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

[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
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Now, the other question for this thread:
*Can I rewrite the original statement more clearly as:
$$\forall \left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} \subset \mathbb{Q}{\text{ where}}\;k_1 < k_2 < \ldots < k_n ,\;\exists \, \varepsilon > 0\;{\text{such that}}\;\forall n \in \mathbb{N},$$
$$\left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} \subseteq k_1 + \varepsilon \left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{k_n - k_1 }}{\varepsilon }} \right\rfloor } \right\}$$

 

[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:
$$\forall \left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} \subset \mathbb{Q}{\text{ where}}\;k_1 < k_2 < \ldots < k_n ,\;\exists \, \varepsilon > 0\;{\text{such that}}\;\forall n \in \mathbb{N},$$
$$\left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} \subseteq k_1 + \varepsilon \left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{k_n - k_1 }}{\varepsilon }} \right\rfloor } \right\}$$

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:,
$$\left\{ {\frac{1}{{11}},\frac{1}{3},\frac{1}{2}} \right\} \not\subset \left\{ {\frac{1}{{11}},\frac{1}{3}} \right\}$$
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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 [itex] \left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} [/tex]. 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!
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*In any case, sorry for skipping lines; I'm still getting used to LaTex

 

[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
$$\{ k_1,k_2,...,k_n \} = \{ \frac{p_1}{q_1},..., \frac{p_n}{q_n} \} = \frac{1}{q_1...q_n} \{ p_1 (q_2...q_n) ,..., p_n (q_1...q_{n-1})$$