The theory of diophantine approximations is a chapter of number theory, which in turn is a part of mathematics. It studies the approximations of real numbers by rational numbers. This article presents an elementary introduction to diophantine approximations, as well as an introduction to number theory via diophantine approximations.
Introduction
In the everyday life our civilization applies mostly (finite) decimal fractions
Decimal fractions are used both as certain values, e.g. $5.85, and as approximations of the real numbers, e.g.
However, the field of all rational numbers is much richer than the ring of the decimal fractions (or of the binary fractions
which are used in the computer science). For instance, the famous approximation
has denominator 113 much smaller than 105 but it provides a better approximation than the decimal one, which has five digits after the decimal point.
How well can real numbers (all of them or the special ones) be approximated by rational numbers? A typical Diophantine approximation result states:
Theorem Let
be an arbitrary real number. Then
is rational
there exists a real number C > 0 such that
![{\displaystyle \left|a-{\frac {x}{y}}\right|>{\frac {C}{y}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8b205bec4983072395a17d91b95220f096d531a)
for arbitrary integers
such that
and
- (Adolph Hurwitz)
is irrational
there exist infinitely many pairs of integers
such that
and
![{\displaystyle \left|a-{\frac {x}{y}}\right|<{\frac {1}{{\sqrt {5}}\cdot y^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00823164291ad408445f0bcf4820d0496a1bd196)
Remark Implication
of the first part of the theorem is a simple and satisfaction bringing exercise.
Notation
— "equivalent by definition" (i.e. "if and only if");
— "equals by definition";
— "there exists";
— "for all";
— "
is an element of set
";
— the semiring of the natural numbers;
— the semiring of the non-negative integers;
— the ring of integers;
— the field of rational numbers;
— the field of real numbers;
![{\displaystyle (x;y)\ :=\{t\in \mathbb {R} :\,x<t<y\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db22bd42f42fe4bc3f7ba9bf02389c3172049eba)
![{\displaystyle |(x;y)|\ :=y-x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f745495722cc70b5694b51ead5aee96784af78e)
![{\displaystyle {\mathit {Span}}(x,y)\ :=\ \left(\min \left(x,y\right);\,\max \left(x,y\right)\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7e4f650a3e3d60a61a1cd9be1cb743085c4c0b5)
— "
divides
" (i.e.
);
— the greatest common divisor of integers
and ![{\displaystyle \ b.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3658546c70c97d710581246f2c4952e8d6997cb2)
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
and
with integer numerators and natural denominators, are called neighbors (in the given order)
![{\displaystyle {\frac {a}{c}}-{\frac {b}{d}}\ =\ {\frac {1}{c\cdot d}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6eb510c0a5f1c8e177ec9ec4f7d94437a99c6b3)
Fraction
is called the top neighbor of the other,
is called the bottom neighbor, and the interval
is called neighborhood; thus a neighborhood is an open interval such that its endpoints are neighbors.
- If
and
are neighbors then
( i.e.
).
- Let
Fractions
and
are neighbors
fractions
and
are neighbors
fractions
and
are neighbors.
Examples:
- Fractions
and
are neighbors for every positive integer ![{\displaystyle \ n.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91200b76008a0b9e4fcfd5aa0d156bf78b66b5b1)
- Fractions
and
are neighbors for every positive integer ![{\displaystyle \ n.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91200b76008a0b9e4fcfd5aa0d156bf78b66b5b1)
Thus it easily follows that for every positive irrational number
there exists a pair of neighbors
and
with positive numerators and denominators, such that:
Definition A pair of neighboring fractions
with integer numerators and natural denominators, is called a top pair
Otherwise it is called a bottom pair.
Thus now we have notions of top/bottom neighbors and of top/bottom pairs of neighbors.
- Let
be a pair of neighbors. Then
is a top pair of neighbors, and
is a bottom pair of neighbors.
First results
Theorem Let fractions
and
with integer numerators and natural denominators, be neighbors. Then
- if integers
and
are such that
then ![{\displaystyle t\geq c+d;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f5ddd6b4843b740f7a0e453dd8531de8df4769c)
- the median
is a bottom neighbor of
and a top neighbor of
- let
be an irrational number such that
then
![{\displaystyle 0\ <\ \max({\frac {a}{c}}-x,\,x-{\frac {b}{d}})\ <\ {\frac {1}{c\cdot d}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6de6139326e2fe9047ce7ac310431ab18cc733d)
- and
or ![{\displaystyle 0\ <\ x-{\frac {b}{d}}\ <\ {\frac {1}{2\cdot d^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e615c3aa51b61fd3d013c18dc4befe2e08ec96e1)
Proof Let
then
and ![{\displaystyle {\frac {s}{t}}-{\frac {b}{d}}\ \geq \ {\frac {1}{d\cdot t}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f22e2dc5c6ca1fa9f6fe5afb77ca4f75eb61bfe9)
and
![{\displaystyle {\frac {1}{c\cdot t}}+{\frac {1}{d\cdot t}}\ \leq {\frac {1}{c\cdot d}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f8ff460a9af0e780e7f5660c731dc4b0a4ca37c)
Multiplying this inequality by
gives
![{\displaystyle t\geq c+d}](https://wikimedia.org/api/rest_v1/media/math/render/svg/562ceef336c833f9bac203d81bbf99a7d78b475b)
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:
![{\displaystyle \max({\frac {a}{c}}-x,\,x-{\frac {b}{d}})\ <\ ({\frac {a}{c}}-x)+(x-{\frac {b}{d}})\ =\ {\frac {1}{c\cdot d}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7f055c40fac37f8c9f82e94442cfe39d6138279)
Next,
![{\displaystyle \left({\frac {1}{c}}-{\frac {1}{d}}\right)^{2}\ \geq \ 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9b027d170dd6a3927e87cfb91496eefc96c39af)
hence
![{\displaystyle {\frac {1}{2\cdot c^{2}}}+{\frac {1}{2\cdot d^{2}}}\ \geq \ {\frac {1}{c\cdot d}}\ =\ {\frac {a}{c}}-{\frac {b}{d}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1dd77a1f6b44839eb0681d407c062dbbfcfe26db)
![{\displaystyle =\ \left|{\mathit {Span}}({\frac {a}{c}},{\frac {b}{d}})\right|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f9bb4e139bf3560320429fd91d51caad88a0d3c)
and
![{\displaystyle {\frac {b}{d}}+{\frac {1}{2\cdot d^{2}}}\ \geq \ {\frac {a}{c}}-{\frac {1}{2\cdot c^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2a2efe3f693be019796aaaf2139681442a05e60)
i.e.
End of proof
Corollary Let fractions
and
with integer numerators and natural denominators, be neighbors. Then, if integers
and
are such that
then either
![{\displaystyle s=a+b\quad \land \quad t=c+d;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c70fda7ad356ace439c001b92357260b851d9fd)
or
![{\displaystyle t\ \geq \ c+d+\min(c,d)\ >\ c+d}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ecceaf276b1716b00596d615a60be68feb2d4b0a)
Hurwitz theorem
- Let
be an arbitrary irrational number. Then
![{\displaystyle \left|x-{\frac {s}{t}}\right|\ <\ {\frac {1}{{\sqrt {5}}\cdot t^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8081e127d2ec50c1f566bb68fd6b6e390c29656a)
- for infinitely many different
![{\displaystyle \ s,t\in \mathbb {Z} \backslash \{0\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f6f2f07ab3f105fc9309303866acb0f992c124b)
Lemma 1
Let
Let
Then:
![{\displaystyle c^{2}-{\sqrt {5}}\!\cdot \!c\!\cdot \!d+d\,^{2}\ >\ 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15c15756c93c2819c34e3bd0becb9f8d6786f703)
or
![{\displaystyle C\,^{2}-{\sqrt {5}}\cdot \!C\!\cdot \!d+d\,^{2}\ >\ 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0e8ec991acb2f8fb26c6a09af9449468893c40d)
Proof of lemma 1 It's easy to show that
Thus the square of
is positive. Now,
![{\displaystyle {\sqrt {2}}\cdot {\sqrt {3-{\sqrt {5}}}}\ =\ {\sqrt {5}}-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51057215f49b44088c9d7b46654a1db9e5527f22)
which means that we may write
as follows:
![{\displaystyle 2\!\cdot \!c^{2}\ -\ 2\!\cdot \!({\sqrt {5}}-1)\cdot c\!\cdot \!d\ +\ (3-{\sqrt {5}})\!\cdot \!d\,^{2}\ >\ 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1502161282d7099846bc18c37e84832f984c77b)
i.e.
![{\displaystyle (c^{2}-{\sqrt {5}}\cdot c\cdot d+d^{2})\ +\ (C^{2}-{\sqrt {5}}\cdot C\cdot d+d^{2})\ >\ 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63719bddecadee67dc529d7c1e1282a62c3b7cfe)
and lemma 1 follows. End of proof
Lemma 2
Let
and
Let
and
Furthermore, let fractions
be neighbors, and let:
![{\displaystyle {\frac {A}{C}}\ >\ x\ >\ {\frac {b}{d}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0832e0855147f1cbdbc18e80cad062f807f940d9)
where
is real. Then one of the following three inequalities holds:
![{\displaystyle 0\ <\ {\frac {a}{c}}-x\ <\ {\frac {1}{{\sqrt {5}}\cdot c^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6766390741689e6f804366cf504eb7d4dacb4550)
![{\displaystyle 0\ <\ {\frac {A}{C}}-x\ <\ {\frac {1}{{\sqrt {5}}\cdot C^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7447d0a2923ed084099d7af212ac8cc0abbc4740)
![{\displaystyle 0\ <\ x-{\frac {b}{d}}\ <\ {\frac {1}{{\sqrt {5}}\cdot d^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9334ae33f5bcce43352f425de2d6c06c3822257)
Proof There are two cases along the inequalities of Lemma 1. Let's assume the first one, which is equivalent to:
![{\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|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a438f3098b7c7bd3486eb7b99d91bca07cd72269)
Thus
![{\displaystyle {\frac {b}{d}}+{\frac {1}{{\sqrt {5}}\cdot d\,^{2}}}\ >\ {\frac {a}{c}}-{\frac {1}{{\sqrt {5}}\cdot c^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6d51f92db630d6099ff19575f2dc125cd68733c)
which means that
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9b4942647de97b80619a79e34d77b69ec3e8ee6)
![{\displaystyle \supseteq \ {\mathit {Span}}\left({\frac {A}{C}},{\frac {b}{d}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97b0a671492cd7b65e934fdc8233cd9b81a79287)
Thus lemma 2 is proven when the first inequality of lemma 1 holds. By replacing in the above proof the lower case
by the upper case
we obtain the proof when the second inequality of lemma 1 holds.
End of proof (of lemma 2)
Lemma 2'
Let
and
Let
and
Furthermore, let fractions
be neighbors, and let:
![{\displaystyle {\frac {a}{c}}\ >\ x\ >\ {\frac {A}{C}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1e9edcac3dfab4e4d06a2415fd6c910bc6ce3cc)
where
is real. Then one of the following three inequalities holds:
![{\displaystyle 0\ <\ x-{\frac {b}{d}}\ <\ {\frac {1}{{\sqrt {5}}\cdot d\,^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc8671a5676b5a15a490469a4104a4a2e9aaf2b6)
![{\displaystyle 0\ <\ x-{\frac {A}{C}}<\ {\frac {1}{{\sqrt {5}}\cdot C^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c52e02749ba77b8ed04b59b6d3307fb62f8c36e6)
![{\displaystyle 0\ <\ {\frac {a}{c}}\ <\ {\frac {1}{{\sqrt {5}}\cdot x^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f62aabd5c8c19842a4a0e7db0a05f4925a9c4e9b)
Proof It's similar to the proof of lemma 2. Or one may apply lemma 2 to
and
, which would provide us with the respective fraction
Then the required
are given by
End of proof (of lemma 2')
Proof of Hurwitz theorem
When
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
(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
be an irrational number, We may always squeeze it between the extremal neighbours:
![{\displaystyle {\frac {1}{0}}\ >\ x\ >\ {\frac {0}{1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0b08c1e454decd65ee8fe200a1000f9864fc9cf)
But if you don't like infinity (on the left above) then you may do one of the two things:
![{\displaystyle x>1\quad \quad \Rightarrow \quad \quad {\frac {n+1}{1}}>x>{\frac {n}{1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae062497290fe64a676d26863508bf82f232ca9d)
or
![{\displaystyle x<1\quad \quad \Rightarrow \quad \quad {\frac {1}{n}}>x>{\frac {1}{n+1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a92a7411d6e618549da01d1a2ecd2b5c2a966f4)
where in each of these two cases
is a respective unique positive integer.
It was mentioned in the previous section (First results) that if fractions
and
with positive (or non-negative) integer numerators and denominators are neighbors then also the top and the bottom (bot for short) pair:
![{\displaystyle {\mathit {top}}\!\left({\frac {a}{c}},{\frac {b}{d}}\right)\ :=\ \left({\frac {a}{c}},{\frac {a+b}{b+d}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d7890632cae5f355bbf852faff52e01bfd3b352)
and
![{\displaystyle {\mathit {bot}}\!\left({\frac {a}{c}},{\frac {b}{d}}\right)\ :=\ \left({\frac {a+b}{c+d}},{\frac {b}{d}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41bddfe9dc65c8d1763a4bf4b737a805a354b77c)
are both pairs of neighbors.
Let
be a pair of neighbors, and
be an irrational number. Assume that pairs of neighbors
are already defined, and that they squeeze
i.e. that
for each
Then we define
as the one of the two pairs:
or
which squeezes
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
there exist fractions of integers
with arbitrarily large denominators, such that
![{\displaystyle \left|x-{\frac {s}{t}}\right|\ <\ {\frac {1}{{\sqrt {5}}\cdot t^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8081e127d2ec50c1f566bb68fd6b6e390c29656a)
(see section Hurwitz theorem).
If cases top-top and bot-bot happen only finitely many times then starting with an
we get an infinite alternating top-bot-top-bot-... sequence:
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/16d323d4098fef061fe9748c4cafe6cf57b8fd20)
Then the new neighbor of the
pair (i.e. the median of the previous pair
) is equal to
![{\displaystyle e_{k}\ :=\ {\frac {F_{k+1}\cdot a+F_{k}\cdot b}{F_{k+1}\cdot c+F_{k}\cdot d}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c74376ad0a92cb13aced9d4214a778c6eb8eeebd)
for every
where
are the Fibonacci numbers, where
![{\displaystyle \left({\frac {a}{c}},{\frac {b}{d}}\right)\ :=\ A_{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b4b365767fbda8eed6505910c1e8fd81d2db33b)
It is known that
![{\displaystyle \lim _{k\rightarrow \infty }{\frac {F_{k+1}}{F_{k}}}\ =\ \Phi \ :=\ {\frac {{\sqrt {5}}+1}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c21101a5b0193f74d516d25f9efa0ed175eb9797)
hence
![{\displaystyle x\ =\ \lim _{k\rightarrow \infty }e_{k}\ =\ {\frac {\Phi \cdot a+b}{\Phi \cdot c+d}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/470a0f948bc0476d793d3ebaebed5fa58dca17df)
But if our infinite alternation has started with bot :
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/55d68f4f65218037acbc7439e75100294e6b4603)
Then we would have
![{\displaystyle x\ =\ \lim _{k\rightarrow \infty }e_{k}\ =\ {\frac {a+\Phi \cdot b}{c+\Phi \cdot d}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca7103dbb77cd47627057540ef8bc55aefda6a84)
Another proof of Hurwitz Theorem (further insight)
Reduction to x > 0
Since
![{\displaystyle \left|(-x)-{\frac {-s}{t}}\right|\ =\ \left|x-{\frac {s}{t}}\right|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/200a49a10f49f9a04c4d972c32390ad45f923d89)
it is enough to prove Hurwitz theorem for positive irrational numbers only.
Two cases
Consider the sequence
of pairs of neighbors, which squeeze
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
for which
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c0a6f847f8ec03b3144d2752e233a280138058e)
or
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/592223236d5894d8c96d97c229ac419f234e10b5)
holds, where
![{\displaystyle \left({\frac {a}{c}},{\frac {b}{d}}\right)\ :=\ A_{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b4b365767fbda8eed6505910c1e8fd81d2db33b)
The top-top-bot case
Let's consider the latter top-top-bot case. Let
The squeeze by neighbors:
![{\displaystyle {\frac {a}{c}}\ >\ x\ >\ {\frac {2\cdot a+b}{2\cdot c+d}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b070ba22f2f4e6795c96bda252baf342a19f8294)
shows that
![{\displaystyle 0\ <\ {\frac {a}{c}}-x\ <\ {\frac {1}{c\cdot (2\cdot c+d)}}\ =\ {\frac {1}{(2+\xi )}}\cdot {\frac {1}{c^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33fd529270c26d89a5375fc9553b7c39cb911d87)
This inequality provides the first insight (otherwise, we are not going to use it), so it deserves to be written cleanly as an implication:
The relevant neighborhoods
Consider the next two pairs of neighbors, pair
and pair
which squeeze
The relevant neighborhoods are:
![{\displaystyle A\ :=\ {\mathit {Span}}({\frac {a}{c}},\,{\frac {3\cdot a+b}{3\cdot c+d}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3113c0864ad79b4bba3344d642fd985e45cca670)
![{\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}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2779808343bf8039b7a21b9e6cced9b0b178034)
![{\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}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e10db16a1642a06761658614e3ed00b85ff4f58b)
![{\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}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5726ab8493ca99569ad39960421ba3941697c90a)
Neighborhood B
Let
Then
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e73d7f5ec8a53dd015b66317e3bd8c0ba9a074a4)
and
![{\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)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02b782da5f17f4bf4e2a6c5fd5b5cbbf690bb342)
![{\displaystyle =\ {\frac {2}{c\cdot (5\cdot c+2\cdot d)}}\ =\ {\frac {2}{(5+2\cdot \xi )}}\cdot {\frac {1}{c^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c40036dbdb9566de45a4da8897fe9450b2b92b4f)
Conclusion:
Neighborhood C (first C-inequality)
Let
Then
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e5e6807c608f21a14aeef857281d747afa27198)
Thus using the calculation for neighborhood B also for C, we get
![{\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)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7260e0d3597bd57f41515ed9af0ce303fec3bf8f)
![{\displaystyle =\ {\frac {3}{7+3\cdot \xi }}\cdot {\frac {1}{c^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc030bd39bda3498dc02e830d5d195b959694ccd)
First C-conclusion:
Neighborhood D
Let
i.e.
![{\displaystyle {\frac {7\cdot a+3\cdot b}{7\cdot c+3\cdot d}}\ >\ x\ >\ {\frac {2\cdot a+b}{2\cdot c+d}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d57c878e85c4f8a639182316a796a6511c3926c)
Let
and
Then
![{\displaystyle 0\ <\ x-{\frac {\alpha }{\gamma }}\ =\ x-{\frac {2\cdot a+b}{2\cdot c+d}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84e6ec54945fddaa974f1d6e7add9eb6c26f072f)
![{\displaystyle <\ {\frac {1}{(7\cdot c+3\cdot d)\cdot (2\cdot c+d)}}\ =\ {\frac {1}{(7+3\cdot \xi )\cdot c\cdot \gamma }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4e21e108b78aacc6ac3f97aa6129beb113baf11)
![{\displaystyle =\ {\frac {2+\xi }{7+3\cdot \xi }}\cdot {\frac {1}{\gamma ^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20d223559fbc793e4ff77523cfb563acfa52a000)
Conclusion:
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:
![{\displaystyle 0\ <\ {\frac {a}{c}}-x\ <\ {\frac {2}{5}}\cdot {\frac {1}{c^{2}}}\quad \quad \Leftarrow \quad \quad x\in B}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba94e851817a71a2e79d76c9bdd45a77692c670e)
![{\displaystyle 0\ <\ {\frac {a}{c}}-x\ <\ {\frac {3}{7}}\cdot {\frac {1}{c^{2}}}\quad \quad \Leftarrow \quad \quad x\in C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1f7fa587a1793f4d9dc11db9b8021017c9592ad)
![{\displaystyle 0\ <\ x-{\frac {\alpha }{\gamma }}\ <\ {\frac {2}{7}}\cdot {\frac {1}{\gamma ^{2}}}\quad \quad \Leftarrow \quad \quad x\in D}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9d753b06610b049d4561d960643d3aa207746a9)
Since
![{\displaystyle \max({\frac {2}{5}},\,{\frac {3}{7}},\,{\frac {2}{7}})\ =\ {\frac {3}{7}}\ <\ {\frac {1}{\sqrt {5}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4900e57bc72c8322bfe95a11f04bd42575d4a64d)
each occurrence of the top-top-bot subsequence, i.e. of equalities:
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9f8dce3d26308eac3d7e1bd96b94ec8a6d1335d)
provides a fraction
with integer numerator and natural denominator, such that
![{\displaystyle \left|x-{\frac {s}{t}}\right|\ <\ {\frac {3}{7}}\cdot {\frac {1}{t^{2}}}\ <\ {\frac {1}{\sqrt {5}}}\cdot {\frac {1}{t^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00477f9652577212e2a21397409659b91a51351f)
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
by:
for ![{\displaystyle A_{n}=({\frac {a}{c}},{\frac {b}{d}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2ce1a496293a5999676612274535b9709903934)
Let
be a bot-bot-top progression. Then
is a top-top-bot progression which squeezes
Thus
![{\displaystyle \left|(-x)-{\frac {u}{w}}\right|\ <\ {\frac {3}{7}}\cdot {\frac {1}{w^{2}}}\ <\ {\frac {1}{\sqrt {5}}}\cdot {\frac {1}{w^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a61d3fbb8566ca1c7b2c735b8ae56fde7ea1f3b)
for certain
with integer numerator and natural denominator. Then
for
satisfies:
![{\displaystyle \left|x-{\frac {s}{t}}\right|\ <\ {\frac {3}{7}}\cdot {\frac {1}{t^{2}}}\ <\ {\frac {1}{\sqrt {5}}}\cdot {\frac {1}{t^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00477f9652577212e2a21397409659b91a51351f)
When another top-top-bot or bot-bot-top progression starts with a sufficiently large index
then
which means that the respective new approximation
is different. It follows that if there are infinitely many progressions top-top-bot or bot-bot-bot then there are infinitely many fractions
which satisfy the inequality above. Thus we have obtained the following version of Hurwitz theorem:
- Theorem Let
be an arbitrary irrational number. Then inequality
![{\displaystyle \left|x-{\frac {s}{t}}\right|\ <\ {\frac {1}{\sqrt {5}}}\cdot {\frac {1}{t^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92801bdfd53077bd0facadeeadccafab80388472)
- holds for infinitely many fractions
with integer numerator and natural denominator. Furthermore, if the squeezing sequence of
does not eventually alternate top-bot-top-bot-... till infinity, i.e. if it has infinitely many top-top or bot-bot progressions, then
![{\displaystyle \left|x-{\frac {s}{t}}\right|\ <\ {\frac {3}{7}}\cdot {\frac {1}{t^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aefa655fccc908c67cac9fc9b749cb451d1799b9)
- holds for infinitely many fractions
with integer numerator and natural denominator.
Neighborhood C (second C-inequality)
Let
Then
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2c3ac4cbffc8455a59b9eb73bd7aeb0c500804e)
Thus, using the earlier conclusion for neighborhood
also for
we obtain
![{\displaystyle 0\ <\ x-{\frac {\alpha }{\gamma }}\ =\ x-{\frac {2\cdot a+b}{2\cdot c+d}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84e6ec54945fddaa974f1d6e7add9eb6c26f072f)
![{\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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ecbfba5e3a226b46310830b34cee6d57e983e58e)
![{\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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d8b8317c90f3c8b639f189996a1c881fef712f6)
![{\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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10ed9c98b8f4d0e5c81909040632df07c4fdb1ff)
![{\displaystyle =\ {\frac {2+\xi }{5+2\cdot \xi }}\cdot {\frac {1}{\gamma ^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3f437619bc53fdb8e9a6d46653d624bf5449994)
Second C-conclusion:
Neigborhood C (the combined inequality)
Let:
![{\displaystyle \xi _{0}\ :=\ {\frac {{\sqrt {61}}-7}{6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71c9efc95ef4ee6bb5539c63012f7b0e536314d1)
Then
![{\displaystyle {\frac {3}{7+3\cdot \xi _{0}}}\ =\ {\frac {2+\xi _{0}}{5+2\cdot \xi _{0}}}\ =\ {\frac {{\sqrt {61}}-7}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2d1e3bf7b78c149fd988f1217dee923e840cba1)
Thus for
:
![{\displaystyle {\frac {3}{7+3\cdot \xi }}\ \leq \ {\frac {{\sqrt {61}}-7}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5d0ecfa1987f90c5c407d529e371c5e3fc1dbcd)
and for
:
![{\displaystyle {\frac {2+\xi }{5+2\cdot \xi }}\ \leq \ {\frac {{\sqrt {61}}-7}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/032422840f1f827426203359096f883563683392)
It follows that
- for one of the fractions
the following inequality holds for every ![{\displaystyle \ x\in C:}](https://wikimedia.org/api/rest_v1/media/math/render/svg/036f97d937641e3cdb76951e2f3ae8afc3856495)
![{\displaystyle \left|x-{\frac {s}{t}}\right|\ <\ {\frac {{\sqrt {61}}-7}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb1acd3b269e563f221919da07d497e64cf36994)
(the choice of
depends on
).
Divisibility
Definition Integer
is divisible by integer
Symbolically:
![{\displaystyle \exists _{c\in \mathbb {Z} }\ a=b\cdot c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03c81b33594cf83ba2e5b7cd4b889ade2d3c1bd3)
When
is divisible by
then we also say that
is a divisor of
or that
divides
- The only integer divisible by
is
(i.e.
is a divisor only of
).
is divisible by every integer.
is the only positive divisor of ![{\displaystyle \ 1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cadcfff0f7aabf708593c25995db8a10af8f2535)
- Every integer is divisible by
(and by
).
![{\displaystyle a|b\ \Rightarrow (-a|b\ \land \ a|\!-\!\!b)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e30ed473d47351c0ec679fc7510ffbaa8e1decde)
![{\displaystyle a|b>0\ \Rightarrow a\leq b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f646f9cdd4cfe3c175c6c88f7206376027e2cf2b)
![{\displaystyle a|a\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/255a27b35dd78e2626fce6914bdda12257d66807)
![{\displaystyle (a|b\ \land \ b|c)\ \Rightarrow \ a|c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e6c55b5c611b91641fd56fcd938b7d360823d691)
![{\displaystyle (a|b\ \land \ b|a)\ \Rightarrow \ |a|=|b|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1bd7d2f3c0c6a3ac75bdf4fa3f8ac312b04d6a6)
Remark The above three properties show that the relation of divisibility is a partial order in the set of natural number
and also in
—
is its minimal, and
is its maximal element.
![{\displaystyle (a|b\ \land \ a|c)\ \Rightarrow \ (a\,|\,b\!\cdot \!d\ \land \ a\,|\,b\!+\!c\ \land \ a\,|\,b\!-\!c)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/751c84610824a2406fc0e64db460198c3e72070a)
Relatively prime pairs of integers
Definition Integers
and
are relatively prime
is their only common positive divisor.
- Integers
and
are relatively prime ![{\displaystyle \Leftrightarrow \ |x|=1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6063982dda364a20a30bfcce524f4ce54fda8837)
is relatively prime with every integer.
- If
and
are relatively prime then also
and
are relatively prime.
- Theorem 1 If
are such that two of them are relatively prime and
then any two of them are relatively prime.
- Corollary If
and
are relatively prime then also
and
are relatively prime.
Now, let's define inductively a table odd integers:
![{\displaystyle (\nu _{k,n}:k\in \mathbb {Z} _{+},\ 0\leq n\leq 2^{k})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9a78813dedc946a68507b5cc1f7c2315703d629)
as follows:
and ![{\displaystyle \nu _{0,1}:=1\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/63b4021472755c6b53856888a7d0ab52e915a089)
for ![{\displaystyle 0\leq n\leq 2^{k}\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc256387ccca68a6686de17cb969fbef308f082b)
for ![{\displaystyle 0\leq n<2^{k}\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e2287f7be5314bb339a42b66ddbf060a2632322)
for every
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,
and
is relatively prime.
- For every pair of relatively prime, non-negative integers
and
there exist indices
and non-negative
such that:
![{\displaystyle \{a,b\}\ =\ \{\nu _{k,n},\nu _{k,n+1}\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b28a3629723690a749ba0465275ff08a6a245c6)
Proof Of course the pair
![{\displaystyle \{\nu _{0,0},\nu _{0,1}\}\ =\ \{0,1\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/070bcdcf7cb64763fe1196240e7cbe3c8bb890b3)
is relatively prime; and the inductive proof of the first statement of Theorem 2 is now instant thanks to Theorem 1 above.
Now let
and
be a pair of relatively prime, non-negative integers. If
then
and the second part of the theorem holds. Continuing this unductive proof, let's assume that
Then
Thus
![{\displaystyle \ \max(a,b)<a+b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6370fe61a81edcc3fb378de582bb1fd66ba532e6)
But integers
and
are relatively prime (see Corollary above), and
![{\displaystyle c+d\ =\ max(a,b)\ <\ a+b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26e1002e93533dd91eb99a941d992ad280fb4bf2)
hence, by induction,
![{\displaystyle \{c,d\}\ =\ \{\nu _{k,n},\nu _{k,n+1}\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cdfcee9a88c41727a294fe6140cef17100f45e48)
for certain indices
and non-negative
Furthermore:
![{\displaystyle \{a,b\}\ =\ \{\min(a,b),\max(a,b)\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e765d368991ecb27572ded2efcb29bf041f39f88)
It follows that one of the two options holds:
![{\displaystyle \{a,b\}\ =\ \{\nu _{k+1,2\cdot n},\nu _{k+1,2\cdot n+1}\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1531f97030d2cecdd7ce42e3e367b9e9f55af032)
or
![{\displaystyle \{a,b\}\ =\ \{\nu _{k+1,2\cdot n+1},\nu _{k+1,2\cdot n+2}\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a671e28c40bf9ad4b426d62e011a53d84f3b6273)
End of proof
Let's note also, that
![{\displaystyle \max _{\quad 0\leq n\leq 2^{k}}\nu _{k,n}\ =\ F_{k+1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed567fd5338e27b9d3247a1d3599fb2d89df56bc)
where
is the r-th Fibonacci number.
Matrix monoid ![{\displaystyle {\mathit {SO}}(\mathbb {Z} _{+},2)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98d21a7d80bdccf21ae5fa7dfb8bd59cd4ca9b3c)
Definition 1
is the set of all matrices
![{\displaystyle M\ :=\ \left[{\begin{array}{cc}a&b\\c&d\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca0c30ae19dc60ab9029a08ebbd73a7cf895039f)
such that
and
where
Such matrices (and their columns and rows) will be called special.
![{\displaystyle M\ :=\ \left[{\begin{array}{cc}a&b\\c&d\end{array}}\right]\in {\mathit {SO}}(\mathbb {Z} _{+},2)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9746e9f95199d8ff37ecef1284d9701d6fb4e39e)
then
and each of the columns and rows of M, i.e. each of the four pairs
is relatively prime.
Obviously, the identity matrix
![{\displaystyle {\mathcal {I}}\ :=\ \left[{\begin{array}{cc}1&0\\0&1\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67b0aca44b3fd67f0f11cbe7f4270cf375c119b2)
belongs to
Furthermore,
is a monoid with respect to the matrix multiplication.
Example The upper matrix and the lower matrix are defined respectively as follows:
and ![{\displaystyle {\mathcal {L}}\ :=\ \left[{\begin{array}{cc}1&0\\1&1\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4fa566d4a9bcf7c796dea750749473026255271e)
Obviously
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:
![{\displaystyle M\cdot {\mathcal {U}}\ =\ \left[{\begin{array}{cc}a&a+b\\c&c+d\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c669e63c048428fcd22cdd0e425d51710c39c25)
and
![{\displaystyle M\cdot {\mathcal {L}}\ =\ \left[{\begin{array}{cc}a+b&b\\c+d&d\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f8f6f268e0ae90244f08de7ff09e9c93341aefa)
Definition 2 Vectors
and ![{\displaystyle \left[{\begin{array}{c}b\\d\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a0fe8c71508e4980923fe5c7f30c31aa2dc37cd)
where
are called neighbors (in that order)
matrix formed by these vectors
![{\displaystyle M\ :=\ \left[{\begin{array}{cc}a&b\\c&d\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca0c30ae19dc60ab9029a08ebbd73a7cf895039f)
belongs to
Then the left (resp. right) column is called the left (resp. right) neighbor.
Rational representation
With every vector
![{\displaystyle v\ :=\ \left[{\begin{array}{c}a\\c\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa8ce21cd9cf75f49607a9e5b5416ba00d707d53)
such that
let's associate a rational number
![{\displaystyle {\mathit {rat}}(v)\ :=\ {\frac {a}{c}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ea072b8887dbbb5ae7a034e4eb725efd0aa5961)
Also, let
![{\displaystyle {\mathit {rat}}(v_{\infty })\ :=\ \infty }](https://wikimedia.org/api/rest_v1/media/math/render/svg/37e44c015f26f2a6516d494ceb96f687d4e4028d)
for
![{\displaystyle v_{\infty }\ :=\ \left[{\begin{array}{c}1\\0\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b87f8231efb1fd36b78f839d01924ac148aae947)
Furthermore, with every matrix
let's associate the real open interval
![{\displaystyle {\mathit {span}}(M)\ :=\ ({\mathit {rat}}(w);{\mathit {rat}}(v))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/11f977820210dfd95e51b2f528a5aa6ff41917e4)
and its length
![{\displaystyle diam(M)\ :=\ {\mathit {rat}}(v)-{\mathit {rat}}(w)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3199ae7bc0433632b8f45f9393d3a5ba799a0ab7)
where
is the left, and
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.
![{\displaystyle M\ :=\ \left[{\begin{array}{cc}a&b\\c&d\end{array}}\right]\in {\mathit {SO}}(\mathbb {Z} _{+},2)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9746e9f95199d8ff37ecef1284d9701d6fb4e39e)
then