Re: infinity

Hi Theo. Before answering your question, I would like to do some math
on infinite series first and then explain to you why the vase is not
empty at the noon time.

Let S = 1 + 2 + 3 + 4 + 5 + 6 + ... + oo oo = infinity
Then 2*S = 2 + 4 + 6 + 8 + .... + oo
Subtract S from 2*S, we have
2*S - S = (2 + 4 + 6 + ... + oo) - (1 + 2 + 3 + 4 + 5 + 6 + ... +
oo)
==> S = (2 + 4 + 6 + ... + oo) - (1 + 3 + 5 + ... + oo + 2 + 4 + 6 +
.... + oo)
==> S = (2 + 4 + 6 + ... + oo) - (1 + 3 + 5 + ... + oo) - (2 + 4 + 6 +
.... + oo)
==> S = -(1 + 3 + 5 + 7 .... + oo)
What? The summation of all integers is a negative number? Obviously
there must be something wrong. The error is that we cannot rearrange
numbers like that for an infinite series.

Now go back to your question
Let S1 be the balls being put in the vase
Let S2 be the balls being taken out from the vase

At time = noon
S1 = {1, 2, 3, ... , oo} n(S1) = oo
S2 = {1, 2, 3, ... , oo} n(S2) = oo
# of balls in the vase would be n(S1) - n(S2) = oo - oo which is
undefined.
As seen from the previous example, we cannot just cancel out same terms
from each set and come up with result zero. S1 and S2 are infinite
series with different properties. We cannot simply say every number in
the set S1 also exists in S2 then S1 - S2 = 0. If you do that then you
are rearrangeing the numbers in the infinite series.

In mathematics, usually if at 'noon' the question is undefined then we
would take the limit of the time --> noon to see the tendency of the
result
As time --> noon
S1 = {1, 2, 3, ... , 10n) Number of S1 = 10n
S2 = {1, 2, 3, ... , n) Number of S2 = n
The remaining balls are {n+1, n+2, ... , 10n}
# of balls remained in the vase would be n(S1) - n(S2) = 10n - n = 9n >
0
Therefore, the vase is not empty and actually it is growing rapidly.

>>From a physical point of view, the number of balls in the vase are
growing rapidly as the time approaches noon. Even though, right at
the noon time, the problem is undefined but we would expect the vase is
not empty.

.

Relevant Pages

  • Re: infinity
    ... > on infinite series first and then explain to you why the vase is not ... > Let S2 be the balls being taken out from the vase ... > the noon time, the problem is undefined but we would expect the vase is ...
    (sci.math)
  • Re: infinity
    ... Physically: Infinity is undefined physically. ... >> to show that each ball is removed from the vase before noon and is not ... >> the vase is empty before noon. ... The operation of adding or removal of balls is undefined at noon. ...
    (sci.math)
  • Re: An uncountable countable set
    ... To become empty means there is a change of state in the vase, from having balls to not having balls. ... There are always a specific number of balls, if additions and removals occur instantaneously. ...
    (sci.math)
  • Re: An uncountable countable set
    ... Since the vase was empty to start with, it cannot later "become" empty after once having been empty, at least according to that definition. ... Noon does not exist in the experiment, or else you have infinitely numbered balls. ... insertion or removal or location of balls is a function of time. ...
    (sci.math)
  • Re: infinity
    ... >>> Physically or mathematical it is not difficult to prove that the vase ... # of balls in the vase can be ... this statement alone is not sufficient to claim ... >>> the vase is empty before noon. ...
    (sci.math)
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