If your question was regarding these, then I already replied that, yes, these sum to 1.
1/2+1/4+1/8+1/16+...
Because in this, you're missing the infinitesimal remainder.
- something infinitely small, and so the argument is self defeating that infinities can exist but not infinitesimals as each define the other in terms of lengths.
I also most likely already answered your questions in this reply to meBigGuy that noone appears to have responded to:
We're generally always expected to compute a single numeric limit to irrational numbers as few people want an exhaustive list of possible solutions, but this doesn't mean that those solutions don't exist.
And with regard to the Wiki site, we've already through all this before, but here goes again with regarding to the specific .999... thread on the wikipedia site here
http://en.wikipedia.org/wiki/0.999Fractions
In the case of fractions the detachment occurs when a rational number is converted to an irration form because not even an infinite repetition constructs a precise identity and this can be seen in terms of Number Theory that:
for relatively prime n and m,
n*j!=m^k for all integers k and j (no there is no limit that j and k must be finite either)
And for example 1/3!=m/10^n because 3m!=10^n
-------------------------------
Digit manipulation
In this case the attempted proof relies on an equivalent physical eyeballing of magnitude of the number and treats the '...' representation indeterminantly.
We can show how these errors arise in many ways:
10*.9=9!=9+.9=9.9
10*.99=9.9!=9+.99=9.99
10*.999=9.99!=9+.999=9.999
10*.999...=9.99...!=9+.999...=9.999...
The fundamental problem here is that the multiplication by 10 relies upon an assumed identity that infinity-1 is still infinite, but this can't be generally considered true because it constructs the infinite recursive identity that:
infinity=infinity-1
infinity=(infinity-1)-1
infinity=((infinity-1)-1)...
infinity=infinity-infinity
Which is actually indeterminant. The only way to avoid this is to add a context to infinite that only allows a finite number of such operations to exist, but this is not present in representation of .9r.
-----------------------------------------------
The next attempt there is entitled:
Real Analysis
And this attempt relies upon infinitesimal Calculus and geometric summations to construct, but this again fails to prove an identity outside a non-zero error term (in this case the error is arbitrarily small or large depending upon the context).
If we look at the construct of such a summation, we see this sequence:
1=1/2+1/2=1/2+1/4+1/4=1/2+1/4+1/8+1/8=1/2+1/4+1/8+...+1/2^n+1/2^n
and not:
1!=1/2!=1/2+1/4!=1/2+1/4+1/8!=1/2+1/4+1/8+...
So there is a recurring non-zero error of 1/2^n that occurs in this proof and it can be shown that this value cannot precisely equal zero also.
-------------------------------------------------------------
The next failed attempt is to use:
Nested intervals and least upper bounds
In this case a recursive subdivision occurs along a number line to construct divisions such that a number falls into one or more of these categories:
0<=x<=.1,.1<=x<=.2,.2<=x<=.3,...,.9<=x<=1.0,1.0<=x<=1.1,...
Notice that this is not a deterministic process in which a single bin can be selected at each iteration because every decimal allows for the possibility of its existance in two bins.
For example, the decimal number .2... can line over a range that includes .1999... to .3000...
There is never a point in this process where a decimal number can be allocated to a single bin and in this case every number possess a tolerance of the least significant digit such that n-1/10^m=n/10^m, n+1/10^m
In this case we can prove all real numbers are equal by recursively showing that ...=(n-2)/10^m=(n-1)/10^m=n/10^m=(n+1)/10^m=(n+2)/10^m=...
---------------------------------------------------------------
After the above failure, we come to:
Dedekind cuts
In this case the attempt is to show that there is not a constant finite value that can be computed to lie between .9r and 1 and we're working with infinitesimal Calculus here.
In this case the issue becomes one of whether or not irrational numbers, such as e and pi etc. themselves are specific numbers as they're only defined by infinitely long descriptions.
The attempt here is to construct a number by instead being able to compute all computable numbers not equal to it and then defining it as being the remaining element. So effectively the proof goes that .999...=1 because a construction of .999... can be made that computes all numbers from 0 to less than 1 and so the limit is assumed to only be able to be 1 because there is no other number it could be computed to be.
The fault with this logic is in the incorrect assumption and failed proof that rational numbers are "dense" relative to irrational numbers.
Effectively, a virtual number line is constructed that assumes that for every irrational number there is an associated unique rational number and that by computing all rational numbers we're computing correlations with irrational numbers, but notice that irrational numbers don't terminate but instead have an infinite number of possible values (unless we restrict ourselves to defining a precise infinite value in the computations).
And so an irrational number at any point of a computation is both a rational number, plus an iteration number that generated that rational approximation and so an irrational number is a 2 dimensional value (or 3 dimensional in this case as we have both a rational n/m description as well as an additional iteration number) that no specific rational number equals.
In fact irrational numbers are infinitely denser than rational numbers and any finite difference between numbers possesses an infinite number of irrational numbers between them and at no point does a specific decimal representation of a number limit the possible results to only being a finite number of possibilities, yes, even for an infinitely long decimal expansion, because infinity never stops.
I've shown above how infinitesimal differences can result in finite deviations as well because infinity is underspecified.
----------------------------------------------------------------------------
And then we come across the next attempt to prove .9r=1 as:
Cauchy sequences
In this case it appears that they gave up on trying to prove .999...=1 but instead construct a logical paradox that for any possible non-zero difference, c, we can always make it smaller.
This suffers from the logical paradox that c can be made less than c and so we must find a c that solves:
c>0
c<c
Another way I've seen this same paradox expressed was in the "identity" that there exists:
a<b
c>0
a*c<=b*c
In this case the implication is not that a*c<=b*c, but that a*c<b*c, because if a condition exists where:
a*c=b*c
then we can divide by this non-zero value and derive that
a*c/c=b*c/c
a=b
But this is in violation of the prior restriction that a<b!
------------------------------------------------------------------------------
So, yes, there is a concensus on the .999...?=1 issue and that it can do nothing more than approach a value of 1 but cannot equal 1 and that the limit of 1 indicates an upper bound for which .999... is excluded from equalling.
QUOTE (buttershug+)
edit stop with the stupid ruler. If you can't get past physical thinking then why discuss the subeject at all.
Agreed. Though this also means you can't try to compare infinity with going to Las Vegas or kicking a ball an infinite number of times.
QUOTE (buttershug+)
At work I don't weld because I don't have welding equipment. You obviously don't have whatever it takes to move beyond the physical into the pure theorecitcal.
Ok, apparently I'm simply ignorant.
In the meantime, consider that if the 1/x term in this identity truly equalled 0, the result of this would be 1 and not e, ~2.7182818
2.7182818~=e=lim((1+1/x)^x) as x->infinity
PhysBang
18th January 2008 - 08:40 PM
QUOTE (StevenA+Jan 18 2008, 07:32 PM)
Yes the total equals one, but you're always tossing away the last piece.
1/2=1/2=1-1/2
1/2+1/4=3/4=1-1/4
1/2+1/4+1/8=7/8=1-1/8
...
1/2+1/4+1/8+...=((2^n)-1)/(2^n)=1-1/2^n<1
As everyone here and every mathemtatics text on the subject points out, "1/2+1/4+1/8+...=((2^n)-1)/(2^n)=1-1/2^n<1," is incorrect. You might be able to put a limit on the left-hand side of that equation. If you do so, then you have to work out the limit.
QUOTE
So you need to add in that one last infinitesimal piece if you want to retain the whole, and then it equals 1, so .999....+.000...1=1, but .999...<1 and people have been showing this repeatedly.
Nobody has shown this. "0.000...1" is simply an undefined piece of nonsense. It has nothing to do with any theory of infinitesimal numbers.
Why don't you actually use the mathematics of infinitesimal numbers? You talk about it, but you never apply the theory.
For example:
QUOTE (->
| QUOTE |
| So you need to add in that one last infinitesimal piece if you want to retain the whole, and then it equals 1, so .999....+.000...1=1, but .999...<1 and people have been showing this repeatedly. |
Nobody has shown this. "0.000...1" is simply an undefined piece of nonsense. It has nothing to do with any theory of infinitesimal numbers.
Why don't you actually use the mathematics of infinitesimal numbers? You talk about it, but you never apply the theory.
For example:
1/2+1/4+1/8+...+1/2^n+1/2^n=1
And that is not equal to this
1/2+1/4+1/8+1/16+...
Because in this, you're missing the infinitesimal remainder.
I'm quite happy to say that infinitesimal numbers exist. Please show, using the mathematical theory of these numbers, how they have anything to do with this question.
However, you seem to place yourself in a contradiction when you say that 1/2+1/4+1/8+...+1/2^n+1/2^n=1 does not equal 1/2+1/4+1/8+1/16+... because so much of your analysis requires that this is an equality. For example, you claimed that "1/2+1/4+1/8+...=((2^n)-1)/(2^n)=1-1/2^n<1" in one of your arguents.
QUOTE
Well that's a great paradox isn't it? Are we suppose to compute an uncomputable number because it simply goes on forever?
Even if infinity went on forever, a proof can't.
If you claim that computability theory suggests that 0.999... is not equal to 1, then you had better prove it using computability theory. Standard computability theory textbooks use 0.999... as an example of representing the number 1. You will have to prove them wrong.
QUOTE (->
| QUOTE |
Well that's a great paradox isn't it? Are we suppose to compute an uncomputable number because it simply goes on forever?
Even if infinity went on forever, a proof can't. |
If you claim that computability theory suggests that 0.999... is not equal to 1, then you had better prove it using computability theory. Standard computability theory textbooks use 0.999... as an example of representing the number 1. You will have to prove them wrong.
Fractions
In the case of fractions the detachment occurs when a rational number is converted to an irration form because not even an infinite repetition constructs a precise identity and this can be seen in terms of Number Theory that:
for relatively prime n and m,
n*j!=m^k for all integers k and j (no there is no limit that j and k must be finite either)
And for example 1/3!=m/10^n because 3m!=10^n
Again, if we accept that 1/2+1/4+1/8+...+1/2^n+1/2^n=1 does not equal 1/2+1/4+1/8+1/16+... (as we should and you seem to agree), then we cannot accept that 1/3!=m/10^n bears on the question at hand.
QUOTE
Digit manipulation
In this case the attempted proof relies on an equivalent physical eyeballing of magnitude of the number and treats the '...' representation indeterminantly.
Actually, the proof treats the "..." as completely determined, which is part of the symbol's definition.
QUOTE (->
| QUOTE |
Digit manipulation
In this case the attempted proof relies on an equivalent physical eyeballing of magnitude of the number and treats the '...' representation indeterminantly. |
Actually, the proof treats the "..." as completely determined, which is part of the symbol's definition.
The fundamental problem here is that the multiplication by 10 relies upon an assumed identity that infinity-1 is still infinite, but this can't be generally considered true because it constructs the infinite recursive identity that:
infinity=infinity-1
infinity=(infinity-1)-1
infinity=((infinity-1)-1)...
infinity=infinity-infinity
Which is actually indeterminant. The only way to avoid this is to add a context to infinite that only allows a finite number of such operations to exist, but this is not present in representation of .9r.
You will have to actually provide a mathematical proof of your claims here. You will have to provide a robust definition of "infinity-1" and provide proof that it links to a correct application to the proof at hand.
QUOTE
Real Analysis
And this attempt relies upon infinitesimal Calculus and geometric summations to construct, but this again fails to prove an identity outside a non-zero error term (in this case the error is arbitrarily small or large depending upon the context).
Here is it not clear what proof you are discussing. Real analysis does not require geometry, so it is unlikely that a proof requiring geometry is part of real analysis.
The real analysis proof on the wikipedia page has nothing to do with geometry and does not construct a finite series like this, "1=1/2+1/2=1/2+1/4+1/4=1/2+1/4+1/8+1/8=1/2+1/4+1/8+...+1/2^n+1/2^n."
QUOTE (->
| QUOTE |
Real Analysis
And this attempt relies upon infinitesimal Calculus and geometric summations to construct, but this again fails to prove an identity outside a non-zero error term (in this case the error is arbitrarily small or large depending upon the context). |
Here is it not clear what proof you are discussing. Real analysis does not require geometry, so it is unlikely that a proof requiring geometry is part of real analysis.
The real analysis proof on the wikipedia page has nothing to do with geometry and does not construct a finite series like this, "1=1/2+1/2=1/2+1/4+1/4=1/2+1/4+1/8+1/8=1/2+1/4+1/8+...+1/2^n+1/2^n."
The next failed attempt is to use:
Nested intervals and least upper bounds
In this case a recursive subdivision occurs along a number line to construct divisions such that a number falls into one or more of these categories:
0<=x<=.1,.1<=x<=.2,.2<=x<=.3,...,.9<=x<=1.0,1.0<=x<=1.1,...
Notice that this is not a deterministic process in which a single bin can be selected at each iteration because every decimal allows for the possibility of its existance in two bins.
For example, the decimal number .2... can line over a range that includes .1999... to .3000...
This is simply not true, if one looks at the actual mathematics of the nested intervals approach to generating decimal representations. As soon as one looks to the next nested interval beyond 0.3, one cannot find 0.222..., thus one cannot begin a decimal expansion of 0.222... with "0.3".
QUOTE
Dedekind cuts
In this case the attempt is to show that there is not a constant finite value that can be computed to lie between .9r and 1 and we're working with infinitesimal Calculus here.
In this case the issue becomes one of whether or not irrational numbers, such as e and pi etc. themselves are specific numbers as they're only defined by infinitely long descriptions.
The attempt here is to construct a number by instead being able to compute all computable numbers not equal to it and then defining it as being the remaining element. So effectively the proof goes that .999...=1 because a construction of .999... can be made that computes all numbers from 0 to less than 1 and so the limit is assumed to only be able to be 1 because there is no other number it could be computed to be.
Among the errors of this passage are an incorrect use of the term "compute" and an incorrect description of the Dedekind cut proof as a whole. The Dedekind cut proof shows that the real numbers represented by 0.999... and 1 are the same number given the definition of real number. No recourse to constructing all other numbers is required.
QUOTE (->
| QUOTE |
Dedekind cuts
In this case the attempt is to show that there is not a constant finite value that can be computed to lie between .9r and 1 and we're working with infinitesimal Calculus here.
In this case the issue becomes one of whether or not irrational numbers, such as e and pi etc. themselves are specific numbers as they're only defined by infinitely long descriptions.
The attempt here is to construct a number by instead being able to compute all computable numbers not equal to it and then defining it as being the remaining element. So effectively the proof goes that .999...=1 because a construction of .999... can be made that computes all numbers from 0 to less than 1 and so the limit is assumed to only be able to be 1 because there is no other number it could be computed to be. |
Among the errors of this passage are an incorrect use of the term "compute" and an incorrect description of the Dedekind cut proof as a whole. The Dedekind cut proof shows that the real numbers represented by 0.999... and 1 are the same number given the definition of real number. No recourse to constructing all other numbers is required.
Cauchy sequences
In this case it appears that they gave up on trying to prove .999...=1 but instead construct a logical paradox that for any possible non-zero difference, c, we can always make it smaller.
Unsurprisingly, this is a mistake generated by a failure to understand the mathematics involved. The proof again relies on a definition of the real numbers.
QUOTE
So, yes, there is a concensus on the .999...?=1 issue and that it can do nothing more than approach a value of 1 but cannot equal 1 and that the limit of 1 indicates an upper bound for which .999... is excluded from equalling.
This claim, of course, has nothing to to with any mathematical reasoning.
QUOTE (->
| QUOTE |
So, yes, there is a concensus on the .999...?=1 issue and that it can do nothing more than approach a value of 1 but cannot equal 1 and that the limit of 1 indicates an upper bound for which .999... is excluded from equalling. |
This claim, of course, has nothing to to with any mathematical reasoning.
In the meantime, consider that if the 1/x term in this identity truly equalled 0, the result of this would be 1 and not e, ~2.7182818
2.7182818~=e=lim((1+1/x)^x) as x->infinity
This too, has nothing to to with any mathematical reasoning.
StevenA
18th January 2008 - 10:38 PM
QUOTE (PhysBang+)
As everyone here and every mathemtatics text on the subject points out, "1/2+1/4+1/8+...=((2^n)-1)/(2^n)=1-1/2^n<1," is incorrect. You might be able to put a limit on the left-hand side of that equation. If you do so, then you have to work out the limit.
We could compute a limit if we want to know what the limit is equal to, but notice that the summation never equals the limit in this case, which is the real point and as I posted above this quite typical of most computations of a limit (there's little other reason to utilize this computation for anything else other than as an approximation).
QUOTE (PhysBang+)
Nobody has shown this. "0.000...1" is simply an undefined piece of nonsense. It has nothing to do with any theory of infinitesimal numbers.
Why don't you actually use the mathematics of infinitesimal numbers? You talk about it, but you never apply the theory.
Perfect, yes .000...1 is imprecisely defined in the same manner that a recurring decimal .999... is imprecisely defined.
There are many other such examples and paradoxes as well. For example a perfect circle is assumed to be infinitely subdivisible, yet cyclic around the perimeter. So in this case the definition creates the paradox that we can pass through an infinite number of points, and yet still come back to a precise location - imagine instead that we had a figure 8 and that each half of this was also infinitely subdivisible, in this case if we denoted one half as a string of 9s and the other half as a string of 0s, we should theoretically traverse this by seeing a string of digits identical to .999... in that we see an infinite number of 9s, but ultimately we go past that infinite number and encounter an infinite number of 0s and then 9s etc.
Obviously when we applied the limit to such a function it would also reach the same limit of being equal to 1 as the infinite string of 9s would, so the two versions would be considered identical according to the logic that a finite limit is sufficient to distinguish between infinitesimal differences.
There are other problems and paradoxes as well, but I'm happy to see that you recognized this above.
QUOTE (PhysBang+)
I'm quite happy to say that infinitesimal numbers exist. Please show, using the mathematical theory of these numbers, how they have anything to do with this question.
However, you seem to place yourself in a contradiction when you say that 1/2+1/4+1/8+...+1/2^n+1/2^n=1 does not equal 1/2+1/4+1/8+1/16+... because so much of your analysis requires that this is an equality. For example, you claimed that "1/2+1/4+1/8+...=((2^n)-1)/(2^n)=1-1/2^n<1" in one of your arguents.
Infinitesimals and infinities are processes dependent upon a variable term. In this case I specifically included the variable in order to show in greater detail what the summation represented, so I wasn't being hypocritical in this specific example, though I admit there are places that I have used such nomeclature for convenience as well, but the point remains that '...' isn't sufficiently defined, especially with regard to a proof.
For example, if we wanted a rigorous proof that there are numbers larger than one million, we'd really need to actually create one instead of just saying that 1,2,3,... ends up creating numbers greater than 1 million, especially if there was evidence that no such numbers could be created. In this case we simply compute something like (3^(3^3)) and get a number larger than 1 million, but notice that this still doesn't mean that (3^(3^(3^...))) is a specific number either, as the computation again involves '...' which isn't a number itself but a repeating process.
So if we look at reciprocals of 1/1, 1/2, 1/3, and in general 1/n, there is no n for which 1/n=0 because this violates the multiplicative identity for 0, which is 0*n=0 and not a value for which 0*n=1.
If you try to substitute '...' for n to find a number for which 1/...=0, then we have a new identity that 0*...=1, which might be perfectly useful for some purposes (though '...' is already overused), but I'll be able to grab you another example in which 0*...=17.314 or pi etc. and so you'd need to go back and rewrite all the math textbooks that decided to use ... in this context.
But notice that none of this becomes an issue when people read the fine print and they implicitly understand that .999... never did anything more than get arbitrarily close to (or similarly far from) 1.
QUOTE (PhysBang+)
If you claim that computability theory suggests that 0.999... is not equal to 1, then you had better prove it using computability theory. Standard computability theory textbooks use 0.999... as an example of representing the number 1. You will have to prove them wrong.
They must have really fast computers now. I stopped after 5 digits and saw no change in the process that what arose from using 3 digits. We could try 72 digits if you'd like, but we'd still be infinitely shy of approaching 1.
Suggestions?
QUOTE (PhysBang+)
Act
