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May 28th, 2012, 08:09 AM  #1 
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Another interesting fraction [color=beige]. . [/color] [color=beige]. . [/color] [color=beige]. . [/color] [color=beige]. . [/color] [color=beige]. . [/color] [color=beige]. . [/color] [color=beige]. . [/color] [color=beige]. . [/color] [color=beige]. . [/color] [color=beige]. . . . . . [/color] [color=beige]. . [/color] 
May 28th, 2012, 11:38 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Another interesting fraction
It's a geometric series due to the Binet formula. I don't have pencil or paper at the moment but that should provide the answer readily enough.

May 31st, 2012, 04:49 AM  #3 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: Another interesting fraction
I was hoping that this thread would get more traction. My own quickie internet search came up with the fact that 89 is of the form x^2x1 for the x that serves as the base of our decimal system, ie 10. I've had a general interest in the difference between results that are dependent on the base you are working in and those that are not. Divisibility tests are obviously totally base dependent. Want to know whether a large number is divisible by 137? Just translate the base ten numeral into base 137 and see if it ends in zero! Hey, thanks for the tip, eh? This one, however, seems more interesting. 
May 31st, 2012, 07:41 AM  #4  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Another interesting fraction Quote:
 So F_n = (phi^n + phi^n)/sqrt(5) so This can be simplified mechanically (replace phi with radical definition, rationalize denominators three times, multiply through rationalize denominators again, add, reduce denominators by gcd). I get which is close to but not equal to 1/89.  Oops, got the Binet formula wrong, should be F_n = (phi^n  (phi)^n)/sqrt(5). That does give 1/89. The proof is tedious and essentially identical to the above with some signs reversed.  

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