How exactly does 0.999~ = 1?

Hurkyl

The sequence:

0, 1/2, 2/3, 3/4, 4/5, 5/6, ...

has limit 1. Does that mean that "1" is approaching something, but never quite reaching it?


Of course not, that's silly.

The sequence:

0.9, 0.99, 0.999, ...

has limit 0.999~. It is similarly silly to talk about "0.999~" approaching something.

(Of course, this sequence also has limit 1)

Hurkyl

Why, 0.111...01, of course.

(Please nobody construe this as being a serious reply!)

Hurkyl

If you had a line .999~ cm long, then something is seriously wrong, because lines extend infinitely in both directions.

If you had a line segment 0.999~ cm long, then it would have both its endpoints because, by definition, line segments contain both of their endpoints.

russ_watters

That isn't how limits work. A limit is an exact number - what tends to infinity isn't the limit (actually, I'm not sure if it has a name, but it's the "x" below...).

Ie:

lim 1/(1-x) = 1
x-> 0 (sorry - hate latex)

The equals sign is for real - that limit really is exactly equal to 1. The "x" is approaching something (infinity), but the limit itself is exactly equal to 1.


An interesting link, using polygons inscribed in a circle: http://www.coolmath.com/limit1.htm

"n" approaches infinity, but the limit is exactly equal to the circle.

And that's the entire point of limits - to find exact values for things that are otherwise difficult to pin down (such as the slope of a curve at a single point).

I actually enjoyed calculus-I (yeah, I'm a freak), because when I got my arms around this concept (didn't just plug and chug, but actually understood the point of limits), calculus-I became a piece of cake. It also helped that I took it in conjunction with Newtonian physics and they are so related that both help you learn the other.

sameandnot

do you know that "there are twice as many numbers as numbers"?

Mindscrape

Don't you mean zero?

Sir_Deenicus

Word play.

---

I think the so called Weisserstras Rigorous Epsilon-Delta formulation of limits is every bit as artificial as the naive concept. Simply one level of abstraction high enough to delude one's self into acceptance.
---
nate808, on the ruler you would have the point. The best explanation I can think of is that 0.9... is 1 because we are clever enough to jump out of the system or loop of repetetive non stop addition to see what is actually going on with adroit manipulation of intrinsics.

Sir_Deenicus

I hope I didnt make too many mistakes. For the hyperreals , 0.9... <> 1. First I clarify and set up. If A is an *R statement then R --> *R (if true in R then R* for A and *R statement) , the converse, *R --> R is false. Thus simply stating that something is true in R makes it always true in *R is not enough. There is only a gurantee that if A is a properly defined statement then there will be a behaviour in *R that "functions" as expected. But it need not be the only one. An internal statement in *R is not necessarily true in all subsets of *R, especially depending on your language strength. Also, *N is an internal subset of *R but not N. For example, x + y = 3 is only true in *R when we have (x,y) = (1-e, 2 + e) otherwise, x + y <> 3 even though st(x + y)=3. The behaviour of infinitesimals and finites hint that there exists gaps in any cuts we make in *R (if we began with sets containing only finite hyperreals for example).

Note though, that A an R statement --> R <--> *R. But then *R is not Dedekind Complete, it is not possible to define cuts such that there will always be a number in a gap. It not possible to extend the completeness axiom over to *R. The proof for 0.999... = 1 depedends on an ability to define cuts at its heart and Dedekind completeness. I will not go into that but will argue intuitively.

There exists a hyperreal number e that is infinitely near to any finite number r. I use ~= to mean infinitely approximately (infiniproixmitely) close to. Since there exists e infinitely near to all r then there exists r ~= r - e ~= r + e. The real, r = st(r), the standard part of the hypernumber r. Suppose we have two finite hyperreals r ~= 0.99...9 and s ~=1 then you cannot prove that it converges or whatever because there exists hyperreals r,s,t members of some set S where r < t < s with t not in the interval. Which is undefineable anyway. There is no intreval between the two numbers since such a thing is undefined. You can still however, use hyperreals to study convergance and limits and all that lovely stuff since we would actually be operating on reals but using hyperreals in place of infinities, infinitesimals and stuff. Just as rigorous but much more efficient than epsilon-delta. limits.

st(0.9...) = st(s) = 1 ~= 1 - e < 1+e ~= 1 ~= 1 + e.

This hyperreals vs limits remind me of the old fight between quaternions and vectors and grassman's algebras. limits map to quaternions and hypers to Grassman.

matt grime

No idea what that proves. I know what you think it proves, but I'm afraid that 1/n still tends to zero in the hyperreals. Indeed the transfer principle you cite proves it. Arguing intuitively doesn't prove a single thing.

You do not construct hyper reals by taking limits of 'decimals' or real numbers, 0.9... still represents a real number, the limit of the partial sums, that limit is taken in the subset of the reals since the limits are purely objects lying in the reals, hence they are still equal in any *extension* of R.

russ_watters

Dang, me too.... Fixed now, thanks.

Sir_Deenicus

Perhaps we are talking about 2 different things? 0.9... does not tend to 1 on the hyperreal line because the hyperreal line contains infinitesimals!

The picture argument. On the hyperreal line, There exist numbers between every number that disallow the step required for 0.9... to converge at 1. We cannot make the inductive jump. 0.9...9, 0.9...9 + e, 0.9...9 + 2e and so on.

The formal attacks. Consider the first proof given here.
The proof fails since it is possible to choose e > 0 but less than or equal to all *R. Then e cannot equal 10 ^ (-1/d). Indeed the proof ends up saying something meaningless like infinitely approximately less than zero.

Again in wikipedia it states: Since there exist non zero infinitesimals we can only say that the series gets aproximately close to 1 since we are approximately close to 0 (there exists 0 < e ~= 0). But in fact the whole endeavoure is meaningless since the cuts that created the entire decimals should not have been possible! Since we have e, the gaps created should have been empty. There is no concept of inifinite precision, just a line riddled with gaps as we go through all c in our original set. In fact, the proof is not possible since a meaningful upper bound cannot be found in *R due to the existance of infinitesimals.

Suppose we rewrite the decimal number as a sequence, then because there is no completeness [axiom] in *R, it cannot be shown to converge or diverge etc. I end like so:

The archimdean property is equivelant to the statement s = {1/n} converges to 0. Any system which contains infinitesimals is non-archimedean. The hyperreals contain infinitesimal and thus s = {1/n} does not neccessarily converge to 0. This fact, coupled with the lack of gurantee of convergance on monotonic sequences and a lack of least upper bound property all point that the series 9 * sum(n, infinity,1/10^n) over *R, whose properties are accounted for in their totality cannot be shown to converge at 1.

matt grime

So, what you're saying is that the statement

"1/n does not tend to zero"

is provably true in the hyperreals, which, implies that it is true in real analysis and moreover provably in real analysis, hmm, that seems more than a little problematic. perhaps you need to check what convergence means in the hyperreals (which is more than the statement that x_n-x lies between -e and e, something you point out is not meaningful in *R, I believe. something tends to zero if the limit is both infinitesimal and real, and thus zero since thaht is the unique real infinitesimal, so we are saying 1/n tends to zero because it is *real* at all points and the limits is an infinitesimal, hence 0, convergence is a different beast here.)


*R is an extension of the real number system to a large set, results that are statable about elements of *R and that are true in R are true in *R, that is the transfer principle. The statement 1/n converges to zero is statable in *R, restricted to R is true so it is true in *R.

Hurkyl

How can accepting any axiom or definition be a delusion?

I can't even figure out what makes you think the definition is abstract anyways.

(The Weisserstras formulation is superior to the naive concept, of course, simply because it is a formulation -- I can't mathematically prove anything about the naive concept)


There are several problems with this:
(1) One's "cleverness" (or lack thereof) has nothing to do with what is actually deducible.
(2) Even if I grant you the liberty of calling an infinite sum "non-stop addition", there are plenty of formulations of the decimal numbers that never make reference to such a thing.
(3) In many formulations of the decimal numbers, 0.9~ = 1 (and similar equations) are part of the definition, and not some "adroit manipulation of intrinsics".
(4) Your whole response seems to imply that you think 0.9~ = 1 is a "bad" thing -- but you have to remember that the point of the decimal numbers is not so people can have fun pushing numerals around: it's so that they can be a complete ordered field. (i.e. a model of the real numbers) There are plenty of demonstrations why allowing 0.9~ and 1 to be different would mean this goal was not achieved. (Such as the famous 1.0~/3.0~*3.0~=0.9~ calculation)

Wrong. (Possibly due to sloppiness)

If P is an (internal) statement about the reals, and *P is the corresponding statement about the hyperreals, then P is true if and only if *P is true.


Of course it is enough -- this is the direction that allows me, given the truth of the statement

0.9~ = 1

to conclude the truth of the statement

*0.9~ = *1


I find this all confusing; I'm not sure what it's supposed to say. (e.g. x+y=3 is true in the reals if and only if there is an e such that x = 1-e and y = 2+e)


Yes, you can. Dedekind completeness is a true internal statement about the reals. Therefore it is true for the hyperreals.

You're confusing the internal and external statements of Dedekind completeness. It is true (by the transfer principle) that every internal Dedekind cut is generated by an element of *R, and therefore *R is internally Dedekind complete.


No, the classic proof is a basic epsilon-delta proof that
[tex]
\sum_{n = 1}^{+\infty} 9 \cdot 10^{-n} = 1
[/tex]
and this is valid, for example, in the rational numbers, which are also not Dedekind complete.



I'll tell you what is true, and where you went wrong:

Suppose I take the mapping f : N -> R given by:

[tex]f(n) := \sum_{k = 1}^{n + 1} 9 \cdot 10^{-k}[/tex]

Of course, this is simply the sequence of partial sums of 0.999~.

Now, suppose I also take the injection i : R -> *R.

It is true that the sequence (i o f) : N -> *R does not have a limit.

However, this has absolutely nothing to do with *0.9~.

Your mistake is that you've only transferred part of the whole setup. In the hyperreals, the value of *0.9~ is given by the limit of the function

*f : *N -> *R

which is the *-transfer of the previous function f.

If you don't want to use the transfer principle, it is straightforward to show directly that the limit of *f is 1. (The proof is identical to the real case -- you can get a closed form for the sums of hyperfinitely many terms, and then show the closed form goes to 1)



To sum it up, externally speaking, *0.9~ has more digits than 0.9~.


But the hyperreals are hyperarchmedian! The sequence s = {1/n} does converge to zero when you take the hypernaturals as your index set.

Sir_Deenicus

But it is what allows one to deduce something outside the system. See Godel, Escher Bach an Eternal Golden braid for my meaning.

Again, just an example.

No it Isnt! I showed how there can exist gaps in cuts. Check any text on non standard analysis any you will see that there is no completeness axiom, and nor are they dedekind complete or archimedean. I do not think I could misudnerstand that. I think you use the transfer principle too liberally.

I may be wrong but I think you implicitly just created a real number as the sum you have can be taken as a cut made in the rationals.

Oh wow! I didnt know that. I just learned a cool new word, thanks. EDIT: A search on google, I could not find any cases where hyperreals were linked to hyperarchimidean properties. Only multi values algebras (which gets the hyperacrhimedean property when all its elements are archimidean) whose meaning I am unsure of due to my lack of experience with them. But what of my sequence argument? Represent 0.9... as a sequence, in *R you cannot show convergance.

Sir_Deenicus

No, something true in *R does not need to exist in R, for example st() has no R counterpart. This by the way is what I was trying to say earlier Hurkyl, when you said I was wrong.

But what of my attack of the 2 proofs? They use properties (are not statements) only applicable to R an not *R . For example, an ability to create an upperbound set with a maximal element or the assumption of 0 being the only hyperreal etc. ?

matt grime

None of what you've written had indicated that 0.9... is not 1 in the hyperreals, and you've not rejected any of the arguments put forward that show that they are the same, you've only said that they are not necessarily the same for merely the same reason, and there is no arguning about that, and i agree entirely. convergence in *R is not the same as saying things about epsilon; you've not gotten round the argument that not only do 0.9... and 1 differ by a infinitesimal (which we both agree on) but that the difference is also *real* and hence since the only real infinitesimal is 0 then they are the same thing. Now, since the whole point of infinitesimals in the hyperreals is to provide another way of doing the analysis that agrees on R then how can you remotely believe the phrase 0.999... is 1 to be false in *R and true in R?

Hurkyl

I'm not going to try and quote to what I'm responding -- I don't think it would be helpful.


I confess that I made up the word "hyperarchmedian" -- I'm going along with the convention of prefixing the *-transfer of various concepts with the word "hyper". (e.g. hyperreal, hypernatural, hyperfinite)


One statement of Dedekind completeness is given by:

[tex]
\forall S, T \in P(\mathbb{R}):
(S \neq \emptyset \wedge T \neq \emptyset \wedge S \cup T = \mathbb{R} \wedge S \cap T = \emptyset \wedge
\forall s \in S: \forall t \in T: s < t)
\implies
\exists r \in \mathbb{R}:
\forall s \in S: \forall t \in T: s \leq r \leq t
[/tex]

This is a true statement in standard analysis. Therefore, by the transfer principle, the *-transfer of this statement must also be true. In other words:

[tex]
\forall S, T \in P({}^\star \mathbb{R}):
(S \neq \emptyset \wedge T \neq \emptyset \wedge S \cup T = {}^\star \mathbb{R} \wedge S \cap T = \emptyset \wedge
\forall s \in S: \forall t \in T: s < t)
\implies
\exists r \in {}^\star \mathbb{R}:
\forall s \in S: \forall t \in T: s \leq r \leq t
[/tex]

So, internally speaking, the hyperreals are Dedekind complete.

The catch is that P here is the internal power-set function. P(R) contains only the internal subsets of R. Of course, for the standard model, every subset of R is internal.

But in the nonstandard model, P(*R) is "missing" some subsets. For example [itex]\mathbb{N} \notin P({}^\star \mathbb{R})[/itex], because N isn't an internal subset of *R.

(Incidentally, this whole issue about the internal power-set operation was one of my biggest stumbling blocks in understanding this stuff!)

(I actually have a gripe about the usual treatment of second-order logic that relates to this context -- but that's for another thread!)


Incidentally, this fact is useful to prove certain sets are not internal. For example, I can prove that the set of finite hyperreals is not an internal set, because I can use it to construct a Dedekind cut around one of the gaps!



When I'm working in Q, I can still use all the normal definitions of things like continuity, infinite sums, et cetera. Of course, in Q, things like:

[tex]
\sum_{n = 0}^{+\infty} \frac{1}{n!}
[/tex]

do not converge, despite the fact the sequence of partial sums is Cauchy.

But, it can still be shown (using the exact same proof as one would use in the reals) that the sum

[tex]
\sum_{n = 1}^{+\infty} 9 \cdot 10^{-n}
[/tex]

converges to 1.



A sequence, incidentally is nothing more than a function written with different notation. The domain of this function is called the "index" set.

In general, if we do not say otherwise, we assume that all infinite sequences are indexed by N. (and finite sequences indexed by something of the form {a, a+1, a+2, ..., b-1, b}, usually with a=0 or a=1)

But in the nonstandard case, it becomes more appropriate to index sequences by *N, and not by N. In particular, this means that when we speak about a hyperdecimal, its digits should be indexed by *N and not by N.

You are correct in observing that the N-indexed sequence {1 / (n+1)} does not converge in the hyperreals. However, the *N-indexed sequence {1 / (n+1)} does converge to zero, which can be shown either through the transfer principle, or directly with an epsilon-N argument.



I would also like to point out that st() does not necessarily exist in a nonstandard model of R! It is another example of something external -- a function we can define set-theoretically, but have absolutely no reason to think it is a part of the model we're studying.


(Actually, models of R typically have no functions whatsoever -- what I mean to say is "model of real analysis", but that's too wordy. )