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# Talk:Irrational numbers

See User_talk:Alonso_del_Arte#Irrational_numbers. — Daniel Forgues 16:27, 16 March 2016 (UTC)

## The higher the degree of an algebraic number, the faster the sequence of convergents converges

Is that because the higher the degree
 n
of an algebraic number
 ξ
, the larger the denominator
 q
needs to be, such that
${\begin{array}{l}\displaystyle {\left\vert \xi -{\frac {p}{q}}\right\vert <\epsilon ,\quad (p,q)=1.}\end{array}}$ Since

${\begin{array}{l}\displaystyle {\left\vert \xi -{\frac {p}{q}}\right\vert =\left\vert \xi -\left({\frac {r+s}{q}}\right)\right\vert =\left\vert \xi -\left({\frac {r}{q}}+{\frac {s}{q}}\right)\right\vert =\left\vert \xi -\left({\frac {r^{\prime }}{q^{\prime }}}+{\frac {s}{q}}\right)\right\vert ,\quad r+s=p,\quad (s,q)=1,}\end{array}}$ and

${\begin{array}{l}\displaystyle {\left\vert \xi -{\frac {r^{\prime }}{q^{\prime }}}\right\vert <\epsilon +\left\vert {\frac {s}{q}}\right\vert ,\quad (s,q)=1,}\end{array}}$ where
 r′ q′
is the reduced form of
 r q
for which
 q′
is much smaller than
 q
for some small
 s
coprime to
 q
, would that explain why? — Daniel Forgues 22:18, 16 March 2016 (UTC)
For transcendental numbers, the denominator
 q
would need to be larger than for any algebraic number, which would explain why they are the easiest to approximate with rational numbers. — Daniel Forgues 22:18, 16 March 2016 (UTC)

All the following statements are true (am I correct?)

• rational numbers are everywhere dense but have density 0 among the up to [real] quadratic numbers of the real line;
• up to quadratic numbers are everywhere dense but have density 0 among the up to cubic numbers of the complex plane;
• up to cubic numbers are everywhere dense but have density 0 among the up to quartic numbers of the complex plane;
• up to quartic numbers are everywhere dense but have density 0 among the up to quintic numbers of the complex plane;
• (...)
• algebraic numbers are everywhere dense but have density 0 (worst than that, they are a countable subset with cardinality  ℵ0
of an uncountable set with cardinality  2 ℵ0
) among the complex numbers.

Am I correct? — Daniel Forgues 22:18, 16 March 2016 (UTC)

## Marking up the numbers of the complex plane

You start with a complex plane with all numbers unmarked, and then (am I correct?)

• mark all the rational numbers of the complex plane (thus an everywhere dense subset with cardinality  ℵ0
of the real line is marked);
• mark all the quadratic numbers of the complex plane (thus an everywhere dense subset with cardinality  ℵ0
of the complex plane is marked);
• mark all the cubic numbers of the complex plane (thus between any pair of up to quadratic numbers, you marked a countable infinity  ℵ0
of cubic numbers);
• mark all the quartic numbers of the complex plane (thus between any pair of up to cubic numbers, you marked a countable infinity  ℵ0
of quartic numbers);
• mark all the quintic numbers of the complex plane (thus between any pair of up to quartic numbers, you marked a countable infinity  ℵ0
of quintic numbers);
• (...)
• finally (can you?)*, mark all the transcendental numbers of the complex plane (thus between any pair of those  ℵ0
algebraic numbers, you marked an uncountable infinity  2 ℵ0
of transcendental numbers) and now all the  2 ℵ0
numbers of the complex plane are marked.

* I don't think you can: the transcendental numbers are a non denumerable set (you can't go about marking a first one, a second one, ... because this will only mark a countable infinity
 ℵ0
of numbers!)

Am I correct? — Daniel Forgues 22:18, 16 March 2016 (UTC)