Talk:Ostrowski's theorem

The definition of equivalence of absolute values as given here doesn't seem to go very well with the definition given on Norm_(mathematics). Ncik 00:49, 13 December 2005 (UTC)


 * I think that's because they're not the same thing. A norm is a function on a VS, and it depends on an absolute value on the field for it's definition. Sometimes one has a field extension L/K and one has an absolute value (A) on K and an absolute value (B) on L which agrees with (A) where they are both defined, i.e. on K. Then (B) will certainly be a norm on L, considered as a vector space over (K,(A)). But in general they are not comparable, and I think that's why the equivalence relation on norms doesn't look like the equivalence relation on absolute values. Please correct me if I'm wrong. Owen Jones 12:29, 2 April 2006 (UTC)

Since the only field extensions of the real numbers are the real numbers and the complex numbers - this is wrong, e.g., R(T). Also, the theorem that the only complete Archimedean valued fields are R and C cannot easily be deduced from the theorem here, but is an interesting theorem in its own right.--128.176.188.101 16:04, 27 June 2007 (UTC)

Topological isomorphism
I removed a sentence This is stronger than saying that the two absolute-value structures are topologically isomorphic. It seems unclear exactly what this is supposed to mean. Are there in fact valued fields that are homeomorphic but for which the valuations are inequivalent? Ostrowki's theorem implies that this is not in fact the case for Q. In any event, a reference to a reliable source is needed. Deltahedron (talk) 08:05, 21 August 2012 (UTC)

|a|∞ or |a|∗?
wiped out. $|a|_{∞}$ is a perfectly standard notation, and Wikipedia does not want various originalities with asterisks. Especially if such originalities embed mathmode into section headers that leads to glitches. Especially if one does so because is not skilled enough to make subscripts with HTML. Incnis Mrsi (talk) 15:11, 11 February 2014 (UTC)


 * You have not addressed why you believe what I changed is not a mistake in the logical progression of the proof - did you even check? Notwithstanding that, let me justify them in turn:


 * 1) To change |a|∞ to |a|∗∗ at the end of the first paragraph: this is a mistake - read the logical progression here, we are trying to justify only considering integers - facts about the real absolute value do not come into it at this section.
 * 2) In the bolded section, I changed |n| > 1 to |n|∗ > 1: again, this is a mistake, |n| doesn't mean anything, just read through the proof! The absolute values that we are considering in the proof are |n|∞,|n|∗,|n|∗∗.
 * 3) I changed to latex in the bolded part: this was a typographical thing, I don't really care if you revert this, but if you are going to revert it at least make it accurate - i.e. change the |n| > 1 to |n|∗>1.


 * RubberTyres (talk) 16:25, 11 February 2014 (UTC)
 * Yes, realized you merely tried to uphold a consistency of notation. My [rollback] button pushing was silly. Are you satisfied with ? In any case  have one more question. Incnis Mrsi (talk) 17:33, 11 February 2014 (UTC)
 * also made several unrelated tweaks. For example, strongly disliked “normal absolute value” and replaced it with “standard absolute value”. Does anybody object? Incnis Mrsi (talk) 17:46, 11 February 2014 (UTC)
 * That seems fine to me. RubberTyres (talk) 17:51, 11 February 2014 (UTC)

|n|∗ and |n|∗∗
fail to understand For if we find some $c ∈ R^{+}$ for which $|n|_{∗} = |n|^{c}_{∗∗}$ for all naturals greater than one; then this relation… Is “for if” a correct grammar? Why should we find certain $c$? How this long exercise with $c$th power of non-negative rational numbers is related to further proof? Incnis Mrsi (talk) 17:33, 11 February 2014 (UTC)


 * We should find a certain c, because that is the definition of equivalent absolute values. That is, an absolute value |.|∗ is equivalent to an absolute value |.|∗∗ if and only if there exists a positive real number c such that $|n|_{∗} = |n|^{c}_{∗∗}$ for all values of n. This `long exercise' means that we may assume that we are proving the theorem over the positive natural numbers greater than 1, even though the theorem is a statement about all rational numbers - this exercise shows us that if we prove it for positive natural numbers greater than 1, then it also holds over the rational numbers - in effect, it simplifies the proof significantly. I can't comment about the for if, I was never very good when it came to grammar, I think it might be okay though. RubberTyres (talk) 17:51, 11 February 2014 (UTC)

This is a wrong proof
In the proof, author takes b such that |b| > 1 so that b now is not arbitrary. Indeed, all b > 1, |b| > 1 but you must prove that. With that fact, b becomes arbitrary. Ref:

http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/ostrowskiQ.pdf — Preceding unsigned comment added by 61.28.208.253 (talk) 03:59, 5 November 2018 (UTC)


 * I've tried to fix the existing symmetry argument along these lines, but don't have a reliable source to pin it to. I don't think the PDF is enough, though it adds insight (for me anyway) where it says that the theorem explains why p-adic numbers are somehow natural and it highlights the role of $b^{n}$ in the proof. These may be worth adding as part of the article. NeilOnWiki (talk) 12:57, 26 April 2021 (UTC)

Quesstion about a proof
Perhaps I'm incredibly stupid, but in the first case of the proof where does "Each term $$\left| c_i b^i\right|_*$$ is smaller than $$ (b-1)\left|b\right|^i_*$$ (by the multiplicativity property and the triangle inequality)" come from? Because from $$ c_i \le b-1 $$ in general it doesn't follow that $$ \left|c_i\right|_* \le \left|b-1 \right|_* $$, and much less $$ \left|c_i\right|_* \le b-1 $$ Serpens 2 (talk) 17:05, 15 November 2023 (UTC)


 * This tripped me up at first also -- the multiplicative property gives $$|c_i b^i|_* = |c_i|_* |b|_*^i$$, then using the fact that $$c_i$$ is a digit, write $$c_i = 1 + 1 + \cdots + 1$$ so $$|c_i| \le |1| + |1| + \cdots + |1| = c_i \le b - 1.$$ Caleb Stanford (talk) 04:46, 16 November 2023 (UTC)
 * My bad. I won't delete it in case anyone else would stumble upon the same question. Thanks! Serpens 2 (talk) 07:50, 16 November 2023 (UTC)
 * Not your bad! It was a good idea to clarify & very helpful to point out. I fixed it in the article. Caleb Stanford (talk) 17:09, 16 November 2023 (UTC)