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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.

Did
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Michael
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    Hint: If $N$ is an integer such that $|\beta_{j}| < N$ for all $j$, then $N!a_j$ is an integer for all $j$. – Alex Zorn Sep 11 '18 at 05:29

1 Answers1

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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.
Martin R
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    Another version is that the set $R_n$ of rational numbers in the interval $(\alpha-1,\alpha+1)$ with denominator at most $n$, is finite hence the distance between $\alpha$ and $R_n$ is $\delta_n>0$, which forbids that a sequence with values in $R_n$ converges to $\alpha$. Your version might be simpler to grasp -- provided the OP knows how to prove the second point, that the sequence of rational numbers is eventually constant (and by "prove", I mean "prove", not wave one's hands...). – Did Sep 11 '18 at 06:51
  • @Did: Yes (and it was meant as hints, expecting that OP tells if (s)he has problems making a rigorous proof.) – But actually I just noticed that I am guilty of not seeing https://math.stackexchange.com/q/1465812 in the “Related” section: The question is slightly different, but the answer is the same. Should this be closed as a duplicate? – Martin R Sep 11 '18 at 07:00
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    On second reading, I am nearly sure that anybody asking the question would have trouble explaining why your second point holds. Yes it is natural to "expect(...) that OP tells if (s)he has problems making a rigorous proof", unfortunately... – Did Sep 11 '18 at 07:51