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#76 2006-11-19 02:42:12

Ricky
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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 non-empty.  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..."

#77 2006-11-19 03:37:21

Dross
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Re: 0.9999....(recurring) = 1?

George,Y wrote:

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.

No, I do not admit (2) above - posting as such indicates that you clearly have not been reading my posts. At all. How are we supposed to have what could be described as an inteligent, informed discussion when only one of us is reading the other's posts?

In this thread, or the other of a similar name, you will see that I have written down a rigorous, non-"intuitive" or non-"call to reason" proof that 0.999... = 1. Read it, and see if you can find a hole in it. Don't just dismiss it.

There are many holes in your arguments (a fairly loose definition of the word accepted), but to be honest I simply cannot be bothered to sit at my computer and shoot fish in a barrel.

Since you:

1) Clearly do not make any attempt to actually read what I've written

2) Point-blank refuse to have anything to do with arguments that could even vaguely be called formal or structured

I will not be posting any more in reply to you - unless you convince me that it is worth doing so.

Last edited by Dross (2006-11-19 03:38:42)

#78 2006-11-19 08:29:09

Ricky
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Re: 0.9999....(recurring) = 1?

No, I do not admit (2) above - posting as such indicates that you clearly have not been reading my posts. At all. How are we supposed to have what could be described as an inteligent, informed discussion when only one of us is reading the other's posts?

Well, you have to base it off of some definitions.  Without definitions, we don't even have any numbers.

But I know this is not what you meant.

However, by your proof, I'm assuming that you mean your infinite summation.  This relies on the definition of an infinite summation.  The way mathematicians choose to define it is a limit approaching.  And this only makes prefect common sense.  If we can get arbitrarily close to a real number, that is the same as being at the real number since there are no non-zero infinitesimals in the reals.  But none the less, I'm fairly certain it's still considered a definition.

You can disagree with a definition all you want.  But at soon as you do, I'm talking about apples and you're talking about oranges (you being George), and anything we wish to talk about will not make sense.  Luckily for me though, every other mathematician is talking apples as well.

When life gives you oranges... look for a lemonade stand

Edit: And what the heck is finite logic, and why do you think finite ______ would ever apply to infinite ______?

"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..."

#79 2006-11-19 09:05:25

Dross
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Re: 0.9999....(recurring) = 1?

Okay... either you're misunderstanding me, or I misunderstood it when George wrote:

George wrote:

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.

I meant I didn't think that the first bit (Either 0.999...=1 or 0.111...=1/9 cannot be proven) was true. Of course we need to define what we're talking about so that we're all talking the same language.

#80 2006-11-19 09:56:42

Ricky
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Re: 0.9999....(recurring) = 1?

Well, he said finite logic.  I still have no idea what that is, so it's hard to say what you can prove with it.

Then there is of course my "Argument from destroying calculus" which George has yet to respond to as well.

"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..."

#81 2006-11-19 15:19:16

George,Y
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Re: 0.9999....(recurring) = 1?

First to Dross, I am sorry to mistaken your stand as Ricky's. I apologize.
But in regard of your limit proof promoted in Pst 26, you, Ricky and I discussed it thoroughly from Post 51 to Post 61. And Ricky actually denied a limit proof in Post 59, and said it requires a defination. Further in Post 61, he admitted that the defination replaces the proof needed to fill in the gap from closing in limit framework to the purpose to prove 0.999... equals 1.
Since so far you haven't refuted him, I supposed that you agreedwith him.

About"in real system any number has a infinite expansion", I want to clarify that reals are defined as infinite sets, for which you can consult Georg Cantor's definatio of reals, read Russels opposition to it and the Barber's story, or consult the wiki page you provided, saying 1 is defined as all the rationals before it-notice all the rationals implies infinite amount of them, and thereby I exclude Reals from finite.

To Ricky, I will challenge the Reals later, which is also infinte based.

X'(y-Xβ)=0

#82 2006-11-19 15:31:15

George,Y
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Re: 0.9999....(recurring) = 1?

Finite logic is addition, substraction, multiplication, division, mathematical deduction and Cauchy's defination of limit(e-N or Delta system), particularly when used in trying to prove 0.999... equates 1, or to prove 0.111...=1/9.

Can either of you prove either of the equities using those method? But keeping in mind: Don't make a defination whenever you Fail to prove.

On Ricky's words in Post 80, you haven't explicitly pointed out HOW I will tear the base of Calculus down-or put in other words, you haven't supported your assertion, so in no way  I can refute or reply you. But don't think in this way you are right because you have not prove the assertion yet.

Last edited by George,Y (2006-11-19 15:37:33)

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#83 2006-11-19 15:36:31

Ricky
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Re: 0.9999....(recurring) = 1?

Further in Post 61, he admitted that the defination replaces the proof needed to fill in the gap from closing in limit framework to the purpose to prove 0.999... equals 1.

No, this definition has the side effect of allowing such a proof.  There are many other ways to prove 0.999... = 1 and if you read the wiki article that was posted, you would know that.

Weird how this definition agrees with all other (more stringent) methods of proof, isn't it?  You know, it's kind of like it all fits together without contradiction.  Weird...

"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..."

#84 2006-11-19 15:42:38

Ricky
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Re: 0.9999....(recurring) = 1?

Finite logic is addition, substraction, multiplication, division, mathematical deduction and Cauchy's defination of limit(e-N or Delta system), particularly when used in trying to prove 0.999... equates 1, or to prove 0.111...=1/9.

Can either of you prove either of the equities using those method? But keeping in mind: don't make a defination whenever you fail to prove.

What about completeness of the reals?  Or the Archimedian principle?  Well ordering of the real numbers?

The following are used in my proof in post #76:

Well ordering of the real numbers
Cauchy e-N definition of limits
Nested Interval property - easily proved with Archimedian principle and completeness.

Please point out the error you believe involved.

"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..."

#85 2006-11-19 15:46:45

George,Y
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Re: 0.9999....(recurring) = 1?

Sure-reals and other similar proofsinvolves a infinity concept. And the most of proofs wiki provided have been refuted in this thread and the other one discussing the same topic. Also, the page has some false proofs like using a calculator to fool elementary students. Do I need to respond to this kind of proofs? I would be too torlarant if I do so when you haven't asked.

Would you be kind enough to verify the defination you have mentioned- is the defination to denote a sum of infinite entries as the limit of a sum of Finite entries?

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#86 2006-11-19 15:54:05

George,Y
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Re: 0.9999....(recurring) = 1?

Archimedian principle and completeness- I will challenge this principle quite soon. Sorry I haven't aquired the knowledge of interval proof, but they should be  collarabotively false with Archimedian principle equalavent to Reals completeness.

X'(y-Xβ)=0

#87 2006-11-19 15:58:45

Ricky
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Re: 0.9999....(recurring) = 1?

Would you be kind enough to verify the defination you have mentioned- is the defination to denote a sum of infinite entries as the limit of a sum of Finite entries?

I'm not certain which definition you are talking about.  Can you clarify?

Archimedian principle and completeness- I will challenge this principle quite soon. Sorry I haven't aquired the knowledge of interval proof, but they should be  collarabotively false with Archimedian principle equalavent to Reals completeness.

Another basic principle which real analysis relies heavily upon.  Sigh...

Do you wish me to post a proof of the nested interval property?

"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..."

#88 2006-11-19 16:05:23

George,Y
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Re: 0.9999....(recurring) = 1?

I have denied 0.999... -how can I not disprove your proof on Post 76?

and on Post 74 I just pointed out the fallacy of 0.999... because the last 9, or the 9 after infinite digits. Note my disproof is not targeted at 0.9+0.09+0.009+... alone but on the structure of 0.999... also, since the later can be directly decomponented to the former.

The defination:
"No, this definition" on Post 83

Last edited by George,Y (2006-11-19 16:10:55)

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#89 2006-11-19 16:48:04

Ricky
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Re: 0.9999....(recurring) = 1?

I have denied 0.999...

What, and why?

Definition:

"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..."

#90 2006-11-19 16:58:23

Ricky
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Re: 0.9999....(recurring) = 1?

Then comes the more convincing argument: that the last 9 is quantitiless, infinitesimal that equals to 0, or quasi-0.

Are you seriously talking about the "last 9" in 0.999...?  That's a joke, right?

All infinitesimals in the reals equal 0.  What's quasi-0?  Got a formal definition?

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.

Too bad we aren't dealing with the real world, we're dealing with numbers.  And we are fairly certain that we can't break down distances in the real world infinitely anyways.  Ever hear of a plank length?

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.

Why should I let this assumption in?  We never reach an infinitesimal since we never reach 0.  Non-zero infinitesimals don't exist in the reals.

the last 9 is quantitiless, the second last one is also quantitiless, the N last one is too.

Ack!  Again with this "last nine"!  You're the one that's been always saying there is no last 9.

Even Entry 1/1,000,000,000 of infinity should be a quantitiless.

What?!?  Are you really talking about 1/1,000,000,000 of infinity?  Infinity doesn't exist in the reals, so how the heck are you performing multiplication on it?

And when does the entry turn out to be a quantity from a quantitiless? Which entry?

No quantitiless entry, thus, no need for a "turn".

"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..."

#91 2006-11-19 17:49:06

George,Y
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Re: 0.9999....(recurring) = 1?

First, Ricky, I would like to appreciate you for post 89, the post that express the defination explicitly.

Second, my post from 72 to 74 is mostly targetted at Dross's proposition "Imagine you add them all together". This opposition will remain strong if revised a little bit.

Let us say 0 instead of quasi-zero or quantitiless to describe 9/10, which is embeded in you summation on Post 89. That is just fine, and which entry becomes 0? Can you name one? If you admit the ∞ symbol in the summation in Post 89 cannot substitute n in bn, you mean the summation may never be completed, hence you deny the all-added-up proposal and category 0.999... as a GROWING variable.

A variable cannot be a number. You are too familiar with numbers to let a variable be one of them, aren't you?

In brief, how do you interpret blablabla in 0.999..., particularly in Post 76?

Last edited by George,Y (2006-11-19 22:20:15)

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#92 2006-11-19 17:55:39

George,Y
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Re: 0.9999....(recurring) = 1?

Dross refuted my proof that the infiniteth 9 is unreachable by stating sth like" imagine you add all of them together". So long as he had said the "all", I found it appropriate and necessary to imagine the end in the "all" as well, which was a temp concilatory and did not mean I gave up my stand.

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#93 2006-11-20 05:08:03

Ricky
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Re: 0.9999....(recurring) = 1?

Let us say 0 instead of quasi-zero or quantitiless to describe 9/10∞, which is embeded in you summation on Post 89. That is just fine, and which entry becomes 0? Can you name one? If you admit the ∞ symbol in the summation in Post 89 cannot substitute n in bn, you mean the summation may never be completed, hence you deny the all-added-up proposal and category 0.999... as a GROWING variable.

None of them are 0.  Do you understand what is meant by a limit?  Do you understand how a limit can converge to a real number?  And finally, do you understand why when we can get arbitrarily close to a real number, we consider that the same as being at that real number?

From the above quote, it seems as if you understand none of these.

"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..."

#94 2006-11-20 05:10:54

Ricky
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Re: 0.9999....(recurring) = 1?

In brief, how do you interpret blablabla in 0.999..., particularly in Post 76?

What do you mean interpret?  It is a 0. and then an infinite number of 9's after it.

Please point out an error in my post #76.

"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..."

#95 2006-11-20 13:04:55

George,Y
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Re: 0.9999....(recurring) = 1?

Ricky wrote:

Let us say 0 instead of quasi-zero or quantitiless to describe 9/10∞, which is embeded in you summation on Post 89. That is just fine, and which entry becomes 0? Can you name one? If you admit the ∞ symbol in the summation in Post 89 cannot substitute n in bn, you mean the summation may never be completed, hence you deny the all-added-up proposal and category 0.999... as a GROWING variable.

None of them are 0.  Do you understand what is meant by a limit?  Do you understand how a limit can converge to a real number?  And finally, do you understand why when we can get arbitrarily close to a real number, we consider that the same as being at that real number?

From the above quote, it seems as if you understand none of these.

Uhah! It is the quite common trick that one who failed to refute can employ-  he or she will just state that his/her stand is among the "advanced", or "higher" knowledge the challenger hasn't aquired. But we are too familiar with this trick, Ricky, since we've been coaxed in this way ever since elementary school.("We" refers to everyone of us, who have been told something like" In real numbers, this wouldn't be a problem".)

I said I would deal with the Reals later, but I don't want to do it now cuz I need no further idea to disprove you:

Last edited by George,Y (2006-11-20 14:37:20)

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#96 2006-11-20 13:19:12

George,Y
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Re: 0.9999....(recurring) = 1?

Ricky wrote:

In brief, how do you interpret blablabla in 0.999..., particularly in Post 76?

What do you mean interpret?  It is a 0. and then an infinite number of 9's after it.

Please point out an error in my post #76.

It is a 0. and then an infinite number of 9's after it.

So you have interpreted 0.999...

Here, you have used infinite 9s , and it wouldn't be appropriate for me to further it by there are infinite digits after the digit point, would it?

Well, now what does the 9 on Digit Infinite, or the Infiniteth digit stand for?

Does it stand for nonsense?
No

Does it stand for any finite quantity?
No

Does it stand for a 0, or an infinitesimal or a quantitiless?
Yes, reasonably. Because literally it stands for

Also this is on the wiki page.

Now, let's say it is 0 because you once said so.

The digit prior to it represents  also 0. When it does not, the digit after it will represent one tenth of it.

The digit N digits prior to the Infiniteth digit is 0, too, no matter how large the N is.

Let us see what the 1/2∞th digit stands for. It must be 0, because if it is any x>0, Digit ∞ will stand for sth in the x^2 scale.

So it is ok to say that Digit
represents still 0.

Hence what does 0.999... mean now?

From left, each 9 stands for a positive quatity, no matter how small but not equal to 0, from the right, or from middle-anywhere after ∞-1 9s,  each 9 stands for 0, please do tell me when 9 starts to represent 0?

Last edited by George,Y (2006-11-20 13:27:25)

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#97 2006-11-20 14:05:03

Ricky
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Re: 0.9999....(recurring) = 1?

Infinity isn't part of the real numbers, yet you always try to include it as so.  Why?

"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..."

#98 2006-11-20 14:05:59

Ricky
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Re: 0.9999....(recurring) = 1?

Well, now what does the 9 on Digit Infinite, or the Infiniteth digit stand for?

This question has no meaning.  I might as well attempt to name the continents on the moon.

"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..."

#99 2006-11-20 14:14:50

Ricky
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Re: 0.9999....(recurring) = 1?

George, the digits in a real number are countable.  This means that every digit in any real number has a finite position.

You talk about infinite position but this term is just utter nonsense.

"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..."

#100 2006-11-20 14:17:09

Ricky
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Re: 0.9999....(recurring) = 1?

Ricky wrote:

Let us say 0 instead of quasi-zero or quantitiless to describe 9/10∞, which is embeded in you summation on Post 89. That is just fine, and which entry becomes 0? Can you name one? If you admit the ∞ symbol in the summation in Post 89 cannot substitute n in bn, you mean the summation may never be completed, hence you deny the all-added-up proposal and category 0.999... as a GROWING variable.

None of them are 0.  Do you understand what is meant by a limit?  Do you understand how a limit can converge to a real number?  And finally, do you understand why when we can get arbitrarily close to a real number, we consider that the same as being at that real number?

From the above quote, it seems as if you understand none of these.

Uhah! It is the quite common trick that one who failed to refute can employ-  he or she will just state that his/her stand is among the "advanced", or "higher" knowledge the challenger hasn't aquired. But we are too familiar with this trick, Ricky, since we've been coxed in this way ever since elementary school.("We" refers to everyone of us, who have been told something like" In real numbers, this wouldn't be a problem".)

That isn't what I said George, and you know it.  Which of the above do you not understand and accept?  Because if there isn't a single one that you don't accept, then you should have no problem concluding that 0.999... = 1.  You do have that problem, so obviously there is one which you don't.

"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..."