[NAFV_P]

From wikipedia on algebraic numbers:

> "In mathematics, an algebraic number is a number that is a root of a non-zero polynomial in one variable with rational coefficients (or equivalently—by clearing denominators—with integer coefficients)."

I almost forgot, the set of algebraic numbers also includes complex numbers.

[xyzzyz]

If you mean that some complex numbers are algebraic, then yes -- the imaginary unit i itself is algebraic, as it satisfies an polynomial equation x^2 + 1 = 0 with rational (in fact integer) coefficients. The set of algebraic numbers does not include all complex numbers, though.

[surement]

Fun fact: the set of algebraic numbers is countable. If you recall that the set of rational numbers is also countable, then you get that the reals are uncountable only "because" of transcendental numbers (pi, e, Chapernowne's number, etc.).

[pash]

As Alonzo Church and Alan Turing showed, the computable numbers [0] are countable too. The computable numbers include all the algebraic numbers and some transcendental numbers (including π and e), so the reals are uncountable "because" of those other transcendentals, the uncomputable numbers. Put differently, almost all reals are uncomputable.

0. https://en.wikipedia.org/wiki/Computable_number

[NAFV_P]

Like Omega:

http://en.wikipedia.org/wiki/Chaitin's_constant

[GregBuchholz]

Can't mention Omega without linking to:

Meta Math!

http://arxiv.org/abs/math/0404335

[NAFV_P]

Good book, I read it a few years ago.

His own proof that there are infinitely many primes is a good head spinner.

[Ind007]

Such a good read.Thanks for sharing this.

[JonnieCache]

Check him out on youtube too, he's a fun speaker.

[jsbgir]

This is really interesting. We could take it further and say that, given that some uncomputable reals have a finite definition (e.g. "the probability that a random algorithm halts"), there is a countable number of definable reals (by assigning a Godel number to each definition), so the uncountability of the reals is strictly due to indefinable numbers!

[surement]

I've only ever heard of one uncomputable number (Chaitin's constant), so this is rather mind-blowing.

[nardi]

Actually, because of irrational numbers. There are uncountably infinite irrational numbers in between any pair of rational numbers.

[xyzzyz]

There are also infinitely many rationals between any pair of irrational numbers, but that doesn't make rationals uncountable.

[nardi]

I said uncountably infinite. The parent seemed to be under the impression that a few specific transcendental numbers were the reason that the reals were uncountable. I was pointing out that there are uncountably infinite irrational numbers between any pair of rational numbers. (There being more irrational numbers than real transcendental numbers, as not all irrational numbers are transcendental.)

[ronaldx]

You seem to have missed the point.

You are saying rational numbers are countable but irrational numbers are not countable (uncountably infinite)

This is true.

The parent is saying that algebraic numbers are countable and transcendental numbers are not countable.

The parent's statement is a level deeper and more remarkable. Another way to say this, is that "almost all" numbers are transcendental.

The parent was not suggesting that his short list of examples represented the extent of transcendental numbers: despite almost all numbers being transcendental, they are individually hard for us to define, since the set is normally defined by excluding sets of numbers that we can easily define.

To clarify: transcendental numbers are a subset/a type of irrational numbers. https://en.wikipedia.org/wiki/Transcendental_number If you take out transcendental numbers from the irrationals, those that remain are algebraic and therefore countable. You got it :)

[nardi]

The parent says, and I quote: "...then you get that the reals are uncountable only 'because' of transcendental numbers."

But this is untrue, is it not? Because even if you removed all of the real transcendentals, the real numbers would still be uncountable because of the irrationals.

(Edit: Or are the non-transcendental irrationals countable? I can see how that might be true?)

[auvrw]

more generally, from Lang:

> Let F be a subfield of a field E. An element \alpha of E is said to be {algebraic} over F if there exist elements a_0, ..., a_n (n >= 1) of F, not all equal to 0, such that a_0 + a_1\alpha^n + ... + a_n\alpha^n = 0.

point being, even a sub-par undergrad knows how to generalize algebraic numbers using some machinery that was just beginning to be built in Ramanujan's time (and of course was completely unavailable to the man himself), and now these guys have plugged fields into something that i'd only heard of in the context of group theory and proved something useful. way to go. but at the same time, if Ramanujan saw this, i have to imagine he'd be thinking something along the lines of, "the game ain't the same." (not that it's a game.)