A lot of balls

I agree with Palynka, there should be no balls left in the bag at time 0. The most intuitive (while possibly not the most rigorous) way to think about it is to ask the question "is there any ball I can place in the bag that my friend will not be able to take out?". Since your friend takes out each ball in order, the answer is "no". Therefore, despite the fact that balls are being placed in the bag at twice the rate they are being removed, each ball will eventually be removed in order leaving no balls in the bag.

I think the problem with mtthw's solution is that subtracting infinities leads to funny answers. At first glance, it appears that the number of balls in the bag at any step in the process can be described in either of the two following equivalent forms:

N = 2 + (2i) - (i)

N = 2 + (2i - i) = 2 + i

This method works fine for any finite step "i" in the process. However, I don't think they remain equivalent if the number of steps is infinite since "i" is then no longer an element of the set on natural numbers but rather the set of natural numbers itself. As I understand it, simple arithmetic does not apply to sets and so the equations above cease to have meaning. Palynka's example equating the cardinality of the set of natural numbers with the set of even natural numbers is just one example of the counter-intuitive properties that arise when dealing with infinity, sets and the operations carried out on them.

Contrast this with the type of limit equation found in many calculus text books:

lim x-->inf (2x - x) = lim x-->inf (x)

Here we have a real equivalence since the symbol "x" always represents an element of the set of real numbers, even as "x" increases without bound. In this case, the limit does not exist since both expressions increase without bound as "x" increases without bound. However, the number of balls in the bag should still converge to 0.

I note that the number of balls in the bag increases by one on every step (2 in, one out)

So on step N there are N balls.

As on each step, we are 60*(1-1/2^n) seconds away from zero, we need an infinite number of steps to reach zero. And as, at time zero there have been infinite steps, there are infinite balls by then.

This does assume time is not quantum.....

Then how do you address Palynka's proof by contradiction that there is no smallest numbered ball in the bag, and the problems with using arithmetic on finite and infinite numbers I pointed out above?

Originally posted by PBE6
Then how do you address Palynka's proof by contradiction that there is no smallest numbered ball in the bag, and the problems with using arithmetic on finite and infinite numbers I pointed out above?

Maybe the smallest number is infinitely big.

Originally posted by Thomaster
Maybe the smallest number is infinitely big.

Touche Cecil, hehe...

Actually, I think Palynka's argument works even if the smallest number were infinitely big.

Originally posted by Thomaster
Maybe the smallest number is infinitely big.

Infinity is not a number. Take any ball from the bag (there's infinite of them, that should be easy), its number N must be finite (there is no "infinite number" ). If it's not the smallest, then N-1 must exist. Iterate. Contradiction when N=1.

Originally posted by iamatiger
I note that the number of balls in the bag increases by one on every step (2 in, one out)

So on step N there are N balls.

As on each step, we are 60*(1-1/2^n) seconds away from zero, we need an infinite number of steps to reach zero. And as, at time zero there have been infinite steps, there are infinite balls by then.

This does assume time is not quantum.....

But there is no step "Infinity", so it's not necessarily true that the extrapolation is correct.

Originally posted by Palynka
But there is no step "Infinity", so it's not necessarily true that the extrapolation is correct.

Nevertheless, 0 is reached. Which step is it reached on?

Originally posted by iamatiger
Nevertheless, 0 is reached. Which step is it reached on?

There's no step numbered "infinity".

Originally posted by Palynka
There's no step numbered "infinity".

So every step has a number? What is the last step called?

Originally posted by Thomaster
So every step has a number? What is the last step called?

Jesus Christ.

There. Is. No. Last. Step. There is an infinite number of steps.

Originally posted by Palynka
There's no step numbered "infinity".

If you read what I actually said, you should see that I didn't say there was.

Originally posted by iamatiger
If you read what I actually said, you should see that I didn't say there was.

It's a limit result and so it is reached as the number of steps approaches infinity. So, yes, your question about "which step" implies that it does.

Originally posted by PBE6
Then how do you address Palynka's proof by contradiction that there is no smallest numbered ball in the bag, and the problems with using arithmetic on finite and infinite numbers I pointed out above?

step 1, bag has 1, 2

step 2, bag has 2,3,4

step 3, bag has 3,4,5,6

step 4, bag has 4,5,6,7,8

Even though your friend takes your numbers out, he takes them out too slowly to catch up. Lets do a simple transform that does not change the number of the balls:

The converted ball number on step N is ball_number-n+1

Now:

Step 1 converts to balls 1,2
Step 2 converts to balls 1,2,3
step 3 converts to balls 1,2,3,4
step 4 converts to balls 1,2,3,4,5

This shows your friend is getting nowhere. He may as well stop removing balls, it makes no difference.