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Topic: Proof 0.999... is not equal to one.
Replies: 194   Last Post: Feb 16, 2017 5:56 PM

 Messages: [ Previous | Next ]
 hagman Posts: 1,923 Registered: 1/29/05
Re: Proof 0.999... is not equal to one.
Posted: Jun 2, 2007 3:20 AM

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.

>
>

> > 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.

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)

>
>
>

> > 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".
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) ?

>
> > 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

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