Re: infinity

Dave Seaman wrote:
> > Physically or mathematical it is not difficult to prove that the vase
> > is not empty at noon. Mathematically, # of balls in the vase can be
> > expressed by the equation f(t) = 9+9*log(1/t)/log(2) where t is the
> > time in minute before noon. 't' can be 1, 1/2, 1/4, 1/8, ... 1/OO.
> > f(1) = 9
> > f(1/2) = 18
> > f(1/4) = 27
> > f(1/8) = 36
> > Number of balls in the vase at noon is f(0) = OO. Try it :)
>
> Incorrect. You are confusing f(0) with lim_{t->0-} f(t). The former is
> 0, but the latter is +oo. The former is what the problem asks for. Once
> you get your notation straight, the paradox disappears.
>

Hi Dave, thanks for your response,
According to the fomula f(t) = 9+9*log(1/t)/log(2),
f(0) = 9 + 9 log(1/0)/log(2)
= 9 + 9 log(OO)/log(2)
= 9 + 9 * OO / log(2)
= OO not 0.

In mathematical term, usually we refer infinity as undefined
e.g 1/0 = OO (infinity) but very often we say 1/0 is undefined.

I think you do not mean t->0- above. It should be t->0+, right? f(t)
is undefined as t->0-.
f(0+) = 9 + 9 log(1/0+)/log(2)
= 9 + 9 log(OO)/log(2)
= 9 + 9 * OO / log(2)
= OO

f(0-) = 9 + 9 log(1/0-)/log(2)
= 9 + 9 log(-OO)/log(2)
Cannot proceed because log(-n) is not valid

> > The statement: "All that is necessary to show that the vase is empty is
> > to show that each ball is removed from the vase before noon and is not
> > subsequently replaced." is a logical argument. Unfortunately, if
> > infinity gets involved, this statement alone is not sufficient to claim
> > the vase is empty before noon. The vase can only be claimed to be
> > empty at a particular time t is when all the balls are removed at that
> > time.
>
> Incorrect. Change "at that time" to "at or before that time". Then the
> time t=noon fills the bill.
>
> > At t = 1, ball 1 is removed but ball 10 (the last ball) is added
> > to the vase. At t = 1/2, ball 2 is removed but ball 20 (the new last
> > ball) is in the vase. There is no such time that all balls are
> > removed. Therefore, the vase cannot be empty.
>
> All balls are removed before noon.
Again this argument alone is not sufficient to claim that the vase is
empty at noon. Do you have any other argument?

> > If the problem is changed to remove the last ball each time then the
> > argument would be much simplier - ball 1 is always in the base.
>
> If the problem is changed so that all the balls are in the vase at 11:00
> and from then on balls are only removed, never added, then do you agree
> that it is possible for the vase to be empty at noon?

If 'all the balls' refers to only a finite number of balls, say a
million, a billion, a trillion or a google balls then yes the vase will
be empty before noon. Let n be the total number of ball at 11:59
t minute be the time before noon
f(t) = n - log(1/t)/log(2)
f(1) = n
f(1/2) = n - 1
f(1/4) = n - 2
....
f(1/2^n) = n - n = 0

You can see that one of the requirements for the vase to be empty at
noon is the ball withdrawal rate is greater than the adding rate. In
this case the adding rate is 0 but the removal rate is greater than 0.

The last ball (ball n) is to be removed at a time 1/2^n minute before
noon. By the time the last ball is being removed, no other ball is
left in the vase. This is exactly the criteria for the vase to be
empty.

Unfortunately, 'all the balls' in this problem refers to an infinite
number of balls.
f(t) = OO - log(1/t)/log(2)
f(1) = OO
f(1/2) = OO - 1 = OO
f(1/4) = OO - 2 = OO
....

There is no such a time that the vase would be empty.
---------------------------------------------------------
Case 1. If the removal rate is 10 balls/each time and the adding rate
is 1 ball/each time then the vase will be empty at noon provide that
there are not infinitely number of balls in the vase initially.

Case 2. If the removal rate is 1 ball/each time and the adding rate is
10 balls/each time then the vase will not be empty at noon.

Case 3. If the removal rate is 1 ball/each time and the adding rate is
1 ball/each time then the vase will be empty at noon only if there is
no ball in the vase initially.

I hope you can see that the argument "All balls are removed before
noon." alone is not sufficient to claim that the vase is empty at noon.

.

Relevant Pages

  • 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)
  • Re: infinity
    ... Number of balls in the vase at noon is f= OO. ... Unfortunately, if infinity gets involved, this statement alone is not sufficient to claim the vase is empty before noon. ... then the sum becomes 0. ...
    (sci.math)
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