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#526 2007-03-06 04:56:02

Anthony.R.Brown
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Registered: 2006-11-16
Posts: 516

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

To Maelwys
Quote:

" Those aren't infinite numbers, those are real numbers (you're just multiplying 0.9 x 1.1, no infinite-long recursions involved in that). You can't apply normal rules for finite numbers into infinitely recursive "

A.R.B

How many times do I have to explain it to you! they are the start! unless you can show how Infinite Numbers start in a Different way! then you have no argument!!

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#527 2007-03-06 04:59:46

Anthony.R.Brown
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Registered: 2006-11-16
Posts: 516

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

manner. Moderator action has been taken against those who are not civilized by Ricky and Zach. ]

The above obviously means nothing! has no Values! no standards! and no morals!

A.R.B

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#528 2007-03-06 05:00:34

Maelwys
Member
Registered: 2007-02-02
Posts: 161

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

Anthony.R.Brown wrote:

How many times do I have to explain it to you! they are the start! unless you can show how Infinite Numbers start in a Different way! then you have no argument!!

Infinite numbers don't "start" in any way. 0.999... is an infinitely recursive number. So is 0.111... and so is 0.010101.... These numbers each have several ways of being represented (infinitely recursive decimals or fractions), but they are still numbers, independantly of the formula that you might've used to achieve that number.

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#529 2007-03-06 07:04:10

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

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

Maelwys wrote:
Anthony.R.Brown wrote:

To mathsyperson

This is the Sister thread! the other one got no response! lets keep it fare!

A.R.B

By "sister thread" I think he meant that he'd also close the "proof why 0.999... = 1" thread that you created your thread in retaliation to. Now both are closed, and all of the discussion about this should continue on this thread alone.

Maelwys has it exactly right. Your topic was the one containing proofs that 0.999...<>1.
The related topic to that was the one containing proofs that 0.999...=1. Both of those have been closed.

This topic is unbiased to either conclusion. It merely poses the question and invites discussion, and so leaving it open would not be unfair.


Why did the vector cross the road?
It wanted to be normal.

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#530 2007-03-06 09:03:26

lightning
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Registered: 2007-02-26
Posts: 2,060

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

hmmm i never thought of it like that.....oh well


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#531 2007-03-06 19:04:35

George,Y
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Registered: 2006-03-12
Posts: 1,379

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

Some words too stupid too bear:
my 1 divided by 3 = 0.333... is a repeating value
read my Post 513 and you'll know your repeating is still in finite stage.

Ricky wrote:

Because infinity has been banned from being included with the real numbers.

So you know maths, I don't. So all the things on textbooks are right, and I am wrong, OK? Go back to Post 513 and check my apple quiz.

Ricky wrote:

You are "banned" from talking about the infiniteth digit because it doesn't freakin' exist

Again, go back to Post 513 and check the second structure.

Can' you read?! Are you Blind? Do you call sharp shooting only a few words of your opponent without understanding his or their whole passages as "civilized manner"? Well, put on your "civilized" manner, as the way as the king put on his new suit. Carry on, Ruler!


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#532 2007-03-06 19:36:57

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

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

how exactly is 0.333....  in a finite stage? that makes no sense,


The Beginning Of All Things To End.
The End Of All Things To Come.

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#533 2007-03-06 21:22:59

Dross
Member
Registered: 2006-08-24
Posts: 325

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

George,Y wrote:

Can' you read?! Are you Blind? Do you call sharp shooting only a few words of your opponent without understanding his or their whole passages as "civilized manner"? Well, put on your "civilized" manner, as the way as the king put on his new suit. Carry on, Ruler!

To be fair George, you have been guilty of this yourself. And sometimes you don't make a whole lot of sense. Also, what has your "apples" example got to do with anything? Could you please tell all how it is relevant?

George,Y, in post 513 wrote:

Now, let's say there is another structure of 0.999..., infinite amount of 9's but not endless, this doesn't satisfy you either, am I right?

How can there be an infinite amount of digits after the decimal place and they aren't endless (i.e. have no end) - these two are mutually exclusive. Have one, fine - have the other, great - but you can't have them both simply by their very nature.


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#534 2007-03-07 00:17:34

George,Y
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Registered: 2006-03-12
Posts: 1,379

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

luca-deltodesco wrote:

how exactly is 0.333....  in a finite stage? that makes no sense,

no sense? Are you judging me no sense? Or simply you don't understand what I am saying? Okay, I explain, but please, don't you ask anyone for more explanation by just judging "that doesn't make sense".

Maelways has just said the 9's after 0 are just repeating, so naturally I can interpret the 0.999... in his repeating view is still adding digits, and each digit place should repeat the 9. Interpreting the verb "repeating" by adding the same looking 9 isn't too far, is it? So 0.999... of Maelways(I mean his repeating 0.999... not the one in your perception) is still an addition carrying on and not over yet. But at each stage the addition is finite, regarding both the amount of elements added and the poor summation that has not equate to 1 yet.

How about them not repeating? The last 9 shall be present. Not repeating any more means a stop, so the 9 that stops reproducing more 9 is just the last 9.

If he meant only a bunch of repetitive monotonous 9's already. That goes to apple pile problem.

To Dross, Static Apple Pile is relevant to your second question. Finish the second question first, then any new question ask me . Substitution is not that hard. If you type your answer to the second question out, much appreciated.

Last edited by George,Y (2007-03-07 01:11:53)


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#535 2007-03-07 02:03:34

Dross
Member
Registered: 2006-08-24
Posts: 325

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

George,Y wrote:
luca-deltodesco wrote:

how exactly is 0.333....  in a finite stage? that makes no sense,

no sense? Are you judging me no sense? Or simply you don't understand what I am saying? Okay, I explain, but please, don't you ask anyone for more explanation by just judging "that doesn't make sense".

Maelways has just said the 9's after 0 are just repeating, so naturally I can interpret the 0.999... in his repeating view is still adding digits, and each digit place should repeat the 9. Interpreting the verb "repeating" by adding the same looking 9 isn't too far, is it? So 0.999... of Maelways(I mean his repeating 0.999... not the one in your perception) is still an addition carrying on and not over yet. But at each stage the addition is finite, regarding both the amount of elements added and the poor summation that has not equate to 1 yet.

How about them not repeating? The last 9 shall be present. Not repeating any more means a stop, so the 9 that stops reproducing more 9 is just the last 9.

If he meant only a bunch of repetitive monotonous 9's already. That goes to apple pile problem.

The above shows a serious misunderstanding of what we're talking about - just because the 9's in 0.9999... are said to be "repeating" does not in any way mean that they are "still" being put on the end. This, in turn, does not mean that if you take a "snapshot" of 0.9999... at any given moment, it will have some finite number of 9's in it.

The digits do not have to be added at any point - they were always there, they took no time to add and there was never an instant of time in which they weren't all there.

You have some sort of belief that this number undergoes change over time - it does not, it is timeless, just like 0.1 or 0.52 or π.

George,Y wrote:

To Dross, Static Apple Pile is relevant to your second question. Finish the second question first, then any new question ask me . Substitution is not that hard. If you type your answer to the second question out, much appreciated.

What???

"you can get any positive integer amount of 9's out of it, or in other words, you can count to any positive integerth (i.e. 2nd, 1millionth) 9 in it. The question is, listen carefully, how many 9's does the number have?
Infinite!"

You actually haven't been reading what we've been saying. I know this because this is the (supposedly) amazing result you wanted me to "find out" - but we all agree with this, George. Yes, there are an infinite number of 9's after the decimal place in 0.9999... - and the point is?


George, you frustrate my every effort to talk to you in a sensible and comprehendible fasion. I'll be abbandoning this thread until you and/or Anthony start acting like adults having adult conversations.


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#536 2007-03-07 02:49:04

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

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

The digits do not have to be added at any point - they were always there, they took no time to add and there was never an instant of time in which they weren't all there.
-that's not for you, Dross. That's for the repeating 9's belief
___________________________________________________________

Generally, the belief around what the elipse mean in 0.999... differs around people, even among those who believe 0.999...=1. One of the naturalistic interpreting of 0.999... is that one 9 after one 9, then another one, endless, not over. This belief naturally means no one has it out of finite steps yet and the finiteth stop is not over(starting from only one 9, then one 9 after another is not over)
Let's put it another way, even the arguer hasn't count out of finite 9's, how can he ask anyone to equate 0.999repeating to 1?

And this belief is often useful in refuting none believers?-How can you find/get to the infiniteth 9?(Because you cannot count to that) So spotting this belief is useful in avoiding double standard. I know you are frank and you insist already infinite 9's regardless counting to or not. But someone may pick up this double standard sometimes, unconciously.
________________________________________________________________

I do not presume the existence of π, and don't tell me the Pi already exists in real world. So far our drawings of circles are only at best polygons, and whether a circle can exist in reality is not testified or testable. If you say Pi or Circle exist in Geometry concept, I have already challenged the concept of point and line segment. The same arguement applies to points and an arc too. (friendly reminder:anything already in maths textbook ain't inborn correct for me, so quoting maths textbook doesn't appeal to me)

Dross wrote:

"you can get any positive integer amount of 9's out of it, or in other words, you can count to any positive integerth (i.e. 2nd, 1millionth) 9 in it. The question is, listen carefully, how many 9's does the number have?
Infinite!"

You actually haven't been reading what we've been saying. I know this because this is the (supposedly) amazing result you wanted me to "find out" - but we all agree with this, George. Yes, there are an infinite number of 9's after the decimal place in 0.9999... - and the point is?

First thank you Dross for doing me a favor. Your answer is the best refutation to Ricky's conception that not mentioning means no existence. (Like the bell story lol)

Second, You have got infinite 9's already, and no more, the 9's are there, static. So can I say there exists a last 9? Or can I say there exists an infiniteth 9? Since the 9's are not increasing, not moving to the right, can I say there is a 9 at the right end? (Don't you say I cannot count to that, do you?) What does it mean?

Or put it in a more complex way: You shall admit each 9 among infinite 9 bunch stands for a different value, don't you? Okay, since you have got a static amount or quazi-amount of 9's, please find out the smallest 9, or the 9 standing for smallest value, and tell me what it means. Finding such a 9 out of a static bunch is not hard, is it? (Don't count, we've already abandoned this primitive method.) What does it mean?

Or you do? That cannot get by counting means no existence? In this way you might eat your own words.

Last edited by George,Y (2007-03-07 18:44:47)


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#537 2007-03-07 04:10:36

Ricky
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Registered: 2005-12-04
Posts: 3,791

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

George, if I don't understand something you have posted because of miscommunication, then you need to rephrase what you have said.  Telling me to go back and read the same thing over again will not help one bit.  You talk about the infiniteth digit, and I have shown repeatedly that there is no such thing.  Please show me how I am wrong.


"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..."

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#538 2007-03-07 05:54:53

lightning
Real Member
Registered: 2007-02-26
Posts: 2,060

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

where


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#539 2007-03-07 17:14:31

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

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

Ricky wrote:

George, if I don't understand something you have posted because of miscommunication, then you need to rephrase what you have said.  Telling me to go back and read the same thing over again will not help one bit.  You talk about the infiniteth digit, and I have shown repeatedly that there is no such thing.  Please show me how I am wrong.

Shown repeatedly??? What's your shown? Your 0.999... containing any integer amount of 9's?

Dross has shown it to you that if 9's amount or quazi-amount is static, then you have infinite 9's in 0.999....

I am not repeating infiniteth 9- I am demanding the 9 at the right end, or the smallest 9.

You may say that is equivalent to saying the infiniteth 9, but let's see why you think the infiniteth 9 cannot exist:

Simply because you cannot count to the infiniteth!

That's typical Double Standards, Ricky. You do not deny the existence of the infinite apple pile despite that you cannot count them over, do you?! But why this time you deny the existence of the infiniteth 9 or simply the 9 at the right end just because no one can count to that?!

Give me a reason, Ricky! See whether you have a way to show your marvelous communication! Your marvelous Double Standards!

If you deny the existence of the apple pile, you can only have your abitary finiteth 9 realized in the first form/structure of 0.999..., as I have mentioned. So please keep counting through all your life and don't tell anyone you have counted 0.999... to 1 !

Last edited by George,Y (2007-03-07 18:29:40)


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#540 2007-03-07 17:20:24

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

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

Don't even try to put your cotton in my ear!

Last edited by George,Y (2007-03-07 18:43:16)


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#541 2007-03-07 18:50:01

Ricky
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Registered: 2005-12-04
Posts: 3,791

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

I am demanding the 9 at the right end, or the smallest 9.

This contradicts the notion of an infinite number of 9's.

A 9 in the kth position has the value of:

And thus, if we assume that there exists a position n such that it contain the smallest 9, then:

And contradiction.  n+1 is actually the smallest 9.  Because of this contradiction, we say that there is no smallest 9.

You do not deny the existence of the infinite apple pile despite that you cannot count them over, do you?! But why this time you deny the existence of the infiniteth 9 or simply the 9 at the right end just because no one can count to that?!

Take those apple pies.  Start removing them one by one, apply a number to each, 1, 2, 3...  How many apple pies must you remove before you get to infiniteth pie, George?  In fact, you will never get to the infiniteth pie, and it's the same exact way with the 9's.  It's not just that there are an infinite amount of 9's.  There are a countable amount of 9's.


"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..."

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#542 2007-03-07 18:55:52

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

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

Ha ha! You got it, Ricky! Because of contradiction! See you cannot get a smallest 9 even from a static pile!

The second, Ricky, how many times do I need to communicate it to you?
The removing is just another equivalent saying of counting (they are both one then one), haven't you seen?!
Just making word puzzle doesn't help you out. Just describing how beautiful the cloth was didn't prevent the king from naked either.


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#543 2007-03-07 19:00:00

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

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

Let me tell you why there is a contradiction:

Because 9/10[sup]∞[/sup]=9/10[sup]∞-1[/sup] contradiction already, you only use finite 9's so as to avoid mentioning this contradiction, no wonder you cannot get it.


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#544 2007-03-07 19:03:01

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

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

Ricky wrote:

In fact, you will never get to the infiniteth pie, and it's the same exact way with the 9's.

I admit I cannot count to the infiniteth pie. But can you count the apple pile over?

Stop Claiming "countable" "contable" from a textbook, yes countable from the begining, but you cannot count over forever.


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#545 2007-03-07 19:04:12

Ricky
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Registered: 2005-12-04
Posts: 3,791

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

Ha ha! You got it, Ricky! Because of contradiction! See you cannot get a smallest 9 even from a static pile!

What do you mean static pile?

Because 9/10^∞=9/10^(∞-1) contradiction already, you only use finite 9's so as to avoid mentioning this contradiction, no wonder you cannot get it.

10^infinity does not make sense as infinity is not a real number.  Sorry, you can't raise something to the infinity power before you define exponents with infinity.  I know of no such definitions, please define the operation before you use it.

Let me try it this way.  Start with 0.9.  Write another 9 after it, and then another.  If you keep on writing 9's forever, will you ever reach a 9 which is at the infiniteth position?  No.  Are there a finite amount of 9's?  No, because you keep writing them forever.


"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..."

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#546 2007-03-07 19:06:37

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

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

Stop Claiming "countable" "contable" from a textbook, yes countable from the begining, but you cannot count over forever.

Why 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..."

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#547 2007-03-07 19:26:21

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

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

Ricky wrote:

Why not?

substitute "write" by "count" in the following, and see it youself.

Ricky wrote:

Let me try it this way.  Start with 0.9.  Write another 9 after it, and then another.  If you keep on writing 9's forever, will you ever reach a 9 which is at the infiniteth position?  No.  Are there a finite amount of 9's?  No, because you keep writing them forever.

"write" and "count" are the same by nature that is one then one; one, then another.


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#548 2007-03-07 19:29:28

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

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

Let me try it this way.  Start with 0.9.  Write another 9 after it, and then another.  If you keep on writing 9's forever, will you ever reach a 9 which is at the infiniteth position?  No.  Are there a finite amount of 9's?  No, because you keep writing them forever.

I interpret this paragraph as your concept of 0.99..., that is 0.99repeating, you can check my first structure or my other posts on this page and see if you can write 0.99repeating to 1.


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#549 2007-03-07 19:31:59

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

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

George, you restrict yourself from the wonders of math by sticking with what can only be done in reality.  And before you say that math should be based on reality again, I ask you once again, where is infinity in reality?

Just because you can't physically write them out doesn't mean you can't talk about what it would be if you could.


"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..."

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#550 2007-03-07 19:32:29

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

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

Plus, you can write out only a finite amount of 9's forever. And your 9's are always growing, any questions check my first structure or the following:

George,Y wrote:

If you deny the existence of the apple pile, you can only have your abitary finiteth 9 realized in the first form/structure of 0.999..., as I have mentioned. So please keep counting through all your life and don't tell anyone you have counted 0.999... to 1 !


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