Algebraic and Transcendent numbers

[Anonymous No.4800903]

Question /sci/:

I've been reading about number theory and stumbled upon something you may find interesting:
-Algebraic numbers are solutions of a polynomial equation, while transcendent numbers are not.
-Transcendent numbers are also irrational numbers, but irrational numbers can be ( ! can, but not always ! ) solutions of a polynomial (like square root two, etc).
-Euler-number (e) and Pi are transcendent numbers, and these two are limits of certain equations.

ITT: number theory, set theory in general.

I feel if I talk about it, I get to understand it even more. So shall we?

[Anonymous No.4800908]

>>4800903

>Question /sci/

disregard that, the thread started as a question.

op.

[Anonymous No.4800914]

1 + 1 = 2

[Anonymous No.4800917]

algebraic irrationals are harder to approximate with rationals than transcendentals

[Anonymous No.4800921]

>>4800903
This is wrong. Consider the equation
<div class="math">x-\pi=0</div>

[Anonymous No.4800924]

>>4800921
mhm, yes quite.

[Anonymous No.4800933]

>>4800921
As I'm not an autistic pedant, I assumed he meant polynomial with integer coefficients.

[Anonymous No.4800994]

>>4800921

sorry but I don't get what you want to point out.
If you take a polynomial and subtract pi from it, the polynomial basically becomes a transcendent (?).

We ought to define polynomials, since I'm actually kinda puzzled.

[Anonymous No.4801011]

>>4800921
That's not a polynomial.
sqrt(5)^sqrt(7) is another example of transcendent number.

[Anonymous No.4801018]

>>4801011
it is, just not one with rational coefficients

here's another polynomial

pi x^2 + e x - sqrt(5)

[Anonymous No.4801019]

>>4801011
proof?

[Anonymous No.4801036]

my mind is full of fuck.

- op

[Anonymous No.4801042]

>>4801011
>>4800921
>>4800994
Not sure if trolling, misunderstanding, autistic or retarded.
A transcendent number is a number that isn't the root of a polynomial with integer coefficients, and that was OP meant, even though he didn't specify it and anyone without autism could have understood.
X - Pi is a polynomial but it doesn't have integer coefficients, hence the fact that Pi is a root of this polynomial doesn't prove that Pi isn't transcendent. The actual proof that Pi is transcendent requires quite a lot of work.

[Anonymous No.4801048]

>-Algebraic numbers are solutions of a polynomial equation, while transcendent numbers are not.
Wrong, see >>4800921
>-Transcendent numbers are also irrational numbers, but irrational numbers can be ( ! can, but not always ! ) solutions of a polynomial (like square root two, etc).
Wrong, the irrational numbers are only the real numbers which aren't rational numbers but there are
algebraic numbers which are not real like i.
>-Euler-number (e) and Pi are transcendent numbers, and these two are limits of certain equations.
Limits of certain equations doesn't make any sense.

[Anonymous No.4801062]

>>4801048
so what's your definition of a transcendental number?

[Anonymous No.4801064]

>>4801062
Numbers that aren't the root of a polynomial with integer coefficients. This is the only definition.

[Anonymous No.4801072]

>>4801042
>"The actual proof that Pi is transcendent requires quite a lot of work."
true, I just checked the proof and I didn't understand a word of it.

>>4801048

1.: not true.
2.: that's true, but it doesn't have anything to do with what I said. Most of the real- and complex numbers are transcendent, since non-transcendent numbers are countable unlike transcendent numbers which are not-countable.
3.: "Limits of certain equations doesn't make any sense."
Maybe it does not for you, but I think for most of /sci/ it does.

-op was here.

[Anonymous No.4801075]

>>4801064
so you are just being pedantic about him not saying "integer coefficients"

inb4 rigor isn't pedantry

[Anonymous No.4801079]

>>4801075
Actually this needed to be clarified since autists such as >>4800921 showed up.

[Anonymous No.4801080]

>>4801064
>Numbers that aren't the root of a polynomial with integer coefficients. This is the only definition.
It is better to define thef algebraic numbers as the algebraic closure of the integers or rationals and to define the transcedent number as its complement.

[Anonymous No.4801082]

>>4801064

Wiki says:
"In mathematics, a transcendental number is a (possibly complex) number that is not algebraic—that is, it is not a root of a non-constant polynomial equation with rational coefficients."

as I said, Pi is irrational and transcendent, while sqrt(2) is for example irrational but not transcendent (since it's a solution of an integer coefficient polynomial)

-op

[Anonymous No.4801095]

>>4801082
every rational coefficient polynomial can be written as an integer coefficient polynomial, so you usually say integer coefficients

[Anonymous No.4801102]

Every algebraic closed field with countable infinite elements has a subfield which is isomorphic to the field of algebraic numbers.

[Anonymous No.4801105]

>>4801102

oh wow that helps a lot.

2+1=1+2

[Anonymous No.4801106]

>>4801082
To be clear,
Sqrt(2) is A solution to the equation
x^2 = 2;

Or, in ancient greek terms, it is the length of the side of a square that has an area of 2.

[Anonymous No.4801118]

>>4800917
Wrong, most transcendentals are impossible to approximate at all, as there are only a countable number of approximation algorithms.

[Anonymous No.4801119]

What OP means is that transcendentals are never solutions to polynomial equations with rational coefficients. The field of algebraic numbers is the smallest algebraically closed field extension of the rational numbers. Also the set of algebraic numbers has cardinality of aleph null, while complex numbers have cardinality aleph one.

[Anonymous No.4801121]

>>4801119
> complex numbers have cardinality aleph one.
LOLOLOLOLOLOLOL
Prove it.

[Anonymous No.4801139]

>>4801121
This trivially follows from the continuum hypothesis.

[Anonymous No.4801147]

>>4801139
which is undecidable, I think that was the joke >>4801121 implied

[Anonymous No.4801160]

>>4801119

Complex? I think you mean something else. Complex refers to numbers of the form

a + bi, a,b E R.

For any two numbers, a1a2a3..., b1b2b3 we can construct a new number by interleaving the digits:
a1b1a2b2a3b3.... which is a real number.

Therefor, the cardinality of the complex numbers is the same as the cardinality of the real numbers. (similarly for R^2, R^3 ... )

So it's got cardinality of 2^ Aleph Null, which may or may not be equal to Aleph 1, but not probably so.

[Anonymous No.4801162]

>>4801139

Which is dependent on your choice of axioms.

[Anonymous No.4801163]

I might have not used the proper statement. I meant that the set of algebraic numbers has the same cardinality as the integers, and that the set of complex numbers has the same cardinality of the power set of the integers.

[Anonymous No.4801169]

There was a thread on here a while ago about developing an algorithm that can take any algebraic number and return the polynomial of lowest degree that has this as a solution. I think I solved the problem, but I haven't wrote the algorithm yet. I'll probably write it in Mathematica if I do because it already has built in handling for algebraic numbers.