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 Dross
 Power Member
Re: 0.9999....(recurring) = 1?
There is a welldefined concept of what a limit is. A human's inability to calculate a number using an infinite number of decimal places is very much not what is at issue here. Just because you could never actually add them by writing on a piece of paper does not mean anything as far as the decimal expansion is concerned. the sum of 0.9(recuring) does not have to "reach" it's "last" digit (even though we might define it in a way that might suggest that it does, this is again our shortcoming)  it is not "in time", so to speak. Every digit has already been added, it always has been and always will be. A human must add them one at a time, but I suppose you could think of adding them all at once. I really don't know how to express it another way  I've given you a sound argument which basically says that 0.9(recuring) = 1, simply by the way we define the number 0.9(recuring). But I am not sure about what it will be after infinite, larger than any "many", additions. So who is sure? and how?
I am sure. Again, we have a rigorous concept of limits to show that this is the case.
Bad speling makes me [sic]
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
"I am sure. Again, we have a rigorous concept of limits to show that this is the case."
A rigorous limit concept to show as closer as we can, then imagine the reaching part??
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 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
Instead of quoting the closing sentence of Dross's post and distorting it with rhetoric, how about you try to find something actually wrong with it. As far as I have seen, you have not made an attempt to do so. Is it because you can not?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
Ricky, sorry I miss the formula in latex and the defination. But you know people like you within the system have tons of capacity to make definations or rules.
So I need to disprove the defination of 0.999... promoted in post #51, all right.
But first allow me to distinguish some common mistake among some, though not all, young mathematicians. The point is that a ∞ simbol does not necessarily mean a infinity in practice.
For example, 0/0 and ∞/∞ problems. In fact they are not, 0/0 has no defination, so does ∞/∞. 0 should stand for an infinitesimal a shrinking variable, as well as ∞ should represent a growing variable. And some textbooks express them with quotation marks.
Another instance might be when you say f(x) goes to infinity as x goes to infinity, you mean f(x) diverges.
∞ in the expression " limit(n>∞) An" does not mean real infinity, the reached state by its clear defination in the majority of textbooks, which can only provide a finite Nepsilon proof.
So if you interpret the infinity under a limit symbol as real, you've already make a closerreaching imaginary leap.
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 Dross
 Power Member
Re: 0.9999....(recurring) = 1?
George,Y wrote:∞ in the expression " limit(n>∞) An" does not mean real infinity, the reached state by its clear defination in the majority of textbooks, which can only provide a finite Nepsilon proof.
Granted, there are textbooks out there that don't provide such a definition  my main calculus textbook from first year didn't Here's one anyhow, see if it makes sense to you.
Last edited by Dross (20061021 07:12:57)
Bad speling makes me [sic]
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
The defination you've provided is not about a series but about a function.Thank you for this defination, for it does not contradict my argument.
then do you want to conclude that ? I doubt if you've find a book with this conclusion, and your replacement of N by x does not garantee the difference being eliminated to 0.
Last edited by George,Y (20061021 13:08:27)
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 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
I guess the real question George is why you think we (or anyone) would ever conclude that or even think of concluding it.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
Because Dross want to prove 0.9+0.09+0.009+...=1 And if he use a limit proof, he is supposed to get 10.90.09...=0. Only by this way can Dross get a reaching backup by limit theory. epsilon means still difference, not 0 though can be any small, doesn't it?
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 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
By adding a finite number of terms in that sequence we can get arbitrarily close to 1. If you define epsilon to be 10^100,000,000, we can find a delta which suffices. If you define epsilon to be 100,000,000,000,000^100,000,000,000,000,000,000, again, we can find a delta. We can get as abitrarily close to 1 as we want.
Now go back to real analysis. Write out each s_n, s_n being the sum of all terms up to n.
s_0 = 0 s_1 = 0.9 s_2 = 0.99 s_3 = 0.999 s_4 = 0.9999
And so on. We define the summation of an infinite series to be the limit of this series as it approaches infinity.
If you don't accept that definition, that's fine. You don't have to. However, by not accepting it, you throw most, if not all, of real analysis out the window.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
I am glad that you finally realised there has to be a defination to fill the gap between finite logic to infinite framework. And that's the point I've been standing at.
Hence any proof of 0.999...=1 is a somewhat equivalent of defining 0.9+0.09+0.009...=1 when by finite logic we know the addition approaches it, and they lose their use. We all know perfectly 0.999... (growing) approaches 1, then why just state honestly we can define it as 1 as it imaginarily come to infinite digits?
Still, I am happy we get a consensus.
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 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
Hence any proof of 0.999...=1 is a somewhat equivalent of defining 0.9+0.09+0.009...=1 when by finite logic we know the addition approaches it, and they lose their use. We all know perfectly 0.999... (growing) approaches 1, then why just state honestly we can define it as 1 as it imaginarily come to infinite digits?
Not quite. We do define what we mean when we say "sum of an infinite convergent sequence." However, it is not because of the problem 0.999... = 1 do we define it so. We define it so because large parts of real analysis need sums of infinite series and this definition fits with everything we need.
It is only an side effect that this means 0.999... = 1.
Now I know you disagree with the above definition. But what I'm not sure is if you understand how big of a hole this makes in real analysis. Reimann sums depend upon this definition to work, and reimann sums is how we arrive at a definition for an integral. By not allowing such a definition, you eliminate all possible single variable, multivariable, and vector calculus.
By denying this defintion, you destroy all of integral calculus.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
Yeah, I know that anything relating to limit have something to do with this defination, derivativestangents, magnitudes(with nabla notations),and all kinds of integrals.
But the defination only assume their accuracies instead of their moreorless nature. I bet a electrodynamic professor will advice you to use integrals wisely because there are no infinitesimals in practice. Rather, there are sufficiently small masses of electrons and small scanties of electrons. Thus the theoretical results produced via infinitesimals may be a little bit different from a reality consisting of discrete particles. And in economics you will meet a great trouble if you interpret the Average Fixed Cost =Fixed Cost/Quantity as infinity when no product is produced. The solution is to realise the quantity should be integers instead of reals, which have been adopted only for the convenience to apply calculus.
Hence the exactness might be some pure imagination in mathematicians' mind, and the difference between "exactly the same" and "just a little bit different" can be too small to test. And when it can be tested, they just deny the reality. For example, they blame the reality cannot give a perfect circle(a polygon with infinity sides).
And I doubt the importance you've stated, for before Georg Cantor the calculus had been on a "closing" basis for so many years and it had been applied so well.
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 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
I bet a electrodynamic professor will advice you to use integrals wisely because there are no infinitesimals in practice.
What does the nonexistence of infinitesimals (to our knowledge) in nature have to do with integrals? Integrals are done on the assumption that there are no infinitesimals.
And in economics you will meet a great trouble if you interpret the Average Fixed Cost =Fixed Cost/Quantity as infinity when no product is produced. The solution is to realise the quantity should be integers instead of reals, which have been adopted only for the convenience to apply calculus.
0 is not an integer?
Hence the exactness might be some pure imagination in mathematicians' mind and the difference between "exactly the same" and "just a little bit different" can be too small to test.
We can not infinitely measure anything. Thus, any statements about the exactness of the universe is pure speculation and nothing but.
You continue to be talking about applicative mathimatics. But real analysis is pure mathimatics. Do you not see the difference?
And I doubt the importance you've stated, for before Georg Cantor the calculus had been on a "closing" basis for so many years and it had been applied so well.
"closing" basis? Can you expand on this? And to imagine that real analysis has not changed since Cantor was around is amazing.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
"Integrals are done on the assumption that there are no infinitesimals." The sum will never reach the amount to which the limit approaches unless the pieces being divided into fulldeveloped infinitesimals.
"0 is not an integer?" First, having to be Integers but not all of them. When applying integers, you are not dellusioned to foolishly devide an amount by 0, and you know perfectly the denominater is an integer Except 0. But common economic books give a continueous curve to plot the function, with "AVC" beside vertical lines on the left, which is definately misguiding. If a discrete step by step, no one will make that mistake.
"closing" basis? yes, N∈ or Delta∈ defination by Cauchy, which does not state "reaching" in any conditions. Besides, do you know where epsilon comes from? One book I've read say it is from the ancient greek word " error".
"We can not infinitely measure anything. Thus, any statements about the exactness of the universe is pure speculation and nothing but." Actually infinity concept is fundamentally flawed proven by simple reasoning and have no reason to exist, which I shall illustrate later.
Last edited by George,Y (20061025 15:33:35)
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 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
I've said this before George, but I'll repeat it again. When dealing with real numbers, whenever we talk about infinity, we are talking about it within the concept of limits. This includes sums to infinity. This is because infinity is not a real number.
But you ignored that before so I'll assume you're going to ignore it again. How's that for induction?
"Integrals are done on the assumption that there are no infinitesimals." The sum will never reach the amount to which the limit approaches unless the pieces being divided into fulldeveloped infinitesimals.
We define the sum of an infinite series to be the limit that the sum approaches. It is this definition of "sum of an infinite series" which Reimann used and it is Reimann Sums which are used to define integrals.
No infinitesimals even mentioned.
yes, N∈ or Delta∈ defination by Cauchy, which does not state "reaching" in any conditions. Besides, do you know where epsilon comes from? One book I've read say it is from the ancient greek word " error".
Sum of an infinite series is defined to be a limit. That you think it shouldn't be is where the disagreement is. We have no disagreement on limits.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
Ricky wrote:We define the sum of an infinite series to be the limit that the sum approaches. It is this definition of "sum of an infinite series" which Reimann used and it is Reimann Sums which are used to define integrals.
No infinitesimals even mentioned.
Can you explain Reimann Sum a little bit further? how about the entry of the infinite series? If the series has infinite entries, each entry should be a particular infinitesimal to enture the sum to be a number, I suppose. And this involves a quasizero*infinity trick.
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 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
And what do you mean by infinite series? Can I think of it having infinite entries? So what does "infinite" stand for when you deny infinity as a number?
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 Dross
 Power Member
Re: 0.9999....(recurring) = 1?
Ah, the irony! I opened up Wikipedia to find that the subject of this thread is also the topic of their article of the day
Have a gander at the main page, or the article itself if you're not reading this today.
And yes, I know  "just because it's on the internet doesn't make it true". But seriously, if you disagree with 0.999... = 1, have a look at the article. A REAL look and really READ it  it may just clarify something you didn't even think was important, or didn't think to think about at all. Or it may just be explained in a way that you "get".
Bad speling makes me [sic]
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
Yes, thank you for your surf. I will read it through.
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 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
Under the "skepticism" heading:
Intuition and ambiguous teaching lead students to think of the limit of a sequence as a kind of infinite process rather than a fixed value, since the sequence never reaches its limit. Where students accept the difference between a sequence of numbers and its limit, they might read "0.999…" as meaning the sequence rather than its limit.
Which I believe nails Georges thoughts/opinions/arguments right on the head.
If the series has infinite entries, each entry should be a particular infinitesimal to enture the sum to be a number, I suppose.
I believe I see your reasoning. However, I believe you forget that the sequence we sum is (normally) monotonely decreasing. The above quote would be true if we were doing:
Where c is some constant. Then c must be an infinitesimal, I believe. I'm not very familiar with them. However, we aren't summing constants. Note that the same reasoning applies if: c must still be an infinitesimal. However, if we do: Now we have an infinite sum just as above, but are we summing infinitesimals? Absolutely not. Each term in that sum is not only a real number, it is rational as well. And what do you mean by infinite series? Can I think of it having infinite entries? So what does "infinite" stand for when you deny infinity as a number?
It has infinite entries. But don't think of infinity here as a number, as I have stated. Think of it as a property. Specifically, infinity means nonfinite. And when given this, as I have said before, we must interpret it as a limit as that is the only way we know how to deal with infinity.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Re: 0.9999....(recurring) = 1?
Its impossible for Infinite 0.9 to ever = 1
Beacause there will always be a difference of Infinite 0.1
A.R.B
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
Until now in this topic, having debated so long, Ricky, Dross and I all admit that: 1) 0.999...=0.9+0.09+0.009+...; 2) Either 0.999...=1 or 0.111...=1/9 cannot be proven by any finite logic, and either a defination or an imagination is needed.
And Dross insists the imagination or intuition method, while Ricky favors a defination he thinks useful and essential. I, certainly, deny both.
One day, Dross found a long passage written by a professor on wikipedia, which tries to prove 0.999...=1. Since it looked a big deal and I lacked time, I asked for a pause and promised to challange the 0.999...=1 thing later.
Now it's been too long for a pause, thus I complete the challenge below.
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 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
First of all, let us see what need not be proven. 10.9=0.1, and 10.99=0.01 for sure. And everyone knows the entry or the sum will get closer and closer to 1 so long as the number of digits in the entry of {0.9, 0.99, 0.999,...} get bigger and bigger, or as more and more entries being added in 0.9+0.09+0.009+.... It's foolproof that it "grows" to 1 but so far in any finite step it is still not close enough to reach 1.
So is there a somewhere where it finally, utimately reaches 1? "When the series comes to the infinitth entry or the summation has infinite entries", some mathematicians would reply. Such guess seems by no means to be wild, but by developing it logically, one could easily discover its discontents.
The first one is the famous and smart proof using 10 multiplier: 10×0.999...=9.999... 0.999...=0.999... \9×0.999...=9 Nice substraction! But one thing is puzzling  0.999... has infinite 9 s, and multiplied by 10 doesn't change the amount of 9s by algebraic rules, if any in infinite state. Hence how does the substraction deal with the last 9? Is it substracted as well?
Some one would argue that infinity is docile, and the 9.999... naturally grows one digit to substract the last 9 in 0.999.... It sounds of some sense, but what if this proof: 100×0.999...=9.999... 0.999...=0.999... \99×0.999...=99 Infinity is indeed "docile" previously infinity+1=infinity, and now infinity+2=infinity. But one thing we should pay attention is that the amount of digits now begin to grow or vary now, hence whether the infinite numbers is a final state is doubtful how can a final something begin to change again? And if 0.999... with infinite 9s is 1, what are 0.999... with 1, 2, or 10, 1000 more than infinite 9s?
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 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
Then comes the more convincing argument: that the last 9 is quantitiless, infinitesimal that equals to 0, or quasi0.
It, however, is also doubtful. The basic doubt is that when and how it is gotten? Zeno used to question the quantitiless concept by analyzing a moving arrow: The arrow should reach the mid point before reaching the end, but should also reach the quater point before reaching the mid point,... Each step is finite, and it seems no hope to get an infintesimal or a quantitiless by reducing the quantity, in number.
Still, let us say somehow the quantitiless or the infinitesimal is reached, then how about add the sum up this way? From the last, backwards.
the last 9 is quantitiless, the second last one is also quantitiless, the N last one is too.
Going further, we can say that the halfinfiniteth 9 is quantitiless It has to be otherwise the last 9 is not. (think of any digit number then double the collective 0s after the digit point) Even Entry 1/1,000,000,000 of infinity should be a quantitiless.
And when does the entry turn out to be a quantity from a quantitiless? Which entry? This is the reverse question of when quantities come to a quantitless state.
Altogether, can anyone distinguish which entry separate the quantities on the left and the quantitiless on the right? Or, can anyone illustrate how the quantities grow to the quantitiless gradually?
To be continued...
X'(yXβ)=0
 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
George, the reals are defined not to contain infinitesimals.
You have yet to answer my challenge. How do we deal with infinities in the reals without limits? You seem to be claiming there is some way as you don't accept the limit definition of infinite summation. But you never proposed another way.
If you can get arbitrarily close to a number, as the limit definition provides, in the reals, that is good enough to consider equal. This has everything to do with the fact that there are no infinitesimals.
But it also seems you have entirely hand waved off the proofs provided by the wiki. Let me go through my favorite:
Let a = 0.999... throughout.
a is in A_0 = [0.9, 1] a is in A_1 = [0.99, 1] a is in A_2 = [0.999, 1] ... a is in A_n = [0.999...n...9, 1] ...
Note that because of the infinite number of 9's, this statement will be true for any number of A_n.
But, A_0 is a subset of A_1 is a subset of A_2 is a subset of ... A_n is a subset of A_n+1 ...
Also, note that with any finite amount of 0.999..., we may get arbitrarily close to 1. Thus, for any e (epsilon) greater than 0, we may find a finite amount of 9's such that 1  e < 0.999...
This means that any real number less than 1 is not in one of the A_n sets listed above. Also, any real number greater than 1 is obviously not.
However, by the nested interval property, and infinite series of intersections of closed intervals is nonempty. That is:
A_0 intersected with A_1 intersected with A_2 ....
Is not empty. But it contains no numbers less than 1 or greater than 1. Thus, it contains 1 and only 1. But remember that a is in every set A_0, ..., A_n, ... and thus, a = 1.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Topic closed
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