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> Point Nine Recurring Equals One (or not), What do you think?
 
Does point nine recurring equal one?
Yes [ 102 ]  [54.26%]
No [ 68 ]  [36.17%]
Don't ask me or I'll hurt you [ 18 ]  [9.57%]
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gonegahgah
Posted on Mar 6 2007, 04:32 PM


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Let's get to the proofs.

1.
x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - x
10x - x = 9.9999... - 0.9999...
9x = 9.0000...
9x / 9 = 9.0000... / 9
x = 1.0000...

In fact this should be:
x = 0.9999...
10x = 9.999... (notice still only 4 nines unlike example above)
10x - x = 9.999... - x
10x - x = 9.999... - 0.9999...
9x = 8.999...1 (yes this is nine recurring with a one at the end of the number)
9x / 9 = 8.999...1 / 9
x = 0.9999...

2.
1/3 = 0.3333...
1/3 * 3 = 0.3333... * 3
1 = 0.9999....

In fact this should be:
1/3 = 0.3333... remainder 0.0...1 (yes this is 0 recurring with a one at the end)
1/3 x 3 = (0.3333... remainder 0.0...1) x 3
1 = 1

They try to get past the first one by saying that you can not stick a 1 at the end of a recurring sequence of nines. Nonsense.

They try to use the second one because we automatically assume that 1/3 = 0.3333... which it doesn't. No matter how far you work this sum out you always have to carry the one so there will always be a remainder of 0.0...1. You try working it out on paper without always having to carry the one.
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StevenA
Posted on Mar 6 2007, 04:46 PM


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Let's do it like this:

1-.9=.1
1-.99=.01
1-.999=.001
...
1-.999...999=.000...001

At no step of this recursion are the two values equal, so unless you can pick some magic number at infinity, they're never precisely equal

.000...001 != .000...000

A number ending in 1 never exactly equals the same number ending in zero, unless you're willing to weight the digits down to the point where you can close your eyes and ignore the difference. So, yes, for real world applications with successly smaller and smaller weighted digits it might be close enough but technically they aren't identical and can't be manipulated arbitrarily and see the same result apply.

If we could treat them indentically then we couldn't do this:

x=10^y

1-.9r = (1-.9)/x (where y is the number of 9 digits) = 0

But:

x*(1-.9)/x = x*0

1-.9=0

Is obviously incorrect.

Another way to see this is that it's a process continually approaching 1, but never reaching it. If it could actually reach 1 then no more 9s would be needed ... but that violates the repetition.

QUOTE (gonegahgah)
1.
x = 0.9999... (here, you're assuming x=0.9r)
10x = 9.9999...
.....


And there it is. In the second line of the "proof" there's an assumption made that 9*0.9r=9 or that 0.9r=1, but that's cheating because that's what you're trying to prove and you can't use it yet until you prove it so the "proof" dies here.

If instead you're trying to do this via. multiplication then you should add a multiplicative error term and see if it remains the same:

So you should rewrite it as:

x=0.9999...+e
10x=10*(0.9999...+e)
10x=9.999...+10e

The error increased by 10 when you multiplied it and will continue to erase the validity of the result as you go along.

This post has been edited by StevenA on Mar 6 2007, 05:32 PM
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AlphaNumeric
Posted on Mar 6 2007, 05:54 PM


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Steven, you are completely ignoring the fact 0.9r is the LIMIT of the sequence 0.9, 0.99, 0.999,.... While none of the elements a finite length along the sequence are equal to 1 their limit is.

How do I prove this? Well suppose they limit to something which isn't 1. Call is 1-X where X is some small, non infinitesimal number belonging to the Reals. It has to be non-infinitesimal because there's only one infinitesimal in the Reals and that's zero and that would make 1=0.9r if X=0.

As you point out, 1-0.999.....9 = (0.1)^n for n=(number of 9s in the decimal expansion). You'll also agree that (0.1)^(n) > (0.1)^(n+1), therefore the sequence 0.1, 0.01, 0.001, etc is always decreasing.

Now the question is "Can I find an n such that (0.1)^n < X for any non-zero X in the Reals?". The answer is a quite obvious yes. Give me any non-zero member of the Reals and I'll find you an n such that (0.1)^n < X.

But what does this mean? It means that 1-0.9r < X for ANY X>0. I don't think anyone claims that 0.9r > 1 and it's obviously not. That, by triality (ie a number is either positive, negative or zero), means that 0.9r=1.

Other proofs not so far mentioned include :

If 0.9r is not equal to 1, give me a number between them. If two numbers are not equal, then their average is betwee them and equal to neither of them, that being if A < B, then A < 0.5(A+B ) < B. So if by your claim 0.9r < 1, then what's the number (or numbers) between them?

If you say "There doesn't have to be a number between them" then you are ignoring the fact the Reals are a field under multiplication and addition and therefore for any A,B,C, C*(A+B ) is a Real number too. C=0.5, B=0.9r and A=1 means that 0.5(0.9r+1) must be in the Reals too. If it's not between 0.9r and 1, it must be equal to both of them, ie they are equal.

Remember that physics has nothing to do with this, before someone brings up "But what if space is quantised". We are not discussing anything which is contrained by physics or reality here but the algebraic structure of the field of Real numbers.

Anyone whose sat an introductory course in analysis or set theory will know of several ways to prove 0.9r=1 and in maths that's concrete (and you only need 1 proof anyway). It's not like physics where someone comes along later and disproves a theory, a correct proof in maths is true for all time. Someone proved 2+2=4 and noone will ever come along and from the same axioms prove 2+2=5, that's the nature of maths.

And if you don't accept my word, how about Professor Tim Gowers, Fields Medalist in Combinatorics and professor of maths at Cambridge uni who specifically states 0.9r=1 on that page I just linked to.

No doubt a bunch of muppets will now give their two cents worth with irrelevent comments about how they don't accept the proofs, it doesn't 'feel' right and Zephir will just have to post a pointless picture and mumble something about aether and how AWT explains all of this rolleyes.gif

/cue the idiots....

This post has been edited by AlphaNumeric on Mar 6 2007, 05:57 PM


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gonegahgah
Posted on Mar 6 2007, 06:17 PM


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LOL
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rpenner
Posted on Mar 6 2007, 06:39 PM


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Is it true the Fields medal is given out only once every 4 years?


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AlphaNumeric
Posted on Mar 6 2007, 06:44 PM


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Yep, but they give it to a number of people. Gowers got it late 90s I think. Weils just missed out on it and the guy who proved the Poincare Conjecture might have turned it down recently.


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StevenA
Posted on Mar 6 2007, 07:04 PM


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Alpha,

If we take two lines that are parrallel and one foot apart and mark them at exponentially increasing distances of 1 foot, 10 feet, 100 feet etc., we get the observed series of slopes 1,.1,.01,.001 ... etc.

Just because this value is forever approaching zero doesn't mean it truly reaches it. The lines are always parallel and never touch even though they seem to merge in the distance.

The existance of 0.9r relies upon an assumption of infinite precision. That assumption is self defeating and the result can be interpreted either way depending upon where you decide to make the value real (and at some point you need a finite proof, so you can prove it either way depending upon where you "clip" off the infinitesimal).

Likely this is the same over simplification that makes black holes appear as paradoxes or why people might imagine perfect circles can exist etc.

People can prove anything they want but those proofs rely on assumptions and not everyone will agree upon the underlying assumptions.

Yes, you can round an infinitesimal down to zero in some cases without any noticeable change in the result, but other times you can't and not all infinities are created equally (they just look a lot alike upon course inspection).

QUOTE (Alphanumeric)
While none of the elements a finite length along the sequence are equal to 1 their limit is.
...
We are not discussing anything which is contrained by physics or reality here but the algebraic structure of the field of Real numbers.


I change my mind and have to actually disgree with even this as well because consider the following:

What's (1-.9r)/(1-(0.9r)^2)?

Now if you want to try to calculate the value in the limit, you actually have mutliple ways of doing it and getting different results (again, the problem is that people assume all infinities are equal)

(1-.9)/(1-.9^2)= .1/.19 ~= .526
(1-.99)/(1-.99^2)= .01/.0199 ~= .5025
...
So here we have (1-.9r)/(1-(0.9r)^2)=0.5

But if we instead mistakenly assume 0.9r is equal to 1, then we can get a range of results:

(1-.9r)/(1-1*.9r) = (1-0.9r)/(1-0.9r)

or

(1-0.9r)/(1-0.9r)=1/2!

But wait, that can't be because:
(1-.9)/(1-.9)= .1/.1 = 1
(1-.99)/(1-.99)= .01/.01 = 1
(1-.999)/(1-.999)= .001/.001 = 1
...
So (1-.9r)/(1-.9r)=1 or 1/2?!

The problem remains that the underlying process of recursion that the r represents is removed when you assign them all the same representation as a single letter and the differences can easily go unnoticed.

You can get the recursive function to do all sorts of twists and turns along the way before it reaches infinity (in the complex plane you have it draw a smiley face before circling forever!) or even make it ballon up to some value and appear to be approaching infinity when you'd assume it should end up as some finite value or produce an irrational sequence of jumps along the way etc.. It all depends upon the manner in which the recursion is applied which isn't specified by the letter r alone.

Now I have to admit that it may not be possible to do all this by remaining entirely within the realm of real numbers but that still doesn't negate the fact that you can't calculate the limit of a value unless you have the process being used to generate that limit precisely defined.

QUOTE (Alphanumeric)
Steven, you are completely ignoring the fact 0.9r is the LIMIT of the sequence 0.9, 0.99, 0.999,.... While none of the elements a finite length along the sequence are equal to 1 their limit is.
...
There doesn't have to be a number between them" then you are ignoring the fact the Reals are a field under multiplication and addition and therefore for any A,B,C, C*(A+B ) is a Real number too. C=0.5, B=0.9r and A=1 means that 0.5(0.9r+1) must be in the Reals too. If it's not between 0.9r and 1, it must be equal to both of them, ie they are equal.


Ok, and I just treated 0.9r as if it lay within an infinite field of numbers, and used the value in the limit and even assumed 0.9r was equal to 1 and still got an indeterminant answer after doing all this. Now make it harder and try to define a set of rules by which these infinite sequences can be deterministically compressed in a real expression and always act like they approach the same limit. I doubt you can pack all that into the letter r.

This post has been edited by StevenA on Mar 6 2007, 08:02 PM
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Dr Obvious
Posted on Mar 6 2007, 08:12 PM


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Simple proof

1/9=0.1111...
2/9=0.2222...
....
7/9=0.7777...
8/9=0.8888...
9/9=0.9999...

So

9/9=1=0.9999

Youu need more than your caculator.


PS
I agree /set 'idiots' = 92 '%'/


-Dr O
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AlphaNumeric
Posted on Mar 6 2007, 08:40 PM


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QUOTE (StevenA @ Mar 6 2007, 08:04 PM)
Just because this value is forever approaching zero doesn't mean it truly reaches it. The lines are always parallel and never touch even though they seem to merge in the distance.

You assume physical requirements there. There's no 'limit' here on time or a process. Does it take 'mathematical time' to add together 2 and 2? Does it need 1 second for me to do that? Me, yes, it takes me a short time to do mental arithmetic, but that's because I'm a physical entity, not a mathematical system. 2+2=4, it's not a matter of "Wait, I have to do it", it's 4.

Lets do another thing like that. You mention e^x. What's that? It's an infinite series

e^x = 1 + x + (1/2)x^2 + (1/3!)x^3 + ...

Yet does the fact I cannot physically add together infinitely many terms mean that e^x doesn't exist? Bollocks it does.

log(1 +x ) = x + ... (another infinite series)

Can I therefore not say with absolute certainty that e^(log x) = x because both log and exp involve infinite series? Of course I can say it with certainty! I am talking about mathematical entities and systems not some monkey with a calculator doing it by hand!
QUOTE (StevenA @ Mar 6 2007, 08:04 PM)
The existance of 0.9r relies upon an assumption of infinite precision. That assumption is self defeating and the result can be interpreted either way depending upon where you decide to make the value real (and at some point you need a finite proof, so you can prove it either way depending upon where you "clip" off the infinitesimal).
The existence of 0.9r requires a concept known as 'completeness'. I suggest you look it up along with related things like convergences, compactness and Cauchy sequences. You'll find that ALL the requirements and framework to completely justify and uniquely define 0.9r exist.

It's not self defeating, it's an extremely common notion in plenty of mathematical systems. Infact, it's in plenty of physical systems. You do NOT require infinitesimals to perfectly define a Real number, by definition! If you required an infinitesimal (other than 0) to define a Real number, X, X would not be Real but something like HyperReal.

I REALLY suggest you get your hands on a book of mathematical analysis (such as the truely excellent Burkhill - Mathematical Analysis) because it would explain rigorously the utter validity of what I'm saying within mathematics.

For instance, there's a theorem which says "If an infinite sequence is monotonic and bounded, then it converges". In other words, if you have a sequence A, B, C such that A>B>C> .... or A<B<C<.... then the limit of this sequence is a set number. This seperates such sequences from oscillating ones like 0, 1, 0, 1 etc which don't converge despite being bounded (but aren't monotonic).

Is the sequence 0.1, 0.01, 0.001,. ... monotonic? Yes. Is it bounded below? Yes, by -1, -2, -3 or any set negative number. Therefore it converges. And as I demonstrated in my last post, it converges to 0, because it doesn't converge to a negative number and it doesn't converge to any positive number, so it must converge to 0. Therefore 1-0.9r = 0. Alternatively you can consider the sequence 0.9, 0.99, 0.999, ... which converges to 1.
QUOTE (StevenA @ Mar 6 2007, 08:04 PM)
Likely this is the same over simplification that makes black holes appear as paradoxes or why people might imagine perfect circles can exist etc.
Black holes are (supposedly) physical objects. Their existence relies on our assumption that our current understanding of the universe is valid, such as the strong energy condition. In maths WE DEFINE OUR RULES, our axioms, the rules of the game and so we are CERTAIN that our base conditions are right because we define them to be and we see what structure we develop from those base rules.

No offence but you're coming at this like someone whose done very little maths, thinks of maths as nothing more than a tool of physics and hence bound by the same rules as physics and also as someone who doesn't understand how the logic of maths works.
QUOTE (StevenA @ Mar 6 2007, 08:04 PM)
People can prove anything they want but those proofs rely on assumptions and not everyone will agree upon the underlying assumptions.
I suggest you look up the axioms of maths and have a go rewriting them then. ALL the maths you've ever seen comes from the same set of axioms. You're welcome to try to rewrite them but it'll be ahell of a job and I don't think many people would want to try it from the ground up.

It took two of the greatest logicistic mathematicians of the early 20th century a decade to write out 3 volumes of maths from axioms and they didn't even get to geometry. It took 360 pages to get from the axioms to defining the number 1. The axioms are that basic!
QUOTE (StevenA @ Mar 6 2007, 08:04 PM)
Yes, you can round an infinitesimal down to zero in some cases without any noticeable change in the result, but other times you can't and not all infinities are created equally (they just look a lot alike upon course inspection).
The sequence 0.9, 0.99, 0.999, .... is a sequence of countably many terms. There are extensions within HyperReals which have uncountably many terms in the sequence but the end result is the same. 0.9r* = 1*, where * signifies the HyperReal version of the number.

And again, no offence, but if you're struggling with this level of analysis, I don't think you're really up to talking about various types of cardinality in infinite sets unless you just skipped all the foundations which lead to that result.

And besides, if you accept such notions of cardinality along with basic arithmetic, calculus, geometry, set theory and everything else then you must accept 0.9r=1 because they ALL come from the same axioms.
QUOTE (StevenA @ Mar 6 2007, 08:04 PM)
What's (1-.9r)/(1-(0.9r)^2)?
It's undefined because you've asked me "What's 0/0", which is undefined.

Now you can define sequences which limit to that, but to simply ask what the number is is undefined.
QUOTE (StevenA @ Mar 6 2007, 08:04 PM)
(1-.9)/(1-.9^2)= .1/.19 ~= .526
You are playing with numbers and actually missing the fact you're simply demonstrating a well known factorisation.

What is (1-x)/(1-x^2) ? I can factorise the bottom to give me (1-x)/(1-x)(1+x) = 1/(1+x)

Therefore if you put in x->1, you get (1-x)/(1-x^2) -> 1/2

So infact you've proved nothing about your own claim, simply that you shouldn't directly evaluate (1-x)/(1-x^2) at x=1 but should consider a limiting process x->1. Well done, that's high school algebra sorted for you.

You mistakenly thought that (1-x)/(1-x^2) = 1 if x=1 since the top and bottom end up being the same value. WRONG. You have ignored a basic result which I remember teaching to 1st year geographers last term, never mind 1st year mathematicians! If f(x) = g(x)/h(x) and f(a) = 0/0, then you must FIRST factorise g(x) and h(x) and cancel the common factors of (x-a) (which must exist by the fundamental theorem of algebra) and THEN evaluate f(a), if you can.

If you don't, you can prove some otherwise seemingly off things. Lets consider f(x) = (2-x)/(4-x^2). By your initial logic f(2) = 1, but if I put in x->2 I get 1/4. Oh no!! Somehow I've destroyed algebra as we know it!!!

No, I've just been sloppy and made a mistake a 1st year should know better of.
QUOTE (StevenA @ Mar 6 2007, 08:04 PM)
You can get the recursive function to do all sorts of twists and turns along the way before it reaches infinity (in the complex plane you have it draw a smiley face before circling forever!) or even make it ballon up to some value and appear to be approaching infinity when you'd assume it should end up as some finite value or produce an irrational sequence of jumps along the way etc.. It all depends upon the manner in which the recursion is applied which isn't specified by the letter r alone.
Unless you care to rephrase that, I'm going to have to say that's bollocks. The sequence 0.9, 0.99, 0.999, .... is well behaved (monotonic and bounded!), has nothing to do with the complex plane and it sounds distinctly like you're trying to BS your way past me.
QUOTE (StevenA @ Mar 6 2007, 08:04 PM)
Now I have to admit that it may not be possible to do all this by remaining entirely within the realm of real numbers but that still doesn't negate the fact that you can't calculate the limit of a value unless you have the process being used to generate that limit precisely defined.
It can and is done entirely within the Real numbers and the sequence is well defined, it's properties understood and by various results in basic analysis it's known to convert to exactly 1.
QUOTE (StevenA @ Mar 6 2007, 08:04 PM)
Ok, and I just treated 0.9r as if it lay within an infinite field of numbers, and used the value in the limit and even assumed 0.9r was equal to 1 and still got an indeterminant answer after doing all this.
I don't follow what you're saying because you're just being too vague. I notice you failed to give me a number between 0.9r and 1.
QUOTE (StevenA @ Mar 6 2007, 08:04 PM)
I doubt you can pack all that into the letter r.
Probably because what you said was nonsense.

If you're going to talk about all that stuff, at least learn some mathematical analysis first. Learn what 'complete' means, what 'compact' means. Why a monotonic, bounded function always converges. I mean, this isn't like physics where someone can slip up or misinterpret data, it's logic and once a maths proof is done, it's true forever, that's the essence of proof.

I just get the feeling you're trying to bluff your way past me by throwing nonsense my way but all the time you didn't counter any of my proofs, pointing out the flaw in the logic. Burkhill's book would explain this nicely. Infact....

/checked book on shelf

No, it's not specifically mentioned in there, but things like a geometric convergent series within infinitely many terms are, which 0.9r is an example of.

And just to let you know, I did maths as an undergrad and this stuff is 1st year work so short of decending into gibberish (which wouldn't help your case anyway), you're going to have to do better than that wink.gif

This post has been edited by AlphaNumeric on Mar 6 2007, 08:42 PM


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Eric England
Posted on Mar 6 2007, 08:45 PM


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QUOTE
Steven, you are completely ignoring the fact 0.9r is the LIMIT of the sequence 0.9, 0.99, 0.999,.... While none of the elements a finite length along the sequence are equal to 1 their limit is.

Recursiveness has a limit? Based on a finite derived from infinity?

I'll give you a 1, but only as "representing" something, that you can't "describe" any greater than .9r.

This post has been edited by Eric England on Mar 6 2007, 08:47 PM


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StevenA
Posted on Mar 6 2007, 08:48 PM


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QUOTE (Dr. Obvious)
1/9=0.1111...


But why?

If you perform each step of the long division you'll find that there never is a point where 0.1111... is exactly equal to 1/9.

If someone made the claim that 1/9!=0.1111... then the proof you presented would instead say that 1!=0.9999...

Is faith enough to make something true? That may very well end up being the case but then someone could assume 0.1111... is not equal to 1/9 and be able to prove it as much as someone else claiming they are equal, so the only way to resolve this paradox is to not treat them identically or things become confusing.

The best way to do this is to show in what cases .9r=1 and in what cases it doesn't equal 1. Then you're covered on both sides and can see much more accurately when and where the division between 0 and infinitesimals lies.

This post has been edited by StevenA on Mar 6 2007, 09:00 PM
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AlphaNumeric
Posted on Mar 6 2007, 09:26 PM


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QUOTE (Eric England @ Mar 6 2007, 09:45 PM)
I'll give you a 1, but only as "representing" something, that you can't "describe" any greater than .9r.

That's known as 'word salad' Eric. Rather than using logic to refute what I've said you throw out pseudophilisophical crap.
QUOTE (StevenA @ Mar 6 2007, 09:48 PM)
Is faith enough to make something true?
No, rock solid logic is. You seem to be taking your faith over logic about the logical results of statements.
QUOTE (StevenA @ Mar 6 2007, 09:48 PM)
The best way to do this is to show in what cases .9r=1 and in what cases it doesn't equal 1.
Sorry, do you actually think what you just said is valid?! blink.gif

That's like saying "Lets find when 2+2=4 and when is doesn't". Why would a number's value change? Does the value of 2 ever change? Are there cases when 2 is actually 3? But then what's 2? 1? 7? -3?

Numbers don't change, 0.9r=1 and once you've proved it (and I have, several times in several ways), it's an unchanging fact. Or do you need a course in the basic workings of maths?
QUOTE (StevenA @ Mar 6 2007, 09:48 PM)
Then you're covered on both sides and can see much more accurately when and where the division between 0 and infinitesimals lies.
Might I suggest learning what the definitions and properties of fields containing infinitesimals and their relation to Reals or extensions of the Reals before making such comments because it seems very much like you're just stumbling in the dark, trying to hide behind pseudo-sounding nonsense the fact you can't logically prove your claim so if you just go round and round in circles with vague comments then maybe you'll somehow come out right?

The problem is, all the people who claim 0.9r isn't 1 haven't done maths. They have little to no idea the machinary which underpins the result and so when I say something like that theorem about bounded monotonic sequences, it means nothing to them. But rather than think "Hey, perhaps the guy with the maths degree and the guy who lectures this stuff at Cambridge know a little bit about this area and might be right?", it's back to the crank logic where you can't possibly be wrong and to hell with the people who know about this stuff. Only this time, it's not a matter of debate like physics can be, it's a cold hard, irrefutable (but people try!) fact that the axioms of maths undeniably result in the proof that 0.9r=1.

To claim otherwise is just to show you haven't actually got the knowledge to understand how the result is proved. It also means you don't accept 1+1=2, 4/2 = 2 and 1+0 = 1, along with ALL other maths, because they are all interconnected with one another.

But hey, don't let a little thing like knowledge and logic get in your way, most cranks seem to manage well without either.


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StevenA
Posted on Mar 6 2007, 09:26 PM


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QUOTE (Alphanumeric)
You are playing with numbers and actually missing the fact you're simply demonstrating a well known factorisation.

What is (1-x)/(1-x^2) ? I can factorise the bottom to give me (1-x)/(1-x)(1+x) = 1/(1+x)

Therefore if you put in x->1, you get (1-x)/(1-x^2) -> 1/2

So infact you've proved nothing about your own claim, simply that you shouldn't directly evaluate (1-x)/(1-x^2) at x=1 but should consider a limiting process x->1. Well done, that's high school algebra sorted for you.


I DID use the limit of x->1 and got 1/2 as you said. That was my point! If 0.9r can only be equal to one, then I should be able to do this simplification:

(1-x)/(1-x^2) = (1-x)/(1-1*x) = (1-x)/(1-x)

but that gives a different result in the limit. So analyzing a repeating value in the limit isn't reliable and simplifying it to a finite expression removes the ratios of infinity underlying it.

You're repeating what I said and then ignoring the paradox of it. You can't say it's only ok to evaluate the limit of an expression approaching zero in some cases and not others. The reason why this is no paradox to me is because I know why it works in some cases and gets certain results versus when it gets a different result.

0.9r=1 is based upon an assumption that's not always true and again, not all infinities are created equal.

What's x/x?

What's x^2/x or x/x^2?

In all these cases x can't be equal to zero, but if you know what's generating this then you can attempt to calculate a limit instead, but that's only possible by knowing what process is generating the limit!. Otherwise zeroes cannot be treated as infinitesimals, because you end up creating an artificial structure where it didn't exist before or on the other hand you take a structure and attempt to replace it with a symbolic and then detach it from an ability to verify whether it's true or not.

In the case of the equation 1-0.9r=0, you're doing the exact same thing but it's arithmetic so the error falls off by a factor of 10^-y (where y is the resolution of the error), but anything more sensitive than that (and many functions are) will cause it to explode into an infinite error instead of an infinitesimal so you have to know what's causing the repetition to occur in order to determine if the relationship can be evaluated. It's not about simple symbolics - there are processes behind these that the symbolics are intended to represent and separating the two means bogus results.

This post has been edited by StevenA on Mar 6 2007, 09:31 PM
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Precursor562
Posted on Mar 6 2007, 10:06 PM


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Edit.

Nevermind lol. I should have seen that sooner. 1 - .9 = .1, 1 - .99 = .01 etc. so the n doesn't represent the number of 1's after the decimal it represents the number of zeros between the decimal and the one.



This post has been edited by Precursor562 on Mar 6 2007, 10:15 PM


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AlphaNumeric
Posted on Mar 6 2007, 10:16 PM


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QUOTE (StevenA @ Mar 6 2007, 10:26 PM)
I DID use the limit of x->1 and got 1/2 as you said.  That was my point!  If 0.9r can only be equal to one, then I should be able to do this simplification:

(1-x)/(1-x^2) = (1-x)/(1-1*x) = (1-x)/(1-x)

but that gives a different result in the limit.

That's not valid algebra. That's like saying :

1-x^2 = 0. I know x=1 is a solution so I can say 1-1*x = 0, therefore 1-x=0, so x=1 is the only solution.

You can't just sub back in SOME of the variable terms with ONE possible value. It's complete nonsense and hence not suprising you get nonsense as a result. As you'll notice from my example, by doing that you completely missed x=-1 is a valid solution to.
QUOTE (StevenA @ Mar 6 2007, 10:26 PM)
You're repeating what I said and then ignoring the paradox of it. You can't say it's only ok to evaluate the limit of an expression approaching zero in some cases and not others. The reason why this is no paradox to me is because I know why it works in some cases and gets certain results versus when it gets a different result.
I'm not ignoring the paradox of anything and you don't know why it works that way because it doesn't.
QUOTE (StevenA @ Mar 6 2007, 10:26 PM)
0.9r=1 is based upon an assumption that's not always true and again, not all infinities are created equal.
No it isn't! It doesn't even require you to know anything about cardinalities of infinite sets! 0.9r=1 was proved hundreds of years before Cantor did his work on infinite cardinality.

You're just making unfounded claims about somethign you haven't bothered to learn about.
QUOTE (StevenA @ Mar 6 2007, 10:26 PM)
In the case of the equation 1-0.9r=0, you're doing the exact same thing but it's arithmetic so the error falls off by a factor of 10^-y (where y is the resolution of the error), but anything more sensitive than that (and many functions are) will cause it to explode into an infinite error instead of an infinitesimal so you have to know what's causing the repetition to occur in order to determine if the relationship can be evaluated. It's not about simple symbolics - there are processes behind these that the symbolics are intended to represent and separating the two means bogus results.
You talk about the processes behind these things but it's clear you've never bothered to learn such processes.

You are arguing about this kind of maths from the point of view of someone whose done high school maths. High school maths is devoid of any analysis (at least it is in the UK) and the notion of rigorous proof is hence lost on plenty of people. You therefore try to argue by armwaving. That doesn't work.

Stop arm waving and rambling (which boils down to 'Because I don't like it') and instead try to disprove using only maths and only logic the proofs people have given. There's plenty of results in maths which people 'don't like' because they seem counter intuitive, but that's because logic and intuition are not always the same thing, particularly in abstract systems.

Consider the sequence : (0.9, 0.99, 0.999, .....)

If that doesn't limit to 1, what does it limit to? Be precise, no arm waving, use mathematical statements.
QUOTE (Precursor562 @ Mar 6 2007, 10:26 PM)
(0.1)^(n) where n=5
0.11111

(0.1)^(n+1) where n=5
0.111111
You misunderstand the notation. (0.1)^n = 0.1*0.1*...*0.1 n times, so 0.1^n = 0.000....01 with n-1 zeros.

(0.1)^(n+1) = 0.1*(0.1)^n < 0.1^n smile.gif

QUOTE (Precursor562 @ Mar 6 2007, 10:26 PM)
(0.1)^(n) > (0.1)^(n-1)
That's actually exactly what I said, just both sides divided by 0.1.

This post has been edited by AlphaNumeric on Mar 6 2007, 10:33 PM


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The views in the above post are those of its author and not those of the people who educated him through a degree and masters, supervised him or collaborated with him during his PhD, paid him to teach and mark undergraduate mathematics and physics courses or who pay him to do research now.

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