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Throwing rocks in a hole

  1. Feb 29, 2012 #1
    Say it's 11:00 and you throw 100 rocks in a hole and at the exact same second a rabbit throws one rock out of the hole. Then at 11:30 you throw a 100 rocks in and the rabbit throws one rock out. At 11:45 you throw 100 rocks in and the rabbit throws one rock out and so on, with the time increments halving each time. How many rocks will be in the hole at 12:00? Well one thought is that you've thrown a countably infinite number of rocks in and the rabbit has thrown countably infinite number out so the answer is 0, but say you threw 99 rocks in the hole each time and there was no rabbit then obviously the answer is infinity. Where is the contradiction?
     
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  3. Feb 29, 2012 #2

    DaveC426913

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    You assume countably-infinite-X minus countably-infinite-Y is equal to zero, if X=Y.

    Is that a valid assumption?
     
  4. Feb 29, 2012 #3
    I would have to assume that to get an answer of zero.
     
  5. Mar 1, 2012 #4

    Office_Shredder

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    It depends on the rabbit's algorithm for throwing rocks out of the hole.

    Suppose that at each step he throws out the rock most recently thrown into the hole. Then at each step you throw 99 rocks into the hole which are never removed by the rabbit, and you will have infinite rocks in the hole at the end.

    On the other hand suppose that at each step he takes the rock which is on the bottom of the pile (the rock which was thrown in before any other rock currently in the hole) and removes that one. Then every rock which is thrown into the hole is removed by the rabbit, and you will end up with no rocks in the hole
     
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