Does 0.999~ Equal 1?

[Sir_Deenicus]

Hurkyl your post is long and I was only able to skim but will read it more thorougly later. (it also got a bit disconnected as you went on :P). I have no problems with the internal sets agreeing, I see now that it is obvious. But I extended my issue not with them but to P(*R), which you later say is not fair game (exclude finites etc.) but state there exists an isomorphic statement to dedekind completeness in *R. I have to read more on this but from what i know DC is not fair game for extension. Sorry to seem so stubborn but I am quite certain of that correctness.

Another thing, if *R is totally disconnected then you can't define a concept of a topology for it! This is something my intuition strongly rebels agaisnt. Surely you must have made a mistatement that needs clarification there?

 

[matt grime]

Of course we can define a topology for it; if we couldn't then the statement 'is totally disconnected' would be meaningless since it is a statement about topological spaces (the cantor set is another such, as is any discrete topological space). If your intuition rebels against this and you're prepared to make such a strong statement of assertiveness then I will automatically dismiss any other suggestions you make.

 

[Hurkyl]

If it helps you internalize things, you should observe that Q is also a totally disconnected topological space.

("totally disconnected" means that every open set is disconnected)

 

[moose]

If there is no law stating whether .999999...=1 or not, than it's neither... I guess it depends on if you think it does or not... There is no way to prove that .9999...=1 without anything speculitive... It depends on how you view it, seriously...

 

[matt grime]

No, it depends upon how we have chosen to define the terms. There is no law stating anything if we choose not to accept definitions as being true, I am not even compelled to accept that 4/2=2 if I choose to accept different meanings from those commonly accepted. (No one is saying anything is absolutely true, only true within in the meanings we give things, by the way, and if you ignore those meanings that is your problem and puts you outside the realm of mathematics.)

 

[Sir_Deenicus]

("totally disconnected" means that every open set is disconnected)

Thanks, that one helped alot. My knowledge of topology is pretty cursory and gotten from a CS angle. I am soon to undertake a proper math treatment on it and its algebraic counterpart however. On NSA however, I am more confident on, having read more (first Keisler, then numbers, then an interest in proof systems) and so will continue to contend.

By the way, I've managed to find something that states explicitly what I've been arguing. More than just arguments from what may possibly be my misintepretation...

http://www.motionmountain.net/CAppendix.pdf" Apparantly the surreals have this property as well!

 

[Hurkyl]

I hate to take an authoritative stance, but that reference simply has it wrong.

I can only imagine three things the notation 0.99\bar{9} can possibly mean with regards to the surreals:

(1) It is a "hyperdecimal" -- that is, a *Z-indexed sequence of digits -- for which every place to the left of the decimal point is zero, and every place to the right of the decimal point is 9. This "hyperdecimal" is exactly equal to 1.

(2) It is trying to denote an "ordinary" decimal expansion of a hyperreal number -- but any nonterminating "ordinary" decimal does not converge. As I mentioned earlier, a countable sequence converges if and only if it is eventually constant. A strictly increasing sequence, such as 0.9, 0.99, 0.999, ... (that is, \{ 1 - 10^{-n} \}_{n \in \mathbb{Z}^+}), does not converge in *R, so it cannot denote any hyperreal number at all.

(3) The reals are embedded in the hyperreals. 0.99\bar{9} denotes a real number, which in turn is a hyperreal number. But this approach also clearly leads to 0.99\bar{9} = 1.


The surreal case is even stickier -- the surreals are far "too big" for any sequence to converge in the ordinary sense... no matter how long the sequence was... unless it is eventually constant. So, I don't really have any idea what the author could possibly mean by trying to refer to a surreal number via a decimal expansion.

(of course, the real numbers can be embedded in the surreals -- so you denoting a surreal by a decimal expansion can make sense if you say that the decimal expansion denotes a real number, and then that real number is in the surreals... but then clearly that would say 0.99\bar{9} = 1 in the surreals)



I suspect the author is simply confused by the fact that the sequence \{ 1 - 10^{-n} \}_{n \in \mathbb{Z}^+} does not converge to 1 in either the hyperreals or the surreals. This is easy to see in the hyperreals because, for any positive infinitessimal e, we have that (1-e) is greater than every term in that sequence. (But, that's not true for the sequence \{ 1 - 10^{-n} \}_{n \in {}^\star \mathbb{Z}^+})

Similarly for the surreals. We can construct a number "between" that sequence and 1 as being the surreal number:

s = { 0.9, 0.99, 0.999, ... :: 1 }

This defines a surreal number, which is strictly greater than each number in { 0.9, 0.99, 0.999, ... }, and yet strictly less than each number in { 1 }.

And then, I could look at { 0.9, 0.99, 0.999, ... :: s }, which is strictly less than s, and so forth.



However, this is a true statement in the surreals:
1 = {0.9, 0.99, 0.999, ... :: }

But this is misleading, because this is also true:

1 = {0.4, 0.49, 0.499, 0.4999, ... :: }


In both cases, 1 is the "simplest" number greater than everything in the left set, which gives us the equality.

 

[Maxwell]

Quite impressive. I think people here keep thinking I'm like 40 or so and have a PhD in mathematics because i may be able to understand limits one day but this...I doubt it.

What?

 

[Sir_Deenicus]

I hate to take an authoritative stance, but that reference simply has it wrong.

(2) It is trying to denote an "ordinary" decimal expansion of a hyperreal number -- but any nonterminating "ordinary" decimal does not converge. As I mentioned earlier, a countable sequence converges if and only if it is eventually constant. A strictly increasing sequence, such as 0.9, 0.99, 0.999, ... (that is, \{ 1 - 10^{-n} \}_{n \in \mathbb{Z}^+}), does not converge in *R, so it cannot denote any hyperreal number at all.

But it is your word vs his... Nonetheless I have no knowledge of surreal numbers beyond that they have to do with something called Conway Games. However on hyperreals, your second point, point 2, is exactly what I have been arguing!

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.
...

But then, since the hyperreals is not complete, a cut in *R would not allow one one to create a decimal concept would it? 9/10^n + 9/10^(n + 1) + 9 /10^(n + 2) + ... where all the numbers are not necessarily [simply] reals but [also] finite hyperreals, the series would not converge since there would be infinitesimals in the way [of precision]. Thus the question precludes itself.

Why am I still so certain of incompleteness of *R? I have 4 texts which claim so, (one by Keisler, , one a Springer Verlag Graduate text on Numbers, one on NSA with Automated Formal Systems) and http://www.cs.uiowa.edu/~stroyan/InfsmlCalculus/FoundInfsmlCalc.pdf" . Pages 11 and 14.

 

[Hurkyl]

I can go through the proof that a convergent countable sequence of hyperreals must be eventually constant from the ultrapdroduct construction, if you like. It's an ugly mess, though!


I don't think it really matters though -- I get the suspicion you're so busy trying to refute "*R is complete" that there's a subtle difference between that and what I'm saying! (And as I mentioned, it took me a while to pin down this subtlety myself!)


You are correct in saying *R is not (externally) Dedekind complete.

But I'm saying *R is internally Dedekind complete.


You are correct in saying the N-indexed sequence {1 - 10^(-n)} does not converge to 1 in *R. (it doesn't converge at all).

But I'm saying the *N-indexed sequence {1 - 10^(-n)} does converge to 1.


Also, I'm saying that the correct notion of "decimal" when looking at the hyperreal numbers involves having *N-many decimal places (not N-many).

 

[Sir_Deenicus]

I can go through the proof that a convergent countable sequence of hyperreals must be eventually constant from the ultrapdroduct construction, if you like. It's an ugly mess, though!
I don't think it really matters though -- I get the suspicion you're so busy trying to refute "*R is complete" that there's a subtle difference between that and what I'm saying! (And as I mentioned, it took me a while to pin down this subtlety myself!)
You are correct in saying *R is not (externally) Dedekind complete.
But I'm saying *R is internally Dedekind complete.
You are correct in saying the N-indexed sequence {1 - 10^(-n)} does not converge to 1 in *R. (it doesn't converge at all).
But I'm saying the *N-indexed sequence {1 - 10^(-n)} does converge to 1.
Also, I'm saying that the correct notion of "decimal" when looking at the hyperreal numbers involves having *N-many decimal places (not N-many).

I agree with all that! And can also truthfully say that I did not (fully) before this exchange. I have much to learn yet and wish I had already the knowledge so that a more interesting conversation would have been possible. I nonetheless do not think this a wasted endeavour since I at least, gained from it. :P

Perhaps I shall return to this with more to offer months hence and with a robust construction and some practical results proven in Isabella in tow! Thanks. EDIT: except for that last bit. Although true, my argument truly did not rest on that but on the cuts across all *R and not just internal subsets. I sort of feel that focusing on internal sets is boring since its kinda obvious, it is after all what makes analysis possible and useful with hyperreals. I much like the intricacies involved with P(*R) and creating a consistent treatment across all of it.

 

[HallsofIvy]

There is a lot of difference between something tending to 1 and something being 1. we approximate it to 1, but need not be that it is actually 1. The explanation given by Nagzun was really fantastic. Something tending to 1 being written as 1 when we don't know the actual value is a different case. This is what happens in GP. We don't know the actual value and so the formula given is a/1-x wherew x<1. Here a Mathematician is sue that no one tommorow will give a unique perfect answer to this question about summation about GP to infinite values, So he approximated the formula. We write root-2 as 1.414, But 1.414 is a unique rational number. However root-2 is not really 1.414, it only tends to that value. We have no use of finding root-2 to hundred decimals to solve numericals. So we neglect the remaining. We can be sure about the fact that 1 is something very near to the number 0.999~. However we cannot tell, that both are equal.

You seem to have an unfortunate concept of limits. The sequence 0.9, 0.99, 0.999, 0.9999, ... "tends" to 1. However the definition of 0.9999... is that it is the limit of that sequence which is 1. Let me repeat that for clarity: the sequence "tends" to 1 so the limit IS 1. Yes, we can say that 0.999... is equal to 1.

I'm not sure where you got your information about geometric progressions (and 0.999... is the limit of a geometric progression. But note that it is the limit of the geometric progression, not the progression itself.) but I hope no teacher actually ever told you that " We don't know the actual value and so the formula given is a/1-x where x<1.Here a Mathematician is sue that no one tommorow will give a unique perfect answer to this question about summation about GP to infinite values, So he approximated the formula."
If mathematicians use approximations, they say so! We do know, for a fact, that the sum of the infinite geometric progression \sum_{n=0}^\infty ax^n is exactly \frac{a}{1-x} as long as -1< x< 1.

And though it was probably just a matter of imprecise wording, I feel compelled to point out that ("We write root-2 as 1.414, But 1.414 is a unique rational number. However root-2 is not really 1.414, it only tends to that value.") root-2 does NOT tend to 1.414. It does not "tend" to anything- it is a specific number (close to but not equal to 1.414) and numbers stay put!

Finally, when I saw "Here a Mathematician is sue", I was sorely tempted to respond that I know one mathematician who is sue- though she prefers to spell it with a capital S! Fortunately, I restrained myself. Of course, I nevr mak typographzcal errirs!

 

[mombogumbo]

I still don't undertstand how .999~ = 1.

Below is my reasoning of why .999~ does not equal 1:

+You say the partial sums = to 1 and 1 is the limit. I still don't see how .999~ = 1. I see two separate individual numbers. Knowing .999~ is not a process it is not "extending" "tending" "traveling" (all of which require time) to infinity it is already in the state of being .999~. If the partial sums of this number 9/10+ 9/100 +9/1000 +9/10000 and we continue on with this progression into infinity we will still be short of 1 by
.1^-Nth . Therefore with my reasoning i conclude that .99999 does not equal to 1.

+Why do we even speak of the "partial sums of the number = 1" how does this relate to .9999~ = 1 .9999~ is an individual number. Take a real world example. There exists 1 cat named tom and 1 dog name spark, two separate individuals Tom = Tom True Spark = Spark True Tom = Spark False just the same as .99999~ (Tom) Does not equal 1(Spark).

+"if you let it go to infinity it will equal 1" it will not it will continue on with the same progression forever .999...

+Acums Razor : .9999~ simply by looking at it and by the definition of equality cannot in any way other then appoximation be equal to 1.

+no two objects can exists in the same place at the same time.
.9999~ = 1 (Two separate objects occupying same space) False 1 = 1 (same object occupying same location) True.

+"they'd need to accept that 3*1/3 is not only 1 but also 3*1/3 = 3*0.3~ = 0.9~."
I accept 3 * 1 / 3 = 1 I also accept 3/3 = 1 & 3*0.3 = .9~
I don't follow your logic 3*1/3 is 1 but ALSO
3*1/3 = 3*0.3~ = 0.9~
well i just said 3 * 1/3 = 1 and 3*0.3 does not equate to 1 (it quates to 0.9~) . 0.9 ~ does not qual 3*1/3 it equates to 1 . What you are saying is 1 = .9~ = .9~
same as saying Bush = Clinton = Clinton False

But then again

+ .9~ is in the state of being the number not in the process of "extending" or "traveling" to infinity it is static. therefore taking the partail sums of the number.9~ will not result 1. There is no end to infinity .999~ the sequence (9/10 + 9/100..) will continue (timeless, it is already in the state) and it wil never equate to 1.

Please feel free to comment on my reasonings maybe you can make me understand.

 

[HallsofIvy]

I still don't undertstand how .999~ = 1.
Below is my reasoning of why .999~ does not equal 1:
+You say the partial sums = to 1 and 1 is the limit. I still don't see how .999~ = 1. I see two separate individual numbers. Knowing .999~ is not a process it is not "extending" "tending" "traveling" (all of which require time) to infinity it is already in the state of being .999~. If the partial sums of this number 9/10+ 9/100 +9/1000 +9/10000 and we continue on with this progression into infinity we will still be short of 1 by
.1^-Nth . Therefore with my reasoning i conclude that .99999 does not equal to 1.
+Why do we even speak of the "partial sums of the number = 1" how does this relate to .9999~ = 1 .9999~ is an individual number. Take a real world example. There exists 1 cat named tom and 1 dog name spark, two separate individuals Tom = Tom True Spark = Spark True Tom = Spark False just the same as .99999~ (Tom) Does not equal 1(Spark).
+"if you let it go to infinity it will equal 1" it will not it will continue on with the same progression forever .999...
+Acums Razor

Actually, it's "Ockham's Razor" named for William of Ockham, a franciscan friar, born in Ockham, Surrey, England, about 1280.

: .9999~ simply by looking at it and by the definition of equality cannot in any way other then appoximation be equal to 1.
+no two objects can exists in the same place at the same time.
.9999~ = 1 (Two separate objects occupying same space) False 1 = 1 (same object occupying same location) True.
+"they'd need to accept that 3*1/3 is not only 1 but also 3*1/3 = 3*0.3~ = 0.9~."
I accept 3 * 1 / 3 = 1 I also accept 3/3 = 1 & 3*0.3 = .9~
I don't follow your logic 3*1/3 is 1 but ALSO
3*1/3 = 3*0.3~ = 0.9~
well i just said 3 * 1/3 = 1 and 3*0.3 does not equate to 1 (it quates to 0.9~) . 0.9 ~ does not qual 3*1/3 it equates to 1 . What you are saying is 1 = .9~ = .9~
same as saying Bush = Clinton = Clinton False
But then again
+ .9~ is in the state of being the number not in the process of "extending" or "traveling" to infinity it is static. therefore taking the partail sums of the number.9~ will not result 1. There is no end to infinity .999~ the sequence (9/10 + 9/100..) will continue (timeless, it is already in the state) and it wil never equate to 1.
Please feel free to comment on my reasonings maybe you can make me understand.

Do you see what you are doing?

"I see two separate individual numbers"
"There exists 1 cat named tom and 1 dog name spark, two separate individuals"
"Bush = Clinton = Clinton False"

In everyone of these you start from the assumption that there are two separate things. You can't use that to then prove that there are two separate things.

I accept 3 * 1 / 3 = 1 I also accept 3/3 = 1 & 3*0.3 = .9~
I don't follow your logic 3*1/3 is 1 but ALSO
3*1/3 = 3*0.3~ = 0.9~
well i just said 3 * 1/3 = 1 and 3*0.3 does not equate to 1 (it quates to 0.9~) . 0.9 ~ does not qual 3*1/3 it equates to 1 . What you are saying is 1 = .9~ = .9~

You are leaving out an important step: do you accept that 1/3= 0.3333~? If you do then 3(1/3)= 1= 3(0.3333~)= 0.9999~. If you don't accept that 1/3= 0.3333~ then, of course, we are back where we started. The point of the person you quoted here is that many people who will not accept 1= 0.9999~ will accept that 1/3= 0.3333~
(If you do not accept that 1/3= 0.3333~ are you saying that 1/3 cannot be written in decimal form?)

.9~ is in the state of being the number not in the process of "extending" or "traveling" to infinity it is static. therefore taking the partail sums of the number.9~ will not result 1. There is no end to infinity .999~ the sequence (9/10 + 9/100..) will continue (timeless, it is already in the state) and it wil never equate to 1."

Okay, so you accept that we are talking about a single number, not a "process". But then it makes no sense to say it "continues".

Once again, the definition of the "decimal fraction" 0.a_1a_2a_3... is that it is the limit of the infinite sequence 0.a_1, 0.a_1a_2, 0.a_1a_2a_3... or (same thing) the sum of the infinite series
\frac{a_1}{10}+ \frac{a_2}{100}+ \frac{a_3}{1000}+....
(Those are the same because an infinite sum is, by definition, the limit of the sequence of partial sums.)

In particular, the value of the infinite repeating decimal, 0.aaaa..., is, by definition, \frac{a}{10}+ \frac{a}{100}+ \frac{a}{1000}+ ...= <br /> \frac{a}{10}\left(1+ \frac{1}{10}+ \frac{1}{100}+ ... which is a sum of a "geometric progression". It's easy to show that its sum is \frac{a}{10}\left(\frac{1}{1- \frac{1}{10}}\right)= \frac{a}{9}.

 

[benorin]

So, if x=0.\bar{9},
then 10x=9.\bar{9},
thus 10x-x=9.\bar{9}-0.\bar{9},
that is 9x=9,
so x=1. End of debate?

 

[Treadstone 71]

You could just do 1-0.\bar{9}=0.

 

[benorin]

You could just do 1-0.\bar{9}=0.

Yeah, but that would assert the truth of that which was to be proven.

 

[matt grime]

End of debate?

No (and that isn't a proof anyway, just a justification) as you'll see if you stick around. It's quite amazing how many people really don't get it when it's explained, and even more amazing how they will often then refuse to get it.

 

[selfAdjoint]

No (and that isn't a proof anyway, just a justification) as you'll see if you stick around. It's quite amazing how many people really don't get it when it's explained, and even more amazing how they will often then refuse to get it.

As Harry Turtledove likes to have his characters say, Ain't that the sad and sorry truth!

 

[benorin]

A humble contribution...

A humble contribution... Principles Of Mathematical Analysis - Rudin (1976), pg. 11, section 1.22 Decimals.

 

[AlphaNumeric]

+no two objects can exists in the same place at the same time.
.9999~ = 1 (Two separate objects occupying same space) False 1 = 1 (same object occupying same location) True.

So by that reasoning you can't have 0.5 = 1/2. Or 3/6 = 1.2/2.4. By that logic, if a number is expressed in a different way, it is a different number. Is 3/6 different from 1/2?

 

[Integral]

Ok, OK, we will go ahead and repeat the information already posted in the https://www.physicsforums.com/showthread.php?t=5513"
Several people found http://home.comcast.net/~integral50/Math/proof2a.pdf" of interest.

 

[mathwonk]

coming next to this same theater, the exciting sequel:
why does 1.9999... = 2?

 

[Moo Of Doom]

coming next to this same theater, the exciting sequel:
why does 1.9999... = 2?

Good God! I finally get my head around the fact at discussion, and you throw this at me! I see why 0.999~=1 now, but I just won't believe that 1.999~=2! No way!

 

[rhj23]

Several people found http://home.comcast.net/~integral50/Math/proof2a.pdf" of interest.

that proof uses the nested interval theorem, and doesn't quite quote it correctly. you also need that the length of the interval tends to zero (although this is clearly still satisfied)
[:shy: sorry for being pedantic, just wanted to make it rigorous!]

 

[HallsofIvy]

To expand on what rhj23 said: the proof Integral refers to asserts:
"I now apply the nested interval theorem, which states that the infinite intersection of a set of nested intervals contains a single real number"- which is not correct. The "nested interval theorem" (most notably used to prove that any closed and bounded interval in the real numbers is compact) says:
"The intersection of a sequence, {I_n}, of closed, bounded, nested intervals is non-empty. If, in addition, the length of I_n goes to 0 as n goes to infinity, then the intersection contains a single number".

The intervals used in the cited proof are not closed, so the intersection is not necessarily non-empty. (Example: Let I_n= (0, 1/n). The intersection of that infinite sequence of intervals is the empty set.)

 

[rhj23]

presumably we could just take the closure of these sets though, and then the conditions are satisfied..
i.e. 1 - \frac{1}{n} \leq 1 \leq 1 + \frac{1}{n}