Re: infinity


Dave Seaman wrote:
> 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.
>

It is such a simple equation and simple definition. I don't expect you
get confused and claim that they are wrong. Please take look at them
again carefully. My definition of t is the time in minute before noon.
So 11:59am is t=1 and 11:59:30am is t=1/2 and so on.

I don't care what convention you want to use for t. But If you use my
equation, please use also my definition of t. Please do change my
definition of t and then claim my equation is wrong and then say my
definition is wrong. This is your problem.


> >> > 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?

It would be easier for you to see if the operation add only 1 ball (not
10) and then take out 1 ball. Assuming that we have 1 ball in the vase
initially. You can see that no matter how many times there is always
one ball in the vase. The number on the ball in the vase always
changes.
At 11:59 Ball 1 is in the vase
At 11:59:30 Ball 2 is added and ball 1 is removed. Ball 2 remains in
the vase.
At 11:59:45 Ball 3 is added and ball 2 is removed. Ball 3 leaves in
the vase.

All ball comes into the vase will eventually moved out from the vase
but at no time the vase is empty. Therefore, the vase can never be
empty. Are you telling me after an infinite steps the vase will become
empty? How? My suggestion would be after an infinite number of steps,
there would still be 1 ball left in the vase. Can you give me a
counter example?

>
> 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?
>
Not be able to name the number of a ball remains in the vase does not
mean that the vase is empty. You can keep all the balls and only take
away whatever ball I name at the very last step. It does not prove
anything, right? I can prove that the number of balls grow
indefinitely, but you cannot prove that the vase is empty.

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

.

Relevant Pages

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