# Sequence of rationals with an irrational limit have denominators going to infinity

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?

• Not tending to infinity does not mean bounded, $1,1,1,2,1,3,1,4,1,5,\dots$. – André Nicolas Oct 5 '15 at 17:20
• What is $\alpha_j$? – user237392 Oct 5 '15 at 17:58
• It is just used to denote the numerator for the $j^{th}$ term. Nothing special. – LKSR Oct 5 '15 at 17:59
• I just asked my professor, he said that $\beta_j$ should be bounded... – LKSR Oct 5 '15 at 23:25

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 $\lim\limits_{j\to\infty}|\beta_j|=+\infty.$

• 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 5 '15 at 23:35
• @taiweic 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 5 '15 at 23:44
• 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 5 '15 at 23:49
• So the constant subsequence $\{ \beta_{j_k}\} = \{b\} = \{M\}$ ? – LKSR Oct 5 '15 at 23:58
• Oh I see I was being stupid. Thank you! – LKSR Oct 6 '15 at 0:05

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.

• As noted in the comment above, $\beta_j$ is not bounded. I was wrong, sorry. – LKSR Oct 5 '15 at 17:36
• @taiweic I am not saying that $\beta_j$ is bounded. In fact, I'm not necessarily using contradiction here. – Omnomnomnom Oct 5 '15 at 17:39
• Finitely many fractions with denominator less than $M$? – zhw. Oct 5 '15 at 17:56
• An interval of radius $\frac{1}{M}$ clearly suffice. But how do we use this fact to argue that it does not converge? Also, how do we rigorously aruge that there are finitely many fractions with denominator less than $M$? – LKSR Oct 5 '15 at 18:01
• @zhw fixed it ${}$ – Omnomnomnom Oct 5 '15 at 18:05