Re: infinity

On 5 Aug 2005 23:31:30 -0700, snapdragon31 wrote:
> 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),

That formula does not apply to the case t=0. You also have a sign error
if the formula is intended to apply to times before noon (t<0).

> f(0) = 9 + 9 log(1/0)/log(2)
> = 9 + 9 log(OO)/log(2)
> = 9 + 9 * OO / log(2)
> = OO not 0.

Your math is faulty. Division by 0 is not allowed, and oo is not a
number. Your formula can be made to say something about the limit as
t->noon, but the limit is not what the problem asks for.

> 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

I mean t->0-. I use t<0 for times before noon, and t>0 for times after
noon. The left-hand limit is +oo, but the f(0) and the right-hand limit
are both 0.

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

That's why I pointed out that you had a sign error in your formula. It
should be log(-1/t), not log(1/t), in order to make sense for times
before noon.

>> > 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?

My other argument is that no ball is returned to the vase after it is
removed. Which ball do you think is in the vase at noon?

>> > 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?

[... irrelevant finite case snipped ...]

> 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.

Wrong. Noon is such a time. If you disagree, name a ball that is in the
vase at noon.

> 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.

You keep asserting this, but you have never given any hint as to what it
is that makes you think such a thing. What is the reasoning behind your
outlandish claim?

You also keep ignoring my question. If you think the vase is not empty
at noon, then what is the number of a ball that remains in the vase?

As long as you keep ignoring my question and refuse to provide any
supporting argument of your own, we will never make any progress.



--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
.

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
    ... 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)
  • Re: infinity
    ... >>> the vase keeps growing as you approach noon. ... the algorithm which describes the filling of the vase with balls ... Start with an empty vase. ... we try to connect some logical reasoning (about putting ...
    (sci.math)