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September 14

Square Root of Two

I've been trying to prove that the decimal expansion of root two contains more than two digits infinitally often, but can't seem to get past the extension of multiplication to an infinite decimal. Has this ever been proven before, (I'm pretty sure its Normality has not) and if it has, any pointers would be helpful. Thanks :) --Gilderien Chat|List of good deeds 20:52, 14 September 2013 (UTC)[reply]

I have no answers: only questions... But sometimes asking the right questions can be just as helpful: so here goes nothing: If a number is normal in a certain base, then is it normal in all other bases ? I'm asking this because -if this were true- you could probably simplify your work a lot by expressing these numbers in binary, since it only uses two digits, and the lack of any of tne two would imply the number to be rational... and it's also probably way easier to check the frequency of only two possible digits, as opposed to ten, for instance... Do you think such an approach would prove itself helpful ? — 79.113.226.248 (talk) 21:03, 14 September 2013 (UTC)[reply]

Normality in one base does not imply normality in all other bases. This is mentioned at normal number, and at the reference: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pjm/1103038420 Staecker (talk) 16:43, 15 September 2013 (UTC)[reply]

(edit conflict) I don't know the answer, but I have suggestions. In our article Square root of 2, in the section Series and product representations, for each of the series given: for each decimal position only finitely many terms control what that digit is. If you could characterize each term's decimal expansion, you might be able demonstrate it this way. Related to that, the section Computation algorithms spells out a method that, again by characterizing the terms of the sequence involved, could give you some results about the decimal expansion. I tried poking about for results from others on this question, but I kept getting flooded with results about computations of the decimal expansion, I'll dig about more later. :-)Phoenixia1177 (talk) 21:08, 14 September 2013 (UTC)[reply]

If the nth root of a positive integer isn't an integer then it is irrational. Gardner's Workout, by Martin Gardner, page 10, says that all tests so far are consistent with all such irrationals being normal in every base. Of course, such tests don't prove that a number is normal. Bubba73 ^{You talkin' to me?} 00:59, 18 September 2013 (UTC)[reply]

Starting with the Continued fraction [2; 2, 2, 2,...] for 1 + sqrt(2), you get a sequence a(n)/b(n). a(n+1) and b(n+1) depend linearly on a(n) and b(n), so you can carry out iteration using powers of a matrix acting on the vector with entries a(0) = 2, b(0) = 1. To simplify the process, we find the eigenvalues of the matrix and diagonalize, the end result is the sequence of terms sqrt(2)(a₊^N + a_-^N)/((a₊^N - a_-^N)) ,where we have a_+/- = 1 +/- sqrt(2), converging to sqrt(2). To deal with the sqrt(2) terms, you can expand all the powers out using the Binomial theorem, then you get a ratio of two series. This is as far as I got, but I imagine you could do some analysis on these to get some ideas about the decimal expansion. By the way, you'll notice I don't include many actual details, the reason for this is that you can easily extract them from Silver ratio and Pell number- I didn't realize this till working it out from scratch...then just happening upon them. Was fun nonetheless:-)Phoenixia1177 (talk) 05:36, 19 September 2013 (UTC)[reply]