Number theory/Signed Articles/Elementary diophantine approximations: Difference between revisions
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How well can real numbers (all of them or the special ones) be approximated by rational numbers? A typical Diophantine approximation result states: | 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 <math>\ a</math> be an arbitrary real number. Then | '''Theorem''' Let <math>\ a</math> be an arbitrary real number. Then | ||
* <math>\ a</math> is rational | * <math>\ a</math> is rational <math>\Leftrightarrow</math> there exists a real number C > 0 such that | ||
:::<math>|a - \frac{x}{y}| > \frac{C}{y}</math> | :::<math>\left|a - \frac{x}{y}\right| > \frac{C}{y}</math> | ||
for arbitrary integers <math>\ (x,y)</math> such that <math>\ y>0</math> and <math>\ a\ne \frac{x}{y};</math> | for arbitrary integers <math>\ (x,y)</math> such that <math>\ y>0</math> and <math>\ a\ne \frac{x}{y};</math> | ||
* <math>\ a</math> is irrational | * ([[Adolph Hurwitz]]) <math>\ a</math> is irrational <math>\Leftrightarrow</math> there exist infinitely many pairs of integers <math>\ (x,y)</math> such that <math>\ y>0</math> and | ||
:::<math>\left|a - \frac{x}{y}\right| < \frac{1}{\sqrt{5}\cdot y^2}.</math> | |||
'''Remark''' Implication <math>\Leftarrow</math> of the first part of the theorem is a simple and satisfaction bringing exercise. | |||
== Notation == | == Notation == | ||
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* <math>\mathbb{R}</math> — the field of real numbers; | * <math>\mathbb{R}</math> — the field of real numbers; | ||
| | ||
*<math>x|y\ </math> — "<math>x\ </math> divides <math>\ y</math>"; | * <math>(x;y)\ := \{t\in\mathbb{R} :\, x<t<y\}</math> | ||
* <math>|(x;y)|\ := y-x</math> | |||
* <math>\mathit{Span}(x,y)\ :=\ \left(\min\left(x,y\right);\,\max\left(x,y\right)\right)</math> | |||
| |||
*<math>x|y\ </math> — "<math>x\ </math> divides <math>\ y</math>" (i.e. <math>\frac{y}{x}\in\mathbb{Z}</math>); | |||
* <math>\gcd(a, b)\ </math> — the greatest common divisor of integers <math>\ a</math> and <math>\ b.</math> | * <math>\gcd(a, b)\ </math> — the greatest common divisor of integers <math>\ a</math> and <math>\ b.</math> | ||
| |||
== 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 <math>\frac{a}{c}</math> and <math>\frac{b}{d},</math> with integer numerators and natural denominators, are called '''neighbors''' (in the given order) <math>\Leftarrow:\Rightarrow</math> | |||
:::<math>\frac{a}{c}-\frac{b}{d}\ =\ \frac{1}{c\cdot d}</math> | |||
Fraction <math>\frac{a}{c}</math> is called the '''top neighbor''' of the other, <math>\frac{b}{d}</math> is called the '''bottom neighbor''', and the interval <math>\ \mathit{Span}(\frac{a}{c},\frac{b}{d})</math> is called '''neighborhood'''; thus a neighborhood is an open interval such that its endpoints are neighbors. | |||
* If <math>\frac{a}{c}</math> and <math>\frac{b}{d}</math> are neighbors then <math>\ a\cdot b\ge 0</math> ( i.e. <math>\frac{a}{c}\cdot \frac{b}{d}\ \ge 0</math> ). | |||
* Let <math>\ a,b,c,d\in\mathbb{N}.</math> Fractions <math>\frac{a}{c}</math> and <math>\frac{b}{d}</math> are neighbors <math>\Leftrightarrow</math> fractions <math>\frac{d}{b}</math> and <math>\frac{c}{a}</math> are neighbors <math>\Leftrightarrow</math> fractions <math>\frac{-b}{d}</math> and <math>\frac{-a}{c}</math> are neighbors. | |||
'''Examples:''' | |||
* Fractions <math>\frac{n+1}{1}</math> and <math>\frac{n}{1}</math> are neighbors for every positive integer <math>\ n.</math> | |||
* Fractions <math>\frac{1}{n}</math> and <math>\frac{1}{n+1}</math> are neighbors for every positive integer <math>\ n.</math> | |||
Thus it easily follows that for every positive irrational number <math>\ x</math> there exists a pair of neighbors <math>\frac{a}{c}</math> and <math>\frac{b}{d},</math> with positive numerators and denominators, such that: | |||
::: <math>\frac{a}{c}\ >\ x\ >\ \frac{b}{d}</math> | |||
'''Definition''' A pair of neighboring fractions <math>\left(\frac{a}{c},\frac{b}{d}\right),</math> with integer numerators and natural denominators, is called a ''top pair'' <math>\Leftarrow:\Rightarrow\ c > d.</math> Otherwise it is called a ''bottom pair''. | |||
Thus now we have notions of top/bottom neighbors and of top/bottom pairs of neighbors. | |||
* Let <math>\left(\frac{a}{c},\frac{b}{d}\right),</math> be a pair of neighbors. Then <math>\left(\frac{a}{c},\frac{a+b}{c+d}\right)</math> is a top pair of neighbors, and <math>\left(\frac{a+b}{c+d},\frac{b}{d}\right)</math> is a bottom pair of neighbors. | |||
| |||
=== First results === | |||
'''Theorem''' Let fractions <math>\frac{a}{c}</math> and <math>\frac{b}{d},</math> with integer numerators and natural denominators, be neighbors. Then | |||
:* if integers <math>\ t>0</math> and <math>\ s</math> are such that <math>\ \frac{a}{c} > \frac{s}{t} > \frac{b}{d},</math> then <math>t \ge c+d;</math> | |||
:* the '''median''' <math>\frac{a+b}{c+d}</math> is a bottom neighbor of <math>\frac{a}{c},</math> and a top neighbor of <math>\frac{b}{d};</math> | |||
:* let <math>\ x</math> be an irrational number such that <math>\frac{a}{c}\ >\ x\ >\ \frac{b}{d};</math> then | |||
:::<math>0\ <\ \max(\frac{a}{c}-x,\, x-\frac{b}{d})\ <\ \frac{1}{c\cdot d}</math> | |||
:and | |||
:<math>0\ <\ \frac{a}{c}-x\ <\ \frac{1}{2\cdot c^2}</math> or <math>0\ <\ x-\frac{b}{d}\ <\ \frac{1}{2\cdot d^2}</math> | |||
'''Proof''' Let <math>\ \frac{a}{c} > \frac{s}{t} > \frac{b}{d};</math> then | |||
:<math>\frac{a}{c} - \frac{s}{t}\ \ge\ \frac{1}{c\cdot t}</math> and <math>\frac{s}{t} - \frac{b}{d}\ \ge\ \frac{1}{d\cdot t}</math> | |||
and | |||
:<math>\frac{1}{c\cdot t} + \frac{1}{d\cdot t}\ \le \frac{1}{c\cdot d}</math> | |||
Multiplying this inequality by <math>c\cdot d\cdot t\ </math> gives | |||
:::<math>t \ge c+d</math> | |||
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: | |||
:<math>\max(\frac{a}{c}-x,\, x-\frac{b}{d})\ <\ (\frac{a}{c}-x)+(x-\frac{b}{d})\ =\ \frac{1}{c\cdot d}</math> | |||
Next, | |||
:::<math>\left(\frac{1}{c}-\frac{1}{d}\right)^2\ \ge\ 0</math> | |||
hence | |||
::<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> | |||
::::<math> =\ \left|\mathit{Span}(\frac{a}{c},\frac{b}{d})\right|</math> | |||
and | |||
::<math>\frac{b}{d}+\frac{1}{2\cdot d^2}\ \ge\ \frac{a}{c}-\frac{1}{2\cdot c^2}</math> | |||
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)\ \backslash\ \mathbb{Q}\ \supseteq\ \mathit{Span}\left(\frac{a}{c},\frac{b}{d}\right)\ \backslash\ \mathbb{Q}</math> | |||
'''End of proof''' | |||
'''Corollary''' Let fractions <math>\frac{a}{c}</math> and <math>\frac{b}{d},</math> with integer numerators and natural denominators, be neighbors. Then, if integers <math>\ t>0</math> and <math>\ s</math> are such that <math>\ \frac{a}{c} > \frac{s}{t} > \frac{b}{d},</math> 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> | |||
| |||
=== Hurwitz theorem === | |||
:Let <math>\ x\in \mathbb{R}</math> be an arbitrary irrational number. Then | |||
::::<math>\left|x-\frac{s}{t}\right|\ <\ \frac{1}{\sqrt{5}\cdot t^2}</math> | |||
:for infinitely many different <math>\ s,t\in\mathbb{Z} \backslash \{0\}.</math> | |||
==== Lemma 1 ==== | |||
Let <math>\ c,d\in\mathbb{N}.</math> Let <math>\ C:=c+d.</math> Then: | |||
:::<math>c^2-\sqrt{5}\!\cdot\! c\!\cdot\! d+d\,^2\ >\ 0</math> | |||
or | |||
:::<math>C\,^2-\sqrt{5}\cdot\! C\!\cdot\! d+d\,^2\ >\ 0</math> | |||
'''Proof of lemma 1''' It's easy to show that <math>\sqrt{2}\cdot c \ne \sqrt{3-\sqrt{5}}\cdot d.</math> Thus the square of <math>\ X := \sqrt{2}\cdot c - \sqrt{3-\sqrt{5}}\cdot d</math> is positive. Now, | |||
:::<math>\sqrt{2}\cdot \sqrt{3-\sqrt{5}}\ =\ \sqrt{5}-1</math> | |||
which means that we may write <math>\ X^2</math> as follows: | |||
:<math>2\!\cdot\! c^2\ -\ 2\!\cdot\!(\sqrt{5}-1)\cdot c\!\cdot\! d\ +\ (3-\sqrt{5})\!\cdot\! d\,^2\ >\ 0</math> | |||
i.e. | |||
:<math>(c^2-\sqrt{5}\cdot c\cdot d+d^2)\ +\ (C^2-\sqrt{5}\cdot C\cdot d+d^2)\ >\ 0</math> | |||
and lemma 1 follows. '''End of proof''' | |||
==== Lemma 2 ==== | |||
Let <math>\ a,b\in\mathbb{Z}</math> and <math>\ c,d\in\mathbb{N}.</math> Let <math>\ A:=a+b</math> and <math>\ C:=c+d.</math> Furthermore, let fractions <math>\frac{a}{c},\frac{b}{d},</math> be neighbors, and let: | |||
:::<math>\frac{A}{C}\ >\ x\ >\ \frac{b}{d}</math> | |||
where <math>\ x</math> is real. Then one of the following three inequalities holds: | |||
:::* <math>0\ <\ \frac{a}{c}-x\ <\ \frac{1}{\sqrt{5}\cdot c^2}</math> | |||
:::* <math>0\ <\ \frac{A}{C}-x\ <\ \frac{1}{\sqrt{5}\cdot C^2}</math> | |||
:::* <math>0\ <\ x-\frac{b}{d}\ <\ \frac{1}{\sqrt{5}\cdot d^2}</math> | |||
'''Proof''' There are two cases along the inequalities of Lemma 1. Let's assume the first one, which is equivalent to: | |||
:<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> | |||
Thus | |||
::<math>\frac{b}{d}+\frac{1}{\sqrt{5}\cdot d\,^2}\ >\ \frac{a}{c}-\frac{1}{\sqrt{5}\cdot c^2}</math> | |||
which means that | |||
::<math>\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)</math> | |||
::::<math>\supseteq\ \mathit{Span}\left(\frac{A}{C},\frac{b}{d}\right)</math> | |||
Thus lemma 2 is proven when the first inequality of lemma 1 holds. By replacing in the above proof the lower case <math>\ a,c,</math> by the upper case <math>\ A,C,</math> we obtain the proof when the second inequality of lemma 1 holds. | |||
'''End of proof''' (of lemma 2) | |||
==== Lemma 2' ==== | |||
Let <math>\ a,b\in\mathbb{Z}</math> and <math>\ c,d\in\mathbb{N}.</math> Let <math>\ A:=a+b</math> and <math>\ C:=c+d.</math> Furthermore, let fractions <math>\frac{a}{c},\frac{b}{d},</math> be neighbors, and let: | |||
:::<math>\frac{a}{c}\ >\ x\ >\ \frac{A}{C}</math> | |||
where <math>\ x</math> is real. Then one of the following three inequalities holds: | |||
:::* <math>0\ <\ x - \frac{b}{d}\ <\ \frac{1}{\sqrt{5}\cdot d\,^2}</math> | |||
:::* <math>0\ <\ x-\frac{A}{C} <\ \frac{1}{\sqrt{5}\cdot C^2}</math> | |||
:::* <math>0\ <\ \frac{a}{c}\ <\ \frac{1}{\sqrt{5}\cdot x^2}</math> | |||
'''Proof''' It's similar to the proof of lemma 2. Or one may apply lemma 2 to <math>\ x':=-x,\ a':=-b,\ b':=-a,</math> and <math>\ c':=d,\ d':=c</math>, which would provide us with the respective fraction <math>\frac{s'}{t'}.</math> Then the required <math>\ s,t,</math> are given by <math>s:=-s',\ t:=t'.</math> | |||
'''End of proof''' (of lemma 2') | |||
| |||
==== Proof of Hurwitz theorem ==== | |||
When <math>\ x</math> 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 <math>\frac{s}{t}</math> (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 <math>\ x > 0</math> be an irrational number, We may always squeeze it between the extremal neighbours: | |||
:::<math>\frac{1}{0}\ >\ x\ >\ \frac{0}{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> | |||
or | |||
:<math>x < 1\quad\quad\Rightarrow\quad\quad\frac{1}{n} > x > \frac{1}{n+1}</math> | |||
where in each of these two cases <math>\ n\in \mathbb{N}</math> is a respective unique positive integer. | |||
It was mentioned in the previous section ('''''First results''''') that if fractions <math>\frac{a}{c}</math> and <math>\frac{b}{d},</math> with positive (or non-negative) integer numerators and denominators are neighbors then also the top and the bottom (''bot'' 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 | |||
*<math>\mathit{bot}\!\left(\frac{a}{c},\frac{b}{d}\right)\ :=\ \left(\frac{a+b}{c+d},\frac{b}{d}\right)</math> | |||
are both pairs of neighbors. | |||
Let <math>\ A_0</math> 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> are already defined, and that they squeeze <math>\ x,</math> i.e. that <math>\ x\in\mathit{Span}(A_k)</math> for each <math>\ k=0,\dots,n-1.</math> Then we define <math>\ A_n</math> as the one of the two pairs: <math>\ \mathit{top}(A_{n-1})</math> or <math>\ \mathit{bot}(A_{n-1}),</math> which squeezes <math>\ x.</math> 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> there exist fractions of integers <math>\ \frac{s}{t},</math> with arbitrarily large denominators, such that | |||
:::<math>\left|x - \frac{s}{t}\right|\ <\ \frac{1}{\sqrt{5}\cdot t^2}</math> | |||
(see section '''''Hurwitz theorem'''''). | |||
If cases top-top and bot-bot happen only finitely many times then starting with an <math>\ n</math> we get an infinite alternating top-bot-top-bot-... sequence: | |||
:<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> | |||
Then the new neighbor of the <math>\ A_{n+k}</math> pair (i.e. the median of the previous pair <math>\ A_{n+k-1}</math>) is equal to | |||
::<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> where <math>\ F_{t}</math> are the [[Fibonacci number]]s, where | |||
:::<math>\left(\frac{a}{c},\frac{b}{d}\right)\ :=\ A_n</math> | |||
It is known that | |||
:::<math>\lim_{k\rightarrow\infty}\frac{F_{k+1}}{F_k}\ =\ \Phi\ :=\ \frac{\sqrt{5}+1}{2}</math> | |||
hence | |||
::<math>x\ =\ \lim_{k\rightarrow\infty}e_k\ =\ \frac{\Phi\cdot a+b}{\Phi\cdot c+d}</math> | |||
But if our infinite alternation has started with ''bot'' : | |||
:<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> | |||
Then we would have | |||
::<math>x\ =\ \lim_{k\rightarrow\infty}e_k\ =\ \frac{a+\Phi\cdot b}{c+\Phi\cdot d}</math> | |||
| |||
=== Another proof of Hurwitz Theorem (further insight) === | |||
==== Reduction to x > 0 ==== | |||
Since | |||
:::<math>\left|(-x)-\frac{-s}{t}\right|\ =\ \left|x-\frac{s}{t}\right|</math> | |||
it is enough to prove Hurwitz theorem for positive irrational numbers only. | |||
==== Two cases ==== | |||
Consider the sequence <math>\ \left(A_n\right)</math> of pairs of neighbors, which squeeze <math>\ x>0,</math> 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 <math>\ n</math> for which | |||
:<math>\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}</math> | |||
or | |||
:<math>\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}</math> | |||
holds, where | |||
::<math>\left(\frac{a}{c},\frac{b}{d}\right)\ :=\ A_n</math> | |||
| |||
==== The top-top-bot case ==== | |||
Let's consider the latter top-top-bot case. Let <math>\ \xi := \frac{d}{c}.</math> The squeeze by neighbors: | |||
:::<math>\frac{a}{c}\ >\ x\ >\ \frac{2\cdot a+b}{2\cdot c+d}</math> | |||
shows that | |||
:<math>0\ <\ \frac{a}{c}-x\ <\ \frac{1}{c\cdot(2\cdot c+d)}\ =\ \frac{1}{(2+\xi)}\cdot\frac{1}{c^2}</math> | |||
This inequality provides the first insight (otherwise, we are not going to use it), so it deserves to be written cleanly as an implication: | |||
<math>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}</math> | |||
==== The relevant neighborhoods ==== | |||
Consider the next two pairs of neighbors, pair <math>\ A_{n+4}</math> and pair <math>\ A_{n+5},</math> which squeeze <math>\ x.</math> The relevant neighborhoods are: | |||
*<math>A\ :=\ \mathit{Span}(\frac{a}{c},\,\frac{3\cdot a+b}{3\cdot c+d})</math> | |||
*<math>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})</math> | |||
*<math>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})</math> | |||
*<math>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})</math> | |||
| |||
==== Neighborhood B ==== | |||
Let <math>\ x\in B.</math> Then | |||
:<math>\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}</math> | |||
and | |||
:<math>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)}</math> | |||
::<math>=\ \frac{2}{c\cdot(5\cdot c+2\cdot d)}\ =\ \frac{2}{(5+2\cdot\xi)}\cdot\frac{1}{c^2}</math> | |||
Conclusion: | |||
<math>0\ <\ \frac{a}{c}-x\ <\ \frac{2}{5+2\cdot\xi}\cdot\frac{1}{c^2}\quad\quad\Leftarrow\quad\quad x\in B</math> | |||
==== Neighborhood C (first C-inequality) ==== | |||
Let <math>\ x\in C.</math> Then | |||
:<math>\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}</math> | |||
Thus using the calculation for neighborhood B also for C, we get | |||
:<math>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)}</math> | |||
:<math>=\ \frac{3}{7+3\cdot\xi}\cdot\frac{1}{c^2}</math> | |||
First C-conclusion: | |||
<math>0\ <\ \frac{a}{c}-x\ <\ \frac{3}{7+3\cdot\xi}\cdot\frac{1}{c^2}\quad\quad\Leftarrow\quad\quad x\in C</math> | |||
| |||
==== Neighborhood D ==== | |||
Let <math>\ x\in D,</math> i.e. | |||
:<math>\frac{7\cdot a+3\cdot b}{7\cdot c+3\cdot d}\ >\ x\ >\ \frac{2\cdot a+b}{2\cdot c+d}</math> | |||
Let <math>\alpha := 2\cdot a+b</math> and <math>\gamma := 2\cdot c+d = (2+\xi)\cdot c.</math> Then | |||
:<math>0\ <\ x-\frac{\alpha}{\gamma}\ =\ x-\frac{2\cdot a+b}{2\cdot c+d}</math> | |||
:<math><\ \frac{1}{(7\cdot c+3\cdot d)\cdot(2\cdot c+d)}\ =\ \frac{1}{(7+3\cdot\xi)\cdot c\cdot\gamma}</math> | |||
:<math>=\ \frac{2+\xi}{7+3\cdot\xi}\cdot\frac{1}{\gamma^2}</math> | |||
Conclusion: | |||
<math>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</math> | |||
==== 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: | |||
* <math>0\ <\ \frac{a}{c}-x\ <\ \frac{2}{5}\cdot\frac{1}{c^2}\quad\quad\Leftarrow\quad\quad x\in B</math> | |||
* <math>0\ <\ \frac{a}{c}-x\ <\ \frac{3}{7}\cdot\frac{1}{c^2}\quad\quad\Leftarrow\quad\quad x\in C</math> | |||
* <math>0\ <\ x-\frac{\alpha}{\gamma}\ <\ \frac{2}{7}\cdot\frac{1}{\gamma^2}\quad\quad\Leftarrow\quad\quad x\in D</math> | |||
Since | |||
:::<math>\max(\frac{2}{5},\,\frac{3}{7},\,\frac{2}{7})\ =\ \frac{3}{7}\ <\ \frac{1}{\sqrt{5}},</math> | |||
each occurrence of the top-top-bot subsequence, i.e. of equalities: | |||
:<math>A_{n+1} = \mathit{top}(A_n),\quad A_{n+2} = \mathit{top}(A_{n+1}),\quad A_{n+3} = \mathit{bot}(A_{n+2})\quad\dots</math> | |||
provides a fraction <math>\frac{s}{t}\in \mathit{Span}(A_{n+3}),</math> with integer numerator and natural denominator, such that | |||
:<math>\left|x-\frac{s}{t}\right|\ <\ \frac{3}{7}\cdot\frac{1}{t^2}\ <\ \frac{1}{\sqrt{5}}\cdot\frac{1}{t^2}</math> | |||
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 <math>\ -x</math> by: | |||
::<math>B_n := (\frac{-b}{d},\frac{-a}{c})</math> for <math>A_n = (\frac{a}{c},\frac{b}{d})</math> | |||
Let <math>\left(A_n,A_{n+1},A_{n+2},A_{n+3}\right)</math> be a bot-bot-top progression. Then <math>\left(B_n,B_{n+1},B_{n+2},B_{n+3}\right)</math> is a top-top-bot progression which squeezes <math>\ -x.</math> Thus | |||
:<math>\left|(-x)-\frac{u}{w}\right|\ <\ \frac{3}{7}\cdot\frac{1}{w^2}\ <\ \frac{1}{\sqrt{5}}\cdot\frac{1}{w^2}</math> | |||
for certain <math>\frac{u}{w}\in \mathit{Span}(B_{n+3}),</math> with integer numerator and natural denominator. Then <math>\frac{s}{t}\in\mathit{Span}(A_{n+3}),</math> for <math>\ (s,t):=(-u,w),</math> satisfies: | |||
:<math>\left|x-\frac{s}{t}\right|\ <\ \frac{3}{7}\cdot\frac{1}{t^2}\ <\ \frac{1}{\sqrt{5}}\cdot\frac{1}{t^2}</math> | |||
When another top-top-bot or bot-bot-top progression starts with a sufficiently large index <math>\ n',</math> then <math>\left|\mathit{Span}(A_{n'+3})\right| < \left|x-\frac{s}{t}\right|,</math> which means that the respective new approximation <math>\frac{s'}{t'}\in\mathit{Span}(A_{n'+3})</math> is different. It follows that if there are infinitely many progressions top-top-bot or bot-bot-bot then there are infinitely many fractions <math>\ \frac{s}{t},</math> which satisfy the inequality above. Thus we have obtained the following version of '''Hurwitz theorem''': | |||
:'''Theorem''' Let <math>x\in\mathbb{R}\backslash\mathbb{Q}</math> be an arbitrary irrational number. Then inequality | |||
:*<math>\left|x-\frac{s}{t}\right|\ <\ \frac{1}{\sqrt{5}}\cdot\frac{1}{t^2}</math> | |||
:holds for infinitely many fractions <math>\ \frac{s}{t},</math> with integer numerator and natural denominator. Furthermore, if the squeezing sequence of <math>\ x</math> does not eventually alternate top-bot-top-bot-... till infinity, i.e. if it has infinitely many top-top or bot-bot progressions, then | |||
:*<math>\left|x-\frac{s}{t}\right|\ <\ \frac{3}{7 | |||
}\cdot\frac{1}{t^2}</math> | |||
:holds for infinitely many fractions <math>\ \frac{s}{t},</math> with integer numerator and natural denominator. | |||
| |||
==== Neighborhood C (second C-inequality) ==== | |||
Let <math>\ x\in C.</math> Then | |||
:<math>\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}</math> | |||
Thus, using the earlier conclusion for neighborhood <math>\ D</math> also for <math>\ C,</math> we obtain | |||
:<math>0\ <\ x-\frac{\alpha}{\gamma}\ =\ x-\frac{2\cdot a+b}{2\cdot c+d}</math> | |||
:<math><\ \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}</math> | |||
:<math>=\ \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}</math> | |||
:<math>=\ \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}</math> | |||
:<math>=\ \frac{2+\xi}{5+2\cdot\xi}\cdot\frac{1}{\gamma^2}</math> | |||
Second C-conclusion: | |||
<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 <math>\ \xi\ge\xi_0</math>: | |||
:<math>\frac{3}{7+3\cdot \xi}\ \le\ \frac{\sqrt{61}-7}{2}</math> | |||
and for <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 <math>\frac{s}{t}\in\left\{\frac{a}{c},\,\frac{2\cdot a+b}{2\cdot c+d}\right\}</math> the following inequality holds for every <math>\ x\in C:</math> | |||
:::<math>\left|x-\frac{s}{t}\right|\ <\ \frac{\sqrt{61}-7}{2}</math> | |||
(the choice of <math>\frac{s}{t}</math> depends on <math>\xi := \frac{d}{c}</math> ). | |||
| | ||
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Let's note also, that | |||
:::<math>\max_{\quad 0\le n \le 2^k}\nu_{k,n}\ =\ F_{k+1}</math> | |||
where <math>\ F_r</math> is the r-th [[Fibonacci number]]. | where <math>\ F_r</math> is the r-th [[Fibonacci number]]. | ||
== Matrix monoid <math>\mathit{SO}(\mathbb{Z}_+, 2)</math> == | |||
'''Definition 1''' <math>\mathit{SO}(\mathbb{Z}_+, 2)</math> is the set of all matrices | |||
:::<math>M\ :=\ \left[\begin{array}{cc}a & b\\ c & d\end{array}\right]</math> | |||
such that <math>\ a,b,c,d\in\mathbb{Z}_+,</math> and <math>\ \det(M) = 1,</math> where <math>\ \det(M) := a\cdot d - b\cdot c.</math> Such matrices (and their columns and rows) will be called '''special'''. | |||
* If | |||
:::<math>M\ :=\ \left[\begin{array}{cc}a & b\\ c & d\end{array}\right]\in\mathit{SO}(\mathbb{Z}_+, 2)</math> | |||
''' | then <math>\ a,d > 0,</math> and each of the columns and rows of M, i.e. each of the four pairs <math>(x,y) \in \{a,d\}\times \{b,c\},\ </math> is relatively prime. | ||
Obviously, the identity matrix | |||
:::<math>\mathcal{I}\ :=\ \left[\begin{array}{cc}1 & 0\\ 0 & 1\end{array}\right]</math> | |||
belongs to <math>\mathit{SO}(\mathbb{Z}_+, 2).</math> Furthermore, <math>\mathit{SO}(\mathbb{Z}_+, 2)</math> is a [[monoid]] with respect to the matrix multiplication. | |||
'''Example''' The ''upper matrix'' and the ''lower matrix'' are defined respectively as follows: | |||
:<math>\mathcal{U}\ :=\ \left[\begin{array}{cc}1 & 1\\ 0 & 1\end{array}\right]</math> and <math>\mathcal{L}\ :=\ \left[\begin{array}{cc}1 & 0\\ 1 & 1\end{array}\right]</math> | |||
Obviously <math>\ \mathcal{U},\mathcal{L}\in \mathit{SO}(\mathbb{Z}_+,2).</math> 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: | |||
:::<math>M\cdot \mathcal{U}\ =\ \left[\begin{array}{cc}a & a+b\\ c & c+d\end{array}\right]</math> | |||
and | |||
:::<math>M\cdot \mathcal{L}\ =\ \left[\begin{array}{cc}a+b & b\\ c+d & d\end{array}\right]</math> | |||
'''Definition 2''' Vectors | |||
:::<math>\left[\begin{array}{c}a\\ c\end{array}\right]</math> and <math>\left[\begin{array}{c}b\\ d\end{array}\right]</math> | |||
where <math>\ a,b,c,d\in\mathbb{Z}_+,</math> are called '''neighbors''' (in that order) <math>\Leftarrow:\Rightarrow</math> matrix formed by these vectors | |||
:::<math>M\ :=\ \left[\begin{array}{cc}a & b\\ c & d\end{array}\right]</math> | :::<math>M\ :=\ \left[\begin{array}{cc}a & b\\ c & d\end{array}\right]</math> | ||
such that <math>\ a, | belongs to <math>\mathit{SO}(\mathbb{Z}_+, 2).</math> Then the left (resp. right) column is called the left (resp. right) neighbor. | ||
== Rational representation == | |||
With every vector | |||
:::<math>v\ :=\ \left[\begin{array}{c}a\\ c\end{array}\right]</math> | |||
such that <math>\ c\ne 0,</math> let's associate a rational number | |||
:::<math>\mathit{rat}(v)\ :=\ \frac{a}{c}</math> | |||
Also, let | |||
:::<math>\mathit{rat}(v_{\infty})\ :=\ \infty</math> | |||
for | |||
:::<math>v_{\infty}\ :=\ \left[\begin{array}{c}1\\ 0\end{array}\right]</math> | |||
Furthermore, with every matrix <math>M\in \mathit{SO}(\mathbb{Z}_+,2),</math> let's associate the real open interval | |||
:::<math>\mathit{span}(M)\ :=\ (\mathit{rat}(w);\mathit{rat}(v))</math> | |||
and its length | |||
:::<math>diam(M)\ :=\ \mathit{rat}(v)-\mathit{rat}(w)</math> | |||
where <math>\ v</math> is the left, and <math>\ w</math> 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 | |||
:::<math>M\ :=\ \left[\begin{array}{cc}a & b\\ c & d\end{array}\right]\in\mathit{SO}(\mathbb{Z}_+, 2)</math> | |||
then <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 Failed to parse (SVG (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{k}{10^n}.} Decimal fractions are used both as certain values, e.g. $5.85, and as approximations of the real numbers, e.g. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \pi \approx 3.14159.} However, the field of all rational numbers is much richer than the ring of the decimal fractions (or of the binary 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{k}{2^n},} which are used in the computer science). For instance, the famous 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 \ \pi \approx \frac{355}{113}} 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 Failed to parse (SVG (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}
be an arbitrary real 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 \ a} is rational Failed to parse (SVG (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} there exists a real number C > 0 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|a - \frac{x}{y}\right| > \frac{C}{y}}
for arbitrary 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,y)} 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 \ y>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 \ a\ne \frac{x}{y};}
- (Adolph Hurwitz) Failed to parse (SVG (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 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 \Leftrightarrow} there exist infinitely many pairs 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 \ (x,y)} 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 \ y>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 \left|a - \frac{x}{y}\right| < \frac{1}{\sqrt{5}\cdot y^2}.}
Remark 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 \Leftarrow}
of the first part of the theorem is a simple and satisfaction bringing exercise.
Notation
- Failed to parse (SVG (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\ } — "equivalent by definition" (i.e. "if and only 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 :=\ } — "equals by definition";
- Failed to parse (SVG (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} — "there exists";
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \forall} — "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\ } — "Failed to parse (SVG (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 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 \ 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 \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;
- Failed to parse (SVG (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{R}} — 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 \frac{y}{x}\in\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 \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 Failed to parse (SVG (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.}
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}
- Failed to parse (SVG (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}\ =\ \frac{1}{c\cdot d}}
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{a}{c}} is called the top neighbor of the other, Failed to parse (SVG (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}} 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 Failed to parse (SVG (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 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\cdot b\ge 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 \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}}
- Failed to parse (SVG (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}}
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 Failed to parse (SVG (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:=-s',\ t:=t'.}
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
- Failed to parse (SVG (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}}
(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 Failed to parse (SVG (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+5},} 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.} 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 Failed to parse (SVG (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 B.} 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
- Failed to parse (SVG (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{2}{5},\,\frac{3}{7},\,\frac{2}{7})\ =\ \frac{3}{7}\ <\ \frac{1}{\sqrt{5}},}
each occurrence of the top-top-bot subsequence, i.e. of equalities:
- Failed to parse (SVG (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{top}(A_{n+1}),\quad A_{n+3} = \mathit{bot}(A_{n+2})\quad\dots}
provides a 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}\in \mathit{Span}(A_{n+3}),} 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}
:
- Failed to parse (SVG (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}\ \le\ \frac{\sqrt{61}-7}{2}}
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 \ \xi\le\xi_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 \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 Failed to parse (SVG (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 Failed to parse (SVG (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} (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 Failed to parse (SVG (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.}
- 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 Failed to parse (SVG (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 \ |a-b\,|} 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 Failed to parse (SVG (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_{0,1} :=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 \nu_{k+1,2\cdot n}\ :=\ \nu_{k,n}} 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 Failed to parse (SVG (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 := \min(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 \ d := |a-b|} 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
- Failed to parse (SVG (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+1},\nu_{k+1,2\cdot n+2}\}}
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:
- Failed to parse (SVG (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{U}\ =\ \left[\begin{array}{cc}a & a+b\\ c & c+d\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 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}}