prove that sequence of rationals goes to infinity

Question

This question is part of my attempts to prove that the pop-corn function is continious. (here)

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It is known that every rational $q \in \mathbb{Q}$ have a unique representation of the form $\frac{a}{b} $ with $\gcd (a,b)=1$

Now concider a real number $r\in \mathbb{R}$.

It is also known that we could find a sequence of rationals $(q_n)_{n=1}^{\infty}$ such that $$\lim_{n \to \infty} q_n= r$$

So there are two sequences of integers $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ such that $\forall n \in \mathbb{N} \Rightarrow$ $$ \frac{a_n}{b_n}= q_n$$ and $$\gcd(a_n,b_n)=1$$

Prove that $$\lim_{n \to \infty} \frac{1}{b_n}=0$$

The statement seems true.

If we try to approximate $\sqrt{2}$ we should get the folowing sequence $$1, \quad 1.4 =\frac{14 }{ 10}, \quad 1.41=\frac{ 141}{ 100}, \quad 1.4142 =\frac{14142 }{10000 }, \dots$$

It is obvious that as the proximity increases , as the $q_n$ approaches $r$ so the values of numerators and denominators increase.

But i need some help to write down the rigorous proof, if it exists.

Edit

As it pointed out by many users my statement is not entirely correct.

• If $r$ is a rational number $\implies r = \frac{a}{b}$. Then, from the sequence $$(q_n=\frac{a}{b})_{n=1}^{\infty}$$ we do not get $$\lim_{n \to \infty} \frac{1}{b_n}=0$$

• Moreover even if the $r$ is irrational, as user астон вілла олоф мэллбэрг illustrated in his example, we can find a sequence that approximates $r$ but do not fulfil the statement

So let me refine my statament

Consider a irrational number $r$ and construct the sequences $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ as before,

prove that there is a subsequence of $(b_n)_{n=1}^{\infty}$ that tends to $\infty$.

Or more formally,

Prove that $\limsup b_n=\infty$

Answer 1

Your claim is false if $r$ is rational.

However, for irrational $r$ it is true. Hint: If there is some constant $m$ such that infinitely many of the denominators are all less than $m$, then they cannot get arbitrarily close to the true value since you can assume that those denominators are always $m!$. Therefore for any $m$ we only have finitely many denominators less than $m$ and hence beyond a certain point they would be all more than $m$.

(Thanks to Rob Arthan for pointing out my blurness in my previous answer!)

Answer 2

EDITED

The following statement is true : let $p_n = \frac{a_n}{b_n}$, with $a_n,b_n \in \mathbb Z, b_n > 0, (a_n,b_n) = 1$ be a sequence of rational numbers converging to $q \notin \mathbb Q$. Then $b_n \to \infty$.

The best way of seeing this, is that if $b_n \not \to \infty$, then there exists $C > 0$ such that $\forall N$ there is $n > N$ such that $b_{n} < C$. This gives rise to a subsequence $0 < b_{n_k} < C$ for all $k$. Every bounded sequence has a convergent subsequence, and every convergent sequence of integers is eventually constant. That is, there is a constant subsequence of $b_{n_k}$. The upshot : there is a subsequence of $b_n$ which is constant. Let this constant be $d$, and let the subsequence be $b_{n_t}$ i.e. $b_{n_t} = l$ for all $t$. Note $l > 0$.

Note that $\frac{a_{n_t}}{b_{n_t}} \to q$ since every subsequence of a convergent sequence also has the same limit. However, this means that $a_{n_t} \to lq$, since $b_{n_t} = l$ for all $t$. However, $a_{n_t}$ is also a sequence of integers, therefore eventually constant. Let $a_{n_t} = a$ after some $t$, then $\frac{a_{n_t}}{b_{n_t}} = \frac al$ after some time. Consequently, this sequence has two limits : $\frac al$ and $q$. This forces $\frac al = q$, a contradiction since $q$ is irrational.

Hence, $b_n \to \infty$.

Answer 3

I think it can be done as follows: suppose we have

$$p_n\in\Bbb Z\,,\;q_n\in\Bbb N\;,\;\;\lim_{n\to\infty}\frac{p_n}{q_n}=x\in\Bbb R\setminus\Bbb Q$$

Observe we can always choose the denominator of any rationa fraction to be positive. This is just to avoid using absolute value and make things slightly simpler.

Suppose $\;q_n\rlap{\;\;\;/}\longrightarrow\infty\;$, so that there exists $\;R\in\Bbb R^+\;$ such that any $\;m\in\Bbb N\;$ there exists $\;n_m>m\;$ with $\;q_{n_m}<R\;$ . Observe now the subsequence

$$\left\{\,\frac{p_{n_m}}{q_{n_m}}\,\right\}_{m=1}^\infty.\;\;\text{Obviously we still have}\;\;\;\;\frac{p_{n_m}}{q_{n_m}}\xrightarrow[m\to\infty]{}x$$

but since there are only a finite number of possible denominators $\;q_{n_m}\;$ in that sequence we can find a subsequence of (a sub-subsequence of the original one) with constant denominators and still

$$\left\{\,\frac{p_{n_{m_k}}}{q_{n_{m_k}}}\,\right\}_{k=1}^\infty\xrightarrow[k\to\infty]{}x\;,\;\;q_{n_{m_k}}=q=\,\text{a constant}$$

But then the above means

$$x=\lim_{k\to\infty}\frac{p_{n_{m_k}}}{q=q_{n_{m_k}}}=\frac{\lim\limits_{k\to\infty}p_{n_{m_k}}}q=\frac pq$$

as a convergent sequence of integers must eventually be constant, and thus we got $\;x\in\Bbb Q\;$ , contradiction.

Answer 4

By contradiction.Suppose $\neg (\lim_{n\to \infty}b_n=0)$. Then there is a bounded subsequence $(b_{f(n)})_{n\in N},$ where $f:N\to N$ is strictly increasing.

Let $b\leq \max \{b_{f(n)}:n\in N\},$ such that the set $f^{-1}\{b\}=\{n\in N: f(n)=b\}$ is infinite.

Let $m\in Z$ such that $\frac {m}{b}<r<\frac {1+m}{b}.$

Let $e=\min (r-\frac {m}{b},\; \frac {1+m}{b}-r).$ Then $\min \{|r-\frac {z}{b}|:z\in Z\}=e>0.$

Therefore $$\inf \{|r-\frac {a_{f(n)}}{b_{f(n)}}|: n\in f^{-1}\{b\}\}=$$ $$=\inf \{|r-\frac {a_{f(n)}}{b}|: n\in f^{-1}\{b\}\}\geq$$ $$\geq \min \{|r-\frac {z}{b}|:z\in Z\}=e>0.$$ But $\lim_{m\to \infty}|r-\frac {a_m}{b_m}|=0$ and $f^{-1}\{b\}$ is an infinite set, so $$\inf \{|r-\frac {a_{f(n)}}{b_{f(n)}}|:n\in f^{-1}\{b\}\}=0,$$ a contradiction.