Author |
Topic: Transcendental binary game (Read 1764 times) |
|
Aryabhatta
Uberpuzzler
Gender: 
Posts: 1321
|
 |
Transcendental binary game
« on: Jul 26th, 2004, 12:49am » |
 Quote  Modify
|
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.
|
|
 IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Transcendental binary game
« Reply #1 on: Jul 26th, 2004, 2:13am » |
Quote Modify
|
::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 
|
| « Last Edit: Jul 26th, 2004, 2:14am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7526
|
|
Re: Transcendental binary game
« Reply #2 on: Jul 26th, 2004, 7:01am » |
Quote Modify
|
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?
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Transcendental binary game
« Reply #3 on: Jul 26th, 2004, 7:13am » |
Quote Modify
|
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..
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Transcendental binary game
« Reply #4 on: Jul 26th, 2004, 8:48am » |
Quote Modify
|
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? |
|
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.
|
| « Last Edit: Jul 26th, 2004, 9:07am by Eigenray » |
IP Logged |
|
|
|
Aryabhatta
Uberpuzzler
Gender:
Posts: 1321
|
|
Re: Transcendental binary game
« Reply #5 on: Jul 26th, 2004, 11:46pm » |
Quote Modify
|
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 |
|
Not if you play this game with Zeno. 
|
|
IP Logged |
|
|
|
Aryabhatta
Uberpuzzler
Gender:
Posts: 1321
|
|
Re: Transcendental binary game
« Reply #6 on: Jul 26th, 2004, 11:51pm » |
Quote Modify
|
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
|
|
IP Logged |
|
|
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Transcendental binary game
« Reply #7 on: Jul 27th, 2004, 5:11am » |
Quote Modify
|
Whoops. I guess I was thrown off by seeing the word 'decimal' in the problem statement. Oh well, back to the drawing board...
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print