And you want to do so, because you want to be able to do calculus.

you don't need real-line continuity from 3 to pi in order to do calculus. in fact, that statement makes no sense, because it's impossible to use undefinable numbers, so you couldn't use them in calculus even if you wanted to.

why don't you get all your real-line loving friends to join the current live game, it's not even gunboat, so you can all join up and beat me and take my points! i'd rather that then not play at all, we're having trouble recruiting. gameID=81853

than

lol, I'm doing math. ;-)
And yes, you do. Without a continuous real line, you can't prove the basic theorems of calculus. That's why it was derived. You need the numbers' existence, even if you can't name them.
Also, they are not undefineable. They are perfectly well defined. Every real number is the limit of a sequence of rationals.

oh thank god, i stumbled on the article i wanted. you assholes can all study this since you clearly don't know what it means for a set to be countable

" However, most real numbers are not definable: the set of all definable numbers is countably infinite (because the set of all logical formulas is) while the set of real numbers is uncountably infinite (see Cantor's diagonal argument). As a result, most real numbers have no description (in the same sense of "most" as 'most real numbers are not rational')."

http://en.wikipedia.org/wiki/Definable_real_number

Hey, I wasn't here last night. I didn't even read all that mess lol. But I'm perfectly aware of what it means for a set to be countable.

seems like you don't, since you think you need undefinable reals in order to do calculus.

Um, yes you do. You need a complete metric space for reasonable calculus. The definable numbers are not complete, as the page you linked pointed out.

A set is countable if and only if there exists a bijection between it and the natural numbers (or a subset thereof). Examples of countable sets include the integers, the rationals, and the algebraics. None of these, however, is analytically complete, nor is any other countable subset of the reals. You need analytic completeness to do calculus.

you don't know what you're talking about because you need even less than definable numbers for calculus. all you need are the computable numbers. but it would be beyond the scope of this thread for me to explain why.

if you took every definable number, and even added in the things you think are missing, which are number pointed to by functions but can't be explicitly defined, you add all that together, you still only have a countable set. which takes up 0% (exactly) of the real number line.

Well, it depends what you mean by calculus. The Mean Value Theorem, for example, requires a complete space. That means the fundamental theorem of calculus does, too. Not much calculus without those.

@bolshoi: "you don't know what you're talking about"

Yeah I know. And you can't do calculus with 0% of the real number line.

you know what? why don't you disprove me? give me an example of something in calculus that requires undefinable numbers.

presumably calculus involves taking numbers, then applying algorithms to them to get results. do you agree with that, at least? and that the algorithm has to be finite? and produces a result?

I already did, bolshoi.
And what I meant was that to PROVE the theorems in calculus, you need all the reals. Sure, you can define algorithms, but you can't prove they do a dang thing.

can you do integer arithmetic with 0% of the real line?

your proofs are based on algorithms dumbass

Statement: "your proofs are based on algorithms dumbass"
Response: "you don't know what you're talking about"

"The question naturally arises of whether it is possible to dispose of the full set of reals and use computable numbers for all of mathematics. This idea is appealing from a constructivist point of view, and has been pursued by what Bishop and Richman call the Russian school of constructive mathematics."

spoiler alert: you can

No, bolshoi, they're based on axioms and definitions, among which are a construction of the entire, analytically complete real line.

I recommend you read David Berlinski's "A Tour of the Calculus" for a fine exposition of the importance of the continuum to calculus.

i don't understand, do you know english? maybe you need to reread that post.

Source, bolshoi?

oh like you could not have guessed
http://en.wikipedia.org/wiki/Computable_number

http://en.wikipedia.org/wiki/Computable_number#Can_computable_numbers_be_used_instead_of_the_reals.3F

I see. I like how you added "Spoiler alert: the answer is yes" even though the article does not say that (in any form).

but that one is more of a debate. what's not a debate is that every number, every symbol, every variable ever imaginable forms a countable set.

Every symbol, yes. Every number? No. That's obviously what we're arguing about.
(Unless you mean _individually_ imaginable. In that case, it forms a FINITE set).
The point remains, you can't do calculus without the reals. (Your constructivist friends have an impressive list of partial results, but they work incredibly hard for them and end up with inferior results).