be a sequence of rational numbers converging to $\alpha$. Suppose that each $a_{j}$ is expressed in lowest terms: $a_{j} = \frac{\alpha_{j}}{\beta_{j}}$. Prove that the $\beta_{j}$ are unbounded.

I would like some advice on where to start. I have a suspicion that this is likely a contradiction (So we should assume $|\beta_{j}| < M \in \mathbb{R^+}$), and that the definition of a converging sequence will be helpful, but I don't know where to go from there.

Answer 1

Assume that $a_{j} = \frac{\alpha_{j}}{\beta_{j}} \to \alpha$ with integers $\alpha_j, \beta_j$, and $(\beta_{j})$ is bounded.

• First show that $(\alpha_{j})$ must be bounded as well.
• So there are only finitely many different $a_j$, which means that the sequence $(a_j)$ is eventually constant. Hint:

  Let $\epsilon > 0$ be the minimal difference between distinct $a_j$. Choose $j_0 \in \Bbb N$ such that $|a_j - \alpha | < \frac 12 \epsilon$ for $j \ge j_0$. Conclude that $a_j = a_{j_0}$ for $j \ge j_0$.

• Finally conclude that $\alpha$ is equal to some $a_j$, contradicting the assumption that it is irrational.