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Topic: Transcendental binary game (Read 1796 times) |
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Aryabhatta
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Transcendental binary game
« on: Jul 26th, 2004, 12:49am » |
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This problem if from the book: "A problem Seminar", by Donald J Newman.
We alternate writing down binary digits after a decimal point therby producing a real number in the interval [0,1]. You win if this number is transcendental. Can you force a win?
I don't think this appeared before in this forum. If it has, my apologies.
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towr
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Re: Transcendental binary game
« Reply #1 on: Jul 26th, 2004, 2:13am » |
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::No you can't win, a transcendental number has an infinite number of digits. So it would take an infinite amount of time to write down the number, and thus an infinite amount of time before you could win. The universe will come to an end before you can win and will take you (being a part of said universe) with it..::
But I'm sure that's not the answer you were looking for 
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| « Last Edit: Jul 26th, 2004, 2:14am by towr » |
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Grimbal
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Re: Transcendental binary game
« Reply #2 on: Jul 26th, 2004, 7:01am » |
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I think if you just write down the binary digits of pi, whatever the other person writes, you end up with a transcendental number. No proof, though.
Only that non-transcendental number are countable. So, the set of odd digits belonging to a transcendental number is countable, while the set of odd digits you could write is not countable. So, by just writing random digits, you have 100% chance to end up with a transcendental digit. And pi digits are kind of random, aren't they?
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towr
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Re: Transcendental binary game
« Reply #3 on: Jul 26th, 2004, 7:13am » |
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Depends on what you call random, there is a formula to determine the nth digit (in hexadecimal) of pi without first calculating the first n-1 digits..
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Eigenray
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Re: Transcendental binary game
« Reply #4 on: Jul 26th, 2004, 8:48am » |
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on Jul 26th, 2004, 7:01am, Grimbal wrote:| So, by just writing random digits, you have 100% chance to end up with a transcendental digit. And pi digits are kind of random, aren't they? |
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Just because the probability of something happening is 0, it doesn't mean that it can't happen, just that it won't.
And the digits of pi are certainly not random. Calculate the 17th digit of [pi] as many times as you want, you (should) get 3 every time.
But the idea is correct. To make it a bit more rigorous:
Define an independent strategy to be a map t:[bbn][to]{0,1,...,9}. Let IS be the set of all independent strategies, except the one that sends everything to 9, for convenience. Note IS is uncountable.
For t [in] IS, define
St = { x [in] [0,1) | [forall]k[in][bbn] d2k(x) = t(k) },
where dn(x) = floor(10n*x) mod 10 is the n-th decimal digit of x. That is, St is the set of all the possible outcomes when I play strategy t, picking the decimal digits for the even places.
Let A be the set of algebraic numbers in [0,1) (if I play a strategy in IS, the outcome can't be exactly 1).
Now, [cup]t St = [0,1), so [cup]t (St[cap]A) = A.
Since the RHS is countable, and the union is of uncountably many disjoint sets, there must exist t' [in] IS such that St' [cap] A = [emptyset]. Thus, if I play strategy t', the number formed when we play will be transcendental, no matter what your strategy is.
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| « Last Edit: Jul 26th, 2004, 9:07am by Eigenray » |
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Aryabhatta
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Re: Transcendental binary game
« Reply #5 on: Jul 26th, 2004, 11:46pm » |
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on Jul 26th, 2004, 2:13am, towr wrote:::No you can't win, a transcendental number has an infinite number of digits. So it would take an infinite amount of time to write down the number, and thus an infinite amount of time before you could win. The universe will come to an end before you can win and will take you (being a part of said universe) with it..::
But I'm sure that's not the answer you were looking for |
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Not if you play this game with Zeno. 
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Aryabhatta
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Re: Transcendental binary game
« Reply #6 on: Jul 26th, 2004, 11:51pm » |
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Doesn't it remind you of the Sink the Sub puzzle? (Found in the hard riddles)
bw, Eigenray, the problem talked about binary representation, so your solution makes no sense
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Eigenray
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Re: Transcendental binary game
« Reply #7 on: Jul 27th, 2004, 5:11am » |
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Whoops. I guess I was thrown off by seeing the word 'decimal' in the problem statement. Oh well, back to the drawing board...
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