List all Rationals in [a,b]: a + lim x->0 x{n}_0^(floor((b-a)/x))

[rachmaninoff]

I know you are technically correct. But I cannot get rid of the feeling that your objection is like saying 1 LaTeX graphic is being generated. Reload this page in a moment. 0.999... for the reason that "one would never get to the last 9." I know intuition can give you the wrong answer, this must be one of those times.

Yes, it's one of those times.

 

[bomba923]

*As for intuition, not quite sure...
but here's the short history of that rational sequence problem:
--I was browsing aimlessly around a math office in some university,
and saw some chart with sets/comics on it. The next thing I know,
I'm helping someone with unrelated power series questions. Before leaving,
the person asked me where I was born, and later handed me a small
(poorly handwritten) tag, with the statement:
**\
\forall \left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} \subset \mathbb{Q}\,,\;\exists \,\varepsilon > 0\;{\text{such that }}\forall n \in \mathbb{N}\,,
\left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} \subseteq \min \left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} + \varepsilon \left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{\operatorname{range} \left\{ {k_1 ,k_2 , \ldots ,k_n } \right\}}}<br /> {\varepsilon }} \right\rfloor } \right\}
(where the "range" is just the range of this set, max-min that is)
-------------------------------------------------------------------
*Anyway, the notation was a bit funky (or poorly written?)...but then I got this "idea"
(well, actually , I thought of both naturals and rationals being \aleph _0, and then got this
crazy idea represented below, using the n \in \mathbb{N})
-Oh well, here goes :
*Let
\begin{gathered}<br /> a = \min \left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} \hfill \\ b = \max \left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} \hfill \\ \end{gathered}
In addition, let \left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} represent all of the rationals in \left[ {a,b} \right].

!--So basically, the idea was to figure out if the statement would still apply, in the above case.
(I think you can figure out what mean by this "statement application")
-------------------------------------------------------------------
The next thing to do was to post it on PF! (hey, I even recommended the guy to sign up, but he doesn't know me and I think got sort of scared...but that doesn't matter)
(*In conclusion, I'm quite surprised that this thread got almost 700 views. ..well, a surprise only for me )
So um, what say you guys?

 

[rachmaninoff]

It's already been shown that that statement doesn't hold in the infinite case. The finite case is essentially just a way of saying every finite set of integers has a common multiple (so that the 1/n 's have a common denominator).

 

[EnumaElish]

It's already been shown that that statement doesn't hold in the infinite case.

Well, since that was in a math dept. I guess I'd have to agree with rachmaninoff here. Had you said CS dept., for example, and put the question as

"can one generate progressively bigger sets of rationals in any interval using this algorithm, all the way to infinity?"

then the answer would be affirmative. Would you get to infinity? No. With an infinite amount of time, would a parallel quantum super bio-computer get there? No.

 

[bomba923]

Well, since that was in a math dept. I guess I'd have to agree with rachmaninoff here. Had you said CS dept., for example, and put the question as

"can one generate progressively bigger sets of rationals in any interval using this algorithm, all the way to infinity?"

then the answer would be affirmative.

Yes, !Indeed!
** That should have been the question!
hehe / (I'm referring to the "all the way to infinity" part!)

(*-And, I am sorry )
Oh, what's the 'CS'-department?

 

[EnumaElish]

Computer science -- or any engineering or applied science dept., I guess. What are u sorry abt?

 

[bomba923]

Computer science -- or any engineering or applied science dept., I guess. What are u sorry abt?

Well, for two things:
1) I mistyped the general statement, forgot to add \min \left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} to the
\varepsilon \left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{\operatorname{range} \left\{ {k_1 ,k_2 , \ldots ,k_n } \right\}}}{\varepsilon }} \right\rfloor } \right\}
2) My original question was an unnecessary "leap"/stretch from the general statement
-------------------------------------------------------
Now, the other question for this thread:
*Can I rewrite the original statement more clearly as:
\forall \left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} \subset \mathbb{Q}{\text{ where}}\;k_1 &lt; k_2 &lt; \ldots &lt; k_n ,\;\exists \, \varepsilon &gt; 0\;{\text{such that}}\;\forall n \in \mathbb{N},
\left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} \subseteq k_1 + \varepsilon \left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{k_n - k_1 }}{\varepsilon }} \right\rfloor } \right\}

 

[EnumaElish]

Oh... Don't be, in these Forums almost anything goes, I dare guess. What I mean is, there aren't bad questions, only unclear answers.

 

[bomba923]

Whoa-->900 views for this thread! (Compared to my other threads this is by far the most I ever had :shy:)

Btw, I do have one other question:
*Using the LaTex code, this is my second question:
{\text{Does}}\;\exists \,k \ne 0 {\text{ such that}}\;\left( {k\sqrt 2 ,k\sqrt 3 } \right) \in \mathbb{Z}^2 \, ?
Currently I believe there is NO such k. Am I correct?

(I was trying to find a proof why it is/isn't correct, but I found nothing;
Btw, \mathbb{Z} represents the set of all integers)

 

[rachmaninoff]

Trivially - you're wrong: try k = 0.

Nontrivially:

\mbox{Assume } k \sqrt{2} = \frac{m}{n}, \, k \sqrt{3} = \frac{p}{q}, \, m...q \in \mathbb{N}
\sqrt{2} = \frac{m}{n k} = \left( \frac{mq}{pn} \right) \frac{p}{q k} = \left( \frac{mq}{pn} \right) \sqrt{3}
\frac{\sqrt{2}}{\sqrt{3}} = \frac{mq}{pn} \in \mathbb{Q}

which is a contradiction, since root(2)/root(3) is irrational.

 

[EnumaElish]

... Can I rewrite the original statement more clearly as:
\forall \left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} \subset \mathbb{Q}{\text{ where}}\;k_1 &lt; k_2 &lt; \ldots &lt; k_n ,\;\exists \, \varepsilon &gt; 0\;{\text{such that}}\;\forall n \in \mathbb{N},
\left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} \subseteq k_1 + \varepsilon \left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{k_n - k_1 }}{\varepsilon }} \right\rfloor } \right\}

Hmph, I can't readily think of a counterexample to this. Closest I can get to a proof is, if you define \Delta_i = k_i - k_{i-1} then it looks to me like one can select \varepsilon=\min\left\{\Delta_i\right\}_1^n to get the result . All one needs is to select the minimum step size that would produce all elements of the subset, starting with k₁.

 

[bomba923]

Hmph, I can't readily think of a counterexample to this. Closest I can get to a proof is, if you define \Delta_i = k_i - k_{i-1} then it looks to me like one can select \varepsilon=\min\left\{\Delta_i\right\}_1^n to get the result . All one needs is to select the minimum step size that would produce all elements of the subset, starting with k₁.

*Counterexample to proof (informal ), on the "floor" technicality:
For \left\{ {\frac{1}{11},\frac{1}{3},\frac{1}{{2}}} \right\},
\varepsilon = \min\left\{\Delta_i\right\}_1^n = \frac{1}{3} - \frac{1}{{11}} = \frac{8}{{33}}
Then, {\left\lfloor {\frac{{k_n - k_1 }}{\varepsilon }} \right\rfloor } = \left\lfloor {\frac{{27}}{{16}}} \right\rfloor = 1
And so
\varepsilon \left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{k_n - k_1 }}{\varepsilon }} \right\rfloor } \right\} = \frac{8}{33} \left\{ {0,1} \right\} = \left\{ {\frac{1}{{11}},\frac{1}{3}} \right\}
*But:,
\left\{ {\frac{1}{{11}},\frac{1}{3},\frac{1}{2}} \right\} \not\subset \left\{ {\frac{1}{{11}},\frac{1}{3}} \right\}
------------------------------------------------------
However, if :shy:
\varepsilon = \frac{1}{{2 \cdot 3 \cdot 11}} = \frac{1}{{66}}
then
\varepsilon \left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{k_n - k_1 }}{\varepsilon }} \right\rfloor } \right\} = \frac{1}{{66}}\left\{ {0,1, \ldots ,27} \right\}
and
\left\{ {\frac{1}{{11}},\frac{1}{3},\frac{1}{2}} \right\} \subset \frac{1}{{66}}\left\{ {0,1, \ldots ,27} \right\}
*So basically, the greatest value of \varepsilon will be greatest common factor of all the elements in \left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} [/tex]. I&#039;ll update later with a mathematical description of the previous sentence, from reading <a href="http://mathworld.wolfram.com/GreatestCommonDivisor.html" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://mathworld.wolfram.com/GreatestCommonDivisor.html</a> . Later, I&#039;ll use this to develop a proof of the general statement,...unless you guys want to crack it!<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f600.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":biggrin:" title="Big Grin :biggrin:" data-smilie="8"data-shortname=":biggrin:" /><br /> -------------------------------------------------------------------<br /> *In any case, sorry for skipping lines; I&#039;m still getting used to LaTex <img src="https://www.physicsforums.com/styles/physicsforums/xenforo/smilies/oldschool/blushing.gif" class="smilie" loading="lazy" alt=":blushing:" title="Blushing :blushing:" data-shortname=":blushing:" />

 

[rachmaninoff]

*So basically, the greatest value of will be greatest common factor of all the elements in . I'll update later with a mathematical description of the previous sentence, from reading http://mathworld.wolfram.com/GreatestCommonDivisor.html . Later, I'll use this to develop a proof of the general statement,...unless you guys want to crack it!

You've already trivially proved it. The easiest notation to see this in would be
\{ k_1,k_2,...,k_n \} = \{ \frac{p_1}{q_1},..., \frac{p_n}{q_n} \} = \frac{1}{q_1...q_n} \{ p_1 (q_2...q_n) ,..., p_n (q_1...q_{n-1})