Sorry fo the inconvenience. Sometimes, I forget others have different online experience. I have all sorts of ad blocks installed so the only thing I see there is the logo and the link, which is a design I like.

Ok, I think I have devised a formal proof to 0.999... = 1.

It is obvious that 0.999...(where there are k '9's) = 1 - 10^(-k)
For a given epsilon (e1), which is 1- (1-10^(-k)) = 10^(-k), I can choose an integer k+1.
Then e2 = 10^(-k-1) < 10^(-k).
Since there is always a k+1 for every k, by the formal definition of limits, lim (k -> inf, 1-10^(-k)) -> 1.
Since 0.999.... has an infinite no. of 9s, k -> inf, so 0.999... = 1.

If you do not believe in the formal definition of limits, which underscores all of calculus, then unfortunately I cannot do anything.
It is a bit hard to grasp the fact that the little difference of 10^(-k) is equal to 0 when k is infinitely large, but you have to get that in order for any differentiation or integration to make sense.

likewise, can't seem to use the link....

ah-ha... what a terrible experience. Anyway, onto the data
(also @ let's vote: no, I refuse to take part in any legitimizing of the 'democratic maths' movement)

I prefer googlefusion tables to be honest...

i think zultar's proof at the beginning is simpler. mainly that 10*(0.99...)-(0.99...)=9. i don't know what you mean by formal, but it seemed legit to me.

i just don't like the idea of thinking of 0.99... as a number, since the number is 1. it's like -0. i don't like -0 as a number either. but i guess this debate was even more drawn out than the global warming one i was having. so i'd better stop talking about it now.

bolshoi, think of it (AND 1) as _representing_ a number, and they both represent the same number.

This issue is actually kind of annoying in certain proofs -- it's irritating to have two different ways of writing some numbers.

Or you can dodge the whole thing by using nonstandard analysis, as somebody pointed out earlier. In some forms of that, 0.9999.... is not a number. But then you have to live with infinitesimals, which for my money, are even weirder than this.

Well, that proof is not formal, in the sense that 0.999.... is not actually rigorously defined, and you cannot just subtract two numbers that are not well defined.

yeah maybe you're right.... really you're subtracting an infinite sum. and i guess i'm taking for granted that subtraction works on infinite sums like it does on typical numbers... maybe.

i don't know, i just don't like real numbers. this whole infinite-digit construction. i mean if you take all rational numbers and all (that exist in our universe) irrational numbers, add them together, you still have a countably infinite set. so the real number line is essentially all filled with numbers that do not and cannot exist in our universe.

My eyes... my poor eyes... they are bleeding....

... sorry, i meant to follow that up with a decent link to the table of imported data... but i can't focus on this right now.

actually if our universe is finite, then the set of integers doesn't even really fully exist. haha... but it feels like they are closer to existing, to me, than the real line.

Well, not all real numbers occur in nature or the universe, but it is useful in mathematics to have a set of such numbers which cover the whole spectrum from negative infinity to positive infinity, and can describe anything within.
The Greeks of Pythagoras' time did not have the concept of irrational numbers. It was impossible for them to express the length of the diagonal of a unit square.
The concept of complex numbers is relatively new. Mathematicians wanted to be able to use the square root of a negative number in calculations (specifically the formula for cubic roots).
Similarly, we do not want to run into a scenario where there is a number we want to use, cannot describe. That is why the real number set exists. For example, pi and e, two very fundamental constants in mathematics, are real numbers but not rational.

By the way, irrational + rational = real :p

'and i guess i'm taking for granted that subtraction works on infinite sums like it does on typical numbers... maybe.'

yes, i'm guessing that you are right, but you don't have a formal proof that it is ok to take this for granted....

That's what two years of college maths made me want to not know.

also, do any numbers exist in the 'real' universe? i mean, they are all just mental constructs...

what is the 1 anyway?

No, bolshoi is incorrect in this matter. You cannot actually do 9.99....-0.999.... = 9.
The correct way I have demonstrated (hopefully correctly).

i think there was nothing wrong with that proof, it's just that he didn't lay out the formal definitions of his subtraction/etc and why it works the way it does.

it's like saying someone adding complex numbers to prove something is making a mistake because you can't just add those things together. if people know how the operation works for the type of number, then there shouldn't be a problem.

The subtraction does not work as such.
0.999... is not a clear definition, there is no number 'infinity' in mathematics so you cannot say 0.999 = a number where there are infinite number of 9's after the leading 0.
Which means you have to work with limits, say 0.999... where the number of '9's is k, and k -> infinity. But if you are working with limits in this way, then obviously 9.999.. has the same number of '9's as the 0.999... so the last 9 in 0.999 is not cancelled out.

"By the way, irrational + rational = real :p"

yes, but i would counter that countable set + countable set = countable set. so if you're talking about the size of the sets, you're never going to get to an uncountable set by adding and multiplying your countable sets together.

the 9.99... - 0.999... is a (more) intuitive way to prove the answer, but it is not a formal proof.

The irrational set is uncountable. There is a second-order infinity of rational numbers, equal to 2 array of all the integer combinations.
The rational set is uncountable. There is a first-order infinity of rational numbers, equal to the number of integers.

And since the no. of integers is infinite (there is always an integer k + 1 greater than k, and there is always an integer k-1 smaller than k, so there are as many values for k as you arbitrarily decide), no set you mentioned is countable in the sense that it is finite.

uhm. what the hell is second-order infinity? i hate to break it to you, but there are no more rational numbers than integers, according to modern math theory. but anyway, this is what i meant by countable http://en.wikipedia.org/wiki/Countable_set

you probably missed it, but a really long time ago in this thread someone else didn't understand what i meant by countable, so i tried to clarify it. i think i'm just using a bad choice of words. anyway who cares, this isn't even math anymore it's philosophy, i think.

countable infinite as opposed to uncountable infinite...

OK I see what you mean now, but there is no problem with rational numbers being a subset of real numbers.

mathematics is mathematics. mathematics is not philosophy. in mathematics there is one correct answer, not multiple. You are either correct or wrong.

i don't know if it has to be that black and white, but i would be satisfied if we used the term "real" to refer to all possible combinations between rational and irrational numbers. and even that they can be thought of as having infinite precision, to an extent. but just not that the number line has every possible combination of infinite digits... cause that's describing a number line that has no basis in reality.