Number theory/Signed Articles/Elementary diophantine approximations: Difference between revisions

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== The method of neighbors and median ==
== The method of neighbors and median ==


In this section we will quickly obtain some results about approximating irrational numbers by rational (for the sake of simplicity only positive numbers will be considered). To this end we will not worry about the details of the difference between a rational number and a fraction (with integer numerator and denominator)—this will not cause any problems; fully crisp notions will be developed in the next sections, they will involve 2-dimensional vectors and 2x2 matrices. This section is still introductory. It is supposed to provide quick insight into the topic.
In this section we will quickly obtain some results about approximating irrational numbers by rational. To this end we will not worry about the details of the difference between a rational number and a fraction (with integer numerator and denominator)—this will not cause any problems; fully crisp notions will be developed in the next sections, they will involve 2-dimensional vectors and 2x2 matrices. This section is still introductory. It is supposed to provide quick insight into the topic.
 
=== Definitions ===
   
   
Fractions <math>\frac{a}{c}</math>&nbsp; and <math>\frac{b}{d},</math>&nbsp; with integer numerators and natural denominators, are called '''neighbors''' (in the given order) &nbsp; <math>\Leftarrow:\Rightarrow</math>
Fractions <math>\frac{a}{c}</math>&nbsp; and <math>\frac{b}{d},</math>&nbsp; with integer numerators and natural denominators, are called '''neighbors''' (in the given order) &nbsp; <math>\Leftarrow:\Rightarrow</math>
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::: <math>\frac{a}{c}\ >\ x\ >\ \frac{b}{d}</math>&nbsp;
::: <math>\frac{a}{c}\ >\ x\ >\ \frac{b}{d}</math>&nbsp;
'''Definition'''&nbsp; A pair of neighboring fractions <math>\left(\frac{a}{c},\frac{b}{d}\right),</math>&nbsp; with integer numerators and natural denominators, is called a ''top pair'' &nbsp; <math>\Leftarrow:\Rightarrow\ c > d.</math> &nbsp; Otherwise it is called a ''bottom pair''.
Thus now we have notions of top/bottom neighbors and of top/bottom pairs of neighbors.
* Let&nbsp; <math>\left(\frac{a}{c},\frac{b}{d}\right),</math>&nbsp; be a pair of neighbors. Then&nbsp; <math>\left(\frac{a}{c},\frac{a+b}{c+d}\right)</math>&nbsp; is a top pair of neighbors, and&nbsp; <math>\left(\frac{a+b}{c+d},\frac{b}{d}\right)</math>&nbsp; is a bottom pair of neighbors.
&nbsp;


=== First results ===
=== First results ===


'''Theorem'''&nbsp; Let fractions <math>\frac{a}{c}</math>&nbsp; and <math>\frac{b}{d},</math>&nbsp; with integer numerators and natural denominators, be neighbors. Then
'''Theorem'''&nbsp; Let fractions <math>\frac{a}{c}</math>&nbsp; and <math>\frac{b}{d},</math>&nbsp; with integer numerators and natural denominators, be neighbors. Then
:* if integers <math>\ t>0</math>&nbsp; and <math>\ s</math>&nbsp; are such that &nbsp; <math>\ \frac{a}{c} > \frac{s}{t} > \frac{b}{d},</math> &nbsp; then &nbsp; <math>t \ge b+d;</math>
:* if integers <math>\ t>0</math>&nbsp; and <math>\ s</math>&nbsp; are such that &nbsp; <math>\ \frac{a}{c} > \frac{s}{t} > \frac{b}{d},</math> &nbsp; then &nbsp; <math>t \ge c+d;</math>


:* the '''median''' &nbsp;<math>\frac{a+b}{c+d}</math>&nbsp; is a bottom neighbor of &nbsp;<math>\frac{a}{c},</math>&nbsp; and a top neighbor of&nbsp; <math>\frac{b}{d};</math>&nbsp;
:* the '''median''' &nbsp;<math>\frac{a+b}{c+d}</math>&nbsp; is a bottom neighbor of &nbsp;<math>\frac{a}{c},</math>&nbsp; and a top neighbor of&nbsp; <math>\frac{b}{d};</math>&nbsp;
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:::<math>\left(\frac{1}{c}-\frac{1}{d}\right)^2\ >\ 0</math>
:::<math>\left(\frac{1}{c}-\frac{1}{d}\right)^2\ \ge\ 0</math>


hence
hence


::<math>\frac{1}{2\cdot c^2}+\frac{1}{2\cdot d^2}\ >\ \frac{1}{c\cdot d}\ =\ \frac{a}{c}-\frac{b}{d}</math>
::<math>\frac{1}{2\cdot c^2}+\frac{1}{2\cdot d^2}\ \ge\ \frac{1}{c\cdot d}\ =\ \frac{a}{c}-\frac{b}{d}</math>




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and
and


::<math>\frac{b}{d}+\frac{1}{2\cdot d^2}\ >\ \frac{a}{c}-\frac{1}{2\cdot c^2}</math>
::<math>\frac{b}{d}+\frac{1}{2\cdot d^2}\ \ge\ \frac{a}{c}-\frac{1}{2\cdot c^2}</math>


i.e.
i.e.
:<math>\mathit{Span}\left(\frac{a}{c},\frac{a}{c}-\frac{1}{2\cdot c^2}\right)\ \cup\ \mathit{Span}\left(\frac{b}{d}+\frac{1}{2\cdot d^2},\frac{b}{d}\right)\ \supseteq\  \mathit{Span}\left(\frac{a}{c},\frac{b}{d}\right)</math>
 
<math>\mathit{Span}\left(\frac{a}{c},\frac{a}{c}-\frac{1}{2\cdot c^2}\right)\ \cup\ \mathit{Span}\left(\frac{b}{d}+\frac{1}{2\cdot d^2},\frac{b}{d}\right)\ \backslash\ \mathbb{Q}\ \supseteq\  \mathit{Span}\left(\frac{a}{c},\frac{b}{d}\right)\ \backslash\ \mathbb{Q}</math>
 


'''End of proof'''
'''End of proof'''
'''Corollary'''&nbsp; Let fractions <math>\frac{a}{c}</math>&nbsp; and <math>\frac{b}{d},</math>&nbsp; with integer numerators and natural denominators, be neighbors. Then, if integers <math>\ t>0</math>&nbsp; and <math>\ s</math>&nbsp; are such that &nbsp; <math>\ \frac{a}{c} > \frac{s}{t} > \frac{b}{d},</math> &nbsp; then either
* <math>s=a+b\quad\and\quad t=c+d;</math>
or
* <math>t\ \ge\ c+d+\min(c,d)\ >\ c+d</math>
&nbsp;


=== Hurwitz theorem ===
=== Hurwitz theorem ===
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'''Proof of lemma 1'''&nbsp; It's easy to show that&nbsp; <math>\sqrt{2}\cdot c \ne \sqrt{3-\sqrt{5}}\cdot d.</math>&nbsp; Thus the square of&nbsp; <math>\ X := \sqrt{2}\cdot c - \sqrt{3-\sqrt{5}}\cdot d.</math>&nbsp; is positive. Now,
'''Proof of lemma 1'''&nbsp; It's easy to show that&nbsp; <math>\sqrt{2}\cdot c \ne \sqrt{3-\sqrt{5}}\cdot d.</math>&nbsp; Thus the square of&nbsp; <math>\ X := \sqrt{2}\cdot c - \sqrt{3-\sqrt{5}}\cdot d</math>&nbsp; is positive. Now,




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and lemma1 follows.&nbsp; '''End of proof'''
and lemma 1 follows.&nbsp; '''End of proof'''
 


==== Lemma 2 ====
==== Lemma 2 ====
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:<math>\frac{1}{\sqrt{5}\cdot c^2} + \frac{1}{\sqrt{5}\cdot d\,^2}\ >\ \frac{1}{c\cdot d}\ =\ \frac{a}{c}-\frac{b}{d}\ =\ \mathit{Span}\left(\frac{a}{c},\frac{b}{d}\right)</math>
:<math>\frac{1}{\sqrt{5}\cdot c^2} + \frac{1}{\sqrt{5}\cdot d\,^2}\ >\ \frac{1}{c\cdot d}\ =\ \frac{a}{c}-\frac{b}{d}\ =\ \left|\mathit{Span}\left(\frac{a}{c},\frac{b}{d}\right)\right|</math>




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Thus lemma 2 is proven when the first inequality of lemma 1 holds. By replacing in the above proof the lower case&nbsp; <math>\ a,c,</math>&nbsp; by the upper case&nbsp; <math>\ A,C,</math>&nbsp; we obtain the proof when the second inequality of lemma 1 holds.
Thus lemma 2 is proven when the first inequality of lemma 1 holds. By replacing in the above proof the lower case&nbsp; <math>\ a,c,</math>&nbsp; by the upper case&nbsp; <math>\ A,C,</math>&nbsp; we obtain the proof when the second inequality of lemma 1 holds.


'''End of proof'''
'''End of proof''' (of lemma 2)


&nbsp;
==== Lemma 2' ====


=== Squeezing irrational numbers between neighbors ===
Let&nbsp; <math>\ a,b\in\mathbb{Z}</math>&nbsp; and&nbsp; <math>\ c,d\in\mathbb{N}.</math>&nbsp; Let&nbsp; <math>\ A:=a+b</math>&nbsp; and&nbsp; <math>\ C:=c+d.</math>&nbsp; Furthermore, let fractions&nbsp; <math>\frac{a}{c},\frac{b}{d},</math>&nbsp; be neighbors, and let:


Let <math>\ x > 0</math>&nbsp; be an irrational number, We may always squeeze it between the extremal neighbours:
:::<math>\frac{a}{c}\ >\ x\ >\ \frac{A}{C}</math>


:::<math>\frac{1}{0}\ >\ x\ >\ \frac{0}{1}</math>
where&nbsp;<math>\ x</math>&nbsp; is real. Then one of the following three inequalities holds:


But if you don't like infinity (on the left above) then you may do one of the two things:
:::* <math>0\ <\ x - \frac{b}{d}\ <\ \frac{1}{\sqrt{5}\cdot d\,^2}</math>


:<math>x > 1\quad\quad\Rightarrow\quad\quad\frac{n+1}{1} > x > \frac{n}{1}</math>


or
:::* <math>0\ <\ x-\frac{A}{C} <\ \frac{1}{\sqrt{5}\cdot C^2}</math>


:<math>x < 1\quad\quad\Rightarrow\quad\quad\frac{1}{n} > x > \frac{1}{n+1}</math>


:::* <math>0\ <\ \frac{a}{c}\ <\ \frac{1}{\sqrt{5}\cdot x^2}</math>


where in each of these two cases&nbsp; <math>\ n\in \mathbb{N}</math> &nbsp; is a respective unique positive integer.


It was mentioned in the previous section ('''''First results''''') that if fractions <math>\frac{a}{c}</math>&nbsp; and <math>\frac{b}{d},</math>&nbsp; with positive (or non-negative) integer numerators and denominators are neighbors then also the top and the bottom (''bot'' &nbsp;for short) pair:
'''Proof'''&nbsp; It's similar to the proof of lemma 2. Or one may apply lemma 2 to&nbsp; <math>\ x':=-x,\ a':=-b,\ b':=-a,</math>&nbsp; and&nbsp; <math>\ c':=d,\ d':=c</math>, which would provide us with the respective fraction&nbsp; <math>\frac{s'}{t'}.</math>&nbsp; Then the required&nbsp; <math>\ s,t,</math>&nbsp; are given by&nbsp; <math>s:=-s',\ t:=t'.</math>


*<math>\mathit{top}\!\left(\frac{a}{c},\frac{b}{d}\right)\ :=\ \left(\frac{a}{c},\frac{a+b}{b+d}\right)</math>
'''End of proof''' (of lemma 2')


and
&nbsp;


*<math>\mathit{bot}\!\left(\frac{a}{c},\frac{b}{d}\right)\ :=\ \left(\frac{a+b}{c+d},\frac{b}{d}\right)</math>
==== Proof of Hurwitz theorem ====


When &nbsp;<math>\ x</math>&nbsp; is irrational, then it is squeezed between infinitely many different pairs of neighbors (see part two of the theorem from the '''''First results''''' section). They provide infinitely many different required fractions&nbsp; <math>\frac{s}{t}</math>&nbsp; (see lemma 2 and 2' above; but it is not claimed, nor true, that different pairs of neighbors provide different required fractions).


are both pairs of neighbors.
'''End of proof'''


Let <math>\ A_0</math>&nbsp; be a pair of neighbors, and <math>\ x\in \mathit{Span}(A_0)</math> &mdash; an irrational number. Assume that pairs of neighbors <math>\ A_0,\dots,A_{n-1}</math>&nbsp; are already defined, and that they squeeze <math>\ x,</math> i.e. that <math>\ x\in\mathit{Span}(A_k)</math>&nbsp; for each <math>\ k=0,\dots,n-1.</math>&nbsp; Then we define <math>\ A_n</math>&nbsp; as the one of the two pairs: <math>\ \mathit{top}(A_{n-1})</math> &nbsp;or&nbsp; <math>\ \mathit{bot}(A_{n-1}),</math> &nbsp;which squeezes <math>\ x.</math>&nbsp; Thus for every positive irrational number we have obtained an infinite sequence of pairs of neighbors, each squeezing the given irrational number more and more. Thus for arbitrary irrational <math>\ x</math>&nbsp; there exist fractions of integers <math>\ \frac{s}{t},</math>&nbsp; with arbitrarily large denominators, such that
&nbsp;


:::<math>\left|x - \frac{s}{t}\right|\ <\ \frac{1}{t\cdot(t+1)}</math>
=== Squeezing irrational numbers between neighbors ===


(apply theorem from section '''''First results'''''). Let's get a sharper result:  
Let <math>\ x > 0</math>&nbsp; be an irrational number, We may always squeeze it between the extremal neighbours:


First of all, since <math>\ x</math> is irrational, the direction top'''/'''bottom of the sequence of pairs of neighbors changes infinitely many times, i.e.
:::<math>\frac{1}{0}\ >\ x\ >\ \frac{0}{1}</math>


:<math>A_{n+1} = \mathit{top}(A_n),\quad\quad A_{n+2} = \mathit{bot}(A_{n+1})</math>
But if you don't like infinity (on the left above) then you may do one of the two things:


:<math>x > 1\quad\quad\Rightarrow\quad\quad\frac{n+1}{1} > x > \frac{n}{1}</math>


for infinitely many values of &nbsp;<math>n\in\mathbb{Z}_+,</math> &nbsp; '''and'''
or


:<math>A_{n+1} = \mathit{bot}(A_n),\quad\quad A_{n+2} = \mathit{top}(A_{n+1})</math>
:<math>x < 1\quad\quad\Rightarrow\quad\quad\frac{1}{n} > x > \frac{1}{n+1}</math>




for infinitely many values of &nbsp;<math>n\in\mathbb{Z}_+</math>&nbsp; as well. Thus there are infinitely many different <math>\ n</math>&nbsp; of each of the two kinds. The other two kinds (which may or may not actually occur) are described by conditions:
where in each of these two cases&nbsp; <math>\ n\in \mathbb{N}</math> &nbsp; is a respective unique positive integer.


:<math>A_{n+1} = \mathit{top}(A_n),\quad\quad A_{n+2} = \mathit{top}(A_{n+1})</math>
It was mentioned in the previous section ('''''First results''''') that if fractions <math>\frac{a}{c}</math>&nbsp; and <math>\frac{b}{d},</math>&nbsp; with positive (or non-negative) integer numerators and denominators are neighbors then also the top and the bottom (''bot'' &nbsp;for short) pair:
 
*<math>\mathit{top}\!\left(\frac{a}{c},\frac{b}{d}\right)\ :=\ \left(\frac{a}{c},\frac{a+b}{b+d}\right)</math>


and
and


:<math>A_{n+1} = \mathit{bot}(A_n),\quad\quad A_{n+2} = \mathit{bot}(A_{n+1})</math>
*<math>\mathit{bot}\!\left(\frac{a}{c},\frac{b}{d}\right)\ :=\ \left(\frac{a+b}{c+d},\frac{b}{d}\right)</math>




Let <math>\ A_n := (\frac{a}{c},\,\frac{b}{d})</math>&nbsp; be a pair of neighbors for which the top-top property above holds, i.e. for which <math>\ x</math>&nbsp; is squeezed between a pair of neighbors as follows:
are both pairs of neighbors.
 
:::<math>\frac{a}{c}\ >\ x\ >\ \frac{2\cdot a+b}{2\cdot c+d}</math>


Then
Let <math>\ A_0</math>&nbsp; be a pair of neighbors, and <math>\ x\in \mathit{Span}(A_0)</math> be an irrational number. Assume that pairs of neighbors <math>\ A_0,\dots,A_{n-1}</math>&nbsp; are already defined, and that they squeeze <math>\ x,</math> i.e. that <math>\ x\in\mathit{Span}(A_k)</math>&nbsp; for each <math>\ k=0,\dots,n-1.</math>&nbsp; Then we define <math>\ A_n</math>&nbsp; as the one of the two pairs: <math>\ \mathit{top}(A_{n-1})</math> &nbsp;or&nbsp; <math>\ \mathit{bot}(A_{n-1}),</math> &nbsp;which squeezes <math>\ x.</math>&nbsp; Thus for every positive irrational number we have obtained an infinite sequence of pairs of neighbors, each squeezing the given irrational number more and more. Thus for arbitrary irrational <math>\ x</math>&nbsp; there exist fractions of integers <math>\ \frac{s}{t},</math>&nbsp; with arbitrarily large denominators, such that


::<math>0\ <\ \frac{a}{c}-x\ <\ \frac{1}{c\cdot(2\cdot c+d)}\ <\ \frac{1}{2\cdot c^2}</math>
:::<math>\left|x - \frac{s}{t}\right|\ <\ \frac{1}{\sqrt{5}\cdot t^2}</math>


(see section '''''Hurwitz theorem''''').


Similarly, in the bot-bot case we get
If cases top-top and bot-bot happen only finitely many times then starting with an <math>\ n</math>&nbsp; we get an infinite alternating top-bot-top-bot-... sequence:


::<math>0\ <\ x-\frac{b}{d}\ <\ \frac{1}{d\cdot(c+2\cdot d)}\ <\ \frac{1}{2\cdot d^2}</math>
:<math>A_{n+1} = \mathit{top}(A_n),\quad A_{n+2} = \mathit{bot}(A_{n+2}),\quad A_{n+3} = \mathit{top}(A_{n+1})\quad\dots</math>




Thus if the top-top or the bot-bot case holds infinitely many times then there exist infinitely many fractions of integers&nbsp; <math>\frac{s}{t}</math>&nbsp; such that:
Then the new neighbor of the <math>\ A_{n+k}</math>&nbsp; pair (i.e. the median of the previous pair <math>\ A_{n+k-1}</math>)&nbsp; is equal to


::: <math>\left|x-\frac{s}{t}\right|\ <\ \frac{1}{2\cdot t^2}</math>
::<math>e_{k}\ :=\ \frac{F_{k+1}\cdot a + F_{k}\cdot b}{F_{k+1}\cdot c + F_{k}\cdot d}</math>


On the other hand, if cases top-top and bot-bot happen only finitely many times then starting with an <math>\ n</math>&nbsp; we get an infinite alternating top-bot-top-bot-... sequence:
for every <math>\ k=1,2,\dots,</math>&nbsp; where <math>\ F_{t}</math>&nbsp; are the [[Fibonacci number]]s, where


:<math>A_{n+1} = \mathit{top}(A_n),\quad A_{n+2} = \mathit{bot}(A_{n+2}),\quad A_{n+3} = \mathit{top}(A_{n+1})\quad\dots</math>
:::<math>\left(\frac{a}{c},\frac{b}{d}\right)\ :=\ A_n</math>


Then the new neighbor of the <math>\ A_{n+k}</math>&nbsp; pair (i.e. the median of the previous pair <math>\ A_{n+k-1}</math>)&nbsp; is equal to
It is known that
 
::<math>e_{k}\ :=\ \frac{F_{k+1}\cdot a + F_{k}\cdot b}{F_{k+1}\cdot c + F_{k}\cdot d}</math>
 
for every <math>\ k=1,2,\dots,</math>&nbsp; where <math>\ F_{t}</math>&nbsp; are the [[Fibonacci number]]s. It is known that


:::<math>\lim_{k\rightarrow\infty}\frac{F_{k+1}}{F_k}\ =\ \Phi\ :=\ \frac{\sqrt{5}+1}{2}</math>
:::<math>\lim_{k\rightarrow\infty}\frac{F_{k+1}}{F_k}\ =\ \Phi\ :=\ \frac{\sqrt{5}+1}{2}</math>
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::<math>x\ =\ \lim_{k\rightarrow\infty}e_k\ =\ \frac{\Phi\cdot a+b}{\Phi\cdot c+d}</math>
::<math>x\ =\ \lim_{k\rightarrow\infty}e_k\ =\ \frac{\Phi\cdot a+b}{\Phi\cdot c+d}</math>


Thus


:::<math>\frac{a}{c}-x\ =\ \frac{1}{c\cdot(\Phi\cdot c+d)}</math>
But if our infinite alternation has started with ''bot'' :


and
:<math>A_{n+1} = \mathit{bot}(A_n),\quad A_{n+2} = \mathit{top}(A_{n+2}),\quad A_{n+3} = \mathit{bot}(A_{n+1})\quad\dots</math>


:::<math>x-\frac{b}{d}\ =\ \frac{\Phi}{(\Phi\cdot c+d)\cdot d}</math>
Then we would have


If &nbsp;<math>\ d > \frac{c}{\Phi}</math>&nbsp; then
::<math>x\ =\ \lim_{k\rightarrow\infty}e_k\ =\ \frac{a+\Phi\cdot b}{c+\Phi\cdot d}</math>
 
::<math>0\ <\ \frac{a}{c} - x\ <\ \frac{1}{c^2}\cdot\frac{1}{\Phi+\frac{1}{\Phi}}\ =\ \frac{1}{\sqrt{5}\cdot c^2}</math>
 
and if &nbsp;<math>\ d < \frac{c}{\Phi},</math>&nbsp; i.e.&nbsp; <math>\ c > \Phi\cdot d,</math>&nbsp; then
 
::<math>0\ <\ x-\frac{b}{d}\ <\ \frac{\Phi}{(\Phi^2+1)\cdot d^2}\ =\ \frac{1}{\sqrt{5}\cdot d^2}</math>
 
Since&nbsp; <math>\sqrt(5)>2,</math> and due to the earlier inequalities which have covered the top-top and the bot-bot cases, we have obtained the following theorem:
 
* for arbitrary irrational number there exist infinitely many different fractions <math>\ \frac{s}{t},</math>&nbsp; with integer numerator and non-zero denominator, such that:
:::<math>\left|x-\frac{s}{t}\right|\ <\ \frac{1}{2\cdot t^2}</math>
 
However, we are close to replacing constant&nbsp; <math>\ 2</math>&nbsp; by &nbsp;<math>\sqrt{5}</math>&nbsp; in the above denominator. Let's do it:


&nbsp;
&nbsp;


=== Another proof of Hurwitz Theorem (further insight) ===
=== Another proof of Hurwitz Theorem (further insight) ===
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==== The top-top-bot case ====
==== The top-top-bot case ====
Let's consider the latter top-top-bot case. Let <math>\ \xi := \frac{d}{c}.</math>&nbsp; The squeeze by neighbors:
Let's consider the latter top-top-bot case. Let&nbsp; <math>\ \xi := \frac{d}{c}.</math>&nbsp; The squeeze by neighbors:


:::<math>\frac{a}{c}\ >\ x\ >\ \frac{2\cdot a+b}{2\cdot c+d}</math>
:::<math>\frac{a}{c}\ >\ x\ >\ \frac{2\cdot a+b}{2\cdot c+d}</math>
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:::<math>\max(\frac{2}{5},\,\frac{3}{7},\,\frac{2}{7})\ =\ \frac{3}{7}\ <\ \frac{1}{\sqrt{5}},</math>
:::<math>\max(\frac{2}{5},\,\frac{3}{7},\,\frac{2}{7})\ =\ \frac{3}{7}\ <\ \frac{1}{\sqrt{5}},</math>


each occurence of the top-top-bot subsequence, i.e. of equalities:
each occurrence of the top-top-bot subsequence, i.e. of equalities:




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The same is holds for every occurence of the bot-bot-top sequence, which can be shown now mechanically by a proof similar to the proof of the top-top-bot case, or as follows: define the squeezing sequence of&nbsp; <math>\ -x</math>&nbsp; by:
The same is holds for every occurrence of the bot-bot-top sequence, which can be shown now mechanically by a proof similar to the proof of the top-top-bot case, or as follows: define the squeezing sequence of&nbsp; <math>\ -x</math>&nbsp; by:


::<math>B_n := (\frac{-b}{d},\frac{-a}{c})</math> &nbsp; &nbsp; for &nbsp; &nbsp; <math>A_n = (\frac{a}{c},\frac{b}{d})</math>
::<math>B_n := (\frac{-b}{d},\frac{-a}{c})</math> &nbsp; &nbsp; for &nbsp; &nbsp; <math>A_n = (\frac{a}{c},\frac{b}{d})</math>
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  <math>0\ <\ x-\frac{\alpha}{\gamma}\ <\ \frac{2+\xi}{5+2\cdot\xi}\cdot\frac{1}{\gamma^2}\quad\quad\Leftarrow\quad\quad x\in C</math>
  <math>0\ <\ x-\frac{\alpha}{\gamma}\ <\ \frac{2+\xi}{5+2\cdot\xi}\cdot\frac{1}{\gamma^2}\quad\quad\Leftarrow\quad\quad x\in C</math>


==== Neigborhood C (the combined inequality) ====
Let:
:::<math>\xi_0\ :=\ \frac{\sqrt{61}-7}{6}</math>
Then
::<math>\frac{3}{7+3\cdot \xi_0}\ =\ \frac{2+\xi_0}{5+2\cdot\xi_0}\ =\ \frac{\sqrt{61}-7}{2}</math>
Thus for &nbsp;<math>\ \xi\ge\xi_0</math>:
:<math>\frac{3}{7+3\cdot \xi}\ \le\ \frac{\sqrt{61}-7}{2}</math>
and  for &nbsp;<math>\ \xi\le\xi_0</math>:
:<math>\frac{2+\xi}{5+2\cdot\xi}\ \le\ \frac{\sqrt{61}-7}{2}</math>
It follows that
* for one of the fractions&nbsp; <math>\frac{s}{t}\in\left\{\frac{a}{c},\,\frac{2\cdot a+b}{2\cdot c+d}\right\}</math>&nbsp;  the following inequality holds for every &nbsp;<math>\ x\in C:</math>
:::<math>\left|x-\frac{s}{t}\right|\ <\ \frac{\sqrt{61}-7}{2}</math>
(the choice of &nbsp;<math>\frac{s}{t}</math>&nbsp; depends on&nbsp; <math>\xi := \frac{d}{c}</math> ).


&nbsp;
&nbsp;
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and its length
and its length


:::<math>|M|\ :=\ \mathit{rat}(v)-\mathit{rat}(w)</math>
:::<math>diam(M)\ :=\ \mathit{rat}(v)-\mathit{rat}(w)</math>


where <math>\ v</math>&nbsp; is the left, and <math>\ w</math>&nbsp; is the right column of matrix M &mdash; observe that the rational representation of the left column of a special matrix is always greater than the rational representation of the right column of the same special matrix.
where <math>\ v</math>&nbsp; is the left, and <math>\ w</math>&nbsp; is the right column of matrix M &mdash; observe that the rational representation of the left column of a special matrix is always greater than the rational representation of the right column of the same special matrix.
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:::<math>M\ :=\ \left[\begin{array}{cc}a & b\\ c & d\end{array}\right]\in\mathit{SO}(\mathbb{Z}_+, 2)</math>
:::<math>M\ :=\ \left[\begin{array}{cc}a & b\\ c & d\end{array}\right]\in\mathit{SO}(\mathbb{Z}_+, 2)</math>


then&nbsp; <math>\ |M| = \frac{1}{c\cdot d}</math>
then&nbsp; <math>\ diam(M) = \frac{1}{c\cdot d}</math>

Latest revision as of 07:40, 15 March 2021

The theory of diophantine approximations is a chapter of number theory, which in turn is a part of mathematics. It studies the approximations of real numbers by rational numbers. This article presents an elementary introduction to diophantine approximations, as well as an introduction to number theory via diophantine approximations.

Introduction

In the everyday life our civilization applies mostly (finite) decimal fractions   Decimal fractions are used both as certain values, e.g. $5.85, and as approximations of the real numbers, e.g.   However, the field of all rational numbers is much richer than the ring of the decimal fractions (or of the binary fractions   which are used in the computer science). For instance, the famous approximation   has denominator 113 much smaller than 105 but it provides a better approximation than the decimal one, which has five digits after the decimal point.

How well can real numbers (all of them or the special ones) be approximated by rational numbers? A typical Diophantine approximation result states:


Theorem  Let   be an arbitrary real number. Then

  •   is rational     there exists a real number C > 0 such that

for arbitrary integers   such that   and

  • (Adolph Hurwitz)     is irrational     there exist infinitely many pairs of integers   such that   and


Remark  Implication     of the first part of the theorem is a simple and satisfaction bringing exercise.

Notation

  •   —   "equivalent by definition" (i.e. "if and only if");
  •   —   "equals by definition";
  •   —   "there exists";
  •   —   "for all";
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a\in A\ }   —   "  is an element of set ";

 

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{N}\ :=\ \{1, 2 \dots\}}  —  the semiring of the natural numbers;
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{Z}_+\ :=\ \{0, 1, \dots\}}  —  the semiring of the non-negative integers;
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{Z}\ :=\ \{-2,-1,1, 2 \dots\}}  —  the ring of integers;
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{Q}}  —  the field of rational numbers;
  •  —  the field of real numbers;

 

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x;y)\ := \{t\in\mathbb{R} :\, x<t<y\}}
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |(x;y)|\ := y-x}
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathit{Span}(x,y)\ :=\ \left(\min\left(x,y\right);\,\max\left(x,y\right)\right)}

 

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x|y\ }   —   "Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\ }   divides Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ y} "   (i.e. );
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \gcd(a, b)\ }  —  the greatest common divisor of integers Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a}   and

 

The method of neighbors and median

In this section we will quickly obtain some results about approximating irrational numbers by rational. To this end we will not worry about the details of the difference between a rational number and a fraction (with integer numerator and denominator)—this will not cause any problems; fully crisp notions will be developed in the next sections, they will involve 2-dimensional vectors and 2x2 matrices. This section is still introductory. It is supposed to provide quick insight into the topic.

Definitions

Fractions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{b}{d},}   with integer numerators and natural denominators, are called neighbors (in the given order)   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Leftarrow:\Rightarrow}

Fraction   is called the top neighbor of the other,   is called the bottom neighbor, and the interval Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \mathit{Span}(\frac{a}{c},\frac{b}{d})}   is called neighborhood; thus a neighborhood is an open interval such that its endpoints are neighbors.

  • If  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}}   and   are neighbors then    ( i.e.  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}\cdot \frac{b}{d}\ \ge 0} ).
  • Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a,b,c,d\in\mathbb{N}.}   Fractions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{b}{d}}   are neighbors   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Leftrightarrow}   fractions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{b}}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{c}{a}}   are neighbors   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Leftrightarrow}   fractions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-b}{d}}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-a}{c}}   are neighbors.


Examples:

  • Fractions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{n+1}{1}}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{n}{1}}   are neighbors for every positive integer Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ n.}


  • Fractions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{n}}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{n+1}}   are neighbors for every positive integer Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ n.}

Thus it easily follows that for every positive irrational number Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x}   there exists a pair of neighbors  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}}   and  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{b}{d},}   with positive numerators and denominators, such that:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}\ >\ x\ >\ \frac{b}{d}}  

Definition  A pair of neighboring fractions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{a}{c},\frac{b}{d}\right),}   with integer numerators and natural denominators, is called a top pair   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Leftarrow:\Rightarrow\ c > d.}   Otherwise it is called a bottom pair.


Thus now we have notions of top/bottom neighbors and of top/bottom pairs of neighbors.

  • Let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{a}{c},\frac{b}{d}\right),}   be a pair of neighbors. Then  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{a}{c},\frac{a+b}{c+d}\right)}   is a top pair of neighbors, and  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{a+b}{c+d},\frac{b}{d}\right)}   is a bottom pair of neighbors.

 

First results

Theorem  Let fractions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{b}{d},}   with integer numerators and natural denominators, be neighbors. Then

  • if integers Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ t>0}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ s}   are such that   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \frac{a}{c} > \frac{s}{t} > \frac{b}{d},}   then   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t \ge c+d;}
  • the median  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a+b}{c+d}}   is a bottom neighbor of  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c},}   and a top neighbor of  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{b}{d};}  
  • let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x}   be an irrational number such that   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}\ >\ x\ >\ \frac{b}{d};}   then
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ \max(\frac{a}{c}-x,\, x-\frac{b}{d})\ <\ \frac{1}{c\cdot d}}
and
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ \frac{a}{c}-x\ <\ \frac{1}{2\cdot c^2}}       or       Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ x-\frac{b}{d}\ <\ \frac{1}{2\cdot d^2}}


Proof   Let   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \frac{a}{c} > \frac{s}{t} > \frac{b}{d};}   then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c} - \frac{s}{t}\ \ge\ \frac{1}{c\cdot t}}     and     Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{s}{t} - \frac{b}{d}\ \ge\ \frac{1}{d\cdot t}}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{c\cdot t} + \frac{1}{d\cdot t}\ \le \frac{1}{c\cdot d}}

Multiplying this inequality by  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c\cdot d\cdot t\ } gives

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t \ge c+d}

which is the first part of our theorem.


The second part of the theorem is obtained by a simple calculation, straight from the definition of the neighbors.


The first inequality of the third part of the theorem is instant:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \max(\frac{a}{c}-x,\, x-\frac{b}{d})\ <\ (\frac{a}{c}-x)+(x-\frac{b}{d})\ =\ \frac{1}{c\cdot d}}

Next,


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{1}{c}-\frac{1}{d}\right)^2\ \ge\ 0}

hence

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2\cdot c^2}+\frac{1}{2\cdot d^2}\ \ge\ \frac{1}{c\cdot d}\ =\ \frac{a}{c}-\frac{b}{d}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\ \left|\mathit{Span}(\frac{a}{c},\frac{b}{d})\right|}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{b}{d}+\frac{1}{2\cdot d^2}\ \ge\ \frac{a}{c}-\frac{1}{2\cdot c^2}}

i.e.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathit{Span}\left(\frac{a}{c},\frac{a}{c}-\frac{1}{2\cdot c^2}\right)\ \cup\ \mathit{Span}\left(\frac{b}{d}+\frac{1}{2\cdot d^2},\frac{b}{d}\right)\ \backslash\ \mathbb{Q}\ \supseteq\ \mathit{Span}\left(\frac{a}{c},\frac{b}{d}\right)\ \backslash\ \mathbb{Q}}


End of proof


Corollary  Let fractions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{b}{d},}   with integer numerators and natural denominators, be neighbors. Then, if integers Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ t>0}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ s}   are such that   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \frac{a}{c} > \frac{s}{t} > \frac{b}{d},}   then either

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s=a+b\quad\and\quad t=c+d;}

or

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t\ \ge\ c+d+\min(c,d)\ >\ c+d}

 

Hurwitz theorem

Let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x\in \mathbb{R}}   be an arbitrary irrational number. Then
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|x-\frac{s}{t}\right|\ <\ \frac{1}{\sqrt{5}\cdot t^2}}
for infinitely many different Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ s,t\in\mathbb{Z} \backslash \{0\}.}


Lemma 1

Let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ c,d\in\mathbb{N}.}   Let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ C:=c+d.}   Then:


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c^2-\sqrt{5}\!\cdot\! c\!\cdot\! d+d\,^2\ >\ 0}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C\,^2-\sqrt{5}\cdot\! C\!\cdot\! d+d\,^2\ >\ 0}


Proof of lemma 1  It's easy to show that  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{2}\cdot c \ne \sqrt{3-\sqrt{5}}\cdot d.}   Thus the square of  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ X := \sqrt{2}\cdot c - \sqrt{3-\sqrt{5}}\cdot d}   is positive. Now,


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{2}\cdot \sqrt{3-\sqrt{5}}\ =\ \sqrt{5}-1}


which means that we may write  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ X^2}   as follows:


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\!\cdot\! c^2\ -\ 2\!\cdot\!(\sqrt{5}-1)\cdot c\!\cdot\! d\ +\ (3-\sqrt{5})\!\cdot\! d\,^2\ >\ 0}

i.e.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (c^2-\sqrt{5}\cdot c\cdot d+d^2)\ +\ (C^2-\sqrt{5}\cdot C\cdot d+d^2)\ >\ 0}


and lemma 1 follows.  End of proof

Lemma 2

Let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a,b\in\mathbb{Z}}   and  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ c,d\in\mathbb{N}.}   Let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ A:=a+b}   and  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ C:=c+d.}   Furthermore, let fractions  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c},\frac{b}{d},}   be neighbors, and let:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{A}{C}\ >\ x\ >\ \frac{b}{d}}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x}   is real. Then one of the following three inequalities holds:

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ \frac{a}{c}-x\ <\ \frac{1}{\sqrt{5}\cdot c^2}}


  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ \frac{A}{C}-x\ <\ \frac{1}{\sqrt{5}\cdot C^2}}



Proof  There are two cases along the inequalities of Lemma 1. Let's assume the first one, which is equivalent to:


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\sqrt{5}\cdot c^2} + \frac{1}{\sqrt{5}\cdot d\,^2}\ >\ \frac{1}{c\cdot d}\ =\ \frac{a}{c}-\frac{b}{d}\ =\ \left|\mathit{Span}\left(\frac{a}{c},\frac{b}{d}\right)\right|}


Thus


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{b}{d}+\frac{1}{\sqrt{5}\cdot d\,^2}\ >\ \frac{a}{c}-\frac{1}{\sqrt{5}\cdot c^2}}


which means that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathit{Span}\left(\frac{a}{c},\frac{a}{c}-\frac{1}{\sqrt{5}\cdot c^2}\right)\ \cup\ \mathit{Span}\left(\frac{b}{d}+\frac{1}{\sqrt{5}\cdot d\,^2},\frac{b}{d}\right)\ \supseteq\ \mathit{Span}\left(\frac{a}{c},\frac{b}{d}\right)}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \supseteq\ \mathit{Span}\left(\frac{A}{C},\frac{b}{d}\right)}

Thus lemma 2 is proven when the first inequality of lemma 1 holds. By replacing in the above proof the lower case  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a,c,}   by the upper case  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ A,C,}   we obtain the proof when the second inequality of lemma 1 holds.

End of proof (of lemma 2)

Lemma 2'

Let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a,b\in\mathbb{Z}}   and  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ c,d\in\mathbb{N}.}   Let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ A:=a+b}   and  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ C:=c+d.}   Furthermore, let fractions  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c},\frac{b}{d},}   be neighbors, and let:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}\ >\ x\ >\ \frac{A}{C}}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x}   is real. Then one of the following three inequalities holds:

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ x - \frac{b}{d}\ <\ \frac{1}{\sqrt{5}\cdot d\,^2}}


  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ x-\frac{A}{C} <\ \frac{1}{\sqrt{5}\cdot C^2}}


  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ \frac{a}{c}\ <\ \frac{1}{\sqrt{5}\cdot x^2}}


Proof  It's similar to the proof of lemma 2. Or one may apply lemma 2 to  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x':=-x,\ a':=-b,\ b':=-a,}   and  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ c':=d,\ d':=c} , which would provide us with the respective fraction  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{s'}{t'}.}   Then the required  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ s,t,}   are given by 

End of proof (of lemma 2')

 

Proof of Hurwitz theorem

When  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x}   is irrational, then it is squeezed between infinitely many different pairs of neighbors (see part two of the theorem from the First results section). They provide infinitely many different required fractions  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{s}{t}}   (see lemma 2 and 2' above; but it is not claimed, nor true, that different pairs of neighbors provide different required fractions).

End of proof

 

Squeezing irrational numbers between neighbors

Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x > 0}   be an irrational number, We may always squeeze it between the extremal neighbours:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{0}\ >\ x\ >\ \frac{0}{1}}

But if you don't like infinity (on the left above) then you may do one of the two things:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x > 1\quad\quad\Rightarrow\quad\quad\frac{n+1}{1} > x > \frac{n}{1}}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x < 1\quad\quad\Rightarrow\quad\quad\frac{1}{n} > x > \frac{1}{n+1}}


where in each of these two cases  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ n\in \mathbb{N}}   is a respective unique positive integer.

It was mentioned in the previous section (First results) that if fractions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{b}{d},}   with positive (or non-negative) integer numerators and denominators are neighbors then also the top and the bottom (bot  for short) pair:

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathit{top}\!\left(\frac{a}{c},\frac{b}{d}\right)\ :=\ \left(\frac{a}{c},\frac{a+b}{b+d}\right)}

and

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathit{bot}\!\left(\frac{a}{c},\frac{b}{d}\right)\ :=\ \left(\frac{a+b}{c+d},\frac{b}{d}\right)}


are both pairs of neighbors.

Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ A_0}   be a pair of neighbors, and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x\in \mathit{Span}(A_0)} be an irrational number. Assume that pairs of neighbors Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ A_0,\dots,A_{n-1}}   are already defined, and that they squeeze Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x,} i.e. that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x\in\mathit{Span}(A_k)}   for each Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ k=0,\dots,n-1.}   Then we define Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ A_n}   as the one of the two pairs: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \mathit{top}(A_{n-1})}  or  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \mathit{bot}(A_{n-1}),}  which squeezes Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x.}   Thus for every positive irrational number we have obtained an infinite sequence of pairs of neighbors, each squeezing the given irrational number more and more. Thus for arbitrary irrational Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x}   there exist fractions of integers Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \frac{s}{t},}   with arbitrarily large denominators, such that

(see section Hurwitz theorem).

If cases top-top and bot-bot happen only finitely many times then starting with an Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ n}   we get an infinite alternating top-bot-top-bot-... sequence:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_{n+1} = \mathit{top}(A_n),\quad A_{n+2} = \mathit{bot}(A_{n+2}),\quad A_{n+3} = \mathit{top}(A_{n+1})\quad\dots}


Then the new neighbor of the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ A_{n+k}}   pair (i.e. the median of the previous pair Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ A_{n+k-1}} )  is equal to

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e_{k}\ :=\ \frac{F_{k+1}\cdot a + F_{k}\cdot b}{F_{k+1}\cdot c + F_{k}\cdot d}}

for every Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ k=1,2,\dots,}   where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ F_{t}}   are the Fibonacci numbers, where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{a}{c},\frac{b}{d}\right)\ :=\ A_n}

It is known that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{k\rightarrow\infty}\frac{F_{k+1}}{F_k}\ =\ \Phi\ :=\ \frac{\sqrt{5}+1}{2}}

hence

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\ =\ \lim_{k\rightarrow\infty}e_k\ =\ \frac{\Phi\cdot a+b}{\Phi\cdot c+d}}


But if our infinite alternation has started with bot :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_{n+1} = \mathit{bot}(A_n),\quad A_{n+2} = \mathit{top}(A_{n+2}),\quad A_{n+3} = \mathit{bot}(A_{n+1})\quad\dots}

Then we would have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\ =\ \lim_{k\rightarrow\infty}e_k\ =\ \frac{a+\Phi\cdot b}{c+\Phi\cdot d}}

 

Another proof of Hurwitz Theorem (further insight)

Reduction to x > 0

Since

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|(-x)-\frac{-s}{t}\right|\ =\ \left|x-\frac{s}{t}\right|}

it is enough to prove Hurwitz theorem for positive irrational numbers only.

Two cases

Consider the sequence Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \left(A_n\right)}   of pairs of neighbors, which squeeze Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x>0,}   from the previous section. The case of the infinite alternation top-bot-top-bot-... has been proved already. In the remaining case the bot-bot-top or top-top-bot progressions appear infinitely many times, i.e. there are infinitely many non-negative integers Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ n}   for which

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}\ >\ \frac{a+b}{c+d}\ >\ \frac{a+2\cdot b}{c+2\cdot d}\ >\ x\ >\ \frac{a+3\cdot b}{c+3\cdot d}\ >\ \frac{b}{d}}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}\ >\ \frac{3\cdot a+b}{3\cdot c+d}\ >\ x\ >\ \frac{2\cdot a+b}{2\cdot c+d}\ >\ \frac{a+b}{c+d}\ >\ \frac{b}{d}}


holds, where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{a}{c},\frac{b}{d}\right)\ :=\ A_n}

 

The top-top-bot case

Let's consider the latter top-top-bot case. Let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \xi := \frac{d}{c}.}   The squeeze by neighbors:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}\ >\ x\ >\ \frac{2\cdot a+b}{2\cdot c+d}}

shows that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ \frac{a}{c}-x\ <\ \frac{1}{c\cdot(2\cdot c+d)}\ =\ \frac{1}{(2+\xi)}\cdot\frac{1}{c^2}}

This inequality provides the first insight (otherwise, we are not going to use it), so it deserves to be written cleanly as an implication: 

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ \frac{a}{c}-x\ <\ \frac{1}{(2+\xi)}\cdot\frac{1}{c^2}\quad\quad \Leftarrow\quad\quad\frac{a}{c}\ >\ x\ >\ \frac{2\cdot a+b}{2\cdot c+d}}

The relevant neighborhoods

Consider the next two pairs of neighbors, pair Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ A_{n+4}}   and pair   which squeeze   The relevant neighborhoods are:

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A\ :=\ \mathit{Span}(\frac{a}{c},\,\frac{3\cdot a+b}{3\cdot c+d})}


  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B\ :=\ \mathit{Span}(\frac{3\cdot a+b}{3\cdot c+d},\,\frac{5\cdot a+2\cdot b}{5\cdot c+2\cdot d})}


  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C\ :=\ \mathit{Span}(\frac{5\cdot a+2\cdot b}{5\cdot c+2\cdot d},\,\frac{7\cdot a+3\cdot b}{7\cdot c+3\cdot d})}


  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D\ :=\ \mathit{Span}(\frac{7\cdot a+3\cdot b}{7\cdot c+3\cdot d},\,\frac{2\cdot a+b}{2\cdot c+d})}

 

Neighborhood B

Let    Then


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}\ >\ \frac{3\cdot a+b}{3\cdot c+d}\ >\ x\ >\ \frac{5\cdot a+2\cdot b}{5\cdot c+2\cdot d}\ >\ \frac{2\cdot a+b}{2\cdot c+d}}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ \frac{a}{c}-x\ <\ \frac{1}{c\cdot(3\cdot c+d)}+\frac{1}{(3\cdot c+d)\cdot(5\cdot c+2\cdot d)}}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\ \frac{2}{c\cdot(5\cdot c+2\cdot d)}\ =\ \frac{2}{(5+2\cdot\xi)}\cdot\frac{1}{c^2}}


Conclusion: 

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ \frac{a}{c}-x\ <\ \frac{2}{5+2\cdot\xi}\cdot\frac{1}{c^2}\quad\quad\Leftarrow\quad\quad x\in B}

Neighborhood C (first C-inequality)

Let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x\in C.}   Then


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{c}\ >\ \frac{3\cdot a+b}{3\cdot c+d}\ >\ \frac{5\cdot a+2\cdot b}{5\cdot c+2\cdot d}\ >\ x\ >\ \frac{7\cdot a+3\cdot b}{7\cdot c+3\cdot d}}


Thus using the calculation for neighborhood B also for C, we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ \frac{a}{c}-x\ <\ \frac{2}{(5+2\cdot\xi)\cdot c^2}+\frac{1}{(5\cdot c+2\cdot d)\cdot(7\cdot c+3\cdot d)}}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\ \frac{3}{7+3\cdot\xi}\cdot\frac{1}{c^2}}


First C-conclusion: 

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ \frac{a}{c}-x\ <\ \frac{3}{7+3\cdot\xi}\cdot\frac{1}{c^2}\quad\quad\Leftarrow\quad\quad x\in C}

 

Neighborhood D

Let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x\in D,}   i.e.


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{7\cdot a+3\cdot b}{7\cdot c+3\cdot d}\ >\ x\ >\ \frac{2\cdot a+b}{2\cdot c+d}}


Let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha := 2\cdot a+b}   and   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \gamma := 2\cdot c+d = (2+\xi)\cdot c.}   Then


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ x-\frac{\alpha}{\gamma}\ =\ x-\frac{2\cdot a+b}{2\cdot c+d}}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <\ \frac{1}{(7\cdot c+3\cdot d)\cdot(2\cdot c+d)}\ =\ \frac{1}{(7+3\cdot\xi)\cdot c\cdot\gamma}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\ \frac{2+\xi}{7+3\cdot\xi}\cdot\frac{1}{\gamma^2}}


Conclusion: 

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ x-\frac{\alpha}{\gamma}\ <\ \frac{2+\xi}{7+3\cdot\xi}\cdot\frac{1}{\gamma^2}\quad\quad\Leftarrow\quad\quad x\in D}


Early yield (Hurwitz Theorem)

Let's impatiently indulge ourselves in already getting some crude results from the above hard work (:-). The above three BCD-conclusions instantly imply:

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ \frac{a}{c}-x\ <\ \frac{2}{5}\cdot\frac{1}{c^2}\quad\quad\Leftarrow\quad\quad x\in B}


  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ \frac{a}{c}-x\ <\ \frac{3}{7}\cdot\frac{1}{c^2}\quad\quad\Leftarrow\quad\quad x\in C}


  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ x-\frac{\alpha}{\gamma}\ <\ \frac{2}{7}\cdot\frac{1}{\gamma^2}\quad\quad\Leftarrow\quad\quad x\in D}


Since

each occurrence of the top-top-bot subsequence, i.e. of equalities:


provides a fraction    with integer numerator and natural denominator, such that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|x-\frac{s}{t}\right|\ <\ \frac{3}{7}\cdot\frac{1}{t^2}\ <\ \frac{1}{\sqrt{5}}\cdot\frac{1}{t^2}}


The same is holds for every occurrence of the bot-bot-top sequence, which can be shown now mechanically by a proof similar to the proof of the top-top-bot case, or as follows: define the squeezing sequence of  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ -x}   by:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_n := (\frac{-b}{d},\frac{-a}{c})}     for     Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_n = (\frac{a}{c},\frac{b}{d})}


Let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(A_n,A_{n+1},A_{n+2},A_{n+3}\right)}   be a bot-bot-top progression. Then  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(B_n,B_{n+1},B_{n+2},B_{n+3}\right)}   is a top-top-bot progression which squeezes  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ -x.} Thus


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|(-x)-\frac{u}{w}\right|\ <\ \frac{3}{7}\cdot\frac{1}{w^2}\ <\ \frac{1}{\sqrt{5}}\cdot\frac{1}{w^2}}


for certain  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{u}{w}\in \mathit{Span}(B_{n+3}),}   with integer numerator and natural denominator. Then  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{s}{t}\in\mathit{Span}(A_{n+3}),}   for  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ (s,t):=(-u,w),}   satisfies:


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|x-\frac{s}{t}\right|\ <\ \frac{3}{7}\cdot\frac{1}{t^2}\ <\ \frac{1}{\sqrt{5}}\cdot\frac{1}{t^2}}


When another top-top-bot or bot-bot-top progression starts with a sufficiently large index  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ n',}   then  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|\mathit{Span}(A_{n'+3})\right| < \left|x-\frac{s}{t}\right|,}   which means that the respective new approximation  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{s'}{t'}\in\mathit{Span}(A_{n'+3})}   is different. It follows that if there are infinitely many progressions top-top-bot or bot-bot-bot then there are infinitely many fractions  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \frac{s}{t},} which satisfy the inequality above. Thus we have obtained the following version of Hurwitz theorem:


Theorem  Let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\in\mathbb{R}\backslash\mathbb{Q}}   be an arbitrary irrational number. Then inequality
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|x-\frac{s}{t}\right|\ <\ \frac{1}{\sqrt{5}}\cdot\frac{1}{t^2}}
holds for infinitely many fractions  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \frac{s}{t},}   with integer numerator and natural denominator. Furthermore, if the squeezing sequence of  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x} does not eventually alternate top-bot-top-bot-... till infinity, i.e. if it has infinitely many top-top or bot-bot progressions, then
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|x-\frac{s}{t}\right|\ <\ \frac{3}{7 }\cdot\frac{1}{t^2}}
holds for infinitely many fractions  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \frac{s}{t},}   with integer numerator and natural denominator.


 

Neighborhood C (second C-inequality)

Let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x\in C.}   Then


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{5\cdot a+2\cdot b}{5\cdot c+2\cdot d}\ >\ x\ >\ \frac{7\cdot a+3\cdot b}{7\cdot c+3\cdot d}\ >\ \frac{2\cdot a+b}{2\cdot c+d}}


Thus, using the earlier conclusion for neighborhood Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ D}   also for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ C,}   we obtain


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ x-\frac{\alpha}{\gamma}\ =\ x-\frac{2\cdot a+b}{2\cdot c+d}}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <\ \frac{1}{(5\cdot c+2\cdot d)\cdot(7\cdot c+3\cdot d)}\ +\ \frac{2+\xi}{(7+3\cdot\xi)\cdot \gamma^2}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\ \frac{1}{(5+2\cdot\xi)\cdot(7+3\cdot\xi)}\cdot\frac{1}{c^2}\ +\ \frac{2+\xi}{(7+3\cdot\xi)\cdot \gamma^2}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\ \frac{(2+\xi)^2}{(5+2\cdot\xi)\cdot(7+3\cdot\xi)}\cdot\frac{1}{\gamma^2}\ +\ \frac{2+\xi}{(7+3\cdot\xi)\cdot \gamma^2}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\ \frac{2+\xi}{5+2\cdot\xi}\cdot\frac{1}{\gamma^2}}


Second C-conclusion: 

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ <\ x-\frac{\alpha}{\gamma}\ <\ \frac{2+\xi}{5+2\cdot\xi}\cdot\frac{1}{\gamma^2}\quad\quad\Leftarrow\quad\quad x\in C}


Neigborhood C (the combined inequality)

Let:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \xi_0\ :=\ \frac{\sqrt{61}-7}{6}}

Then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{3}{7+3\cdot \xi_0}\ =\ \frac{2+\xi_0}{5+2\cdot\xi_0}\ =\ \frac{\sqrt{61}-7}{2}}


Thus for  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \xi\ge\xi_0} :



and for  :


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2+\xi}{5+2\cdot\xi}\ \le\ \frac{\sqrt{61}-7}{2}}


It follows that

  • for one of the fractions  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{s}{t}\in\left\{\frac{a}{c},\,\frac{2\cdot a+b}{2\cdot c+d}\right\}}   the following inequality holds for every  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x\in C:}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|x-\frac{s}{t}\right|\ <\ \frac{\sqrt{61}-7}{2}}

(the choice of  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{s}{t}}   depends on  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \xi := \frac{d}{c}} ).

 

Divisibility

Definition  Integer Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a}   is divisible by integer Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ b}   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Leftarrow:\Rightarrow\ }   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \exists_{c\in\mathbb{Z}}\ a=b\cdot c}

Symbolically:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b|a\ }   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Leftarrow:\Rightarrow\ }   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \exists_{c\in\mathbb{Z}}\ a=b\cdot c}


When Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a}   is divisible by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ b}   then we also say that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ b}   is a divisor of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a,}   or that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ b}   divides Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a.}

  • The only integer divisible by   is   (i.e. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ 0}   is a divisor only of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ 0} ).
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\ }   is divisible by every integer.
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1\ }   is the only positive divisor of
  • Every integer is divisible by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ 1}   (and by  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ -1} ).

 

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a|b\ \Rightarrow (-a|b\ \and\ a|\!-\!\!b)}
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a|b > 0\ \Rightarrow a\le b}

 

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a|a\ }
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a|b\ \and\ b|c)\ \Rightarrow\ a|c}
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a|b\ \and\ b|a)\ \Rightarrow\ |a|=|b|}

Remark  The above three properties show that the relation of divisibility is a partial order in the set of natural number  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{N},}   and also in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{Z}_+} — Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ 1}   is its minimal, and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ 0}   is its maximal element.

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a|b\ \and\ a|c)\ \Rightarrow\ (a\,|\,b\!\cdot\!d\ \and\ a\,|\,b\!+\!c\ \and\ a\,|\,b\!-\!c)}

 

Relatively prime pairs of integers

Definition  Integers Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ b}   are relatively prime   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Leftarrow:\Rightarrow\ }   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ 1}   is their only common positive divisor.

  • Integers Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ x}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ 0}   are relatively prime   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Leftrightarrow\ |x| = 1.}
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1\ }   is relatively prime with every integer.
  • If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ b}   are relatively prime then also Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a}   and  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ -b}   are relatively prime.


Theorem 1  If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a,b,c \in \mathbb{Z}}   are such that two of them are relatively prime and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a+b=c,}   then any two of them are relatively prime.
Corollary  If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ b}   are relatively prime then also   and    are relatively prime.


Now, let's define inductively a table odd integers:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\nu_{k,n} : k\in \mathbb{Z}_+,\ 0\le n\le 2^k)}

as follows:

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nu_{0,0} := 0\ }     and    
  •   for  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\le n\le 2^k\ }
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nu_{k+1,2\cdot n+1}\ :=\ \nu_{k,n}+\nu_{k,n+1}}   for  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\le n < 2^k\ }

for every  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ k=0,1,\dots.}


The top of this table looks as follows:

0 1
0 1 1
0 1 1 2 1
0 1 1 2 1 3 2 3 1
0 1 1 2 1 3 1 3 1 4 3 5 2 5 3 4 1

etc.

Theorem 2
  • Every pair of neighboring elements of the table, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \nu_{k,n}}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \nu_{k,n+1}}   is relatively prime.
  • For every pair of relatively prime, non-negative integers Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ b}   there exist indices Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k\in\mathbb{Z}_+}   and non-negative Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ n<2^k}   such that:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{a,b\}\ =\ \{\nu_{k,n},\nu_{k,n+1}\}}

Proof  Of course the pair

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{\nu_{0,0},\nu_{0,1}\}\ =\ \{0,1\}}

is relatively prime; and the inductive proof of the first statement of Theorem 2 is now instant thanks to Theorem 1 above.

Now let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a}   and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ b}   be a pair of relatively prime, non-negative integers. If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a+b=1}   then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \{a, b\}=\{0,1\}=\{\nu_{0,0},\nu_{0,1}\},}   and the second part of the theorem holds. Continuing this unductive proof, let's assume that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a+b>1.}   Then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \min(a,b) > 0.}   Thus

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \max(a,b) < a+b}

But integers   and   are relatively prime (see Corollary above), and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c+d\ =\ max(a,b)\ <\ a+b}

hence, by induction,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{c,d\}\ =\ \{\nu_{k,n},\nu_{k,n+1}\}}

for certain indices Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k\in\mathbb{Z}_+}   and non-negative Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ n<2^k.}   Furthermore:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{a,b\}\ =\ \{\min(a,b), \max(a,b)\}}

It follows that one of the two options holds:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{a,b\}\ =\ \{\nu_{k+1,2\cdot n},\nu_{k+1,2\cdot n+1}\}}

or

End of proof


Let's note also, that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \max_{\quad 0\le n \le 2^k}\nu_{k,n}\ =\ F_{k+1}}


where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ F_r}   is the r-th Fibonacci number.

Matrix monoid Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathit{SO}(\mathbb{Z}_+, 2)}

Definition 1  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathit{SO}(\mathbb{Z}_+, 2)}   is the set of all matrices

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M\ :=\ \left[\begin{array}{cc}a & b\\ c & d\end{array}\right]}

such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a,b,c,d\in\mathbb{Z}_+,}   and  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \det(M) = 1,}   where  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \det(M) := a\cdot d - b\cdot c.}   Such matrices (and their columns and rows) will be called special.

  • If
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M\ :=\ \left[\begin{array}{cc}a & b\\ c & d\end{array}\right]\in\mathit{SO}(\mathbb{Z}_+, 2)}

then  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a,d > 0,}   and each of the columns and rows of M, i.e. each of the four pairs  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x,y) \in \{a,d\}\times \{b,c\},\ }   is relatively prime.


Obviously, the identity matrix

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{I}\ :=\ \left[\begin{array}{cc}1 & 0\\ 0 & 1\end{array}\right]}

belongs to   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathit{SO}(\mathbb{Z}_+, 2).}   Furthermore,  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathit{SO}(\mathbb{Z}_+, 2)}   is a monoid with respect to the matrix multiplication.

Example  The upper matrix and the lower matrix are defined respectively as follows:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{U}\ :=\ \left[\begin{array}{cc}1 & 1\\ 0 & 1\end{array}\right]}   and   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{L}\ :=\ \left[\begin{array}{cc}1 & 0\\ 1 & 1\end{array}\right]}

Obviously Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \mathcal{U},\mathcal{L}\in \mathit{SO}(\mathbb{Z}_+,2).}   When they act on the right on a matrix M (by multipliplying M by itself), then they leave respectively the left or right column of M intact:

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M\cdot \mathcal{L}\ =\ \left[\begin{array}{cc}a+b & b\\ c+d & d\end{array}\right]}


Definition 2  Vectors

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[\begin{array}{c}a\\ c\end{array}\right]}     and     Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[\begin{array}{c}b\\ d\end{array}\right]}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a,b,c,d\in\mathbb{Z}_+,}   are called neighbors (in that order)   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Leftarrow:\Rightarrow}   matrix formed by these vectors

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M\ :=\ \left[\begin{array}{cc}a & b\\ c & d\end{array}\right]}

belongs to  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathit{SO}(\mathbb{Z}_+, 2).}   Then the left (resp. right) column is called the left (resp. right) neighbor.

Rational representation

With every vector

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v\ :=\ \left[\begin{array}{c}a\\ c\end{array}\right]}

such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ c\ne 0,}   let's associate a rational number

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathit{rat}(v)\ :=\ \frac{a}{c}}

Also, let

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathit{rat}(v_{\infty})\ :=\ \infty}

for

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v_{\infty}\ :=\ \left[\begin{array}{c}1\\ 0\end{array}\right]}

Furthermore, with every matrix  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M\in \mathit{SO}(\mathbb{Z}_+,2),}   let's associate the real open interval

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathit{span}(M)\ :=\ (\mathit{rat}(w);\mathit{rat}(v))}

and its length

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle diam(M)\ :=\ \mathit{rat}(v)-\mathit{rat}(w)}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ v}   is the left, and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ w}   is the right column of matrix M — observe that the rational representation of the left column of a special matrix is always greater than the rational representation of the right column of the same special matrix.

  • If
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M\ :=\ \left[\begin{array}{cc}a & b\\ c & d\end{array}\right]\in\mathit{SO}(\mathbb{Z}_+, 2)}

then  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ diam(M) = \frac{1}{c\cdot d}}

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