Talk:Algebraic number

This is a subject where it is easy to give examples- like sqrt(2) etc. Let me be the first to suggest this. And also what are not algrebraic numbers, i.e. transcendental numbers. I see there is an entry for this. Nevertheless, you could mention them on the algebraic number page and link. RoseParks

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I've added some. -- Jan Hidders---- Lookin good...RoseParks

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What do you call numbers which can be obtained from the integers by a finite series of additions, multiplications, divisions and root extractions? I know they are not algebraic numbers, since solutions to polynomials of degree five or higher cannot be obtained in this way, and yet they are by definition algebraic numbers. -- Simon J Kissane

An equation whose roots are numbers of the form you describe is said to be "solvable by radicals". My guess would be that the class of numbers would be the "radical expressions" or something, but I don't know. -- Carl Witty

I've been asking professors and others the same question and I suggest calling them solvable numbers. It's clearly a subfield. Richard Peterson

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Looking at MathWorld - Algebraic Integer, I think there might be something wrong with the definition of algebraic integers here.

QZ 13:55, 2004 Mar 19 (UTC)

Don't see it. Charles Matthews 15:26, 19 Mar 2004 (UTC)

I think you need to say that the coefficients are all integers. Otherwise, I can say 2/3 is an algebraic integer, because it's a solution of 1*x - 2/3 = 0, which is a monic polynomial. Not entirely sure, so not changed. Jonpin 08:26, Oct 1, 2004 (UTC)

Well, it says all the a_i are integers at the top of the page; still applies. Charles Matthews 18:37, 1 Oct 2004 (UTC)

It doesn't anymore; it was changed on 4 Oct 2004. --RRM 06:43, 8 Mar 2005 (UTC)

Link suggestions

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I added the links that made sense. Edward 00:24, 22 Dec 2004 (UTC)

Ring of Integers redirect

"Ring of Integers" currently redirects to this article, but I'm going to move it so it redirects to Ring (mathematics), which I consider a more suitable target. --Malcohol 09:49, 8 Jun 2005 (UTC)

p-adic algebraic numbers

While I think it's important to point out that p-adic numbers can also be algebraic, I'm a bit concerned if we don't specify precisely what an "algebraic number" is when no field is specified. For instance, statements like "the algebraic numbers form a field" or "the algebraic numbers are countable" are false if we leave it open like that. Also, many other pages that link here and describe properties of algebraic numbers would have to be qualified, since many of these statements don't work for p-adic algebraic numbers.

I'll try to do something about this issue, but other suggestions are welcome. AxelBoldt 18:56, 14 May 2006 (UTC)

As far as I can think of the term "algebraic number" is reserved for a root of a polynomial with coefficients in the rationals. The p-adic numbers are not algebraic, specifically because they are obtained by the analytic process of completing the rational numbers. One can speak of elements that are algebraic over a p-adic number field, but then one enters the more general concept of algebraic elements. For this reason, I will modify the introduction of this article, I'll also make some changes to the rest. RobHar 20:50, 22 December 2006 (UTC)

No, that's not right. The algebraic closure of the rationals in the p-adic field is quite large. All those numbers are algebraic. There is no essential difference, abstractly, between those numbers and the algebraic closure of the rationals in the complex numbers. Charles Matthews 22:53, 22 December 2006 (UTC)

Ah I see, I thought it was being said that the p-adic numbers were algebraic, but what was pointed out was that certain p-adic numbers are algebraic. Actually, any field of characteristic zero contains the rationals, so perhaps I shall replace "complex number" in the definition with something more general. Something like an algebraic number is a root of a polynomial over the rationals, it is often taken in the complex numbers but can be in any field of charateristic zero. Does that sound good? RobHar 00:53, 23 December 2006 (UTC)

... An irrational number may or may not be algebraic. For example ... 31/3/2 (half the cube root of 3) are algebraic because they are the solutions of ... 8x3 − 3 = 0, ... Is this right? Not 2x3 - 3 = 0 ?

No the article states it correctly, since

x = (3/8)^1/3 = 3^1/3/2

--- Please, can some expert clarify "Numbers defined by radicals" Section? Specially the part about n>=5 ---Thanks, AlfC ---

Examles

In the examples section on this page, the statement is made that "an irrational number may or may not be algebraic". Examples of the positive are given, but no counter-examples. Could π be such a conter-example? I think it is, but for the life of me I cannot proove it.

I think a counter-example has to be given to clarify the statement, if someone has one (with proof) please add it. Thanks payxystaxna 15:42, 23 November 2006 (UTC)

If Infinity an algebraic number?

It would seem so from the article, as 1÷0 gives infinity. Worth spelling out? quota 09:50, 12 February 2007 (UTC)

Reread the definition: algebraic numbers are complex numbers. Infinity is not a complex number, so it can't be an algebraic number. --Zundark 13:01, 12 February 2007 (UTC)

A better answer perhaps is simply that infinity is not a root of a polynomial. It could be said to be a solution to 1/x=0 (for example in the context of the Riemann sphere), but 1/x is not a polynomial. RobHar 04:26, 3 May 2007 (UTC)

Contradiction between Algebraic numbers and Transcendental number

The article Transcendental number states that

In mathematics, a transcendental number is a real or complex number which is not algebraic, i.e., not a solution of a non-zero polynomial equation with integer coefficients.

whereas in the article Algebraic numbers the following is stated:

In mathematics, an algebraic number is a complex number that is an algebraic element over the rational numbers. In other words, an algebraic number is a root of a non-zero polynomial with rational (or equivalently, integer) coefficients.

Is it only integer coefficients or rational coefficients? Hakeem.gadi 09:27, 10 April 2007 (UTC)

It makes no difference, since you can change the rationals to integers without changing the roots of the polynomial (just multiply through by the lowest common denominator). --Zundark 10:04, 10 April 2007 (UTC)