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Let $\alpha$ be an irrational real number and let $a_j$ be a sequence of rational numbers converging to $\alpha$. Suppose that each $a_j$ is a fraction expressed in lowest terms: $a_j = \alpha_j / \beta_j$. Prove that the $\beta_j$ tend to $\infty$

Attempt: AFSOC that $\beta_j$ does not tend to infinity, then it is bounded by some $M$. We can also find an interval $\alpha \in (k, k+1)$. Let $\epsilon > 0$, $\exists N_0$ such that for all $N > N_0$ we have $|a_N - \alpha| < \epsilon$. So $|\frac{p_N}{q_N} - \alpha| < \epsilon$. We try to use the fact that $q$ is bounded but fail to derive a contradiction. I do not know how to use the fact that $\alpha$ is between two integers, although it might not be relevant at all.

Ideas?

zhw.
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LKSR
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2 Answers2

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Your statement is not necessarily true, for instance, it is well known that, for each irrational number $x,$ (and in fact $x$ need not be irrational) there exists some rational sequence $\{r_n\}=\{a_n/b_n\}$ ($a_n,b_n\in\mathbb Z$) converging to $x$ and so you can conveniently modify each term of $\{r_n\}$ in such a way that $b_k<0$ for all $k.$ However, the fact that (in your particular case) $\lim\limits_{j\to\infty}|\beta_j|=+\infty$ is true and in fact you can replace the $\alpha$ in your statement for any real number $x$ and add an "additional" condition, which is $a_j\neq x,\;\forall j.$

Now let me help you in the proof that $\lim\limits_{j\to\infty}|\beta_j|=+\infty.$

Suposse, for the sake of contradiction, that $|\beta_j|\not\to+\infty.$ This means that there is some real number $M$ such that no matter how big $N\in\mathbb N$ is, you will always find some natural $n>N$ such that $|\beta_n|<M.$ Therefore, since the sequence $\{|\beta_j|\}$ is of positive integers, it must have a constant subsequence $\left\{|\beta_{j_k}|\right\}=\{b\}.$ Thus, since $\{a_j\}$ converges to $\alpha,$ then $\left\{a_{j_k}\right\}$ converges to $\alpha$ as well and hence $\left\{\alpha_{j_k}\right\}$ converges to $b\alpha$ and it follows that since the subsequence $\left\{\alpha_{j_k}\right\}$ of $\{\alpha_j\}$ is of integers, it must be eventually constant, that is, there is some $a\in\mathbb Z$ such that $\alpha_{j_k}=a$ for sufficiently large $k.$ Thus $a=b\alpha$ and hence $\alpha=a/b,$ which contradicts the fact that $\alpha$ is irrational. Thus $\lim\limits_{j\to\infty}|\beta_j|=+\infty.$

CIJ
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  • What if $\beta_j$ = 1 1 2 1 3 1 4 1 5 1 6 1... Then this does not tend to infinity, but there is no real number $M$ such that you will always find some $n > N$ such that $|\beta_n| < M$ – LKSR Oct 05 '15 at 23:35
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    @LKSR Put $M=2.$ Given any $N\in\mathbb N$ you will always find some $n>N$ such that $\beta_n=1,$ which is certainly $<$ than $M$ – CIJ Oct 05 '15 at 23:44
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    Let me put everything in symbols: $$\lim_{n~\to~\infty}x_n=+\infty;;;\iff;;;\forall M\in\mathbb R;;;\exists N\in\mathbb N;;;\forall n>N;;;x_n\geqslant M$$ and so $$\lim_{n~\to~\infty}x_n\neq+\infty;;;\iff;;;\exists M\in\mathbb R;;;\forall N\in\mathbb N;;;\exists n>N;;;x_n<M$$ – CIJ Oct 05 '15 at 23:49
  • So the constant subsequence ${ \beta_{j_k}} = {b} = {M}$ ? – LKSR Oct 05 '15 at 23:58
  • No. The idea here is that there will be infinitely many subindexes $j$ such that $0<|b_j|<M$ (this is a consequence of the assumption that $b_j\not\to+\infty$) and since the $|b_j|$ are all integers and they all belong to the interval $(0,M)$, then necessarily infinitely many of them must have the same value. Those equal terms form the constant subsequence ${|b_{j_k}|},$ where $|b_{j_k}|=b$ for all $k$ – CIJ Oct 06 '15 at 00:02
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    Oh I see I was being stupid. Thank you! – LKSR Oct 06 '15 at 00:05
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Hint: For any $M > 0$, there are finitely many fractions with denominator less than $M$ that fall within $1$ of $\alpha$. So, there is an interval around $\alpha$ that excludes all such fractions.

Ben Grossmann
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