Talk:Archimedean property

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Why can't 3c/4 be an infinitesimal? (I'm not saying it is...just that it's not obvious that every number less than an infinitesimal is infinitesimal, at least not obvious to those who don't work with infinitesimals...)

Okay, after a moment's thought, it IS obvious, but it still seems this fact should be mentioned (it's not even mentioned, although it's a key point in the proof).

How can anything be an infintesimal. Even the smallest number when summed will eventually exceed 1. Hmm...

As explained in the article, "The non-existence of nonzero infinitesimal real numbers follows from the least upper bound property of the real numbers". So you are absolutely right: There are no real infinitesimals (except 0). Nontheless there are other mathematical structures that contain infinitesimals. See for example Hyperreal number. --SirJective 15:47, 5 August 2005 (UTC)

To say that a small number x is classed as infinitesimal if the inequality

  |x|+...+|x| < 1

always holds, no matter how large the finite number of n terms in this sum, is absolutely rubbish. Why? If x is greater than zero, then there does exist a finite number of terms n such that the sum is equal to or exceeds 1. No such inequality is possible or even remotely realistic. Anonymous 5:30PM Sep. 2005

Again, anonymous, you're missing the point a little bit. For x in the real numbers, you're absolutely right, as the article states (nothing can be a real number and an infinitesimal at the same time). But it can be useful to think of systems where infinitesimals do take part, as in Nonstandard analysis. The idea is that infinitesimals are not the numbers you're familiar with, and do satisfy that inequality, similarly to how imaginary numbers are not real numbers, and do satisfy the inequality x^2 < 0. Andrew Rodland 09:45, 20 November 2005 (UTC)

Oh rubbish! Nonstandard analysis is even a bigger load of hogwash than standard analysis.71.248.131.74 02:55, 6 February 2006 (UTC)

You don't have to get angry with people when you don't understand the explanation. If you have ever attended even a reletively informal lecture on analysis, it would be clear why infinitesimals are an important concept. Take a math class, ask polite questions, or make a cogent counter-argument. Don't pawn your unwillingness to attempt an understanding outside your own little sphere off on other people.66.153.117.118 20:04, 26 February 2006 (UTC)

Here is a very simple example of infinitesimals in an ordered field, comprehensible to any secondary-school pupil. The field contains all rational functions with real coefficients. A rational function is a polynomial over another polyomial. To make this an ordered field, one must assign an ordering compatible with the addition and multiplication operations. Now f > g if and only if f − g > 0, so we only have to say which rational functions are considered positive. Write the rational function in the form of a polynomial plus a remainder over the denominator, where the degree of the remainder is less that the degree of the denominator. Call it positive if either (1) the leading coefficient of the polynomial part is positive, or (2) if polynomial part is zero, the leading coefficient of the remainder is positive (after, if necessary, multiplying the top and bottom of the fractional part by −1 to make the leading coefficient of the bottom part positive).
By this definition, the rational function 1/x is positive but less than the rational function 1. Now if you add
(1/x) + ... + (1/x)
with n terms in the sum, you get n/x, which is positive but still less than 1, no matter how big n is.
I leave it as an exercise to show that this ordering is compatible with the addition and multiplication operations, so that this is indeed an ordered field. Thus 1/x is an infinitesimal in this field.
So, Mr. or Ms. 71.248.131.74, you are mistaken. Michael Hardy 01:10, 27 February 2006 (UTC)

This is such a good example, Michael, I'm going to put it in the article! --Toby Bartels 01:12, 8 April 2006 (UTC)

I've always had a problem with the use of the term "rational function." It implies a function that maps rational numbers onto the rational numbers. The constant function f(x) = pi may be a "rational function, but it is hardly rational! Phrases like "polynomial ratio" or "polynomial quotient" are more self-explanitory.

If the nonarchimedean field of polynomial ratios is modified to define arbitrary relations among such functions to have the same truth as the corresponding relations for all but a finite number of natural values of the variable, the field becomes a model of a hyperreal number system, with all first-order statements about polynomial ratios having the same truth value as the corresponding statement about standard or real numbers. If one adds this coment to the example here, the same section could be added to the article on hyperreals. Alan R. Fisher 21:24, 22 November 2006 (UTC)

Contents

[edit] Problem with the definition

From the article:

If x is a positive number (or positive element of any ordered algebraic structure), then x is infinitesimal if there exists a positive element y such that, for every natural number n, the multiple nx is less than y. That is, the inequality
|x| + \cdots + |x| < y \!

Now assume that the structure in question is an ordered field with z such that 0 < z < 1/n for all natural numbers n (for example, in hyperreals there is such a z). Let x be a non-zero element of the field. Let y = |x|/z. Since z is infinitesimal, 1/z is infinitely large, and hence |x| + \cdots + |x| < y for any finite number of additions. Hence, all elements of the field are infinitesimal.

Is this really the intended definition? The definition seems to give the desired notion if one replaces y with 1.Punainen Nörtti 10:34, 9 June 2006 (UTC)

This is what we need as a definition of Archimedean (if that is to mean: there are no infinitesimals). However, you're right that this is incorrect as a definition of infinitesimal. The problem is that, without an absolute scale (provided by the multiplicative unit 1), there is no absolute meaning of infinitesimal (nor infinite). If you take y = 1, then you get the correct definition of infinitesimal; but this only works if there is such a thing as 1 in the algebraic system.

The real problem is that the idea "Archimedean" = "there are no infinitesimals" is wrong. It is true in fields, but false in rings, and meaningless in groups. (A ring with no infinitesimals is still non-Archimedean if it has infinite elements.) So I've fixed the definition to use only relative notions of infinitesimal (and infinite). But it might be better to rewrite the definition to avoid those terms entirely.

—Toby Bartels 22:54, 20 September 2006 (UTC)

[edit] Capitalized or not?

Should it be "Archimedean" or "archimedean"? Note that archimedean field uses the downcase version. Oleg Alexandrov 01:25, 13 Feb 2005 (UTC)

I'm going to post this to WP:MATH. - grubber 16:02, 18 April 2007 (UTC)

[edit] Clean-up

This article purported to be very general, speaking of arbitrary algebraic structures, but actually, it didn't encompass probably the most important case of non-Archimedean fields (which are usually not ordered!). I am not convinced that the ultimate generality is warranted, and I've begun the clean-up with a slightly more intuitive lead, and a narrower definition in the context of linearly ordered groups, where it can be made precise. I'd like to see natural examples of non-Archimedean objects not related to non-Archimedean fields. The example of the ordered field of of rational functions is nice, but why not just give an example of the ring of formal power series (make it ordered, if you like): it would bring out the same point without the extra complexity. If there are no objections, I would replace the example. Arcfrk 05:56, 13 May 2007 (UTC)

[edit] Archimedean property in constructive analysis

It is also interesting to note that the definition of infinitesimal isn't really constructive. Given positive number, there's is no terminating algorithm that can decide if the number is infinitesimal. Algorithm can only terminate if it finds big enough natural number n. Tlepp (talk) 13:03, 1 April 2008 (UTC)

That is fallacious. You're only proving that ONE PARTICULAR algorithm doesn't terminate, but you're claiming to have proved there IS NO ALGORITHM that terminates, i.e. you're claiming to have proved there is no OTHER algorithm besides that one. Obviously in the example of rational functionss over an ordered field, to claim that certain rational fuunctions are infinitesimal is completely constructive. Michael Hardy (talk) 13:13, 1 April 2008 (UTC)
I not proving nor claiming to have proved anything. I just made comment about the generic definition of infinitesimal in the most generic case of ordered group. I found it interesting that constructivist even care about Archimedean property of reals. Intuitively constructivism and infinitesimals seem to be far apart. Tlepp (talk) 14:19, 1 April 2008 (UTC)
You claimed there is no terminating algorithm. That is wrong. You said an algorithm can only terminate if it finds big enough natural number n. That is wrong. That would apply ONLY to an algorithm that proceeds just by computing nx for n = 1, 2, 3, ... or the like. The fact that that one sort of algorithm won't terminate if x is infinitesimal does not mean that NO OTHER algorithm will terminate. There's nothing at all non-contructive about the example involving rational functions over an ordered field. Michael Hardy (talk) 15:26, 1 April 2008 (UTC)