Despite having often heard the piece of trivia, I had not actually previously been aware of what Goedel’s reasons for such a belief were (other than that they weren’t based on the Proper Forcing Axiom, of course. :)). Looking it up right now, however, it seems to be an interesting story including a surprising amount of error, flip-flopping, and stubborn double-flip-flopping, so to speak, as described here.
What is meant by “a finite number of steps”? I would not say it is standard to say “pi cannot be expressed algebraically”; the standard phrasing is only “pi is not an algebraic number (in the sense that it is not the root of a non-zero polynomial with integer coefficients)”.
I mean, I don’t think there’s anything fundamentally mistaken about using the phrasing “cannot be expressed algebraically” to mean “is not an algebraic number” or whatever, and it does seem like there are some Google hits for this sort of thing (although, one of the top hits for it is this very thread…). But it’s odd phrasing to my ears, and I’d generally not know how to interpret it until the speaker made clear what they meant by it.
There is something fundamentally misguided about considering “a finite number of steps” as any objective notion; it’s all relative to what you consider to be primitive steps. I could produce π in but a handful of steps, if I had 2π, 1, +, and / as primitives… I couldn’t produce sqrt(2) if I had only field operations as primitives; I could easily produce it if I had the square root operation in addition. And so on…
A finite number of algebraic steps involving integers, i.e. addition, subtraction, multplication, division and exponentiation. I.e. taking advantage of the field properties of the real numbers (any expression with an integer exponents can of course be expressed in terms of multplication) as opposed to it’s properties as an ordered field.
It is standard to say this, whilst semantics can be argued the meaning is still clear
shrugs. If you say it’s standard phrasing, fine. I am unaware of that standard. But whatever; despite taking it up, I don’t actually care about the quibble over phrasing.
Regardless, “can be generated by the field operations alone (in the real numbers)” is just a fancy way of saying “is rational”.
I have heard it many times. Arguably it’s not something that’s strictly defined though, but I think the meaning is still clear.
Yes of course the rationals are the field completion of the integers. But algebraic numbers can be represented by expressions using a finite number field operations (of course they don’t represent a completion themselves as clearly not all algebraic expressions represent real algebraic numbers). The real numbers are the Cauchy-completion of the rationals, but the alegbraic numbers themselves can be defined by an algebraic field extension of the rationals and does not rely on it’s properties as an ordered field.
Okay, these are my own observatiosn rather than something laid down in stone, but if something is described as algebriac expression, you would not expect to see a limit in it or any other operation that relied on an order relation.
(Let me clarify one thing: I concur that “algebraic” being used to mean either “related to field structure” or “related to equational logic” is standard (as in “The Fundamental Theorem of Algebra” or “essentially algebraic theory” or such things). It’s only the ambiguities of the context-less phrasing “can be expressed algebraically” (which you above described as “not something that’s strictly defined”) which I was originally commenting upon. But as I said, I don’t really care about the phrasing quibble, so that’s not what this post is about.)
It’s interesting you keep taking pains to note the distinction between field structure and ordered field structure. I’m just being nitpicking and pedantic in this post, but there is hopefully some amusement in it:
Suppose, working in the field of real numbers, I wanted to express the number sqrt(2), by which I mean the positive square root of two. How would I do so with an “algebraic expression” without making any use of the order structure?
Well, I suppose you’d say something like “The square of either of the two real fourth roots of 2” and skate by. So, sorry, let me rephrase. Suppose, in the context of the complex numbers (or any other algebraically closed field), you wanted to uniquely specify 1.414… (or any other particular algebraic irrational number). How would you do so using only the field structure?
Are there algebraic numbers that are not in any radical extension of the rationals? If so, it is rather misleading to describe the algebraic numbers as being those that are “generated algebraically” or “expressed algebraically”. I doubt that anyone who didn’t know about countable/uncountable sets would think that writing down a polynomial and saying “find its roots” was included in being generated/expressed algebraically. Of course, the set of numbers in all radical extensions of the rationals is also countable.
By “radical extensions of the rationals”, you mean numbers that can be expressed in terms of addition, subtraction, multiplication, division, nth roots, and integers? Like sqrt(2+sqrt(3/2)), or something? Then yes, there are algebraic numbers which cannot be expressed that way. There is no generally-applicable equivalent to the quadratic formula for polynomials of degree 5 or higher, so most 5th-order polynomials have roots that, though algebraic, can’t be expressed in that way.
Adding to Chronos’s answer, one particular example of a polynomial whose roots do not lie in a radical extension of the rationals is x[sup]5[/sup] - x - 1, as argued in the Wikipedia article on Galois theory (which was essentially invented to answer David Marcus’s question).
It’s worth noting that, in itself, the lack of a general “quintic formula” only immediately grants us that the generic polynomial a + bx + cx^2 + dx^3 + ex^4 + fx^5 over the field of rational functions in the variables a through f has no solution in a radical extension of this field. It does not (so far as I know) immediately follow that such an insoluble polynomial can be given on just the rationals, although this is of course true as well.
Actually, I read the Wikipedia article before posting my question. It wasn’t obvious to me that being not solvable is the same as the roots not being in any extension of the rationals. It seems that being not solvable means there is no formula for the roots that only involves the coefficients. This sounds like it is weaker than the roots not being in any extension. I confess to being ignorant of almost all of Galois theory. I’m a probabilist by training. Of course, I knew that quintics and higher can’t be solved in general.
If you are a little more specific about what kind of counterexample to the equivalence you are unable as of yet to rule out (i.e., what apparent mismatch causes you to feel the one notion is a priori weaker than the other), I could help clarify why the two notions are exactly the same.
The Abel-Ruffini Theorem from http://www.fact-index.com/a/ab/abel_ruffini_theorem.html: “The content of the theorem is that the solution of a higher-degree equation cannot always be expressed by starting with the coefficients and using only the operations of addition, subtraction, multiplication, division and extracting roots (radicals)”.
Without thinking about it much, I thought the “starting with the coefficients” clause was restrictive. But, we can take any nonzero coefficient and from it express any rational number. So, if the polynomial has rational coefficients and is not solvable, then some root is not in a radical extension of the rationals.
Indeed. It’s worth noting that the constant 1 is the nullary case of multiplication and the constant 0 is the nullary case of addition, and of course every rational can be built from these, so that one needn’t even build the rationals up from some coefficient, as such.
It is hard to tell from the imprecise version of the theorem that I found on the Web and quoted above if that is allowed. That’s the problem with imprecise versions. Thanks.
Yes, there are different infinities. The cardinality of the real numbers is larger than the cardinality of the rationals, for example. But, the cardinality of the rationals is the same as the cardinality of the natural numbers.
It’s sort of like there’s infinity-A, and infinity-B, and infinity-C, and so on, and these guys just proved that infinity-B and infinity-C are “the same,” when it had been thought they were different. It doesn’t say anything about whether infinity-J is the same as infinity-K.
Some of the details, including at least some very important ones, were lost in that article.
That would only mean that p is intermediate between aleph-0 and t, not that it’s intermediate between aleph-0 and c. You’d only get the former if you had also already proven that t ≤ c.
And if they had proven that t ≤ c, then they obviously wouldn’t be able to prove that t ≠ p, since that (as mentioned) would disprove the continuum hypothesis, and it’s already been shown that the continuum hypothesis is undecidable.
I’m guessing that the truth behind the article is that mathematicians expected the question of whether t = p to be undecidable, and that the surprise was just that it is decidable, not that it’s decidable in that particular direction (since being decidable in the other direction would be impossible).
I don’t know how to take this article - as far as I know, the idea that there is an infinity size between integers and reals is not provable or disprovable from the standard axioms of set theory, but the article does seem to be reporting something real - I just don’t know what.
ETA: Now I see that Chronos has answered my question, while I was composing my post. Thanks!