I've re-written this, from Martin Gardner's version in his "

**The Colossal Book of Mathematics**":

Imagine you have access to an infinite supply of ping pong balls, each of which bears a positive integer label on it, which is its 'rank.' And for every integer there are an INFINITE number of such balls available; i.e. an infinite no. of "#1" balls, an infinite no. of "#523" balls, an infinite no. of "#1,356,729" balls, etc. etc. You also have a box that contains some

*FINITE*no. of these very same-type balls. You have as a goal to empty out that box, given the following procedure:

You get to remove one ball at a time, but once you remove it, you

*must*replace it with any finite no. of your choice of balls of

**'lesser**' rank. Thus you can take out a ball labelled (or ranked) #768, and you could replace it with 27 million balls labelled, say #563, just as one of a multitude of examples. The

**sole**exceptions are the #1 balls, because obviously there are no 'ranks' below one, so there are

**NO**replacements for a #1 ball.

Is it possible to empty out the box in a finite no. of steps??? Or posing the question in reverse, as Gardner asks: "Can you not prolong the emptying of the box forever?" And then his answer: "Incredible as it seems at first, there is

**NO WAY**to avoid completing the task." [bold added]

Although completion of the task is "unbounded" (there is no way to predict the number of steps needed to complete it, and indeed it could be a

*large number), the box*

**VERY****MUST**empty out within a

*finite*number of steps!

There are various proofs of this amazing result (which Raymond Smullyan originally published in the "Annals of the New York Academy of Sciences" in 1979, Vol. 321), but it only requires logical induction to see the general reasoning involved:

Once there are only #1 balls left in the box you simply discard them one by one (no replacement allowed) until the box is empty --- that's a given. In the simplest case we can start with only #2 and #1 balls in the box. Every time you remove a #2 ball, you can ONLY replace it with a #1, thus at some point (it could take a long time, but it must come) ONLY #1 balls will remain, and then essentially the task is over. S'pose we start with just #1, #2, and #3 balls in the box... Every time a #3 ball is tossed, it can only be replaced with #1 or #2 balls. Eventually, inevitably, we will be back to the #1 and #2

**only**scenario (all #3 balls removed), and we already know that situation must then terminate. The same logic applies no matter how high up you go (you will always at some point run out of the very 'highest-ranked' balls and then be working on the next rank until they run out, and then the next...); eventually you will of necessity work your way back to the state of just #1 and #2 balls, which then convert to just #1 balls and game over (even if you remove ALL the #1 and #2 balls

*first*, you will eventually work back and be using

*them*as replacements). Of course no human being could live long enough to actually carry out such a procedure, but the process must nonetheless amazingly conclude after some mathematically finite no. of steps. Incredible! (too bad Cantor isn't around to appreciate this intuition-defying problem).

If you wish to read about the problem in Gardner's volume (which is available for free on the Web, BTW) it is near the beginning of his Chapter 34. But again, we have logician Raymond Smullyan to thank for this wonderful thought paradox. I'm just using Gardner as the great explicator that he is.

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