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For those who are interested in connections to more advanced mathematics, there is a sense in which √2 is still an integer, even though it is irrational. Specif
by kevinventullo 2y ago
For those who are interested in connections to more advanced mathematics, there is a sense in which √2 is still an integer, even though it is irrational. Specifically there is the notion of “algebraic integers”, which are the set of all complex numbers expressible as the root of a monic polynomial:
x^n + a_{n-1}x^(n-1) + … + a_1x + a_0.
Here each a_i is a usual integer in ℤ, and monic refers to the leading coefficient being equal to 1.
It turns out the set of such roots is actually closed under multiplication, addition, and subtraction, and there is even an analogue of prime factorization if you squint. Moreover, the intersection of these “algebraic integers” and the rational numbers ℚ are exactly the usual integers ℤ. This is why you sometimes might hear an algebraic number theorist refer to ℤ as the set of “rational integers”.
- gerdesj 2y agoMaths is always a bit boggling. You say that root two can be considered an integer despite being irrational. What does "usual integer" mean?
- kevinventullo 2y agoBy “usual integer”, I mean what people usually refer to as an integer: …, -2, -1, 0, 1, 2, … As opposed to “algebraic integer”, which is a more general notion.
- p-e-w 2y agoIf I wasn't familiar with that concept already, then I would probably assume that math is no more rigorous than psychology after encountering this thread. The exact disciplines are doing themselves a terrible disservice by muddying up established terminology like this (and "algebraic integers" are far from the only such case).
- RGamma 2y agoAh yes, good ol' clopen sets.
- mananaysiempre 2y agoThe red herring principle[1] is unfortunately popular enough in mathematical terminology to have a name and a page about it. Roughly, a fooish bar will frequently be something like a bar except fooish, so not actually a bar. (Algebraic integer, multivalued function, manifold with boundary, etc.) On the other hand, a nonfooish baz when baz is normally fooish often means a not necessarily fooish baz, so a particular one might be fooish but we can’t assume that. (Noncommutative ring, nonassociative algebra, the very field of noncommutative geometry, etc.) [1] https://ncatlab.org/nlab/show/red+herring+principle https://ncatlab.org/nlab/show/red+herring+principle
- Someone 2y ago> The exact disciplines are doing themselves a terrible disservice by muddying up established terminology like this So, what term do propose they use for this? “Hutyfreklop” or “gensym_167336871904” would probably be unique, but wouldn’t tell anything about the subject itself, “roots of polynomials with integer coefficients and a leading coefficient of one” would get cumbersome soon.
- amelius 2y agoYeah, you might as well call any countable set "integers" because you can find a 1:1 mapping between them. This is silly.
- gjm11 2y agoI don't think that's fair. Mathematicians pretty reliably say "algebraic integer" or "integer in [some specific class of numbers that has non-rational integers in it]" when they are talking about the broader notion, and if they're doing something where that broader notion is often relevant they will generally say something like "rational integer" when they mean the narrower notion. So in practice there is seldom any confusion. And algebraic integers really _are_ like ordinary integers in important ways. Inventing a completely new term would not obviously be an improvement. It's not like this sort of thing is unique to mathematics. Once upon a time a "language" was a thing human beings used to communicate with one another. Then along came "programming languages" which are not languages in that sense. And then things like "hypertext markup language" which isn't a language in the programming sense either. (Arguably this is partly mathematicians' fault since I think they were the first to use "language" to refer to purely formal constructs. But I think the use of "language" in computing arose mostly by analogy to human languages.) And it happens plenty outside "the exact disciplines". A republican is someone who favours a mode of government that doesn't have monarchs, but if you call someone a "Republican" in the US you mean something rather more specific and a few "Republicans" would actually quite like a system hard to distinguish from monarchy. A window is a transparent thing placed in a wall to let light in, but a window of opportunity is something quite different. A czar is the absolute ruler of Russia, but when someone says (rightly or wrongly) that Kamala Harris was "border czar" they don't mean that. A star is a gigantic ball of stuff undergoing nuclear fusion and producing unimaginable amounts of energy, but even the most impressive rock stars don't do that, and some people called "rock stars" have never played or sung a note of rock music in their lives.
- skhunted 2y agoTerminology is often times used to encapsulate a lot of information in a single word or phrase. It’s sort of a compression of information to facilitate communication. Things like “roots of f” is a shorter way to say that: “the set of all x such that f(x) = 0”. As you get deeper into a subject the more terminology you encounter. This is why research papers are generally unintelligible to those with no training in the areas that the research is about. To not use terminology would make papers insanely long and far too tedious to read.
- mathgradthrow 2y agoThis seems a bit harsh. Mathematicians tend to be mostly unambiguous when they write. This isn't even on the same planet as the empirical disciplines.
- gowld 2y agoMathematicians (with books in hand) tend to have the ability to disambiguate when confusion arises. But they rarely do, instead relying on context and the intelligence of the reader to derive meaning.
- empath75 2y agohttps://math.stackexchange.com/questions/778004/why-study-integrality https://math.stackexchange.com/questions/778004/why-study-in... There's a good explanation of the motivation for this concept, here.
- cyberax 2y agoPerhaps a translation issue? I've always referred to that set as "algebraic numbers" ( https://en.wikipedia.org/wiki/Algebraic_number https://en.wikipedia.org/wiki/Algebraic_number ). Since they are equipotent with integers, you _can_ call them that, but it's misleading.
- LegionMammal978 2y agoIn general, won't some algebraic numbers' minimal polynomials have a leading coefficient greater than 1, when written with integer coefficients?
- adrian_b 2y agoAs already mentioned by another poster, algebraic numbers are more general than algebraic integers, because the leading coefficient of the polynomial does not have to be one, similarly to the difference between rational numbers and integer numbers, where for the former the denominator does not have to be one, like for the latter.
- Xerox9213 2y agoAhh, that would explain why the intersection of algebraic integers and Q is Z. I wasn’t convinced of that when I had the notion of algebraic numbers in place of algebraic integers. I like teaching this kind of stuff to my grade 9 and 10 advanced math classes. It’s not that hard to understand and yet it gives students a sense of wonder about how math works. I might try to show the grade 10s algebraic integers now.
- kevinventullo 2y agoYou sound like an amazing teacher :)
- bmacho 2y agoNo, algebraic integers are a different set than algebraic numbers. (A subset.) Algebraic integers are much cooler, since there is a number theory on them: https://en.wikipedia.org/wiki/Algebraic_integer https://en.wikipedia.org/wiki/Algebraic_integer . (And also because the most basic facts about it, like that it forms a ring, are not trivial to prove, that's a good sign for a concept to be cool and useful.) These two number sets are more or less in a relationship like regular integers (with a number theory), and rational numbers. In fact A = O/Z where A denotes the set of algebraic numbers, O denotes the set of algebraic integers, and Z denotes the set of integers.
- jl6 2y agoYes, but did God make the algebraic integers? Because this looks suspiciously like the work of man.
- skhunted 2y agoThis depends on your philosophy of mathematics. If you believe god gave us the nonnegative integers then in a very natural way one is led to the notion of algebraic integers. Whether this would be god’s creation or man’s creation depends on your view of who gets the credit in such a situation.
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- rodric 2y ago> the set of such roots is actually closed under multiplication, addition, and subtraction, and there is even an analogue of prime factorization if you squint I did a maths undergrad, but I don’t think I ever studied algebraic integers. That’s something I shall have to remedy now, thanks!
- dhosek 2y agoIf you took abstract algebra (which presumably you did as a math major), you certainly encountered these at least in the exercises as groups of the form ax + b where x is some irrational number (or imaginary) and a and b are integers are a staple of chapter 1–2 proofs. Gaussian integers (ai + b) are a special case that are loads of fun to play with it. They are not unique factorization domains like the integers (e.g., 5 can be expressed as both 1∙5 and (1 - 2i)² where 1, 5 and 1 - 2i are all irreducible).
- rodric 2y agoYou are right, of course. Clearly, I have been away from mathematics too long.
- kevinventullo 2y agoNit: while it is not generally the case that rings of algebraic integers must be unique factorization domains, it is the case for Gaussian integers! In your example, 5 is uniquely factorizable up to units as (1-2i)(1+2i).
- gowld 2y agoIndeed, the integers have the same limitation -- factorization is unique only up to units. 1 = -1 * -1 In elementary mathematics, people wave away "-1" by saying silly things like "positive integers", before Gaussian integers arrive and force us to figure out precisely what we are trying to say without silly ideas from analysis like "ordering". :-)