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Re: Proof 0.999... is not equal to one.
Posted:
Jun 18, 2007 1:22 AM
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On Jun 2, 12:20 am, hagman <goo...@von-eitzen.de> wrote: > On 1 Jun., 22:35, chaja...@mail.com wrote: > > > > > > > On Jun 1, 2:31 am, hagman <goo...@von-eitzen.de> wrote: > > > > On 1 Jun., 08:55, chaja...@mail.com wrote: > > > > > > If you insist, here is the first obvious mistake in the paper: > > > > > > ------------------------------------------- > > > > > Let S be the set of all real numbers in the interval [0,1] > > > > > Let T be the set of all real numbers in the interval (-1,0] > > > > > > If you applied an operator '+' to these two sets that sums the elements > > > > > of both sets, you would get: S + T = 1 > > > > > ------------------------------------------- > > > > > > You are trying to sum an uncountable set of numbers. Please define what > > > > > this means. Also consider: if you pair the numbers differently, can you > > > > > make the sum come out to something else? > > > > > > FInally, I didn't notice anything looking like a proof in your article, > > > > > though I admit I haven't read all of it. > > > > > > -- > > > > > Eric Schmidt > > > > > > -- > > > > > Posted via a free Usenet account fromhttp://www.teranews.com > > > > > Hi Eric, > > > > > what I am attempting to point out in the S + T result is that any > > > > element of the infinite set T corresponds to exactly one element in > > > > the infinite set S of equal value and opposite sign, except for the > > > > positive 1 in set S. Should you hold the infinity of values in your > > > > awareness, all cancel except for one that cannot. > > > > > You would not be able to have a final result upon summing the entire > > > > infinity of elements in S union T other than 1. > > > > > --charlie > > > > Oh, so you want to define sum(X) for certain subsets X of the reals? > > > This should be easy if X is finite: just sum the elements (fortunately > > > addition is associative and commutative). > > > It should also be easy to define sum(X) if sum(Y) is defined > > > where Y = X\(-X): just set sum(X)=sum(Y). > > > I don't understand what the backslash means. > > X is a set (of real numbers). > Then with -X I denote the set { -x | x in X }. > Finally \ denotes set difference, i.e. if A and B are sets then > A\B = { x | x in A and not x in B } > > > > > > One should investigate any further definitions as to whether all that > > > stuff yields a consistent definition of sum. > > > > Your example amounts to obtaining sum( (-1,1] ). > > > With X = (-1,1], I would calculate Y = X\(-X) = {1} and obtain > > > sum(X)=sum(Y)=1. > > > I would agree sum(X) = 1, I don't understand the construction of Y to > > say. > > OK, reread the post with the definition of set difference I gave > above. > Ok, understood and agreed.
> > > > > Nothing has been said yet about sum(S) if S\(-S) is infinite, esp. > > > this has no relevance yet if S is the set of all rationals of the form > > > 9/10^n. > > > > However, we can form the set union of -S and 10 times S, i.e. > > > let C = (-S) u (10*S). > > > It turns out that C\(-C) is {9}. > > > I don't believe this would be accurate. If C were defined as (-S) u (S > > + {9}) I would agree. > > > > We are forced to conclude sum(C)=9. > > > > If t is a non-zero number and sum(S) is defined, can sum(t*S) be > > > anything > > > but defined and equal to t*sum(S)? > > > I wouldn't think so.
By I wouldn't think so, I meant I wouldn't think it could be anything other than what you have stated.
> > Then give *any* definition of sum(A) for as many subsets of the real > numbers as possible. > Here's a suggestion of things that should hold (or specify which you > prefer to remove) > 1) sum({x}) = x for any real x. > 2) A and B disjoint, sum(A),sum(B) defined > => sum(A u B) is defined and is sum(A)+sum(B) > 3) A symmetric with respect to 0 (i.e x in A => -x in A) > => sum(A) is defined and is 0 > 4) sum(A),sum(B) defined, C a subset of the reals, f:C->A, g:C->B > bijective mappings > and for each c in C we have c=f(c)+g(c) > => sum(C) is defined and is sum(A)+sum(B) > 5) A={a_1,a_2,...} with a_1 < a_2 < ..., B={b_1,b_2,...} with b_1 < > b_2 < ..., > C=(a_1+b_1, a_2+b_2, ...}, sum(A), sum(B) defined > => sum(C) is defined and is sum(A)+sum(B) > 5) A subset B, sum(A),sum(B) defined > => sum(B\A) is defined and is sum(B)-sum(A) > 6) sum(A) defined > => sum(A\(-A)) is defined and is sum(A) >
I agree with all of the above.
> > > > If A and B are disjoint and both sum(A) and sum(B) are defined, > > > can sum(A u B) be anything but defined and equal to sum(A)+sum(B) ? > > > I wouldn't think so, and I would tend to use a multiset when doing > > this kind of thing, so I wouldn't require they be disjoint. > > If you don't require them to be disjoint yuo should at least allow > them to be > disjoint. Note that I started with the condition "If A and B are > disjoint".
I would allow disjointness, just not require it.
> Since you disagree, do you think there are disjoint sets such that > sum(A) and sum(B) are disjoint but sum(A u B) is either not defined or > is different from sum(A)+sum(B) ?
Nope. If sum(A) and sum(B) are defined and A and B are disjoint, I would find sum(A u B) always defined and equal to sum(A) + sum(B). I think we miscommunicated somewhere, for the most part I agree with all that preceded unless I specifically say "i disagree" somewhere.
> > > > > > In fact, the last two issues are what motivates the way to > > > calculate sum(X) from sum( X\(-X) ) used above and introduced by you. > > > I'm still not sure what \ means. > > Granted. > > > > > > Thus sum(-S) = -sum(S), sum(10*S)=10 *sum(S) > > > Agreed > > > > and finally > > > sum(-S u 10*S) = 9*sum(S). > > > It should > > > > Hence sum(S) = sum(C)/9 = 1. > > > I disagree sum(C) = 9. I find it strictly less. > > Sigh- Hide quoted text - >
I appeciate your frustration - I should have provided defense.
As I mentioned above, if C were defined as (-S) u (S + {9}) I would agree sum(C) = 9. But we defined C as (-S) u (10*S).
Let's define D = (-S) u (S + {9}).
I say that D <> C, where <> means not equal in every way.
For every element in C, one and exactly one element of equal value exists in D, and vice versa. This one-to-one mapping in respect to value, I propose, is mis-leading.
I here offer an example of two sets with the same one-to-one mapping of value but whose sums are not equal:
sum{n = 1 to infinity}[(-1)^(n+1)*1/n] = ln(2); this is the alternating harmonic series. Let the set of all terms be H, therefore H = {1, -1/2, 1/3, -1/4, ...}.
sum{n = 1 to infinity}[(-1)^(n+1)*1/(2n)] = ln(2)/2; this is simply the above series halved. Let the set of all terms of this series by J therefore J = {1/2, -1/4, 1/6, -1/8, ...}.
Let K = H u J, with all fractions of common denominator combined and reduced to lowest terms. Obviously Sum(K) = Sum(H) + Sum(J) = (3/2)*ln(2)
K and H have a one-to-one explicit mapping of value with no unpairable values. K before combined and reducing = {1, 1/2, -1/2, -1/4, 1/3, 1/6, -1/4, -1/8, 1/5, 1/10, -1/6, -1/12, ...}. Combining like terms we get {1, -2/4, 1/3, -2/8, 1/5, -2/12, ...} and reducing we see this is {1, -1/2, 1/3, -1/4, 1/5, -1/6, ...} which has a one to one mapping with H, but Sum(K) is three-halves times Sum(H) and therefore bigger.
I propose that set D is an even set and set C is odd. I'm sure you understand, and perhaps agree with them, that many are unable to consider an infinite set as having parity. Perhaps in the strictly usual sense it may not, but I find the characteristic to be of extreme importance in it's meaning for infinite sets.
C and D have a one-to-one mapping of value, but not of element. Take into account S and -S have a one-to-one mapping of element. -S is exactly the negative of every single element of S. If we throw a {9} into the union of S and -S, this {9} cannot map to a value element- wise, for S and -S are locked into a one-to-one mapping, they have no element that can spare to map to the {9} without un mapping a pair between them, thereby leaving the entire set of D unpairable or odd. Meanwhile C is simply (10*S) u (-S) and is one-to-one mappable. This is my justicfication that C <> D, and while many may not feel this is necessarily true, I hope to have provided enough example and concept to illustrate that perhaps it it truer than we generally understand or accept.
--charlie
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