Does 0.999~ Equal 1?

[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, $\frac{a}{10}+ \frac{a}{100}+ \frac{a}{1000}+ ...= \frac{a}{10}\left(1+ \frac{1}{10}+ \frac{1}{100}+ ...$ 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. $$1 - \frac{1}{n} \leq 1 \leq 1 + \frac{1}{n}$$