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Home > Paradoxes
Marbles Bonanza (Posted on 2003-09-08) Difficulty: 4 of 5
You have an empty container, and an infinite number of marbles, each numbered with an integer from 1 to infinity.

At the start of the minute, you put marbles 1 - 10 into the container, then remove one of the marbles and throw it away. You do this again after 30 seconds, then again in 15 seconds, and again in 7.5 seconds. You continuosly repeat this process, each time after half as long an interval as the time before, until the minute is over.

Since this means that you repeated the process an infinite number of times, you have "processed" all your marbles.

How many marbles are in the container at the end of the minute if for every repetition (numbered N)

A. You remove the marble numbered (10 * N)

B. You remove the marble numbered (N)

See The Solution Submitted by levik    
Rating: 3.6154 (13 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: more | Comment 36 of 83 |
(In reply to more by Cory Taylor)

Cory,

Like FatBoy, you are making at least a couple of (invalid and unstated) assumptions. In your case, you are likening normal arithmetic operations that work on real numbers to those that apply to infinities. But they do not.

To deal with two infinities of the same Cardinality, you MUST deal with the members of the set. And members of a set ARE named or referred to in some manner.

I understand why you wish to dismiss the "naming" of the elements in a particular set. I am an engineer too, and I agree that it can help to understand problems by assigning numbers/labels.

But it does not change matters.

What if I take an infinite set (the set of all integers), and I remove an infinite set of integers from that (the odd integers). I am left with an infinite set (the even integers). However, if, instead, I remove (the set of all integers), I am left with an empty set. Either way I remove a set of the same size. But that doesn't mean the sets are equivalent.

By this, I think it obvious that the LABELING (or more accurately, which members of the set I remove) DOES in fact make a difference.

So, to paraphrase your statement.... It's very clear that the math you produced is flawed.

--- SK
  Posted by SilverKnight on 2003-09-15 12:14:24

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