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Impish Sprite |
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| Jun2-04, 07:18 AM | #1 |
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Impish Sprite
Suppose there is an infinite number of small balls, all of which have been uniquely numbered from 1 upwards. Furthermore, you have an infinitely large bucket. That is to say, it can be made as large as necessary. You decide to fill the bucket by throwing in all the balls, in order. Starting from 1, every minute you throw in two balls. But every minute, an impish sprite takes one ball back out, always extracting the lowest-numbered ball in the bucket.
For example, 1st minute: You throw in Ball 1 and Ball 2. Impish sprite extracts Ball 1. 2nd minute: You throw in Ball 3 and Ball 4. Impish sprite extracts Ball 2. 3rd minute: You throw in Ball 5 and Ball 6. Impish sprite extracts Ball 3. and so on... QUESTION: After an infinite amount of time has elapsed, how many balls are in the bucket? Argument 1: There is an infinite number of balls in the bucket. After 1 minute there is 1 ball. After 2 minutes there are 2 balls. After 3 minutes there are 3 balls, etc. Argument 2: There are no balls in the bucket. If there are some balls in the bucket, what is the number of the lowest-numbered ball? It can't be Ball 1; that was extracted after 1 minute. Similarly, it can't be Ball 2; that was extracted after 2 minutes. It can't be Ball 3; that was extracted after 3 minutes, etc. (If the phrase 'after an infinite amount of time has elapsed' bothers you, then we can change the problem so that the 1st put-in-and-take-out operation is completed in 1/2 minute, the 2nd operation is completed in 1/4 minute, the 3rd in 1/8 minute, and so on. Now you can ask the question after 60 seconds, and "infinite time" is not longer an issue.) |
| Jun2-04, 07:32 AM | #2 |
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By the way, is this a real brain teaster (i.e is there a definite answer?) or just a paradox? Either way argument 1 seems more sensible to me. The rate of growth of the number of balls in the bucket is 1 per operation, so no matter how many operations you perform there will always be at least 1 ball in the bucket. No? |
| Jun2-04, 07:49 AM | #3 |
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Well the number of balls afetr n minutes is: n(2-1) = n
so as n -> infinity the number of balls tends to infinity too. Of course the number on lowest numbered ball tends to infinity also, but we're only intersted in the number of balls. |
| Jun2-04, 07:59 AM | #4 |
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Impish Sprite
Instead of "infinite", the proprer term is "arbitrarily large". You can't have an infinite number of balls or an infinitely large bucket or an infinite amount of time.
After an arbitrarily long period of time, you will have an arbitrarily large number of balls in the bucket, and an equal amount removed from the bucket (give or take one, depending on the timing). The lowest numbered ball in the bucket will be, you guessed it, arbitrarily large. Njorl |
| Jun2-04, 08:17 AM | #5 |
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Arbitarily large...
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| Jun2-04, 11:23 AM | #6 |
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So the question should be considered purely as a mathematical puzzle. |
| Jun2-04, 03:28 PM | #7 |
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If your question was rhetorical, and in the context of the quote (though I don't see the connection), feel free to ignore this response. |
| Jun2-04, 06:06 PM | #8 |
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His post was not rhetorical, but directly addresses an argument made earlier.
Also note that limx->a f(x) may fail to equal f(a) even when f(a) is defined and the limit exists. For example, f(x) = 0x has this property for a = 0. |
| Jun4-04, 11:16 AM | #9 |
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This is more of a paradox than a brain teaser.
The are half as many balls in the bucket as were originally thrown there, so number of balls equals 1/2 n and n-> infinity. A little like asking which has more points: a circle or a line? |
| Jun4-04, 08:13 PM | #10 |
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There is nothing truly paradoxical about this.
Tell me, what are some of the numbers on the balls left in the bucket? |
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