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

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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 $\ {\frac {k}{10^{n}}}.$ Decimal fractions are used both as certain values, e.g. $5.85, and as approximations of the real numbers, e.g. $\ \pi \approx 3.14159.$ However, the field of all rational numbers is much richer than the ring of the decimal fractions (or of the binary fractions $\ {\frac {k}{2^{n}}},$ which are used in the computer science). For instance, the famous approximation $\ \pi \approx {\frac {355}{113}}$ 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 $\ a$ be an arbitrary real number. Then

$\ a$ is rational $\Leftrightarrow$ there exists a real number C > 0 such that

\left|a-{\frac {x}{y}}\right|>{\frac {C}{y}}

for arbitrary integers $\ (x,y)$ such that $\ y>0$ and $\ a\neq {\frac {x}{y}};$

(Adolph Hurwitz) $\ a$ is irrational $\Leftrightarrow$ there exist infinitely many pairs of integers $\ (x,y)$ such that $\ y>0$ and

\left|a-{\frac {x}{y}}\right|<{\frac {1}{{\sqrt {5}}\cdot y^{2}}}.

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

Notation

$\Leftarrow :\Rightarrow \$ — "equivalent by definition" (i.e. "if and only if");
$:=\$ — "equals by definition";
$\exists$ — "there exists";
$\forall$ — "for all";
$a\in A\$ — " $a\$ is an element of set $\ A$ ";

$\mathbb {N} \ :=\ \{1,2\dots \}$ — the semiring of the natural numbers;
$\mathbb {Z} _{+}\ :=\ \{0,1,\dots \}$ — the semiring of the non-negative integers;
$\mathbb {Z} \ :=\ \{-2,-1,1,2\dots \}$ — the ring of integers;
$\mathbb {Q}$ — the field of rational numbers;
$\mathbb {R}$ — the field of real numbers;

$(x;y)\ :=\{t\in \mathbb {R} :\,x<t<y\}$
$|(x;y)|\ :=y-x$
${\mathit {Span}}(x,y)\ :=\ \left(\min \left(x,y\right);\,\max \left(x,y\right)\right)$

$x|y\$ — " $x\$ divides $\ y$ " (i.e. ${\frac {y}{x}}\in \mathbb {Z}$ );
$\gcd(a,b)\$ — the greatest common divisor of integers $\ a$ and $\ b.$

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 ${\frac {a}{c}}$ and ${\frac {b}{d}},$ with integer numerators and natural denominators, are called neighbors (in the given order) $\Leftarrow :\Rightarrow$

{\frac {a}{c}}-{\frac {b}{d}}\ =\ {\frac {1}{c\cdot d}}

Fraction ${\frac {a}{c}}$ is called the top neighbor of the other, ${\frac {b}{d}}$ is called the bottom neighbor, and the interval $\ {\mathit {Span}}({\frac {a}{c}},{\frac {b}{d}})$ is called neighborhood; thus a neighborhood is an open interval such that its endpoints are neighbors.

If ${\frac {a}{c}}$ and ${\frac {b}{d}}$ are neighbors then $\ a\cdot b\geq 0$ ( i.e. ${\frac {a}{c}}\cdot {\frac {b}{d}}\ \geq 0$ ).

Let $\ a,b,c,d\in \mathbb {N} .$ Fractions ${\frac {a}{c}}$ and ${\frac {b}{d}}$ are neighbors $\Leftrightarrow$ fractions ${\frac {d}{b}}$ and ${\frac {c}{a}}$ are neighbors $\Leftrightarrow$ fractions ${\frac {-b}{d}}$ and ${\frac {-a}{c}}$ are neighbors.

Examples:

Fractions ${\frac {n+1}{1}}$ and ${\frac {n}{1}}$ are neighbors for every positive integer $\ n.$

Fractions ${\frac {1}{n}}$ and ${\frac {1}{n+1}}$ are neighbors for every positive integer $\ n.$

Thus it easily follows that for every positive irrational number $\ x$ there exists a pair of neighbors ${\frac {a}{c}}$ and ${\frac {b}{d}},$ with positive numerators and denominators, such that:

{\frac {a}{c}}\ >\ x\ >\ {\frac {b}{d}}

Definition A pair of neighboring fractions $\left({\frac {a}{c}},{\frac {b}{d}}\right),$ with integer numerators and natural denominators, is called a top pair $\Leftarrow :\Rightarrow \ c>d.$ Otherwise it is called a bottom pair.

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

Let $\left({\frac {a}{c}},{\frac {b}{d}}\right),$ be a pair of neighbors. Then $\left({\frac {a}{c}},{\frac {a+b}{c+d}}\right)$ is a top pair of neighbors, and $\left({\frac {a+b}{c+d}},{\frac {b}{d}}\right)$ is a bottom pair of neighbors.

First results

Theorem Let fractions ${\frac {a}{c}}$ and ${\frac {b}{d}},$ with integer numerators and natural denominators, be neighbors. Then

if integers $\ t>0$ and $\ s$ are such that $\ {\frac {a}{c}}>{\frac {s}{t}}>{\frac {b}{d}},$ then $t\geq c+d;$

the median ${\frac {a+b}{c+d}}$ is a bottom neighbor of ${\frac {a}{c}},$ and a top neighbor of ${\frac {b}{d}};$

let $\ x$ be an irrational number such that ${\frac {a}{c}}\ >\ x\ >\ {\frac {b}{d}};$ then

0\ <\ \max({\frac {a}{c}}-x,\,x-{\frac {b}{d}})\ <\ {\frac {1}{c\cdot d}}

and

0\ <\ {\frac {a}{c}}-x\ <\ {\frac {1}{2\cdot c^{2}}}

or

0\ <\ x-{\frac {b}{d}}\ <\ {\frac {1}{2\cdot d^{2}}}

Proof Let $\ {\frac {a}{c}}>{\frac {s}{t}}>{\frac {b}{d}};$ then

{\frac {a}{c}}-{\frac {s}{t}}\ \geq \ {\frac {1}{c\cdot t}}

and

{\frac {s}{t}}-{\frac {b}{d}}\ \geq \ {\frac {1}{d\cdot t}}

and

{\frac {1}{c\cdot t}}+{\frac {1}{d\cdot t}}\ \leq {\frac {1}{c\cdot d}}

Multiplying this inequality by $c\cdot d\cdot t\$ gives

t\geq 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:

\max({\frac {a}{c}}-x,\,x-{\frac {b}{d}})\ <\ ({\frac {a}{c}}-x)+(x-{\frac {b}{d}})\ =\ {\frac {1}{c\cdot d}}

Next,

\left({\frac {1}{c}}-{\frac {1}{d}}\right)^{2}\ \geq \ 0

hence

{\frac {1}{2\cdot c^{2}}}+{\frac {1}{2\cdot d^{2}}}\ \geq \ {\frac {1}{c\cdot d}}\ =\ {\frac {a}{c}}-{\frac {b}{d}}

=\ \left|{\mathit {Span}}({\frac {a}{c}},{\frac {b}{d}})\right|

and

{\frac {b}{d}}+{\frac {1}{2\cdot d^{2}}}\ \geq \ {\frac {a}{c}}-{\frac {1}{2\cdot c^{2}}}

i.e.

${\mathit {Span}}\left({\frac {a}{c}},{\frac {a}{c}}-{\frac {1}{2\cdot c^{2}}}\right)\ \cup \ {\mathit {Span}}\left({\frac {b}{d}}+{\frac {1}{2\cdot d^{2}}},{\frac {b}{d}}\right)\ \backslash \ \mathbb {Q} \ \supseteq \ {\mathit {Span}}\left({\frac {a}{c}},{\frac {b}{d}}\right)\ \backslash \ \mathbb {Q}$

End of proof

Corollary Let fractions ${\frac {a}{c}}$ and ${\frac {b}{d}},$ with integer numerators and natural denominators, be neighbors. Then, if integers $\ t>0$ and $\ s$ are such that $\ {\frac {a}{c}}>{\frac {s}{t}}>{\frac {b}{d}},$ then either

$s=a+b\quad \land \quad t=c+d;$

or

$t\ \geq \ c+d+\min(c,d)\ >\ c+d$

Hurwitz theorem

Let

\ x\in \mathbb {R}

be an arbitrary irrational number. Then

\left|x-{\frac {s}{t}}\right|\ <\ {\frac {1}{{\sqrt {5}}\cdot t^{2}}}

for infinitely many different

\ s,t\in \mathbb {Z} \backslash \{0\}.

Lemma 1

Let $\ c,d\in \mathbb {N} .$ Let $\ C:=c+d.$ Then:

c^{2}-{\sqrt {5}}\!\cdot \!c\!\cdot \!d+d\,^{2}\ >\ 0

or

C\,^{2}-{\sqrt {5}}\cdot \!C\!\cdot \!d+d\,^{2}\ >\ 0

Proof of lemma 1 It's easy to show that ${\sqrt {2}}\cdot c\neq {\sqrt {3-{\sqrt {5}}}}\cdot d.$ Thus the square of $\ X:={\sqrt {2}}\cdot c-{\sqrt {3-{\sqrt {5}}}}\cdot d$ is positive. Now,

{\sqrt {2}}\cdot {\sqrt {3-{\sqrt {5}}}}\ =\ {\sqrt {5}}-1

which means that we may write $\ X^{2}$ as follows:

2\!\cdot \!c^{2}\ -\ 2\!\cdot \!({\sqrt {5}}-1)\cdot c\!\cdot \!d\ +\ (3-{\sqrt {5}})\!\cdot \!d\,^{2}\ >\ 0

i.e.

(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 $\ a,b\in \mathbb {Z}$ 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 $\ A:=a+b$ 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 $\ x$ 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 $\ n\in \mathbb {N}$ 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:

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

{\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

\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

\ {\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 $\ b$ 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\ }
$\nu _{k+1,2\cdot n}\ :=\ \nu _{k,n}$ 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 $(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}}

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

Latest revision as of 08:40, 15 March 2021

Contents

Introduction

Notation

The method of neighbors and median

Definitions

First results

Hurwitz theorem

Lemma 1

Lemma 2

Lemma 2'

Proof of Hurwitz theorem

Squeezing irrational numbers between neighbors

Another proof of Hurwitz Theorem (further insight)

Reduction to x > 0

Two cases

The top-top-bot case

The relevant neighborhoods

Neighborhood B

Neighborhood C (first C-inequality)

Neighborhood D

Early yield (Hurwitz Theorem)

Neighborhood C (second C-inequality)

Neigborhood C (the combined inequality)

Divisibility

Relatively prime pairs of integers

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)}

Rational representation

Navigation menu

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

Latest revision as of 08:40, 15 March 2021

Introduction

Notation

The method of neighbors and median

Definitions

First results

Hurwitz theorem

Lemma 1

Lemma 2

Lemma 2'

Proof of Hurwitz theorem

Squeezing irrational numbers between neighbors

Another proof of Hurwitz Theorem (further insight)

Reduction to x > 0

Two cases

The top-top-bot case

The relevant neighborhoods

Neighborhood B

Neighborhood C (first C-inequality)

Neighborhood D

Early yield (Hurwitz Theorem)

Neighborhood C (second C-inequality)

Neigborhood C (the combined inequality)

Divisibility

Relatively prime pairs of integers

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)}

Rational representation

Navigation menu

Search