Talk:Algebraic number

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Mathematics rating: Start Class High Priority  Field: Number theory

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


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


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


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 ai 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)

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[edit] 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)

[edit] 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 = 31/3/2


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

[edit] 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)

[edit] 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)

[edit] 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)
Except it does make a difference, in the context of what's written on the page now. It says that if there exists a monic polynomial of degree 1 with x as a root, then that's equivalent to x being rational. That's true if the coefficients of the polynomial are rational, but obviously false if the coefficients are integer.24.91.135.162 23:03, 28 May 2007 (UTC)
Yes, the minimal polynomial is defined to be a polynomial over Q, not Z, since we want it to be monic. I've edited the article to make this clearer. There may be other places in the article where such clarification is needed - my comment above is only about the definition of algebraic numbers. --Zundark 08:59, 29 May 2007 (UTC)

[edit] Non-zero

I don't understand the qualification that it must be a non-zero polynomial. What is a zero polynomial? What would be (wrongly) included in the definition of an algebraic number if this qualifier were dropped? What does it exclude? --DavidConrad 19:48, 31 July 2007 (UTC)

The zero polynomial is 0. Every complex number is a zero of this polynomial. --Zundark 20:07, 31 July 2007 (UTC)

[edit] "Constant" or "real number"?

I changed "The real numbers π and e are not algebraic numbers" to "The constants π and e are not algebraic numbers" and was reverted by with summary "real numbers" is preferable to "constants", this is math, not physics.. The pi and e (mathematical constant) articles begin with "Pi or π is a mathematical constant and a transcendental (and therefore irrational) real number..." and "The mathematical constant e is the unique real number such that...", so I can see a case for both. Opinions? —METS501 (talk) 14:31, 3 August 2007 (UTC)

Hi, I'm the one that reverted it back to real number. I guess I read the change and was sort of surprised to read that. Perhaps the article on algebraic numbers is the one where I would least want pi and e to be called constants since the article discusses in some sense what numbers actually are. That the articles for pi and e refer to them as mathematical constants, I am definitely fine with since people will be looking up those definitions with all sorts of various fields in mind. However, I feel that in this case, pi and e are mentioned halfway through an article on what is definitely pure math (and where no functions are mentioned), and thus within the scope of this article I would rather refer to them as real numbers. Even more specifically, the section that they are mentioned in have a clear sentence pattern rational->irrational->complex->real that seems odd with constant instead of real. But yes I would like to see some discussion on this, as I actually don't know anything about what wikipedia thinks of the scope (and context) of referring to things. Sorry if my revert may have surprised you, I was quite happy with your cleanup, though. RobHar 19:09, 3 August 2007 (UTC)
I've a preference for the real number π, ... as it emphasise the fact that there exist real numbers which are not algebraic. --Salix alba (talk) 19:37, 3 August 2007 (UTC)
OK, no problem. —METS501 (talk) 20:19, 4 August 2007 (UTC)

[edit] Holding place to develop notion of "representation of numbers by decimals"

> Given an algebraic number a0xn + a1xn-1 + a2xn-2+ ... + an-1*x1 + an = 0

It is quite possible for certain ordered collections of coefficients (a0,a1,a2,...an-1,an) that an x satisfying the equality will be a positive whole number, and every coefficient will be less than x. For example:

7*x3 + 3*x2 + 9*x1 - 7390 = 0 is statisfied by x = 10, i.e. the collection of four coefficients (7, 3, 9, 5) is less than 10, and 10 satisfies the equation
(a) 7*103 + 3*102 + 9*101 + 5 - 7390+5 = 0

Any number whatever, given an unbounded sequence of digits (if required), can be expressed as an algebraic number with an integer part and a fractional part.

A rigorous treatment requires the notion of whole numbers (counting numbers, more accurately) as tally marks (such as would be generated by repeated appliction of the successor function), and a process of "division" as defined by e.g. the accumulation in place called Q of the number of successive subtractions of a divisor D until the numerator N (minuend) is less than or equal to the divisor (subtrahend) D; at this point what is left of the numerator N -D -D ... -D = R is defined as the remainder (or residue) and placed in a place called R:

  • Start: (given N, number to be divided, D the divisor, Q is the quotient, R is the residue)
  • 1 empty quotient Q ;i.e. 0 => Q
  • 2 IF N<D THEN step 6
  • 3 N-D => N
  • 4 increment Q
  • 5 goto step 2
  • 6 copy N to R
  • 7 HALT

This presupposes that "subtaction" and "less than" have defintions. But for this we will assume that they do (as primitive recursive functions.)


The first version is customarily written as: M -S*Q -R =0 the tally-count of S-subtractions is Q.

Example: 7395 tally marks in N, divisor is 10 ( |||||||||| tally marks, for instance): After 739 subtractions of 10 tally marks, Q will have 739 tally marks in it, and R will have 5 leftovers in it.

To do a full conversion of a number of unknown "size", the process must repeat until the quotient Q is less than the divisor D. This requires that we "index" our residues R0, R1, R2, ... Rn.

Per our example, the first round starts with 7395 counts, divides these into [Q] = 739 groups of 10 with a residue of 5, so R0 = 5.

This process can proceed again with the quotient 739, and again with 73, and again using 7, until at last the computation arrives at the quotient Q less than the divisor D: 7 < 10.

As before, start with 7395 tally marks in the place called "N" (numerator)

N = 7395 => Q
  • First division of N=173 by D=10 yields (Quotient, Remainder) = (Q, R) = (739, 5). Put "5" (5 tally marks) into place called "R0".
Move the quotient [Q] = 739 into N
  • Second division of N=739 by D=10 yields (Q, R) = (73, 9). Put "9" into place called R1.
Move the quotient [Q]=73 into N
  • Third division of N=73 by D=10 yields (Q, R) = (7, 3). Put "3" into place called R2.
Move the quotient [Q]=7 into N
  • Fourth division of N=7 by D=10 yields (Q, R) = (0, 7). The algorithm escapes the loop. Put "7" into place called R3
HALT

List as an ordered-tuple (R0,R1,R2,R3) = (5,9,3,7). These are the coefficients written backwards, i.e. we can rewrite as follows:

R3*103 + R2*102 + R1*101 + R0*100 - N = 0
a0*103 + a1*102 + a2*101 + a3*100 - N = 0

[?? doesn't look right]] This can be derived from the general form: Nn+1 - D*(Qn - Rn) - Rn+1 = 0, i.e.

Q0 = 0, and N1 - D*(Q0 - R0) - R1 = 0

Example:

  • 7 -0*10 -7 = 0;
  • 73 -10*(0*10 -7) -3 =0
  • 739 -10*(10*(0*10 -7) -3) -9 =0
  • 7395 -10*(10*(10*(0*10 -7) -3) -9) -5 =0
  • 7395 -10*10*10*7 -10*10*3 -10*9 -5 = 0 or 7395 - 103*7 - 102*3 -101*9 -5 = 0

>We reduce any finite number (i.e. one that we can write) to its "mantissa" (basically a string of digits and a decimal point) times its base" to a integer-power in either one of two ways:

1234567.890123456789 = .1234567890123456789 *107
1234567.890123456789 = 1234567890123456789. *10-12

> Theorem 135 p. 111 in Hardy and Wright to the effect that any positive number ξ can be expressed as "a decimal" that extends infinitely to the right. This number will have an "integer part" (customarily written to the left of a "decimal point") and a "fractional part" (customarily written to the right of the decimal point).

> Any postive, whole-number divisor D > 1, aka "base" can be used (even non-whole numbers could be used -- these represent "unit measures" repeatedly subtracted from "the pile" [??])

wvbailey Wvbailey 21:30, 16 October 2007 (UTC)

[edit] The new definition

I don't really like how the definition in the introductory paragraph has changed. Using the fuzzy term "number" is really not preferable. One can't say that numbers that aren't algebraic are transcendental, for example, is aleph_0 transcendental? Also, the writing is cumbersome, with all the parenthetical remarks. I intend on putting the old definition back, and editing many of the recent additions to make the writing style less cumbersome. RobHar 08:06, 14 October 2007 (UTC)

You're just messing with something that comes verbatim from a quoted source. What I recommend you do is try to improve what's there. Cf Hardy and Wright. I will just revert your reversion unless you provide sources. aleph_0 is not a number, it's a 2nd order concept. Bill Wvbailey 16:28, 14 October 2007 (UTC)
Hardy and Wright was written 70 years ago, it will thus phrase things in ways that aren't as good as more recent books. I simply intend on replacing the definition with what it was, and include a reference which actually has that definition verbatim, coincidently. My comment related to aleph_0 is about internal consistency, the wiki article on "Number" mentions all sorts of numbers, such as cardinals. The fact is, you came to this article and made major changes (to things that, in my opinion, didn't need them) without discussing the changes first. Thus the changes are being discussed afterwards. RobHar 19:06, 14 October 2007 (UTC)

RE: your perceived ownership of this article: Read Wikipedia:Be bold.

RE: lead paragraph: This is from the transcendental number article:

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

Bill Wvbailey 20:21, 16 October 2007 (UTC)

I in no way feel that I own this article, and I believe in the "be bold" credo, but if one is bold one must expect people to disagree. I've expressed my disagreement with the replacement of the term "complex number" with just the term "number". I could've easily just reverted your change, but because I do not feel entitled to run this article, I decided to post a discussion about it instead. Basically, your editing of the lead paragraph is essentially the following two actions: 1. replace "complex number" with "number", and 2. replace "with rational coefficients (or equivalently integer coefficients)" with "with integer coefficients". I disagree with these changes and strongly disagree with the first one. You have simply taken away information. Can you tell me what was wrong with the lead paragraph before? RobHar 04:43, 17 October 2007 (UTC)

[edit] Cantor etc.

I removed "The fact that most numbers, indeed almost all, are "transcendental" is proven by use of the Cantor diagonal method.[1]". "Almost all" in math means "all but a finite number", and clearly there are infinitely many rational numbers and infinitely many algebraic numbers. Moreover, Cantor's Diagonal Method proves that the set of the rational numbers is aleph 0, i.e. the same order as the set of the integers. It says nothing about irrational algebraic numbers, e.g. square roots. --Sky Diva 23:11, 1 December 2007 (UTC)

While the term almost all can be used in that sense, it is also frequently used in the sense of almost everywhere, that is, something is true for "almost all" cases if the exceptions form a set of measure zero.
Cantor's diagonal argument does not show that anything is countable. It proves that a set is uncountable. You may be thinking of the argument that the rationals are countable which works by arranging them into a square grid and following diagonals. In contrast, the diagonal argument shows that the set NN of sequences of natural numbers is uncountable by assuming for contradiction that the sequences can be enumerated as the list
α0, α1, α2, …
and constructing a sequence Δ = (αi + 1)i which differs from the ith sequence in the ith place (i.e., on the "diagonal" you'd see if you were to write out all these sequences). Since Δ differs from every αi, it cannot appear anywhere in the list, giving the desired contradiction.
I don't see anything that prevents the diagonal argument from being used to show that there are uncountably many transcendental numbers. After all, it can be used to show that there are uncountably many complex numbers, and it is straightforward to prove that the algebraic closure of the rationals is countable. Michael Slone (talk) 01:35, 2 December 2007 (UTC)
Firstly, the example that was removed was followed by an explicit reference. Looking up that reference, one can see the proof given that indeed uses Cantor diagonalization (in one of its standard uses that shows that the reals are uncountable; having previously shown the algebraic numbers are countable, the result follows). Almost all means for all but a set of measure zero; in the case of counting measure, this is indeed "all but finitely many", but for the standard Lebesgue measure on R, other "bigger" sets have measure zero (e.g. any countable set, or the Cantor set), and looking up almost all on wiki gives this definition. I'm undoing the edit. Thanks for your concern, but please be more careful next time. RobHar 05:13, 2 December 2007 (UTC)