0.9999... != 1

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infinity is a theory, to base it round a particular number is to give it parameters which it defys
....

what he is saying is that 0.999* where the asteric is used to represent an infinite idea , that of never ending 0's the fact that this makes the statement equal to 1 is irrelevant to the 'idea' of infinite 0.999* is still a number, a irrational number expressed through the idea of infinite, another way to express this number rationally is 1.

It is the infinite theory behind 0.999* that makes it equal to 1
 
If we have scales to measure weight and one weight is exactly 1 gram and the other weight 0.9999....! then if 0.9999....! = 1 the scales should be in perfect balance and read 0.

The problem is that you CANNOT demonstrate in reality a weight that = 0.9999....! So while 0.9999...! hypothetically might be equal to 1, practically a 0.99999...! does not even exist.
 
What the f***

9 x (1/9)

Has algebra just been reconstructed. This equals one buddy.

The definition of an irrational number is

A number which cannot be expressed as a fraction for any integers p and q.

0.999... is irrational
 
oh dear lord, for a forum of what i would expect intelligent people your incredibly daft,
0.9999..... is irrational because it cant be express as a fraction, what is your definition of irrational, perhaps if you stated it instead of stupid non-constructive comments like

what the....???

and

are you for real?
 
Originally posted by patty-rick
oh dear lord, for a forum of what i would expect intelligent people your incredibly daft,
0.9999..... is irrational because it cant be express as a fraction, what is your definition of irrational, perhaps if you stated it instead of stupid non-constructive comments like ...

uhh... the statement that 0.99999.... is irrational is so misguided and wrong, that i don t even know how to respond.

well, ok. 0.99999... =1. 1 is a fraction. therefore 0.9999... is rational.

or, equvalently, the decimal expansion of an irrational number never repeats itself. 0.99999... does repeat itself (it s just a sequence of 9s over and over again. that is repetition), and therefore is not irrational, it is instead rational. even if you don t believe that 0.999.... is equal to 1 (if you don t, i m going to cry), you still must conclude that it is rational.

i am a little baffled by ryans statement. if he was not joking, i have to say i am quite surprised. as for you patty-rick, i don t know. please don t make wild aspersions about my intelligence. you don t know me, and are in no place to claim this or that about my intelligence, and i get rather uppity when people make remarks like that, unprovoked.
 
Lethe settle the arguement. I have no doubt that 0.999. is an infinite regression of 9, but does that constitute it being rational, in light of the defintion of an irrational number I gave above. Is this definition correct?

As for your arguement that 0.999...=1, would that not mean that every element in the continous number line would be equal to the next element, and as such, each element equal to every other element. I have constructed a primitive proofusing the sum

0.999....=(9/10)SUM(10^-n) where the sum is from n = 0 to infinity.

Also if 0.999....=1, then this raises the issue of series convergance. Take for example the sMaclaurin series of 1/(1-x) which only converges for values of x less than or greater than 1 or -1 respectively. For 0.999... this series converges, for one it does not.:)
 
Originally posted by ryans
Lethe settle the arguement.
i would very much like to. i am greatly vexed that this issue is still around.

I have no doubt that 0.999. is an infinite regression of 9, but does that constitute it being rational, in light of the defintion of an irrational number I gave above.
this definition: "A number which cannot be expressed as a fraction for any integers p and q."? this implies that 0.9999.... is rational only if you believe that 0.9999.... =1. if you don t believe this fact, then i will have to convince you of that fact first. and frankly, i m not inclined to do so.

the other criterion for irrationality of a number, namely that its decimal expansion be nonrepeating, needs to be proved to be equivalent to your more fundamental definition. to do so requires that you believe in the correctness of the decimal expansion. basically it requires that you believe 0.9999.... =1.



Is this definition correct?
yes.

As for your arguement that 0.999...=1, would that not mean that every element in the continous number line would be equal to the next element,
next element? what does that mean? between any two real numbers, there are an infinite number of other real numbers. how do i define the "next" number after 0, for example? this notion only makes sense for well ordered sets, which the real numbers are not (at least, under the standard definition of <. the axiom of choice claims that the reals are a well ordered set, but such an ordering cannot be constructed explicitly)

let me try to extract what you actually mean with this statement. every number of the form 0.abcdef9999999.... where a, b, c, d, e, and f are digits is equal to 0.abcde(f+1). is that what you mean by every number is equal to the "next" number? if so, then it is true. but rather uninteresting.


and as such, each element equal to every other element.
nonsense. every number is represented by an equivalence class of decimal representations. two decimal representations are equivalent if they converge to the same thing. so 0.99999.... and 1 converge to the same number, but 0.5 and 1 do not. therefore they are different numbers.


I have constructed a primitive proofusing the sum

0.999....=(9/10)SUM(10^-n) where the sum is from n = 0 to infinity.
a proof of what? that every number is equal to every other number? that 0.9999.... is irrational? that 0.99999... != 1? let me see your proof. if you prove something that is false, then your proof is incorrect, i can tell you right now.

Also if 0.999....=1, then this raises the issue of series convergance. Take for example the sMaclaurin series of 1/(1-x) which only converges for values of x less than or greater than 1 or -1 respectively. For 0.999... this series converges, for one it does not.:)

uhh.... for 0.99999.... this series converges to 1. for 1 it also converges. observe: 1=1.000000000.... so the ratio of the series is 0. 1/(1-0)=1. this statement is also nonsense.

since, by definition, a decimal representation carrys a factor of 1/10 (which is less than 1), there are no issues of convergence, in any decimal representation.
 
What have you done

Posted by ryans

Take for example the sMaclaurin series of 1/(1-x) which only converges for values of x less than or greater than 1 or -1 respectively

Posted by Lethe

uhh.... for 0.99999.... this series converges to 1. for 1 it also converges. observe: 1=1.000000000.... so the ratio of the series is 0. 1/(1-0)=1. this statement is also nonsense.

Are you telling me that the Maclaurin series for 1/(1-x) converges for x > 1. I think not. By the looks of your response I think you misread what I have written because you have put down the wrong series in your response above by letting x = 0 For x = 1, this series is undefined, as is its integral. I have been trying to find a way around this for months.
 
let s try this again.

Originally posted by ryans

Are you telling me that the Maclaurin series for 1/(1-x) converges for x > 1.
no.

I think not. By the looks of your response I think you misread what I have written because you have put down the wrong series in your response above by letting x = 0 For x = 1, this series is undefined, as is its integral. I have been trying to find a way around this for months.

i don t know what the hell you re talking about. so try to follow me here:

EVERY REPEATING DECIMAL REPRESENTATION OF A REAL NUMBER IS A CONVERGENT GEMETRIC SERIES WITH RATIO 1/10.

did you get that? let s look at some examples of decimal representations:

0.999999.....

here, this is 9xSUM{1/10^n}. this converges to a value. that value is the number 1.

how about another one?

1.00000000.....

this is also a geometric series. 1+SUM{0}=1.

now, show me a decimal representation of a number that diverges. it doesn t exist. so i don t know why you re talking about convergence. it is irrelevant to the discussion. all decimal represenations converge.
 
patty-rick:

stay out of this then. you probably have no idea how much this has been argued in the past couple of months here , and your comment was very much
stupid non-constructive comments
as much as any other was. If you failed to see the meaning of mine or lethe's comments then you could leave them alone.


and ryans,

I am also suprised you weren't joking about this.
 
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ryans:

All decimal expansions can be represented as geometric series with common factor r=1/10, as lethe says. A geometric series converges provided that |r| &lt; 1, so all decimal expansions converge.

In the case of 0.999..., the series converges to 1.

I, too, find it hard to believe that this thread is still going.
 
EVERY REPEATING DECIMAL REPRESENTATION OF A REAL NUMBER IS A CONVERGENT GEMETRIC SERIES WITH RATIO 1/10.

O.K. then this is what we are saying

0.9999...=9xSUM(1/10^n) converges in the limit as n -> infinity to 1 (n starts at 1)

I agree with that, that is cool. But does this make it rational. Anyway I can see your point, and I conclude that it constitutes a semi-proof, and so yes, an infinite decimal regression of 9's is equal to 1
 
ryans:

Look. It's quite simple:

0.999...
= 9/10 + 9/100 + 9/1000 + 9/10000 + ...
= 9 (1/10 + 1/100 + 1/1000 + ...)
= 9 Sum[n=1 to infinity] (1/10<sup>n</sup>)
= 9 (1/10) / (1 - 1/10)
= 9 (1/10) / (9/10)
= 9 (1/9)
= 1

Agreed?

<i>Anyway I can see your point, and I conclude that it constitutes a semi-proof, and so yes, an infinite decimal regression of 9's is equal to 1</i>

This is not a semi-proof. It is a full proof. If you don't agree, please point out which part is flawed.

Now, presumably you agree that 1 is a rational number? Since we have shown that 0.999... = 1, what follows about the rationality of 0.999... ?
 
Originally posted by James R

This is not a semi-proof. It is a full proof. If you don't agree, please point out which part is flawed.

Now, presumably you agree that 1 is a rational number? Since we have shown that 0.999... = 1, what follows about the rationality of 0.999... ?

well said.
 
Fallacy

Not to beat on a dead horse, but the fallacy in Ben's original post is found in this sentence:
The above proof is not limited to a finite n, n can be infinity, and the proof is still valid.
Nope, it's not. Proofs by induction are valid only for finite n. You can never get to n=infinity by repeatedly applying "if X[n] then also X[n+1]".
 
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StrangeDays:

<i>Proofs by induction are valid only for finite n.</i>

No. They are valid for countable infinities, too.
 
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