Concept of Quantity

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03-06-2011
C_ntua

Quote:

Originally Posted by whiteflags

The whole point is you don't even reach the end of infinity. The whole point of saying anything like "endless nines" is that "Look to the nth place, and there will be a nine there." Furthermore, any real number between 0.999... and 1 would be too big.

You can't really take the difference: 1.000000... - 0.999999... = 0.000000...

You can write that last 1 when you finish writing zero to infinite places, but you will never finish, will you? What's the point of arguing all the rigorous testing done by wolfram? It just seems silly that the only number people can think of between 0.999... and 1 is some number with even more nines. But for any number you can think of, it's part of 0.999... to begin with.

Well, reading a bit more about it I see it is actually a whole debate and there are numerous studies on the subject, so I will agree that at least my arguments are silly compared to those studies :)

I think in the end it is accurate to say it depends if you see it as a number or process as Sang-drax pointed out.
If you see it as a sequence it would depend on how you define the sum of an infinite sequence.

So I guess I am seeing it as a process, since I claimed it is not a number. But if you were to see it as a number then I would agree that it is one. I would say that there is a different logic for a process and a number, thus you would use slightly different arguments. So if I were to "defend" it as a process I would conclude that 1 is its limit. If I was to "defend" it as a number, then it will be equal to 1, since I cannot use a limit for a number as it is something fixed.

As to conclude the mathematical part and officially skip out of the topic

Code:

0.999.... = Σ(x=1...Z)(9*10^-x) lim(Z->inf)0.999.... = lim(Z->inf)(Σ(x=1...Z)(9*10^-x)) 1 - lim(Z->inf)0.999.... = 1 - lim(Z->inf)(Σ(x=1...Z)(9*10^-x)) 1 - 0.999.... = lim(Z->inf)(Σ(x=1...Z)(1 - 9*10^-x)) 1 - 0.999... = lim(Z->inf)(1 -1^-Z) = 0; 1 = 0.999...

if the above or something similar is correct then EVOex is right using limits to prove the point. This assumes that 0.999... is a number and can be used like one.
If it is a process, then the objections would be on

Code:

lim(Z->inf)0.999.... = 0.999....

which is not valid for a process and in a way you would have to conclude to

Code:

lim(Z->inf)0.999.... = 1

which is fine.
So this will prove that if 0.999.... is a number with it is equal to 1 unless I am missing something.

I am guessing similar methods exist and if you are to take 0.999.... it will follow the same algebra and theorems for all numbers and that is why it ends up as 1.

Personally, when infinity is involved I see it as a never ending process.
03-06-2011
EVOEx

Quote:

Originally Posted by C_ntua

Yes, I would disagree on that part even though I agreed at first :)

The "..." is a bit tricky symbol and confuses things. Since you put "x" you assume something like this:

Code:

0.999..... = Σ(x=1...inf)(9*10^-x)

which is fine, but what you do with the above representation is that somehow you use only the last "x". That is how I see it. Where there is a huge difference

x->inf means that x goes to infinity
x=1...inf means that you have a put one by one numbers from 1 to infinity

If you want to use limits you would do:

Code:

lim(Z->inf)(Σ(x=1...Z)(9*10^-x))

You see that there is a separation on the symbols, mixing x with Z here is what makes this assumption wrong.
The symbols "lim" and "->" follow certain rules and those should be used exactly as they are and not just because they seem right. So the starting point is

Code:

0.9999... = 9*10^-1 + 9*10^-2 + 9*10^-3 + ...

and you should add limits if you want by starting there.

Actually, the two are identical:

Code:

lim(Z->inf)(Σ(x=1...Z)(9*10^-x)) 0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )

Except that I was too lazy to get the sigma symbol here, so I simply wrote it out with "+ ... +". Let me proof it again for you using the sigma notation.

But first let me discuss this part of what you said:

Quote:

so you are not getting closer and closer since you will be adding zero. True?
In other words the difference will always be "0.00..001". The more 9s you add the more 0s you can add and you are always on the same point.

But:

Code:

lim(x->inf)(9*10^-x)

Is not a very small number, you have to understand. It's actually 0, per definition of the limit. If x grows towards infinity "9*10^-x" grows smaller. The "9*10^-x" term grows to 0 but never actually becomes 0, meaning the LIMIT is 0.

Now let me proof it using your notation (which is the same except for notational differences). If you still disagree with something please show me exactly what, because I am curious about all this.

Code:

0.9999... = lim[Z->inf]( Σ[x=1...Z](9*10^-x) ) 1 - 0.9999... = 1 - lim[Z->inf]( Σ[x=1...Z](9*10^-x) ) 1 - 0.9999... = lim[Z->inf]( 1 - Σ[x=1...Z](9*10^-x) )

There, same three steps as before. Now we start taking terms from the sigma and combining it with the "1" (notice the change in the "1" in the sum).

Code:

1 - 0.9999... = lim[Z->inf]( (1 - 9*10^-1) - Σ[x=2...Z](9*10^-x) ) 1 - 0.9999... = lim[Z->inf]( 1*10^-1 - Σ[x=2...Z](9*10^-x) )

And then we take another term:

Code:

1 - 0.9999... = lim[Z->inf]( (1*10^-1 - 9*10^-2) - Σ[x=3...Z](9*10^-x) ) 1 - 0.9999... = lim[Z->inf]( 1*10^-2 - Σ[x=3...Z](9*10^-x) )

We repeat this until we get to:

Code:

1 - 0.9999... = lim[Z->inf]( 1*10^-(Z-1) - Σ[x=Z...Z](9*10^-x) ) 1 - 0.9999... = lim[Z->inf]( 1*10^-(Z-1) - 9*10^-Z ) 1 - 0.9999... = lim[Z->inf]( 1*10^-Z )

As said before, that limit is 0! So:

Code:

1 - 0.9999... = 0 1 = 0.9999...

If you still disagree with something, let me know.

Edit: Okay so you already agreed on it by the time I finished writing it... See that it's the same thing I did before except being too lazy to find out how to use the sigma symbol in here.
03-07-2011
Mario F.

I was reading on ancient Greeks today (Tolerance Day over here). It's a recurring point of interest to me. I find certain aspects of their culture the pinnacle of civilization and it's been a downward slope since ;) On many other areas though they were barbaric for sure...

Anyways, I was reminded of the Socratic Problem. In it there's a clear need to reject/replace realism (truth, fact, evidence) for consistency. In no other way would it be possible to write the history of a man that for most modern purposes would have never existed.

It instantly stroke a chord with our debate here. "0.999... = 1" is an unprovable axiom in my view. It can however be demonstrated as being consistent within R. That's the nature of all "proofs" we have seen here, and likewise we will ever see; they are demonstrations of the requirement, not rigorous, undeniable and unquestionable proofs. "0.999... = 1" is accepted for its consistency with any remaining R axioms.

In essence a concept that also borrows from Socrates' saying (again, Socrates) "I know that I know nothing". The idea that one may not know anything with absolute certainty, but can feel confident about certain things. And that alone should be enough, on adequate circumstances, to develop the thinking and the sciences.

This is probably me being a boring sap at this point. But if there is one thing that this exciting debate brought to me was that I became more convict of the fact "0.999... = 1" can't be proved in R (hehe, I know that must feel frustrating to those of you who tried to tell me I'm wrong). However it also taught me the nature of the axiom and to not flat-out deny it as I used to do. For some this could be a paradox. For me, I feel confident living in a world where something may not be proven but be accepted simply on the value of its consistency.
03-07-2011
GReaper

I fill proud when thinking about what our ancestors did, but sick when thinking what we have become!...
Don't you think for a moment that "going far below the bottom" was by chance! They call it "Globalization". I call it "Mind Slavery" to bring "Complete Control".
03-07-2011
Sang-drax

Quote:

Originally Posted by Mario F.

It instantly stroke a chord with our debate here. "0.999... = 1" is an unprovable axiom in my view. It can however be demonstrated as being consistent within R.

Your view is correct. 0.999... is by definition (or: unprovable axiom) equal to lim(n->oo)sum(k=1...N) 9/10^(-k).

I'd be interested in such a demonstration, however. What did you have in mind?
03-08-2011
MacNilly

Quote:

Originally Posted by C_ntua

Yes, but you are adding every time a smaller amount. So in the end you are adding something very close to 0. Which in the case of a truly infinite amount of 9s that will imply that you are actually adding 0 thus you will never get to 1. In other words the amount you are adding can get closer and closer to 0 and in the limiting case it will equal to 0.

Very close, yes, but never equal. When dealing with limits, the difference (epsilon) approaches but never equals 0. If it does, the limit is undefined.

There is no "limiting case," you simply add increasingly smaller amounts forever.

This was an interesting thread, but I have to say there is quite a bit of confusion involving limits and their definition and rationale, especially about the "epsilon," or what some have been calling the "infintesimal." E > 0 always. NOT 0.

Some would be better served reading a calculus textbook (limits and infinite series) than attempting to disprove the calculus. I guess the problem is one of viewpoint; whether or not a (finite) limit produces a single, unique real number or not, and whether this number cannot be used like any other real number for some strange reason.

In other words, its a debate about the validity of the definition of a (finite) limit whether or not 0.999... = 1.

Overall, I have to say that reading this thread has not convinced me that the definition of a limit is flawed.
03-08-2011
grumpy

Quote:

Originally Posted by MacNilly

Very close, yes, but never equal. When dealing with limits, the difference (epsilon) approaches but never equals 0. If it does, the limit is undefined.

There is no "limiting case," you simply add increasingly smaller amounts forever.

That's not true. The limiting case is the value that is continually approached.

If epsilon continually approaches zero, then the limiting case (for epsilon) is zero.

Quote:

Originally Posted by MacNilly

Some would be better served reading a calculus textbook (limits and infinite series) than attempting to disprove the calculus.

While I agree with your comment about people attempting to disprove the theory (of limits and calculus), you might want to read such a textbook more closely yourself.

The notion of a limiting case - whether or not that limiting case can ever be reached - underpins calculus.
03-08-2011
EVOEx

Quote:

Originally Posted by grumpy

That's not true. The limiting case is the value that is continually approached.

If epsilon continually approaches zero, then the limiting case (for epsilon) is zero.

While I agree with your comment about people attempting to disprove the theory (of limits and calculus), you might want to read such a textbook more closely yourself.

The notion of a limiting case - whether or not that limiting case can ever be reached - underpins calculus.

Actually, grumpy, I think he was right. He says there is no limit IF epsilon ever reaches 0, which is right. If epsilon grows closer and closer to 0, then there is a limit.
03-08-2011
User Name:

After spending my 3rd period class, today, reading about hyperreals, I think I've got a unifying statement.

Can we all agree that 1 and .999... are real numbers(and, by inclusion, hyperreal numbers)? And, that 1 - sum(i=0, n).9(10^-i) = 1/10^n?

Quote:

Originally Posted by Mario F.

That's the nature of all "proofs" we have seen here, and likewise we will ever see; they are demonstrations of the requirement, not rigorous, undeniable and unquestionable proofs. "0.999... = 1" is accepted for its consistency with any remaining R axioms.

How is nested interval theorem "not rigorous?"

Both .999... and 1 lie within the intersection of the infinitely nested intervals [.9,1]⊃[.99,1]⊃[.999,1]⊃... therefore, by nested interval theorem, they are equal.

This isn't my proposed unifying statement, just a question.
03-08-2011
MacNilly

Quote:

Originally Posted by grumpy

That's not true. The limiting case is the value that is continually approached.

If epsilon continually approaches zero, then the limiting case (for epsilon) is zero.

While I agree with your comment about people attempting to disprove the theory (of limits and calculus), you might want to read such a textbook more closely yourself.

The notion of a limiting case - whether or not that limiting case can ever be reached - underpins calculus.

Yes, then in that case the "limiting case" is simply the value of the limit itself, L. When I posted that I wasn't so sure the term was being used in that sense, or in the sense that the lower bound (exclusive) on epsilon is always 0 (that never changes). On the other hand, I feel that the term "limiting case" is rather vague. In my calc I and II classes, we never used the term "limiting case," so I wasn't sure what was meant.

I believe its the word "case"... it seems to imply that at some point (ie, a case), we can't decrease epsilon further and we've suddenly reached the limit, which is not how it works.
03-08-2011
Mario F.

Quote:

Originally Posted by MacNilly

Overall, I have to say that reading this thread has not convinced me that the definition of a limit is flawed.

Oh, but no one even got close to mean that. Merely that limits are not enough to establish the identity of a quantity such as a real number with non-terminating decimals.

Also, I don't remember anyone (on any side of the debate) ever attempting to say that an infinitesimal equals 0, although that's the logical counterpart one must accept if they want to establish that 0.999... equals 1.

It's also slightly interesting that you use the term epsilon. Did you know that Cauchy himself called it "error"? Epsilon and Delta = "error" and "distance". Quite the revealing names, when one thinks of using limits as a means to prove the identity of 0.999... Give it a thought.

Quote:

Originally Posted by User Name:

After spending my 3rd period class, today, reading about hyperreals, I think I've got a unifying statement.

Can we all agree that 1 and .999... are real numbers(and, by inclusion, hyperreal numbers). And, that 1 - sum(i=0, n).9(10^-i) = 1/10^n?

I'm actually surprised you propose that. I'm not so sure we should be talking about hyperreals though. I'd rather preferred approaching it through the Internal Set Theory, on account of this option allowing us to remain in R.

But nonetheless; Yes, I fully agree with that.

Quote:

Originally Posted by User Name:

How is nested interval theorem "not rigorous?"

Both .999... and 1 lie within the intersection of the infinitely nested intervals [.9,1]⊃[.99,1]⊃[.999,1]⊃... therefore, by nested interval theorem, they are equal.

This isn't my proposed unifying statement, just a question.

Pretty rigorous, of course. As limits are, for that matter. They aren't however tools that can be used as means of proof to establish the identity of a real with non-terminating digits. See above answer on this post.

To be clear, I feel I may have been careless in my wording sometimes and may have given the feeling I'm somehow taking a jab at establish rules and conventions, when I'm actually not (well, I am in a way certainly. But definitely not attacking calculus as a tool). When I mention the word "rigorous" I'm not attacking limits. I'm merely looking at it in the context of trying to use them to establish the identity of an... actually unknown quantity such as 0.999...

For that purpose they aren't rigorous at all.
03-08-2011
Yarin

Quote:

Originally Posted by ಠ_ಠ

please reread that statement and explain to me why you think it's not completely retarded.

Indeed it is retarded. :p That would make it 0.(3)4, which not only is impossible, but would sum up to 1.(0)2.

Until about half way back in posts of this thread, I was under the impression that 1/infinity = something bigger than 0, specifically 0.(0)1, and likewise that 0.(9) < 1.
The problem in my thinking stemmed from the mistake of trying to tack numbers onto the end of the infinite string - which was really crazy to even try considering the very point of infinity is that there's no end to which anything can be tacked on to in the first place! If I'm not mistaken, this is what Mario was initially trying to make me aware of.

A point that I don't think anyone has brought up on this thread up yet, is the base 3 example. In such a system, 1/3 would actually be represented as 1/10, which would equal 0.1. Consider also that in ternary, 1/2 would actually end up being represented as 0.111... Imagine a society that used ternary, telling us that one couldn't truly accurately represent one-half, we'd think they were crazy! Likewise they'd think we're crazy for saying one-third can't.

I'm glad I started this thread, even with a stupid idea. :)
03-08-2011
User Name:

Quote:

Originally Posted by Mario F.

I'm actually surprised you propose that. I'm not so sure we should be talking about hyperreals though. I'd rather preferred approaching it through the Internal Set Theory, on account of this option allowing us to remain in R.

But nonetheless; Yes, I fully agree with that.

Remind me, who was it who first mentioned *R? I'm just playing along.

Since in *R, it is valid to say 1 - .999... = 1/10^∞, we will start from there.
1 - .999... = 1/10^∞
st(1 - .999...) = st(1/10^∞) // st is order preserving, therefore the equality remains valid
st(1) - st(.999...) = 0 // st(a + b) = st(a) + st(b)
st(1) = st(.999...)
st(1) = 1 = st(.999...) = .999... // st(x) = x iff x is a real number
1 = .999... // transitive property

So, in short, .999... = 1 in R, but not in *R.

Quote:

Originally Posted by Mario F.

Pretty rigorous, of course. As limits are, for that matter. They aren't however tools that can be used as means of proof to establish the identity of a real with non-terminating digits. See above answer on this post.

Last I checked, real numbers didn't have to have terminating decimal expansions. So, following this logic, neither sqrt(2) or 1/7 are real, correct? Nor can we say that, as lim(x->.142857142857142857...) x = 1/7?

Quote:

Originally Posted by Mario F.

To be clear, I feel I may have been careless in my wording sometimes and may have given the feeling I'm somehow taking a jab at establish rules and conventions, when I'm actually not (well, I am in a way certainly. But definitely not attacking calculus as a tool). When I mention the word "rigorous" I'm not attacking limits. I'm merely looking at it in the context of trying to use them to establish the identity of an... actually unknown quantity such as 0.999...

Okay... one more time. This is the very definition of the theorem you have, so far, neglected to read: "Within each infinitely nested nonempty interval there is exactly one real number." Therefore, if two decimal expansions can be shown to exist within the same infinitely nested interval, they must be expansions of the same number. Different representations of the same number is no paradox. It's a simple as (n)/(2n) = 1/2, it's no paradox that there's infinitely many ways to write 1/2. You make it seem paradoxical by injecting ideas of infinitesimals that don't exist in the real numbers.

Quote:

Originally Posted by Mario F.

For that purpose they aren't rigorous at all.

What is the purpose of any rigorous math if you can pick and chose which = means = and which mean ~. = is = whether it seems intuitive to you or not.
03-09-2011
grumpy

Quote:

Originally Posted by MacNilly

I believe its the word "case"... it seems to imply that at some point (ie, a case), we can't decrease epsilon further and we've suddenly reached the limit, which is not how it works.

There are other meanings of the word "case" than that. One meaning of case is "circumstance".

A more complete expansion of "limiting case" would be "theoretical limiting circumstance", reflecting a circumstance in which a limiting circumstance may be continually approached but ultimately unreachable. However, mathematicians are trained to be lazy, so will not more words (or longer words) than necessary. That sometimes introduces ambiguity unless you know exactly what is intended.
03-09-2011
EVOEx

Quote:

Originally Posted by User Name:

After spending my 3rd period class, today, reading about hyperreals, I think I've got a unifying statement.

Can we all agree that 1 and .999... are real numbers(and, by inclusion, hyperreal numbers)? And, that 1 - sum(i=0, n).9(10^-i) = 1/10^n?

I disagree with this. For any integer number n this is true, indeed (actually, there's a small error in there, it should be "i=1, n". For infinity this no longer holds.
That's because "sum(i=1, n) 9(10^-i)" is in itself a limit for n is infinity. It's an implicit limit, but a limit nonetheless. And you can't say that:

Code:

1 - sum(i=1, n) 9(10^-i) = 1/10^n

for n is infinity here, as the implicit limit is disappearing on the right hand side of the equation. Also, "1/10^n" is never actually 0, so that would actually disproof 1 = 0.999... (edit: well, not actually disproof, but it would assume 0 = infinitesimal, which Mario wrongfully accused me of doing).

So it should rather be:

Code:

1 - sum(i=1, n) 9(10^-i) = lim(x->n) 1/10^x

Here, the right hand side isn't "a very small number", but it actually is zero, proving that 1 = 0.9999...
03-09-2011
User Name:

Quote:

Originally Posted by EVOEx

I disagree with this. For any integer number n this is true, indeed (actually, there's a small error in there, it should be "i=1, n". For infinity this no longer holds.
That's because "sum(i=1, n) 9(10^-i)" is in itself a limit for n is infinity. It's an implicit limit, but a limit nonetheless. And you can't say that:

Code:

1 - sum(i=1, n) 9(10^-i) = 1/10^n

for n is infinity here, as the implicit limit is disappearing on the right hand side of the equation. Also, "1/10^n" is never actually 0, so that would actually disproof 1 = 0.999... (edit: well, not actually disproof, but it would assume 0 = infinitesimal, which Mario wrongfully accused me of doing).

So it should rather be:

Code:

1 - sum(i=1, n) 9(10^-i) = lim(x->n) 1/10^x

Here, the right hand side isn't "a very small number", but it actually is zero, proving that 1 = 0.9999...

I was using the hyperreals, in which the pattern that carries for all real n will also carry for the infinite n.

In the reals, you are right. But the hyperreals are a discrete space, and thus the limit is wrong.
03-10-2011
MacNilly

Quote:

Originally Posted by Mario F

It's also slightly interesting that you use the term epsilon. Did you know that Cauchy himself called it "error"? Epsilon and Delta = "error" and "distance". Quite the revealing names, when one thinks of using limits as a means to prove the identity of 0.999... Give it a thought.

To be clear, I feel I may have been careless in my wording sometimes and may have given the feeling I'm somehow taking a jab at establish rules and conventions, when I'm actually not (well, I am in a way certainly. But definitely not attacking calculus as a tool). When I mention the word "rigorous" I'm not attacking limits. I'm merely looking at it in the context of trying to use them to establish the identity of an... actually unknown quantity such as 0.999...

For that purpose they aren't rigorous at all.

Yes, I see your point. Because if there is an error margin, then 0.999... can't be exactly equal to 1. Its an interesting theory, and does raise many questions. In school they brainwash you and usually nobody is smart enough to give a counterargument to the professors. They just accept what they are told.

However, if the error (epsilon) is _arbitrarily small_, and if the function under consideration (0.999... in this case) has a maximum upper bound, we can achieve a value _arbitrarily close_ to that upper bound. Thus, the function is effectively equal to that upper bound (limit). That is, there is no number one can name between 0.999... and 1 (no matter how many 9's you choose, I choose one more 9).

If that isn't proof enough I don't know what could be.

One instance of establishing an unknown quantity via limits is the number pi. Take a circle of fixed radius and inscribe in it a regular polygon composed of N triangles. Limits show that as N -> inf, the area of the polygon approaches the area of the circle, but never increases beyond the area of the circle. A few simple calculations give the area and circumference of a circle, from which pi can be established as a constant factor.

Now if you think pi as derived from a limit proof is valid to use in "regular algebra," I cannot see why you think limits can't prove that 0.999... = 1.
03-10-2011
User Name:

Quote:

Originally Posted by MacNilly

Yes, I see your point. Because if there is an error margin, then 0.999... can't be exactly equal to 1. Its an interesting theory, and does raise many questions. In school they brainwash you and usually nobody is smart enough to give a counterargument to the professors. They just accept what they are told.

However, if the error (epsilon) is _arbitrarily small_, and if the function under consideration (0.999... in this case) has a maximum upper bound, we can achieve a value _arbitrarily close_ to that upper bound. Thus, the function is effectively equal to that upper bound (limit). That is, there is no number one can name between 0.999... and 1 (no matter how many 9's you choose, I choose one more 9).

Or, you can state it as "There are infinitely many numbers between any 2 numbers.[A well known property formally referred to as 'dense'] There are no numbers between .999... and 1, therefore .999... = 1." I'm betting on it working as well for you as it did for me ~1000 pages ago.

There is a problem I see with the lim and sup proofs. Although correct, the result requires the assumption that 1 = .999... For example, if you try proof by contradiction, assuming .999... < 1, then sup(.9, .99, .999, ...) = .999... < 1. I use sup in this case for the ease of notation, and since, in this context, they have the same underlying concept.
03-10-2011
EVOEx

Quote:

Originally Posted by User Name:

There is a problem I see with the lim and sup proofs. Although correct, the result requires the assumption that 1 = .999... For example, if you try proof by contradiction, assuming .999... < 1, then sup(.9, .99, .999, ...) = .999... < 1. I use sup in this case for the ease of notation, and since, in this context, they have the same underlying concept.

Just when I thought we agreed :P. My proof used limits and I never assumed that "1 = .999...". It only uses the definition of "0.999..." and the limit of "10^-x" as x goes to infinity. If you really think that, can you point out where in my proof I assumed any such thing (just out of curiousity).
03-10-2011
Sang-drax

I think many here complicate things too much. A correct proof is:

0.999... = sum(k=1...oo)9 * 10^(-k) = 1.

The first equality by definition and the second one by using the geometric sum formula for the partial sums.
03-10-2011
Mario F.

Quote:

Originally Posted by EVOEx

Just when I thought we agreed :P. My proof used limits and I never assumed that "1 = .999...". It only uses the definition of "0.999..." and the limit of "10^-x" as x goes to infinity. If you really think that, can you point out where in my proof I assumed any such thing (just out of curiousity).

Your proof equalizes an infinitesimal to 0 based on the established rule that, there not being a distance between any two numbers, those two numbers are the same. That's the genesis of the proof. Sang-drax proof just above me uses that rule to prove that 0.999... equals 1 without that requiring establishing the identity of an infinitesimal. Essentially you both prove the same thing though, by slightly different routes.

However I find this argument curious (emphasis mine):

Quote:

Originally Posted by MacNilly

However, if the error (epsilon) is _arbitrarily small_, and if the function under consideration (0.999... in this case) has a maximum upper bound, we can achieve a value _arbitrarily close_ to that upper bound. Thus, the function is effectively equal to that upper bound (limit). That is, there is no number one can name between 0.999... and 1 (no matter how many 9's you choose, I choose one more 9).

But isn't that also an argument that contradicts the whole thing? I can reword the last sentence to an equivalent sentence that puts things in a different perspective: "There is always a real number between 0.999... and 1 (no matter how many 9's you choose, I choose one more 9)."

This is at the core of the concept of infinity. Where you wish to establish you know what happens when we reach infinity, I wish to establish we actually don't. Where you wish to establish you can prove it, I wish to establish you can't. And my own proof for that theory is exactly the contradiction that the proofs we have seen so far present when we consider that for any real number we can't think of, 0.999... is never equal to 1.

This is the basis of my thesis that you can't prove 0.999... equals 1. You can only establish it through an unproven rule that serves the purposes of limits and that gives consistency to the Real number system (that unproven rule being that there is no distance between 0.999... and 1).

I can certainly accept the ruling. But cannot accept that as a proof.
03-10-2011
ಠ_ಠ

Quote:

Originally Posted by Mario F.

I can certainly accept the ruling. But cannot accept that as a proof.

hmm, remind me again why the following isn't good enough to stand as a proof

Code:

1/9 = 0.(1) 9*(1/9) = 9*0.(1) = 0.(9) 9/9 = 1 = 0.(9)
03-10-2011
User Name:

Quote:

Originally Posted by EVOEx

Just when I thought we agreed :P. My proof used limits and I never assumed that "1 = .999...". It only uses the definition of "0.999..." and the limit of "10^-x" as x goes to infinity. If you really think that, can you point out where in my proof I assumed any such thing (just out of curiousity).

I was wrong. The two(lim and sup) are, in fact, not similar, in this case(I was thinking about limits of sequences, which are similar to supremums of subsets of ordered sets. In other words, I was WAY off topic.). I would explain my issue with sup, but it would be an necessary extension to an unnecessarily long thread. (That is, unless you want an explanation.)

After 11+ pages, I'm ready for this thread to just die. :)
03-11-2011
EVOEx

Quote:

Originally Posted by Mario F.

Your proof equalizes an infinitesimal to 0 based on the established rule that, there not being a distance between any two numbers, those two numbers are the same.

I agree with "User Name:" that this thread is ready to die, as everyone but Mario agrees about the proofs I believe.

Let me just try it one more time for you. My proof never uses that infinitesimal equals 0, my proof uses that the LIMIT of something GOING TO infinitesimal equals 0. The two are completely different, and the latter is easily the definition of the limit.
Apparently you fail to see the difference, even with several people explaining and even copying the definitions of limits. And yet you keep insisting on this matter, which I don't get. You're too smart, I believe from all your posts, not to understand it. But for some reason you choose to ignore it.
What is it, some limit-phobia?
03-11-2011
Mario F.

Quote:

Originally Posted by EVOEx

Let me just try it one more time for you. My proof never uses that infinitesimal equals 0, my proof uses that the LIMIT of something GOING TO infinitesimal equals 0. The two are completely different, and the latter is easily the definition of the limit.
Apparently you fail to see the difference, even with several people explaining and even copying the definitions of limits. And yet you keep insisting on this matter, which I don't get. You're too smart, I believe from all your posts, not to understand it. But for some reason you choose to ignore it.
What is it, some limit-phobia?

Maybe you should read my arguments better. Because it's all there. I've explained to you numerous times the problem that what you are saying and what you are actually doing with your proof are not the same thing.

I'll dissect this quote of yours so that once and for all you understand my argument with your proof. And do read the whole thing this time. I'm tired of repeating myself.

>> My proof never uses that infinitesimal equals 0, my proof uses that the LIMIT of something GOING TO infinitesimal equals 0.

Very true. Your proof returns 0 as the L of an infinitesimal. No complains here.

>> The two are completely different, and the latter is easily the definition of the limit.

Very true. They are completely different. Your own words. Completely. Completely different. Again, for it to really sink: Completely different. And I agree. They are completely different. An infinitesimal does not equal zero. The limit of something going to it does.

>> Apparently you fail to see the difference, even with several people explaining and even copying the definitions of limits.

No. I see the difference. There's a difference. They are completely different. I've always said they are different. That's the basis of my whole argumentation. But do YOU see the difference? I doubt it. Because your proof consists of this:

Quote:

Originally Posted by EVOEx

lim(x -> inf) ( 1*10^-x ) = 0

So:
1 - 0.9999... = 0
1 = 0.9999...

If an infinitesimal does not equal zero as you say, how come you bring it to algebra as being 0 in the final part of your proof? It's either different or equal to 0. You have to make up your mind. Because you don't translate the meaning of the equality sign on limits into plain algebra without that resulting into you turning L into not just the limit of zero, but actually 0.

>> And yet you keep insisting on this matter, which I don't get. You're too smart, I believe from all your posts, not to understand it. But for some reason you choose to ignore it.

As I don't get why do you insist in saying that you are not making an infinitesimal equal to 0 when your proof concludes exactly with that. Moreover, how can you claim that 0.999... is 1 without that invariably meaning that an infinitesimal is 0.

More importantly, if you claim that an infinitesimal is not zero (but the limit of something going towards it is what equals 0) then we can conclude that 0.999... does not equal 1, but the limit of something going towards 1 is what equals 1. The only thing you proved was that 0.999... tends towards 1. Not that it equals 1.

Let me say this to you again... you cannot prove the identity of a real with non-terminating decimals. You can only establish its limit.

>> What is it, some limit-phobia?

I've had the opportunity to say a few times (you probably didn't read them, as you probably won't this post) that I have no issues with calculus. Only with the fact people insist they can prove 0.999... equals 1. They can't. They can at best establish why it is useful such a rule to exist within the domain of R (which is what I defend you did and did well). But otherwise that proof is impossible.

More so because the different number systems all have axiomatic consistency. You don't have an axiom that is true for one number system but that proves it false on one that extends it. Instead what you have are certain unspoken rules, universal agreements as to how to represent certain quantities within a limited number system incapable of properly representing them. Not much different from having to be stuck with only the Natural numbers and defining a approximation rule to represent reals. You cannot prove that 1.5 is 2. You just establish it.

Finally one quick explanatory note:
>> even with several people explaining

I suppose you take me for a sensible person? You don't expect me to be insane at least. I'd need to be cuckoo to pursue this debate so long and against the opinions of so many people if I had seen the validity of yours and their arguments, won't you agree? But I'm not crazy. So something else is afoot here.

Besides if you believed I was crazy, you would have already stopped replying. Which you haven't. So I believe you either a) see room for debate, or b) don't like the fact someone is negating the validity of your proof.

I want to believe it's "a)". So with that in mind can you accept that perhaps, just perhaps, no one has actually provided anything concrete, palpable, and unquestionable, that refuted all I have been saying on this post?

Because I also want to believe you know I am man enough to immediately concede and put aside all my arguments. The fact no one did that, the fact no single so-called "proof" i have seen on this thread doesn't immediately raise a number of questions for a cynical (but also unruly and always inquisitive) mind like mine, is what drives me. Not into trying to annoy or troll you and everyone else, but into trying to understand and learn.

Knowledge is what I seek. But don't expect me to accept certain things just because "everyone says it is like so", when I find flaws on them. I will never accept those things until I'm convinced there are no flaws (just my poor judgment). Until then, tough luck; I won't budge. And make no mistake, I question myself all the time. So it's not just you trying to convince me. It's me too. I don't produce convictions and defend them. I attack them, perhaps more violently than you think. Because there's little pleasure with arguing against 10 people for 2 weeks and 13 pages.
03-11-2011
EVOEx

Quote:

Originally Posted by Mario F.

If an infinitesimal does not equal zero as you say, how come you bring it to algebra as being 0 in the final part of your proof? It's either different or equal to 0. You have to make up your mind. Because you don't translate the meaning of the equality sign on limits into plain algebra without that resulting into you turning L into not just the limit of zero, but actually 0.

And this seems to be the problem, here, saying the meaning of the equal sign on limits is different. The equals sign never means anything other than "is equal to", even if a limit is involved:

Code:

lim[x -> infinity] 10^-x = 0

We can even put the limit in a variable:

Code:

t = lim[x -> infinity] 10^-x

And then all the following is true:

Code:

t = lim[x -> infinity] 1/x = lim[x -> infinity] -1/x t = 0 t*5 = 0 t + 1 = 1

Because t is actually 0. Equal to. There is no special meaning to that. And one of the basic rules of equality (again, no matter if limits are involved) says:

Code:

If: a = b b = c Then: a = c

In my proof, I "showed" (the latter I didn't actually show, but those are basic identities) that:

Code:

1 - 0.9999... = lim(x -> inf) ( 1*10^-x ) lim(x -> inf) ( 1*10^-x ) = 0

And again, the equals sign for the limits has exactly the same meaning as for anything else, so we can establish from the basic rules of equality that:

Code:

1 - 0.9999... = lim(x -> inf) ( 1*10^-x ) = 0 1 - 0.9999... = 0

Why am I still arguing? Because I'm hoping I'm missing something, that I'm hoping you'll show me something I have never known. Or maybe the other way around. Though I'm still convinced I'm right.
03-11-2011
Mario F.

Quote:

Originally Posted by EVOEx

And this seems to be the problem, here, saying the meaning of the equal sign on limits is different. The equals sign never means anything other than "is equal to", even if a limit is involved

So saying that the limit of 1 - 0.999... equals 0 is the same as saying that 1 - 0.999... equals 0?

You also said:
My proof never uses that infinitesimal equals 0, my proof uses that the LIMIT of something GOING TO infinitesimal equals 0. The two are completely different, and the latter is easily the definition of the limit.

So... help me here....
03-11-2011
EVOEx

Quote:

Originally Posted by Mario F.

So saying that the limit of 1 - 0.999... equals 0 is the same as saying that 1 - 0.999... equals 0?

Where in my proof do you see the "1 - 0.9999..." part having a limit? The only assumption I make is that "0.9999..." is equal to:

Code:

0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )

That is, the limit of the number as you add more and more 9's. Then:

Code:

1 - 0.9999... = 1 - lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x ) 1 - 0.9999... = lim(x -> inf) ( 1 - 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )

(no limit in that "1 - 0.9999...", though yes, it's equal to a limit)

Then I show that:

Code:

lim(x -> inf) ( 1 - 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x ) = 0

Hence:

Code:

1 - 0.9999... = lim(x -> inf) ( 1 - 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x ) = 0 1 - 0.9999... = 0

There's never a limit any place I don't write it. I didn't even use implicit limits to avoid confusion.
03-11-2011
C_ntua

1)
I want to generalize to the point that whenever you use infinity you have to either prove that the process is valid or define it for infinity.

For example:

Code:

0.9999.... + 0.9.....

To calculate the sum you have to use a very specific method. In this case you would see that it is hard or impossible to sum to decimals with infinite number of decimal digits the way sum is defined.

The same for things like

Code:

0.999... * 10

Conclusion: the important part is to prove the sum or multiplication. Everything else is pure mathematics...

2)
When using sums (Σ) or limits there are some definitions. One definition used by EVOex is that the sum of infinite number of parts is its limit. Its the same thing. So by definition

Code:

Σ(x=1...inf)(9*10^-x) = 1

Using limits on the above as pointed out leaves you at one point with

Code:

lim(...)(0.999... )

which again is not defined directly. You can assume that 0.999... is a number and you go to the field of comment 1) basically.

---------
Concluding, you need to add some rules for infinite decimal digits on a decimal number. Meaning more axioms. To clarify this, you can prove that if "1+1=2" then "1+2=3" with something that everybody would accept. If you try to do something similar involving infinity you would see that not everybody will accept it and you would have disagreements on axioms. Which you cannot prove further by definition.

If somebody posts those rules and the person agrees with them all, then you can go from there and prove your point. So the agreeing and disagreeing is done with conditions that are not clearly stated. If A disagrees and B agrees this means that A might accept some axioms that B doesn't. Or putting this better, A might agree IF the axioms are stated as true.

For me those conditions, those axioms are the interesting part.
03-11-2011
User Name:

I don't remember any axioms or theorems being mentioned to support a dissenting view. But then again, I didn't read the whole thread.
03-12-2011
Sang-drax

Quote:

Originally Posted by Mario F.

Your proof equalizes an infinitesimal to 0 based on the established rule that, there not being a distance between any two numbers, those two numbers are the same. That's the genesis of the proof. Sang-drax proof just above me uses that rule to prove that 0.999... equals 1 without that requiring establishing the identity of an infinitesimal. Essentially you both prove the same thing though, by slightly different routes.

That rule is obviously correct and although I don't use it per se, I use properties of the real numbers to ensure that the limit exists.

0.999... = sum(k=1...oo)9 * 10^(-k) = 1.

is a perfectly valid proof. Here, "oo" should of course not be thought of as a number (it's not), but rather as a convenient form of notation with a well-defined meaning.
03-12-2011
EVOEx

Quote:

Originally Posted by C_ntua

When using sums (Σ) or limits there are some definitions. One definition used by EVOex is that the sum of infinite number of parts is its limit. Its the same thing. So by definition

Code:

Σ(x=1...inf)(9*10^-x) = 1

Using limits on the above as pointed out leaves you at one point with

Code:

lim(...)(0.999... )

which again is not defined directly. You can assume that 0.999... is a number and you go to the field of comment 1) basically.

Acctually, I think you're wrong. See, this:

Code:

Σ(x=1...inf)(9*10^-x)

As I said before, contains an implicit limit. That is to say, it is equal:

Code:

lim[t->inf] Σ(x=1...t)(9*10^-x)

Per definition. I know it's not going to be considered a reliable source, but maybe you believe me if I post a link stating that on wikipedia:
Limit of a function - Wikipedia, the free encyclopedia
See the first in that section. It states that writing the sum of something to infinity is actually saying the limit of that. It's quite obvious, as you can't actually SUM an infinite amount of things, so saying the sum of an infinite amount of non-zero numbers doesn't make sense by itself, which is why we all agreed it's an implicit limit.
03-12-2011
C_ntua

Yes, but in this case you will start with:

Code:

0.999... = lim[t->inf] Σ(x=1...t)( 9*10^-x )

which then will have the argument of differentiating a limit from an actual number.
Meaning that

Code:

x->10

and

Code:

lim[x->10](x) = 10

are not the same.

So that is why I say

Code:

0.999... = Σ(x=1...oo)( 9*10^-x )

because I am implying that oo is truly infinity. A well defined meaning, but you cannot calculate this with "classical" processes. If you use a limit, you are simply changing the meaning of infinity, slightly, but still changing the meaning.
What the article in wikipedia says is that

Code:

Σ(x=1...oo)( 9*10^-x ) IS lim[t->inf] Σ(x=1...t)( 9*10^-x )

not that they are equal, meaning that the meaning of both under the theory of limits is the second. That exactly because you cannot have an infinite SUM, the closest thing is that you have a sum of parts that are NOT infinite, the are t where t->oo. This is certainly not true with 0.999... where it DOES have infinite parts.

If you accept that 0.999... is actually a number, you establish that, you accept that it has a specific quantity then that is a different story. Because a limit of a number is the number itself. But you are just saying that

Code:

0.9999 IS 1.0

that there is actually no difference. It is just a different way to represent it. Yes, you can use the equal (=) sign instead of IS, that is not the point. The point is that unless you define 0.999... to have a specific quantity, not use it as a process where you have to calculate the resulting quantity, then you cannot prove that it is equal to 1.

So if 0.999... is a process, you cannot prove it is 1.
If 0.999... is a number, it is 1, not <1 or >1 but exactly 1.

In th end, since 0.999... is used as a number, you would have to expand the real numbers. When talking about numbers, you have to think of order and difference. Thus the distance needs to be defined.

The thing is that

Code:

0.333... - 0.222....

is something that you can define. You cannot compare all digits, because they are infinite. You cannot compare two infinite things, that is a very fundamental thing about infinity. What you can do is compare infinity with a number. So you can do

Code:

0.34 > 0.3333...... 0.23 > 0.2222...... 0.34 - 0.23 > 0.333... - 0.22222 0 > 0.333.... - 0.2222 0.333.... > 0.2222....

for the first two you just add 0.01 > 0, thus they are true.

Now if you use a similar logic for 0.999.... and 1 you won't succeed. Because you cannot find a decimal number that is greater than 0.999.... and smaller than 1.0. In the end the only way to actually compare 1 and 0.999... is to use a theorem or an axiom. Or just leave 0.9999 undefined or something you cannot really use. Doesn't matter if you can express it or write it, unless you can actually compare it with 1.0 you cannot use it as a number.
This is solved by using a theorem or an axiom which basically will but "trailing 9s" as the next decimal digit being 1. You will just define it like that. You cannot prove that this is not true either so there is absolutely nothing wrong about this.
You can still use 0.999... as a number and prove that "0.3333... * 3 = 0.999... = 1" for example and have a consistence theory. But unless you use an axiom like this I really don't see a proof here.

To be clear this:

Code:

0.999... < 1

is not something you can prove either. Unless you assume that 0.999... is not a number, but a process for the same reasons.

So my conclusion again is that it depends how you see 0.999.... A process or a limit? I would stick with this as the answer and any proof should use one of the two.
03-12-2011
whiteflags

Quote:

Originally Posted by C_ntua

Code:

0.34 > 0.3333...... 0.23 > 0.2222...... 0.34 - 0.23 > 0.333... - 0.22222 0 > 0.333.... - 0.2222 0.333.... > 0.2222....

0.34 > 0.3...
0.23 > 0.2...

I can agree with the above. I cannot agree with anything you said following. Just because a > a0, and b > b0, it does not imply that a - b > a0 - b0. We could try doing some of what examples you have:

0.34 - 0.23 = 0.11
0.3... - 0.2... = 0.1...

Or if you prefer:

34/100 - 23/100 = 11/100
3/9 - 2/9 = 1/9

That's right, isn't it?

0.11 is not greater than 0.1... because 11/100 > 1/9 is contradictory. In the decimal expansion, 0.11 has 0 for the hundredths place, while 0.1... has 1 for that, and all other places. Following through, that means that 0.34 - 0.23 > 0.3... - 0.2... is a false statement. Plus, you've made another mistake. There are numbers between 0.2... and 0.3... 2/9 != 1/3 and any real in between those, like say 0.235, is proof that they aren't the same number. However, there may be no real in between between say 0.23 and 0.229 with 9 recurring, unless you can think of one.

So yeah your whole argument just does not work. You're not wrong because my axioms are different from your axioms either. You're wrong because no one said what you're trying to prove. And I can only guess what that is. Maybe that all non-terminating decimal reals are not equal to their nearest neighbor for some intuitive meaning of "nearest". In fact it's because they have neighbors that they aren't equal. Rather, if 0.9... is equal to 1, then you could write 0.9... as 1, and that has implications for many other reals. 0.2 could be 0.19 with 9 recurring and so on. It's just a different way to write the same number. That is to say that these numbers are equal, which in itself is a comparison.

The reason you can accept various proofs: If you're working with what you think is 0.9... and you get 1 as a result, as long as there is no error in the process equations, then they are equal. And just because you can prove 0.9... = 1 with process equations, it does not make 0.9... a process equation. 0.9... is a number, just like Sang-drax said. 0.9... has a value, (whatever it may be,) and it comes with no operations. Terminating decimals themselves have infinite expansions as well, with 0 filled in for any places after a given place. We do not argue their status as numbers.
03-12-2011
User Name:

What is a process equation?
03-12-2011
whiteflags

Quote:

What is a process equation?

Steps one goes through to arrive at an answer, whatever "answer" may mean in the context of a problem. I sincerely hope this is not where the conversation is going.
03-12-2011
C_ntua

Quote:

Originally Posted by whiteflags

0.34 > 0.3...
0.23 > 0.2...

I can agree with the above. I cannot agree with anything you said following.

There are numbers between 0.2... and 0.3... 2/9 != 1/3 and any real in between those, like say 0.235, is proof that they aren't the same number. However, there may be no real in between between say 0.23 and 0.229 with 9 recurring, unless you can think of one.

Yes, I was completely wrong. I apologize. But what I was getting to is what you are saying, that there is some number in between, unlike 0.23 and 0.22999...

Quote:

Originally Posted by whiteflags

The reason you can accept various proofs: If you're working with what you think is 0.9... and you get 1 as a result, as long as there is no error in the process equations, then they are equal. And just because you can prove 0.9... = 1 with process equations, it does not make 0.9... a process equation. 0.9... is a number, just like Sang-drax said. 0.9... has a value, (whatever it may be,) and it comes with no operations. Terminating decimals themselves have infinite expansions as well, with 0 filled in for any places after a given place. We do not argue their status as numbers.

Yes, but I already claimed why the proofs are not at least complete. So there is an error. Let me use an example:

Code:

x = 0.(9) 10x = 9.(9) 10x - x = 9.(9) - 0.(9) 9x = 9 x = 1

this is a classical proof. But I would argue on the second line. The point is can you prove that the second line is correct??
03-12-2011
whiteflags

Quote:

But I would argue on the second line.

Code:

10x = 9.(9)

?
Argue what? Anything but that isn't multiplication by 10. Does it matter that there are infinite places in x? No. As I said, any number has an infinite expansion, but you still move the . to the right.
03-12-2011
User Name:

sum(i = 0 ... inf){9 * 10^-i) = 9 / (1 - .1) = 10. See here. I quote Euler, therefore I win.
03-13-2011
C_ntua

Quote:

Originally Posted by whiteflags

Code:

10x = 9.(9)

?
Argue what? Anything but that isn't multiplication by 10. Does it matter that there are infinite places in x? No. As I said, any number has an infinite expansion, but you still move the . to the right.

The multiplication exists only as a form of summing a certain times the same number. I believe that is its definition. How do you sum decimals? You start from the last digit and go on. You say there is no last digit, it expands forever. I am stuck.
Or are you telling me that the last digit is 9, therefor the sum has to have a last digit of 8, which is not true since you say it is 9?
03-13-2011
laserlight

Quote:

Originally Posted by C_ntua

How do you sum decimals? You start from the last digit and go on.

You don't have to start from the last digit; you can start from the first corresponding digit pair too, with a little backtracking. Either way, this does not really matter, in my opinion: the fact that we can err... float the decimal point to the right when multiplying by 10 in base 10 should hold even if you cannot fathom how to actually add the numbers on paper using the given representation.
03-13-2011
User Name:

Quote:

Originally Posted by C_ntua

The multiplication exists only as a form of summing a certain times the same number. I believe that is its definition. How do you sum decimals? You start from the last digit and go on. You say there is no last digit, it expands forever. I am stuck.
Or are you telling me that the last digit is 9, therefor the sum has to have a last digit of 8, which is not true since you say it is 9?

Repeated summing is far from the definition of multiplication. How do you do 1/2 * 3/4 with repeated addition? Summing is a special case, not the rule.

If there was a last digit, it wouldn't necessarily be a 9, but there isn't a last digit.

Applying the method of decimal addition I remember from grade school:

Code:

.999... + .999... 0.888... 1.111...

The 1.111... is where I "carried."
03-13-2011
Yarin

Quote:

Originally Posted by C_ntua

The multiplication exists only as a form of summing a certain times the same number. I believe that is its definition.

True...

Quote:

Originally Posted by C_ntua

How do you sum decimals? You start from the last digit and go on. You say there is no last digit, it expands forever. I am stuck.

You don't necessarily have to start at the last digit, if you're willing to do more work, you can start at the first one (which, for simplicity of concept, we can do with numbers that expand in non-zeros)...

1.888888... + 2.333333... =

sum the 1 and 2, we get 3.
Sum the 8 and 3, we get 11, keep the second 1, add the last 1 to the last sum (to be 4).
Sum the 8 and 3, we get 11, keep the second 1, add the last 1 to the last sum (this time to be 2).
Sum the 8 and 3, we get 11, keep the second 1, add the last 1 to the last sum (again to be 2).
Sum the 8 and 3, we get 11, keep the second 1, add the last 1 to the last sum (again to be 2).
Sum the 8 and 3, we get 11, keep the second 1, add the last 1 to the last sum (again to be 2).
And it then keeps going on. We get

4.222222....

There's no problem with this, it's all good. One might say "but it goes on forever, so you can't know what comes out to be", but this isn't true. As soon as things stop changing (that is, we enter the calculation loop of the "..."), we're done, because we at that point know everything we need to know. To prove it, ask me what the value at **any** given index/digit is and I can give you an answer. Try to use numbers that expand in non-zeros with the "normal" numbers that do expand in zeros. It always works without issue. :)

One can't use the argument of "you get stuck" or "you can't know", because we don't ever get stuck, and we do know everything about these numbers that there is to know. They can be used in **any** mathematical operation flawlessly without a problem or "error" arising.
03-13-2011
C_ntua

Quote:

To prove it, ask me what the value at **any** given index/digit is and I can give you an answer.

Lets say index number 5. With your logic
Sum the 8 and 3, we get 11, keep the second 1, add the last 1 to the last sum.
So you have: 4.22221
So how exactly do you get that any digit will be 2? It rather seems that there will always be a digit that will be 1....

Quote:

Repeated summing is far from the definition of multiplication. How do you do 1/2 * 3/4 with repeated addition? Summing is a special case, not the rule.

1/2 * 3/4 = 1*3 / 2*4 = 3 / (4+4)

Quote:

Code:

.999... + .999... 0.888... 1.111...

Wrong, you add one additional 1. If you exclude that 1 you will get 8 as the last digit. In any case, the carrier will NOT apply to the sum that "created" the carrier, so with this logic you always have an 8.

Quote:

float the decimal point to the right when multiplying by 10 in base 10 should hold even if you cannot fathom how to actually add the numbers on paper using the given representation.

And what is exactly the proof that we can do so for a decimal number with infinite digits? When you normally float you end up with one less digit, or the digit 0. Which is not the case here. So this is not exactly consistence, I don't find it really obvious in the sense that it doesn't need further proof or analysis.

I would actually say that all the above rather disprove the point rather than prove it.
Furthermore, if I told you "you have two decimal numbers, one has an integral part of 1 and the other of zero, which one is the bigger?" usually you would answer "the one with integral part 1". And this would be true for all decimal numbers with finite digits. Or actually all decimal number with the exception of trailing 9s. So where I am getting at is that the the infinite 9s doesn't follow anyway all the rules. Why do we assume that it just follows the rule of "multiplying by 10 you just float the digits"?
03-13-2011
User Name:

Quote:

Originally Posted by C_ntua

Wrong, you add one additional 1. If you exclude that 1 you will get 8 as the last digit. In any case, the carrier will NOT apply to the sum that "created" the carrier, so with this logic you always have an 8.

Wrong, you will have an 8 after the last 9. Is there a last 9?
03-13-2011
C_ntua

Quote:

Originally Posted by User Name:

Wrong, you will have an 8 after the last 9. Is there a last 9?

You use the "last 9" to say I am wrong and you ask me if there is a "last 9"? :)
If you are implying that there is not last digit at all then how do you say that you the sum (1.999...) has infinite 9s? How do you know that all the digits are 9?
03-13-2011
User Name:

Quote:

Originally Posted by C_ntua

You use the "last 9" to say I am wrong and you ask me if there is a "last 9"? :)

It's called a "rhetorical question."

Quote:

Originally Posted by C_ntua

If you are implying that there is not last digit at all then how do you say that you the sum (1.999...) has infinite 9s? How do you know that all the digits are 9

It sums to 1.999... because there is no last digit. If there was a last, it would end in 8, but there isn't, so it doesn't.

EDIT: I didn't answer specifically. I know there the 9s continue in 1.999..., because the carry continues as long as the string of 9s in .999... The string of 9s is infinite, so the carry is also infinite. The idea of there being an 8 at the end is incorrect because of the axiom of real numbers that all digits in a decimal expansion must have a definite position. We can say the digit a_x = 9 for all a in 0.a_0a_1a_2a_n..., but, we have no way to specify the index of the 8 "at the end of infinity".
03-13-2011
whiteflags

Quote:

1/2 * 3/4 = 1*3 / 2*4 = 3 / (4+4)

This is not a repeated sum. As proof of the idea that multiplication is repeated sums, I would accept that you added up 1/2 a total of 3/4 times. I'm genuinely interested in you demonstrating this idea as you stated it -- that multiplication is repeated addition, for reals other than integers.

But let's do it your way anyway!

If you want to multiply a number like 0.9... You could imagine multiplying each term in a equivalent series:

9/10 * 10 + 9/100 * 10 + 9/1000 * 10 and continuing. I don't know where you get the idea that there would be a last 8, because all you do when you multiply a number by its base is shift places. It's plenty easy to think of doing this for every 9 in 0.9...
03-13-2011
fronty

Just a quick thought without much thinking behind it:

For a real number m in base-10 with n digits in decimal part and o digits in integer part: m = sum(i = o .. -n)(a_i * 10^i). Isn't 10m = sum(i = o .. -n)(a_i * 10^{i + 1}) and isn't this true even if n = inf?
03-13-2011
Mario F.

Quote:

Originally Posted by C_ntua

And what is exactly the proof that we can do so for a decimal number with infinite digits? When you normally float you end up with one less digit, or the digit 0. Which is not the case here. So this is not exactly consistence, I don't find it really obvious in the sense that it doesn't need further proof or analysis.

There are inconsistencies in the Real Numbers system that have been studied by others. I have not looked yours up in detail. I'm merely picking your quote for truth.

The Real numbers are dubbed by some as self-inconsistent. I'm unsure as to the real meaning of that wording. I suspect it tries to refer to the fact that these inconsistencies are what keeps the number system from falling apart when dealing with infinity. Basically the number system falls apart when dealing with these quantities. However, the rules and axioms that have been established, while inconsistent, keep the number system consistent with the classical approach to calculus that we know as Limits.

Meanwhile, non classical approaches exist for calculus, and for the study of Real numbers, that support the notion of a precise definition for an infinitesimal that is non zero. And this changes everything. What some here seem to keep implying is that this does not happen in R, when it does! While hyperreals do extend the number space, other methods employed by non-standard analysis don't. That is the case of IST. It's not an extension of R. It's an extension to ZFC, the set of axioms governing pretty much all of modern mathematical thinking.

So what can be said at best is that 0.999... equals 1 in classical calculus employing Limits. But definitely not in R where other approaches demonstrate them to be different. This is not a limitation of the number system. This is merely a limitation of the methods employed and their mathematical foundations. And it's also one of the foundations of my claim that 0.999... equals 1 can't be proved.

In fact the importance of an infinitesimal has lead to a sort of friction between renowned investigators. So I find it particularly curious that by the time we reach page 14 of this debate, we are still hearing some students of maths in here fully convinced they can provide proofs on subjects that are at a very minimum minimum arguable and far from being universally accepted. Or that may be subject to change in the future.

Quote:

Originally Posted by C_ntua

Furthermore, if I told you "you have two decimal numbers, one has an integral part of 1 and the other of zero, which one is the bigger?" usually you would answer "the one with integral part 1". And this would be true for all decimal numbers with finite digits. Or actually all decimal number with the exception of trailing 9s. So where I am getting at is that the the infinite 9s doesn't follow anyway all the rules. Why do we assume that it just follows the rule of "multiplying by 10 you just float the digits"?

I can only conclude because they are stuck with classical calculus as their one and only tool to handle real numbers. They see the inconsistency, but they trust their axioms. That's the type of indoctrination that happens at our schools (I apologize if I sound rude or pretentious. That's not what I mean).

In fact, on many places, exposing these inconsistencies or daring to reject established axioms (regardless of still studying them, employing them and showing good knowledge of them) is grounds for a negative note. I wouldn't be surprised if some teachers would flunk you for even daring mentioning non-standard analysis in their classroom.
03-13-2011
C_ntua

Quote:

Originally Posted by whiteflags

This is not a repeated sum. As proof of the idea that multiplication is repeated sums, I would accept that you added up 1/2 a total of 3/4 times. I'm genuinely interested in you demonstrating this idea as you stated it -- that multiplication is repeated addition, for reals other than integers.

But let's do it your way anyway!

If you want to multiply a number like 0.9... You could imagine multiplying each term in a equivalent series:

9/10 * 10 + 9/100 * 10 + 9/1000 * 10 and continuing. I don't know where you get the idea that there would be a last 8, because all you do when you multiply a number by its base is shift places. It's plenty easy to think of doing this for every 9 in 0.9...

Well, I had in mind that that is the way you define multiplication, but the way you use it makes sense (of course) as well.

The point is there that you have one less decimal digit. So if you do that with finite number of digits , lets say N, you get N-1 9s. So you are implying that when N is infinite then N is the same as N-1. In a way having one more 9 is still infinite 9s. Which is true if you say that N->oo but then the number itself doesn't have a specific quantity, it has a quantity always growing to some limit (that is 1). So is it really a number or just some process? I mean, would you define an ever growing quantity as a number?

Putting what I said earlier more clearly maybe
1) A number is a specific quantity, something static. A process changes.
2) Infinity is either the biggest possible number or a process of a quantity going bigger and bigger.

Using infinity in the sense that it is the biggest possible number would result that

Code:

0.99... (N=inf digits)* 10 = 9.999....(X digits) where X = N-1 < inf

if you use infinity as a process going bigger and bigger then 0.99... itself always grows, so I wouldn't consider it a number.

The way we use infinity is normally as something growing bigger without an upper bound, thus from the beginning of this topic I see 0.999... as a process, not really a number.

Of course I use my own definition of number. So my conclusion is that we can see the argument as having two types of quantity
a) Static quantities with a specific valued
b) Growing quantities. They either grow without a limit, which then we call infinity and negative infinity, or they go towards a limit.

Lets get away with the lexical difference and just use both. In this case how would you define the equality between a numberA and a numberB??
The way it is usually defined is that they are equal if there distance is as small as possible. Then how do you define the order? To differentiate you say that one is bigger than the other if there difference is bigger than the smallest possible quantity NOT if its bigger than zero.

Again, not to get subtracted, my thought is that when you use a number as both a) and b) I describe above, that we usually do, then you imply that a quantity very very small is the same as zero. In other words, when you measure a quantity you can just measure its limit, since it might change. So the limit and the quantity itself cannot be distinguished thus we use them the same way. More or less that is the background of all these proofs in my opinion. But this also implies that the equality sign the meaning of two quantities being equal, means something a little bit different, as I phrase above.
03-13-2011
User Name:

Quote:

Originally Posted by Mario F.

But definitely not in R where other approaches demonstrate them to be different.

What approach demonstrates them to be different? (That hasn't been shot down yet.)

Quote:

Originally Posted by Mario F.

This is merely a limitation of the methods employed and their mathematical foundations.

Limitations are limitations, you can't bypass them because you don't like them or because they are unintuitive. You can redefine them, but by doing so you can't necessarily say statements true of your redefinition are true of the reals.
03-13-2011
Mario F.

Quote:

Originally Posted by User Name:

What approach demonstrates them to be different? (That hasn't been shot down yet.)

Haven't I been clear enough already? In non-standard analysis, an infinitesimal is a quantity different than 0. In R, it is defined as a number which absolute value is smaller than any number in the form of 1/n, where n is any natural.

Here there's a clearly a distance between 0.999... and 1.

Quote:

Limitations are limitations, you can't bypass them because you don't like them or because they are unintuitive. You can redefine them, but by doing so you can't necessarily say statements true of your redefinition are true of the reals.

Oh yes I can! If I extend the existing axioms without changing the number space. What do you think the Internal Set Theory is? I've mentioned this to you several times already. How many more times do I need to say it again? Or are you going to confuse it with hyperreals again?

For how long will you keep believing that non-standard analysis is dependent exclusively on extending R?
03-13-2011
User Name:

Quote:

Originally Posted by Mario F.

Haven't I been clear enough already? In non-standard analysis, an infinitesimal is a quantity different than 0. In R, it is defined as a number which absolute value is smaller than any number in the form of 1/n, where n is any natural.

Here there's a clearly a distance between 0.999... and 1.

How hard can it be, to understand that nonstandard analysis is nonstandard?

Quote:

Originally Posted by Mario F.

Oh yes I can! If I extend the existing axioms without changing the number space. What do you think the Internal Set Theory is? I've mentioned this to you several times already. How many more times do I need to say it again? Or are you going to confuse it with hyperreals again?

For how long will you keep believing that non-standard analysis is dependent exclusively on extending R?

I haven't confused IST with anything. I've purposely ignored it, just as you've ignored the 50+ standard analysis proofs I've given you.

IST is, by definition, an nonstandard extension(it adds axioms, what else could it be?), and thereby just as irrelevant as the other tangential counterexample you proposed, the hyperreals.

It's like you're trying to talk naturals with non-Peano axioms. It's irrelevant unless specified.

You can correctly say(assuming it's true, I've not taken the time to study 2 irrelevant subjects) "Under IST, 1 != .999..." But the fact it's true under IST doesn't have any bearing on R.
03-13-2011
C_ntua

Quote:

Originally Posted by Mario F.

Basically the number system falls apart when dealing with these quantities.

That is exactly what I am trying to say, well honestly my phrasing is bad and I am naturally really bad expressing myself, but still the infinity is the issue. And for me the way the system falls apart is the difference between something being static and something actually defined as growing. You can of course use both but then you have to keep in mind that when you say "0" you mean "0+-e" where "e" can be seen as some quantity as small as possible. The reason you have this error is because the quantity you are measuring can be something that is growing, as infinity does. You can see that it is growing towards a limit, 0, but you cannot prove in any possible way that it isn't 0+e or 0-e.

If you don't like the 0+e then you can write it as 0.000.....00001. If you don't like the fact that when you say 1 you mean "1+-e" you can see 1 as 0.999.... or 1 or 1.000...0001. In this case, yes, 0.999... is equal to 1. It is its lower possible limit.

In classical calculus you see numbers as specific quantities where you can order them and even if the have the slightest distance (even e->0) then they are different numbers. Using this terminology and trying to prove that 0.999... = 1 is contradictory for me. You are simply expanding the meaning of numbers or your system as a whole and that expansion has some immediate logical results.

Note that using the "e" is just a way to see it. You can see it differently if you want by changing the meaning of "=". Thus you can say that x=1 even if x is not 1. You see in classical calculus if I tell you x=1 then you can say that x IS 1. Here it is not true, because x can be 1+-e where e->oo or if you prefer x = 0.999... or x = 1.000...0001.

Or lets use "0.999.... = 1" as something proven, since most agree with this. Then I would go and say that every equation you get with a variable you would actually give 3 possible results. One a quantity that is specific and two with the meaning of limits to that quantity.

All of the above are in the same wavelength, the phrasing is different, the logic maybe different, but in all cases I believe that when you use infinity in any way, the infinitesimal has to be defined and it is used either you want it or not. The same way you talk about something going bigger and bigger you talk about something going smaller and smaller. And generalizing, something that goes towards a limit. Lets name that number? OK, but then all the numbers just got a some special qualities. Or, you just re-define the signs of order (=, <, >).
03-13-2011
User Name:

It doesn't fail when dealing with any quantities. It deals with them the way it deals with them, and its way is the correct way for it. For different handling, redefine. But different isn't better, it's just different. Mathematics are the way they are because we define them that way. We define what is correct by axioms, and the axioms of R say, in many different ways, that 1 = .999... It doesn't matter if it is unintuitive.
03-13-2011
Mario F.

Quote:

Originally Posted by User Name:

I haven't confused IST with anything. I've purposely ignored it, just as you've ignored the 50+ standard analysis proofs I've given you.

I haven't ignored them. I've consistently told you you weren't successfully providing a proof.

Quote:

Originally Posted by User Name:

IST is, by definition, an nonstandard extension(it adds axioms, what else could it be?), and thereby just as irrelevant as the other tangential counterexample you proposed, the hyperreals.

To my knowledge there's no such thing as a "nonstandard extension" in mathematical theory. If there is I'd like you to tell me what that is. IST is a conservative extension.

If however you can't see the difference between extending a number space and extending the axioms governing a number space, or that you can't understand the fact different methodologies have been applied to studying R, or even that IST has already had practical uses in R in fields like hydrodynamics and harmonics, there's nothing I can do for you. You simply refuse to accept the fact there is indeed other methods to calculus besides limits.

Quote:

Originally Posted by User Name:

You can correctly say(assuming it's true, I've not taken the time to study 2 irrelevant subjects) "Under IST, 1 != .999..." But the fact it's true under IST doesn't have any bearing on R.

Yes, I believe you can correctly say that you haven't studied these subjects. Because your conclusion that this has no bearing on R, demonstrates that.
03-13-2011
whiteflags

Quote:

The point is there that you have one less decimal digit.

Which means that there is no point. You're telling me that because we have to use a digit from 0.9... to write the answer that somehow this 9.9... is less than other expressions of the same number.

9.9... is ten.

Maybe since we're abusing limits to make a point I can have the same courtesy.

lim x -> oo x - 1 = oo.

Which has nothing to do with the value, but it does communicate to me that 9.9... still has infinite digits. We're taking one digit from infinite digits and there are still infinite digits in the limit. So where you're coming from can't be right either. And when I said it has nothing to do with the value... that also means that 9.9... is a number, not a process. A number expanded to infinity like 10.0... or 9.9... cannot be smaller than a simple 10.
03-13-2011
C_ntua

Quote:

Originally Posted by whiteflags

We're taking one digit from infinite digits and there are still infinite digits in the limit. So where you're coming from can't be right either. And when I said it has nothing to do with the value... that also means that 9.9... is a number, not a process. A number expanded to infinity like 10.0... or 9.9... cannot be smaller than a simple 10.

Why is 9.9... a number or a process when I clearly define a process as a quantity going towards a limit? Isn't 9.9... going towards 10?

And with your logic we have:
9.99.... = 10

So if I add one more 9 the equation is still true. One more nine and the equation is still true. Lets hold this thought. And go to
e>0
x = y
x + e > y

Is the above true? I was taught it in school, it has to be true :)

But the you have the case that
9.999... + e > 10
correct?

Then lets say
e = 0.000....0009
9.999... + 0.000....0009 > 10
so the "adding one more nice" thing isn't true anymore???

Your logic creates a paradox.

So no matter what you do some other theorem will be proven wrong. In order not to change everything you have to simply take away the
e > 0
which is not true if you use the definition of "it is bigger if it differs by an amount greater than e->0" so then you can say that
e = 0.000....0009 -> 0
and there is no paradox.
But without re-defining the ">" sign I don't see how you can avoid such paradoxes!

EDIT:
The use of "..." might be tricky, since there is no actual order, to do 9.999... + 0.000....0009. But if you go down that path then you shouldn't use anything that involves ordering of the digits which includes the multiplication of 9.999... + 10. Either you see the order as something you cannot really touch or play with or you see it as a specific order which continuously grows and there is something that you can imagine grows faster by one digit.
For me all of this doesn't make that sense that is why I would stick of not using the order and the multiplication of "0.999... * 10" is not proven and you cannot use anything that will shift the digits because you are just entering an area where things are not defined...
03-13-2011
whiteflags

Quote:

Isn't 9.9... going towards 10?

Nope! 9.9... has a strict equality with ten. There are infinite digits, and the value is 10.

If you wish to seriously apply limits to this concept, then if 9.9... is 10 at the limit, and we have infinite digits, it is quite contradictory to say it is not 10.
03-13-2011
C_ntua

Quote:

Originally Posted by whiteflags

Nope! 9.9... has a strict equality with ten. There are infinite digits, and the value is 10.

If you wish to seriously apply limits to this concept, then if 9.9... is 10 at the limit, and we have infinite digits, it is quite contradictory to say it is not 10.

So it is a number that creates a paradox? In other words you are just accepting that the system has some paradoxes? Which is not even possible in something theoretical like maths?
03-13-2011
whiteflags

If it is a paradox, I'm not realizing it. It must not be as world crushing as some believe.

This entire discussion is merely the fault of the decimal radix system. People seem to be capable of holding contradictory thoughts in their head.

For instance, there is proof that 0.3... is 1/3 and it is finitely expressed in base 9. But the problem doesn't go away for some numbers by definition and we can imbue the problem in arbitrary numbers.

0.0...

Is it zero or going towards the smallest fraction of 1? Does it matter if infinity has no true end? Aren't you just imagining it's bigger or smaller?

And I feel sorry for circles with a 10pi diameter. Apparently, even though they exist, there is no value for 10pi because it isn't finite.

It's a small and inconsequential 'problem' if you make it one.
03-14-2011
Yarin

Quote:

Originally Posted by C_ntua

Lets say index number 5. With your logic
Sum the 8 and 3, we get 11, keep the second 1, add the last 1 to the last sum.
So you have: 4.22221
So how exactly do you get that any digit will be 2? It rather seems that there will always be a digit that will be 1....

Will there be? You stopped calculating at the digit in question, you can't do that! In order to answer what a certain digit is, you must calculate any remaining digits that will effect the outcome. In this case, any digit beyond the '.' will be a 2 *every time*.

Quote:

Originally Posted by C_ntua

If you are implying that there is not last digit at all then how do you say that you the sum (1.999...) has infinite 9s? How do you know that all the digits are 9?

That's a strange question, the very text "999..." means "9s then continue on forever".

Quote:

Originally Posted by C_ntua

e>0
x = y
x + e > y

Then you would agree that
e = 0
x = y
x + e = y
?

Quote:

Originally Posted by C_ntua

Then lets say
e = 0.000....0009
9.999... + 0.000....0009 > 10

Problem is, 0.000....0009 = 0. ;-) I'm guessing you'd like to say that 1 - 0.000....0009 = 0.999..., which, is true, but, so is this:

0.999... + 0.999... = 1.999...
1.999... - 0.999... = 1
03-14-2011
Sang-drax

Quote:

Originally Posted by Mario F.

Oh yes I can! If I extend the existing axioms without changing the number space. What do you think the Internal Set Theory is? I've mentioned this to you several times already. How many more times do I need to say it again? Or are you going to confuse it with hyperreals again?

You have been reading too much on Wikipedia. No one is using those constructs for anything serious. :)

What's going on here? You started out by asking in post #30

"Isn't it self-evident that an irrational number composed entirely 9s is smaller than its closest higher integer?"
Then a lot of proofs were presented that in fact 0.999... = 1. Then you brought up non-standard analysis.

How many do you think were talking about alternative sets of axioms?
03-14-2011
Mario F.

Quote:

Originally Posted by Sang-drax

What's going on here? You started out by asking in post #30
"Isn't it self-evident that an irrational number composed entirely 9s is smaller than its closest higher integer?"
Then a lot of proofs were presented that in fact 0.999... = 1. Then you brought up non-standard analysis.

How many do you think were talking about alternative sets of axioms?

Actually I've brought non-standard analysis early on in this discussion, not just now. I've been mentioning IST for quite some time.

But to answer your question directly, I no longer question the fact 0.999... equals 1. I've said this repeatedly. This may be true of R and of limits. This has been as much a place of debate for me as of learning and I have had no problem in changing my views as I did.

I just presented on that post you quote what I believe is a corollary of my position: 0.999... equals 1 in classical calculus employing Limits. But this can't be proved. Definitely not in R where other approaches demonstrate them to be different. Quite frankly people may thwart this as they want, but the whole issue of 0.999...l equaling 1 is dependent on how a certain method or axiom understand the figure of an infinitesimal. And an effort to introduce infinitesimals into standard analysis has already been made, which does not affect the number space. If you think this isn't worth of mentioning when I see these "proofs" pretending to prove something for R, I don't know what to say. If that were true, non-standard analysis wouldn't be possible, because at least for IST, ZFC axioms all remain valid. So, proving in ZFC Reals that 0.999... equals 1, would mean that IST (a valid and accepted extension) should have not been formulated, when it fact it was developed exactly to provide a means to make exactly these type of distinctions between real numbers because it is clearly understood as a limitation of classical calculus.

As for you dismissal of non-standard analysis and real-life applications of IST, I find them curious. I presented already other links to them (and no, they weren't wikipedia) many pages behind, so forgive me if I don't go search for them or do another research project on my search engine. But more curious is not so much that you negate any real-life uses, but that you actually seem to pretend this has an effect on the validity of these axioms or that they shouldn't be worth of mention. Particularly because, since your peers in this thread are so deeply interested in being specific about what constitutes a proof on R, IST doesn't change this number space in any way and it is considered a valid set of axioms.
03-14-2011
C_ntua

Quote:

Originally Posted by Yarin

Will there be? You stopped calculating at the digit in question, you can't do that! In order to answer what a certain digit is, you must calculate any remaining digits that will effect the outcome. In this case, any digit beyond the '.' will be a 2 *every time*.
...
That's a strange question, the very text "999..." means "9s then continue on forever".

I don't think we can calculate something that goes forever now can we?
You cannot use it as something that implies time which has nothing to do with a specific quantity, static, irrelevant of time. Unless you have an actual way of calculating what you want. I can very well accept that, as long as you use a logic that is correct. Saying that 0.(3) + 0.(1) = 0.(4) because *all* digits are 1+3=4, that I can accept. Won't be my preferred expression, but still it make sense for me. Saying though that you do something forever, or there is a shift of infinite digits etc etc should not make sense by definition of infinity.

Quote:

Problem is, 0.000....0009 = 0 ;-)
I'm guessing you'd like to say that 1 - 0.000....0009 = 0.999..., which, is true, but, so is this:

0.999... + 0.999... = 1.999...
1.999... - 0.999... = 1

why is 0.000...0009 = 0?

What I like to say is that all this lacks some definition, which more or less is an infinitesimal which is non-zero. Then we can define better decimals with infinite digits as a whole and have no paradoxes.

If we use infinitesimal I would claim that automatically by definition you have some error in your measurement, since you claim that you have a quantity that you cannot distinguish from 0. Then we can define two equalities. One that is absolute and one with having an error a quantity that is an infinitesimal.

Then we can think how we can actually distinguish two quantities of having or not an error. The same logic quantum physics use. There is no real sense declaring something as the "exactly X" if you have no way of measuring exactly anything. It is like asking a blind to speak about colors.

So I would personally conclude that equality has an error, always, if you use infinitesimals. So saying that 0.000...000999... is equal to zero or it -> 0 is the same thing.
03-14-2011
ಠ_ಠ

Quote:

Originally Posted by C_ntua

why is 0.000...0009 = 0?

because 0.0...9, by it's very definition, has an infinite number of zeros, every single digit to the right of the decimal point is a 0
03-14-2011
Mario F.

It's pretty much the same thought behind an infinitesimal being zero with traditional calculus. Simply, you cannot have some entity beyond infinity. So an infinite number of zeroes followed by yet another numeral different than zero makes no sense in this methodology. An infinite quantity by definition excludes the possibility of a "beyond" or a "at the end of".

Of course, this is open for debate since the Compactness Theorem proved that we can have infinitesimals (and by extension such numbers as described here). But still 0.000...9 being equal to zero is pretty much valid in traditional calculus, I suppose.
03-14-2011
NeonBlack

I'm just pointing out that calculus does not use infintesimals. Newton's original formulation used them, but they were abandoned in favor of limits when calculus was formalized. (Probably for some of the same reasons discussed in this thread)
03-14-2011
Mario F.

Indeed.

edit: it's another one of those interesting inconsistencies (limitations?) of Real numbers with the current calculus methodologies. On this case it seems that at infinity numbers break apart. You no longer seem to have a continuum. What comes after 0.888...?
03-14-2011
C_ntua

Quote:

Originally Posted by ಠ_ಠ

because 0.0...9, by it's very definition, has an infinite number of zeros, every single digit to the right of the decimal point is a 0

But 9 is not a 0.

Quote:

It's pretty much the same thought behind an infinitesimal being zero with traditional calculus. Simply, you cannot have some entity beyond infinity. So an infinite number of zeroes followed by yet another numeral different than zero makes no sense in this methodology. An infinite quantity by definition excludes the possibility of a "beyond" or a "at the end of"

Don't disagree, but then the "adding another 9" or "there is always another 9" on 0.999... and any other number make no sense either.

So in the end the definition of infinite decimal digits is still missing. For me is the same thing as trying to figure out if you can have infinite apples in a basket with infinite volume. Doesn't really make that sense. But here we are saying that you can have infinite number of digits in an infinite possible "space" for those digits.

In a way a better definition is needed for decimals with infinite digits or a more complete one.
03-14-2011
Mario F.

Naturally I expect no less than a number like 0.000...9 to exist is some way or another. I think anyone in their sane mind does (including our friends in here). In fact, if we adjust the language a bit, it can be more easily perceived, I suppose: "it's a number that has all 0s for decimals with a 9 at an infinite distance".

There's however very little practical use for it and I expect that to be what Sang-drax meant earlier. The reason infinitesimals were dropped from traditional calculus is probably because we don't find that many uses for them anyways -- or at least didn't at the time -- and their introduction posed serious problems then (it's not up until the last 40 or 50 years that a serious effort was given to reintroduce the infinitesimal in R with valid axioms). Limits were (or are) otherwise good enough for almost anything concerning mathematical analysis. So we stick to them until a new boundary of knowledge is required to again move the field of calculus (limits were not the first, and for sure won't be the last).

Of note however the fact that even with infinitesimals, a number like 0.000...9 still poses serious problems. That is, non-standard analysis doesn't fix all the issues with R either. I cannot find anywhere how a number like that can be treated in R with the current non-standard methods. Maybe I'm missing something, but I suspect there isn't. And if indeed there isn't, guess what's the candidate to "take its place", much the same way 1 took 0.999...'s in traditional calculus?

EDIT: I suppose this is all the holy grail of mathematical analysis. In a world were non-standard methods took hold of calculus we would gain the infinitesimal and all this "nonsense" of 0.999... being equal to 1 would be a thing of the past. Still, we would be no closer to deal with the problem of dealing with a rigorous approach to Reals. 0.000...9 still satisfies the non-standard analysis definition of an infinitesimal. So we would still be "gobbling up" numbers and all sorts of "proofs" would come up by everyone claiming that 0.000...9 = 0.000...1 (or something in those lines). In a distant future we may one day come up with a formulation that indeed once and for all closes the issue of rigorously representing R in all its glory. But that is for now a pipe dream. My only beef?... that we call "proofs" to these stepping stones on that long road, knowing full well we didn't reach the end of it.
03-15-2011
ಠ_ಠ

Quote:

Originally Posted by Mario F.

Naturally I expect no less than a number like 0.000...9 to exist is some way or another. I think anyone in their sane mind does (including our friends in here). In fact, if we adjust the language a bit, it can be more easily perceived, I suppose: "it's a number that has all 0s for decimals with a 9 at an infinite distance".

nope, if you were to ask me "how many how many numbers to the right of the decimal point are 0" I would answer "all of them". The very definition of this number forbids the existence of the 9
03-15-2011
C_ntua

Quote:

Originally Posted by ಠ_ಠ

nope, if you were to ask me "how many how many numbers to the right of the decimal point are 0" I would answer "all of them". The very definition of this number forbids the existence of the 9

Yes, but you can also answer "infinite 0s and then one 9".

Might not make sense to you but that is never enough. Unless we use some more proper definitions about decimals with infinite digits, which we don't really.

You have to answer what is the quantity of digits.
If that quantity is defined as infinite then you can add one more digit as you can do to any quantity.
If you define it as the upper bound of a quantity, that you can never reach, then you cannot have infinite digits.
If you define it as at the upper bound of a quantity, which you can reach (the maximum) then you are OK. But if you have one less digit, you no longer have the maximum number of digits, but exactly one less than the maximum.

I believe infinite digits is just used carelessly. I would suggest that a proper definition to be given to it prior any attempt for proof so we can be on the same page.
03-15-2011
whiteflags

Quote:

Yes, but you can also answer "infinite 0s and then one 9".

I think that you're missing a critical point here. 0.0...9 is not a number.

If we take your interpretation:

Quote:

Yes, but you can also answer "infinite 0s and then one 9".

We would have to surpass infinite places in order for the place where 9 is to exist. Thus 0.0...9 is not a number. For anything not a number, things we know, like

e>0
x = y
x + e > y

Are going to be meaningless.

If you mean to say something else, like eventually there will be a place that definitively makes 0.0... smaller than other expressions of zero, then you could set about showing us. You'd be wrong because any number not zero is obviously not zero, though.
03-15-2011
C_ntua

Quote:

Originally Posted by whiteflags

I think that you're missing a critical point here. 0.0...9 is not a number.
...
We would have to surpass infinite places in order for the place where 9 is to exist

I don't think you read the rest of the post.
You have to define what "infinite places are" in order to use that argument. I gave three possibilities of how you can see infinity. There can be more, but you have to define it first, otherwise we will keep just making circles.
03-15-2011
Clairvoyant1332

I think that in order for a numerical representation to be considered a number, you need to be able to define any given digit.

In the case of 0.0...09, you can't define what digit the 9 occurs at therefore it isn't a number. For something like 0.999... or 0.1666... we can. For example, for 0.166... we can definitively say the 284489286th place after the decimal is 6. For a number like pi, we can say the 4th digit after the decimal is 5. The difference between a rational and an irrational is that for a rational, its finite numerical representation (for example "0.333...") tells us exactly what every digit is, while for an irrational like pi we can't know every digit because it's non-repeating and non-terminating (but with enough calculation we can determine any given digit).
03-15-2011
C_ntua

Quote:

Originally Posted by Clairvoyant1332

I think that in order for a numerical representation to be considered a number, you need to be able to define any given digit
...
In the case of 0.0...09, you can't define what digit the 9 occurs at therefore it isn't a number

I clearly defined it as "it is one digit after infinite digits".
Not to sound crazy, my point is that if you cannot add a something to infinity, you cannot subtract either, thus you cannot "shift" the digits of 0.999...
True?
03-15-2011
ಠ_ಠ

Quote:

Originally Posted by C_ntua

I clearly defined it as "it is one digit after infinite digits".
Not to sound crazy, my point is that if you cannot add a something to infinity, you cannot subtract either, thus you cannot "shift" the digits of 0.999...
True?

you cannot shift the last digit of .999..., which is why 10*0.999... - 0.999... = 9, and not 9.0...9 (which I would argue is also = 9, if it could exist)
03-15-2011
User Name:

Quote:

Originally Posted by C_ntua

I clearly defined it as "it is one digit after infinite digits".
Not to sound crazy, my point is that if you cannot add a something to infinity, you cannot subtract either, thus you cannot "shift" the digits of 0.999...
True?

It's different, because you can say every digit is a 9. You cannot, however, say which digit of 0.000...9 is nine. You can say after infinity, but after infinity doesn't exist. For example, you can write .999... as a sum of digits to powers of 10, but you cannot write 0.000...9 as a similar sum. Or can you? A try wouldn't hurt, I guess. :)

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