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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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