0.999~ = 1

[Durandal]

in real-world terms, it won't always equate to exactly 1 such as travelling at 0.99999 of c. You'll never hit lightspeed, but you'll get damn close and keep getting closer with diminishing returns, like halving a number indefinitely.

0.99999 isn't 0.999..., so I don't see where the contradiction is. It's impossible to reach 0.999...c because it's impossible to reach c. When traveling at such speeds, you're approaching the speed of light.

[Surlethe]

Addendum: I'm not sure in what sense this makes fractional representations are superior as compared to the decimal 0.999~ = 1. Sure, they allow exact representation of all rational numbers, but almost all numbers are not rational, and the fractional representations is no sense unique either, e.g., 1/2 = 2/4. The fact that decimal representations are sometimes non-unique seems to be a poor reason, although of course one can have other reasons depending on the task at hand.

It's a whole lot easier to conceptually grasp a number in a closed form, I think, than in an open form. People are also more likely to confuse and not understand decimal expansions (as Rye did) because of the human mind's inability to really comprehend what it means to "go off to infinity". That certainly doesn't make fractional representations superior, sure, but it would certainly make people more inclined to think in terms of fractions and try to skip over the more difficult non-terminating decimals.

[Darth Servo]

Don't any of you remember from grade school how to mathematically transform repeating decimals into a fractional equivalent? It works for turning 0.333333333333333... into 1/3, 0.5555555555555555... into 5/9, 0.142857142857142857... into 1/7, 0.0909090909090909090... into 1/11.

Try it with .99999999999999999999...

It yields '1'

[Rye]

Because as n increases without bound, this becomes vanishingly small, we can safely say that 0.999 ... /2 = 0.4999 ... = 0.5.

The simplistic reason for this equality is that the terminal 5 is not enough to change where the sequence seems to be headed: as 9s crowd up behind it, they overwhelm the five, until it is so insignificant it doesn't matter. That's what it means for the difference to go to zero as n becomes big (i.e., as you move out further and futher in your sequence): because the 5 is getting pushed out one decimal place each time, the difference between the two sequences decreases by an order of magnitude.

I understand that the smaller the 5 hypothetically goes, the smaller it is, and I guess, once it becomes literally infintesimally small (as in this situation), it "doesn't count" altogether.

That said, I still feel a sense of it dodging it and it still being unresolved to my sense of completion.

I feel like pointing out I could accept that it did equal 1, and subsequently deal with any problems relating to that sort of question (for instance, 1/9), just that conceptually, it doesn't sit well. Plus it seemed like a total ball ache to have infinite numbers moving around in an equation instead of a few numbers that would close everything off. The appreciation for fractions over representations of infinity was purely a personal one when dealing with numbers representable by fractions. It's far easier to deal with the idea of an orange with 9 perfect pieces, rather than 0.1 recurring of an orange.

[Wyrm]

I've read the algebraic proof; I also accept this proof, until something convinces me otherwise. I was wondering about the fraction proof, and by what means we know that 1/3 = .333~.

You mean the other fraction proof? Very easy!

Given that 0.333~ is the decimal representation of 1/3 (setting aside the question of whether 0.333~ = 1/3 applies), the proof is simple:

3 (0.333~) = 3 lim_{n→∞} ∑_{i=1}^n 3 (1/10)^i
= lim_{n→∞} 3 ∑_{i=1}^n 3 (1/10)^i
= lim_{n→∞} ∑_{i=1}^n 9 (1/10)^i
= 0.999~

Ie, the real number named by the decimal representation 0.999~ is three times the real number named by the decimal representation 0.333~. Then the proof is duck soup:

1 = 3 (1/3) = 3 (0.333~) = 0.999~

So much for the fraction proof, now how about the assertion 0.333~ = 1/3? This might not be the most elegant proof, but it's a proof.

The first thing to realize is that 0.333~ is the limit of the sequence S = (0.3, 0.33, 0.333, ...). A limit (in this context) is a real number... a real number characteristic to this sequence, S.

To say that the limit of S means that there exists a unique number, L, such that for every ε>0, there's an N large enough such that whenever n>N, |S_n - L| < ε, where S_n is the n'th member of the sequence S.

By definition, S_n has the decimal representation 0.333... (n times) ...3, and is equal to ∑_{i=1}^n 3 (1/10)^i. We shall prove now that for all n, S_n + 1/3 (1/10)^n = 1/3.

Base step: 1/3 = 10 1/3 (1/10) = (9+1)/3 (1/10) = (3 + 1/3) (1/10) = 3(1/10) + 1/3 (1/10)

Inductive step: Suppose S_n + 1/3 (1/10)^n = 1/3, then by the Base Step, S_n + 1/3 (1/10)^n = S_n + (3(1/10) + 1/3 (1/10)) (1/10)^n = (S_n + 3(1/10)^{n+1}) + 1/3 (1/10)^{n+1} = 1/3, and since S_n + 3(1/10)^{n+1} = S_{n+1}, and therefore if S_n + 1/3 (1/10)^n = 1/3, then so to does S_{n+1} + 1/3 (1/10)^{n+1}.

By induction, therefore, S_n + 1/3 (1/10)^n = 1/3.

Now for the magic: S_n + 1/3 (1/10)^n = 1/3 implies |S_n - 1/3| = 1/3 (1/10)^n. If n>N, then 1/3 (1/10)^n < 1/3 (1/10)^N, and therefore |S_n - 1/3| < 1/3 (1/10)^N. Given a ε>0, then if N > -log_10 3ε, then 1/3 (1/10)^N < ε. Therefore, for all ε>0, then there exists an N > -log_10 3ε such that if n>N, |S_n - 1/3| < ε. Therefore, 1/3 is the limit of S, and therefore is the real number named by the decimal representation 0.333~.

Kuroneko could probably do a much cooler job. Also, I haven't thought too deeply about the correctness of the proof, so I'm prepared for people to rip it to pieces.

[B5B7]

Hmm? If ones halves 0.9999~, one should get 0.4999~, not 0.4999~5. The latter isn't even a real number. And yes, 0.4999~ = 0.5.

I know that's meant to be the answer, and there are probably ways beyond me that show it, and I can accept that ultimately, yes, 0.999... is equal to 1, though I don't intuitively grasp it. The issue I'm having, though, is if you divide 0.9 by 2, you get 0.45, right? So any number in the 0.999... sequence will divide into a variant, 0.9999 / 2 = 0.49995 for instance, and no matter how I try, I can't intuitively see why that won't always be the case, even if we can never get there, the last number should be a 5, due to the way a 9 divides.

Rye - you're not dividing 9 by 2, you're dividing 19 by 2 [except for the initial 9], so any rules about dividing 9 do not apply.

[Kuroneko]

Rye: If you prefer, don't think of the 5 as becoming insignificant, but simply observe the fact that 0.999~/2 doesn't even have a 5 in its decimal expansion.

[Sriad]

Rye: If you prefer, don't think of the 5 as becoming insignificant, but simply observe the fact that 0.999~/2 doesn't even have a 5 in its decimal expansion.

Or that anywhere you could place the ...5, you can divide another 9 to add ...45. That's what it means for something to be infinitely repeating.

The logical proof is my personal favorite btw.

[Mad]

In that case, in what sense does an expression luike "0.4999~5", with infinitely many 9's before the 5, represent a real number? In other words, what would be the exponent of 1/10 corresponding to that digit?

Clearly, it's 10^-inf.

Actually, if we allowed for that, then I had come up with a "proof" a while back that all numbers are the same.

You see, if 1 and 0.999~ are equal (as has been proven), then that means that 1 - 0.000~1 is 0.999~, or 1. Since n + 0.000~1 would be the number adjacent to n, then we can infer that the number adjacent to any number is itself. Since numbers are continuous, then all numbers are adjacent to another number. Therefore, all numbers are equal.

Now that I've disproven math, I need to see my boss about a raise to a much higher equal amount.

(The flaw, of course, is that 0.000~1 would actually be 0 -- or 1/inf -- and 1-0=1, naturally.)

I understand that the smaller the 5 hypothetically goes, the smaller it is, and I guess, once it becomes literally infintesimally small (as in this situation), it "doesn't count" altogether.

That said, I still feel a sense of it dodging it and it still being unresolved to my sense of completion.

Let's assume that your notation of 0.4999~5 is possible. Now, let's try the case of 0.000~1. Since 0.1 is 1 / 10, 0.01 is 1 / 10^2, then we know that 0.000~1 would be 1 / 10^inf, or 1 / inf. Of course, 1 / inf = 0. Therefore, 0.000~1 = 0.

So, 0.4999~5 would be 0.4999~ - 5(0.000~1), or 0.4999~ - 0. Or 0.4999~. And 0.4999~ = 0.5 - 0.000~1 = 0.5.

Or, since we know that infinity has its own funny properties, such as inf + 5 = inf. It tends to "eat" other numbers. We can't compare (inf + 5) and inf, for example. So the 0.4999~ part "eats" the 0.000~5 would never be reached. The same way that "infinity plus one" will never be reached.

[Surlethe]

Aye, that's why we don't really treat infinities as numbers per se.

When we talk about infinities, we're talking about cardinalities of sets -- for example, |N| = inf_0, |R| = inf_1 (the fact there's a difference is given by the famous Cantor diagonal proof). So, to show that inf_0 + 1 = inf_0, you just have to show that there's a bijection f:NU{0} -> N. f(n) = n+1 is bijective (say f(a) = f(b). Then a+1 = b+1, and a=b. Given n in NU{0}, n goes to n+1, which is a postive integer). Since we have the intersection of N and {0} is empty, |NU{0}| = |N|+|{0}| = |N| + 1, and we have |N| + 1 = |N|.

In fact, one definition of an infinite set is as follows: given a set K, we say it has an infinite cardinality iff there exists a bijection from K into a proper subset I in K. For example, we can show the integers are infinite by considering <2> = {... -4, -2, 0, 2, 4, ...}. Let h:Z -> <2> be given by h(n) = 2n. Since h is a bijection (if anyone's up for the exercise, show it), we'll have, since <2> is a proper subset of Z, Z has infinite cardinality.

[Shadow WarChief]

I've read the algebraic proof; I also accept this proof, until something convinces me otherwise. I was wondering about the fraction proof, and by what means we know that 1/3 = .333~.

So if you accept .9999~=1

then .999~/3 = .333~
and
1 divided by 3 = 1/3


so .333~= 1/3

[Admiral Valdemar]

0.99999 isn't 0.999..., so I don't see where the contradiction is. It's impossible to reach 0.999...c because it's impossible to reach c. When traveling at such speeds, you're approaching the speed of light.

The way I saw it was that the 9 was recurring infinitely, but that this doesn't equate to a real-number like c would be, so you're simply getting closer to lightspeed and never reaching it.

I guess I simply misread the article, since I get the impression 0.9 recurring is assumed to be 1, but isn't actually 1. I never did like dealing with infinities.

[Surlethe]

The way I saw it was that the 9 was recurring infinitely, but that this doesn't equate to a real-number like c would be, so you're simply getting closer to lightspeed and never reaching it.

c = c*0.9999... . For the reason why, see below. The idea that you'll never reach c is true, but it's because of relativity, not inherent in the idea that 0.9999 ... < 1.

I guess I simply misread the article, since I get the impression 0.9 recurring is assumed to be 1, but isn't actually 1. I never did like dealing with infinities.

In the real numbers, IIRC a = b iff a-b = 0. Since a real number has a decimal representation (or, more generally, can be represented by any of an equivalence class of sequences of rational numbers), even if you can't evaluate a - b exactly, you can evaluate the decimal sequence of (a_n - b_n) out to the nth term. So, for example, while you can't evaluate 1 - 0.99999999 ..., you can evaluate 1 - 0.9, and 1 - 0.99, and 1 - 0.999, and 1 - 0.9999, and so on.

The idea, then, is that if you find two real numbers a and b such that, given any small number greater than zero, you can go out far enough into their decimals and take a difference which is less than that number, they're equal: as n goes to infinity, the difference goes to zero, and you conclude a - b = 0, so a = b.

That last paragraph is somewhat clunky to read, so this may illustrate it better. Take 1 = 1.00000 ... , and 0.99999 ... . If you give me any small number -- say, 0.000001 -- I can go out far enough into the decimal sequences of 1.0000 ... and 0.9999 ... and take the difference of them to make that difference less than 0.000001. So 1 - 0.9999999 = 0.0000001 < 0.000001. If you say, "Okay, how about finding a difference less than 0.0000000001?" I can just go out a bit further to 1.000000000 and 0.999999999999, and take the difference, and that will be less than 0.00000000001.

That's how we resolve whether or not two real numbers are equal: if we see indicators that the difference of their truncated decimal expansions goes to zero as we go further out into the decimals.

Did that make sense at all?

[Admiral Valdemar]

That made more sense, thanks. Pure theoretical maths is not my strong point, just the application of it to real-world things, so I've never had to properly deal with such brain ticklers.