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A lot of balls

A lot of balls

Posers and Puzzles

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Originally posted by iamatiger
so the lines intersect at infinity?
I don't think I'm saying that (to be honest, geometry is not my strength and I wouldn't know how to define properly intersections at infinity). I'm saying that cardinality relations are not "continuous" in the way you describe (i.e. the cardinality of the limit is not necessarily the limit of cardinalities). The example with natural and even numbers highlights this.

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Originally posted by iamatiger
Possibly the problem lies in the difference between "the same cardinality" and equality?

And the problem with the balls in the bag being zero is that infinity-infinity=infinity?
Yes, I think this is key. But note that "same cardinality" is not enough. The operations with the bags must reflect the one-to-one correspondence (i.e. for any ball we must be sure it is guaranteed to leave the bag at some point).

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Originally posted by Palynka
I don't think I'm saying that (to be honest, geometry is not my strength and I wouldn't know how to define properly intersections at infinity). I'm saying that cardinality relations are not "continuous" in the way you describe (i.e. the cardinality of the limit is not necessarily the limit of cardinalities). The example with natural and even numbers highlights this.
I think parallel lines intersect at infinity.

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Originally posted by AThousandYoung
I think parallel lines intersect at infinity.
Yes, but we have diverging lines here, and these are generally thought to not intersect at infinity.

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Originally posted by adam warlock
But I do have to say that this apparently simple problem really is head spinner. I'll post on some reputable mathematical blogs and see what the big shots answers will be.
Any luck?

What are those blogs, by the way?

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adds 1 and 2 ballcount= 2
adds 3 and 4 ballcount= 4
removes 1 ballcount= 3
adds 5 and 6 ballcount= 5
removes 2 ballcount= 4
adds 7 and 8 ballcount= 6
removes 3 ballcount= 5
adds 9 and 10 ballcount= 7

as you can see, the ballcount approches infinity as the number of actions go up, and the number of actions are related to the time endured. therefore as time goes on the number of balls in the bag will increase and approach infinity as infinite actions are performed

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Originally posted by Klasker
adds 1 and 2 ballcount= 2
adds 3 and 4 ballcount= 4
removes 1 ballcount= 3
adds 5 and 6 ballcount= 5
removes 2 ballcount= 4
adds 7 and 8 ballcount= 6
removes 3 ballcount= 5
adds 9 and 10 ballcount= 7

as you can see, the ballcount approches infinity as the number of actions go up, and the number of actions are related to the time endured. therefore ...[text shortened]... number of balls in the bag will increase and approach infinity as infinite actions are performed
Wow, Klasker can count!

But can he read (the thread)?

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Originally posted by Palynka
Wow, Klasker can count!

But can he read (the thread)?
Klaster is right.

Additionally, because the sets of added balls and removed balls are provably the same order of infinity infinity, and infinity-infinity = infinity (if they are the same order, and one is growing twice as fast as the other), there are an infinite balls in the bag.

http://www.newton.dep.anl.gov/askasci/math99/math99191.htm

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Originally posted by iamatiger
Klaster is right.

Additionally, because the sets of added balls and removed balls are provably the same order of infinity infinity, and infinity-infinity = infinity (if they are the same order, and one is growing twice as fast as the other), there are an infinite balls in the bag.

http://www.newton.dep.anl.gov/askasci/math99/math99191.htm
Infinity is the only answer that leads to a contradiction. You can either chose "None" or "Supertasks are impossible so there is no defined answer". "Infinity" is, however, wrong.

Nice link. No idea why it's relevant, though. At no point is an operation "infinity-infinity" needed.

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