Timeline for answer to What are some counter-intuitive results in mathematics that involve only finite objects? by Daniel R. Collins
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| Feb 20, 2021 at 21:09 | history | edited | Daniel R. Collins | CC BY-SA 4.0 |
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| Feb 22, 2017 at 6:31 | comment | added | D. W. | @DanielMcLaury saying "the integers" and having that finite amount of information point to an infinite object is a bit of smoke and mirrors, imo. Not to get too philosophical/semantic, but it's technically an example of exformation en.m.wikipedia.org/wiki/Exformation. Just saying "the integers" would not represent an infinite thing if you didn't already have the knowledge on what the integers are, and the fact that they are infinite. So technically you still need an infinite amount of information to specify it, you just already have that information, and point to it with "the integers" | |
| Dec 8, 2016 at 11:04 | audit | First posts | |||
| Dec 8, 2016 at 14:59 | |||||
| Dec 7, 2016 at 13:31 | comment | added | Burak | @DanielMcLaury: It seems to me that in order to say "dimension is at most 4" you have to have some existential quantifiers (quantifying over real algebraic numbers) determining the dimension; and of course you have this universal quantifier at the beginning which is supposed to quantify over division rings over real algebraic numbers. Thus, regardless of whether you consider division rings over real algebraic numbers "finite objects" or not, it does not seem to satisfy the requirements... | |
| Dec 6, 2016 at 18:45 | comment | added | Daniel McLaury | @Levent: Just because a set is infinite doesn't mean that it takes an infinite amount of information to specify it. For instance, I can say "the integers" and I've uniquely specified an infinite set using a finite amount of information. Moreover, each real algebraic number can be specified with a finite amount of information (namely, its minimal polynomial together with something specifying which root we're talking about). This is completely different from the situation of the real numbers, where there is no way to specify each real number with a finite amount of information. | |
| Dec 6, 2016 at 18:40 | comment | added | Levent | @Daniel McLaury : It is a fact that any real closed field is infinite. Therefore any real closed field in fact contains 'infinite amount of information'. Also, it is very easy to generate very counter-intuitive facts on real closed fields since it is already very easy to generate counter-intuitive facts on any infinite set. | |
| Dec 5, 2016 at 14:47 | comment | added | Daniel McLaury | @Burak: you can replace "real numbers" with "real closed field," and e.g. the real algebraic numbers (of which there are countable many) are a real closed field. So this example doesn't really have anything to do with the real numbers per se, and in particular is not connected to how it takes an infinite amount of information to specify an arbitrary real number (which was the OP's objection to using the real numbers as examples). | |
| Dec 4, 2016 at 5:27 | comment | added | Burak | @Serge Seredenko: And how is that related to the third proposed example in this post? | |
| Dec 4, 2016 at 5:20 | comment | added | Serge Seredenko | @Burak Polynomials over rational numbers, as well as radicals over rational numbers, are finite constructions. They deal with countable algebraic numbers and do not need complete reals at all. OP allows countability in the question. | |
| Dec 4, 2016 at 5:12 | comment | added | Burak | I downvoted this answer since the second paragraph of the OP explicitly says it does not consider real numbers as finite objects. Therefore statements about division rings over the reals, whether they are finite dimensional or not, is not covered in this question. | |
| Dec 4, 2016 at 2:20 | comment | added | Eric Towers | @IwillnotexistIdonotexist : Not all algebraic numbers can be expressed using algebraic operations (addition, subtraction, multiplication, division, and positive integral roots) applied to rational numbers (or integers, since division makes the distinction not a difference). Only algebraic roots whose minimal polynomial has a solvable Galois group can be (i.e., only those that live in a tower of simple algebraic field extensions of $\Bbb{Q}$) "Most" fifth and higher degree polynomials do not have solvable Galois groups, so cannot be expressed with algebraic operations on rationals. | |
| Dec 4, 2016 at 1:49 | comment | added | Iwillnotexist Idonotexist | @SergeSeredenko Euuh... care to elaborate? That seems impossible, per the definition of an algebraic number: Any complex number that is a root of a non-zero polynomial in one variable with rational coefficients. | |
| S Dec 3, 2016 at 20:57 | history | suggested | Oliphaunt | CC BY-SA 3.0 |
spelling; specify *regular* convex polytopes
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| Dec 3, 2016 at 20:49 | review | Suggested edits | |||
| S Dec 3, 2016 at 20:57 | |||||
| Dec 3, 2016 at 16:48 | comment | added | Kimball | For polynomials you, might need to add "solutions by radicals." There might be closed form solutions for degree 5 in terms of transcendental functions, say. Also, I would not consider division algebras over the reals "finite objects." | |
| Dec 3, 2016 at 8:28 | comment | added | polfosol | This might be interesting as well: en.wikipedia.org/wiki/Exotic_R4 | |
| Dec 3, 2016 at 5:53 | comment | added | Serge Seredenko | I think the fact that some roots of some degree 5+ polynomials cannot be expressed with algebraic operations is even more interesting. | |
| Dec 3, 2016 at 4:02 | history | edited | Daniel R. Collins | CC BY-SA 3.0 |
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| Dec 3, 2016 at 3:55 | history | edited | Daniel R. Collins | CC BY-SA 3.0 |
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| Dec 3, 2016 at 3:25 | history | edited | Daniel R. Collins | CC BY-SA 3.0 |
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| Dec 3, 2016 at 0:07 | history | answered | Daniel R. Collins | CC BY-SA 3.0 |