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


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 $\limsup b_n=\infty$

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    $\begingroup$ Your statement is not true in general (take for instance $r=1$, $a_n=b_n=1$ for all $n$). You also need that $r \notin \mathbb Q$. $\endgroup$ – TonyK Mar 17 '16 at 2:17
  • $\begingroup$ The answer from астон вілла олоф мэллбэрг is all wrong, I'm afraid. Consider unaccepting it. $\endgroup$ – TonyK Mar 28 '16 at 11:40
  • $\begingroup$ could you elaborate? $\endgroup$ – karhas Mar 28 '16 at 17:41
  • $\begingroup$ See my comment to the answer. $\endgroup$ – TonyK Mar 28 '16 at 18:35
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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, $b_n \to \infty$.

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    $\begingroup$ In your second parraph, it is not true that the second sequence, the one you give as counterexample, still tends to $\;\sqrt2\;$ . Some care is needed here, as you seem to write down a sequence with an infinite subsequence that tends to $\;1/2\;$ (In fact, it is a constant sequence. This, I think, renders your whole argument incorrect. $\endgroup$ – DonAntonio Mar 15 '16 at 8:35
  • $\begingroup$ This whole answer is wrong, I'm afraid. If $r$ is rational (e.g. $r=0$), then there is not necessarily a subsequence of $a_m$ that tends to $\infty$. And if $r$ is irrational, then $a_m$ does tend to $\infty$. $\endgroup$ – TonyK Mar 28 '16 at 11:39
  • $\begingroup$ I have made corrections, albeit VERY delayed. Request all to have a look. $\endgroup$ – астон вілла олоф мэллбэрг Sep 20 '18 at 16:25
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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.

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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!)

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    $\begingroup$ The question asks you to show that the denominators tend to infinity not that they are unbounded. And I don't see where $m!$ comes into it. $\endgroup$ – Rob Arthan Mar 14 '16 at 23:55
  • $\begingroup$ @DanielRobert-Nicoud: "unbounded "doesn't "imply tends to infinity". $\endgroup$ – Rob Arthan Mar 15 '16 at 0:40
  • $\begingroup$ @DanielRobert-Nicoud It's the same problem: if for example $\;\{b_n\}=\left\{1,\frac12,2,\frac12,3,...\right\}\;$ then the sequence isn't bounded and also $\;\frac1{b_n}\rlap{\;\,/}\rightarrow 0\;$ and also $\;b_n\rlap{\,\;/}\rightarrow\infty\;$ $\endgroup$ – DonAntonio Mar 15 '16 at 15:26
  • $\begingroup$ @RobArthan: The $m!$ was simply a common denominator if all denominators remain less than $m$. But I was very blur (half awake) when I posted my answer and you are certainly right that unbounded does not mean tend to infinity. However, it so happens that my hint still works for this problem. If it didn't tend to infinity, then there is an $m$ such that infinitely many of the denominators are bounded by $m$. That produces the contradiction in the same way. $\endgroup$ – user21820 Mar 15 '16 at 15:48
  • $\begingroup$ @DanielRobert-Nicoud: Rob Arthan was right. See my fixed answer. $\endgroup$ – user21820 Mar 15 '16 at 15:53
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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.

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