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)

About densities

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)