Decimal Expansion of 1/7 - correction
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From: JEMebius on 17 May 2010 08:50 JEMebius wrote: > Hein wrote: >> Most people look at the first four digits of 1/7 = .1428... and either >> don't notice the appearance of multiples of 7 (2 times 7 and 4 times >> 7) in the decimal representation or think that their existence in the >> representation is a mere coincidence. It's not. Let's compute the >> decimal expansion of 1/7 without doing long division, but using >> instead the identity (***) >> >> x/(1-x) = x + x^2 + x^3 + ... >> >> (Note the identity holds if x is between -1 and 1.) >> >> >> 1/7 >> = 7/49 >> = 7*1/(50-1) >> = 7*1/50/(1-1/50) >> = 7*.02/(1-.02) (now use *** with x=.02) >> = 7*(.02 + .02^2 + .02^3 + ...) >> = 7*(.02 + .0004 + .000008 + .00000016 + .0000000032 + ...) >> = 7*(.02040816326...) >> = 7*(.020408) + 7*(.00000016326...) >> = .142856 + 7*(.00000016) + 7*(.00000000326...) >> = .142856 + (.00000112) + 7*(.00000000326...) >> >> So, 2*7=14 and 4*7=28 are the first 4 digits because 7*.0204 = . >> 1428. >> >> Also, notice that the first 6 digits of 1/7 are >> >> .142856 + .000001= .142857. >> >> For any positive integer j, the decimal representation of 1/j either >> terminates or repeats with at most (j-1) repeating digits and we have >> six (7-1) digits, so those six digits must repeat, so finally 1/7 must >> be >> >> 1/7 = .142857142857142857142857142857142857.... >> >> >> Cheers, >> Hein Hundal > > > Quotation: > "For any positive integer j, the decimal representation of 1/j either > terminates or repeats with at most (j-1) repeating digits (...)" > > Not true in general: > make a careful study of http://en.wikipedia.org/wiki/Repeating_decimal > or still better: spend several feet or yards of scrap paper in doing > several different long divisions. > For starters: 1/13, 1/27, 1/37, 1/41, 1/49, 1/53, 1/73, 1/81, 1/137, > 1/239, 1/243. > In my opinion the nicest are 1/487 and 1/487^2, both of which repeat > after 486 digits. > > I myself discovered in my boyhood years that not only 1/7, but also > 1/13, 1/21, 1/39, 1/63, and ultimately 1/999999 have repeating decimal > expansions of period 6. > A revelation was the factorization 999999 = 3^3 . 7 . 11 . 13 . 37 > With this factorization in mind one easily finds all expansions of 1/Q > of period 6. > > Enjoy! Johan E. Mebius Quotation: "For any positive integer j, the decimal representation of 1/j either terminates or repeats with at most (j-1) repeating digits (...)" This is correct! Again I was too fast and eager in replying! I overlooked the words "at most". Once one has proved that the decimal expansion of a real number A terminates or is periodic if and only if A is rational, it is a piece of cake to prove that the period of the expansion of 1/A (A containing at least one factor other than 2 or 5) is indeed at most A-1 digits long. Around midnight May 15th-16th the Delft region was hit by a power failure. A fire in the regional high-voltage distribution station at Delft. About 250,000 households were without electricity. So I could not reply until now. Nevertheless I hope my post is of some use to some of my fellow newsgroup readers. Ciao: Johan E. Mebius |