
Re: Proof 0.999... is not equal to one.
Posted:
Jun 18, 2007 1:22 AM


On Jun 2, 12:20 am, hagman <goo...@voneitzen.de> wrote: > On 1 Jun., 22:35, chaja...@mail.com wrote: > > > > > > > On Jun 1, 2:31 am, hagman <goo...@voneitzen.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 nonzero 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 onetoone mapping in respect to value, I propose, is misleading.
I here offer an example of two sets with the same onetoone 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 onetoone 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 threehalves 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 onetoone mapping of value, but not of element. Take into account S and S have a onetoone 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 onetoone 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 onetoone 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

