Farey SeriesDate: 10/21/2002 at 09:24:35 From: Clive Subject: Farey's series Dear Dr. Math, I have encountered some difficulties in solving problems about Farey's series. For 3 successive terms in a Farey's series, say a/b, c/d, e/f, how can we show that i) c/d = (a+e)/(b+f) ii) ad-bc = -1 I can prove that i) implies ii) and ii) implies i), but I cannot prove either one independently. Any help will be appreciated. Thank you. Yours sincerely, Clive Date: 12/10/2002 at 01:51:48 From: Doctor Nitrogen Subject: Re: Farey's series Hi, Clive: Here is a suggestion as to how you can prove (i) and (ii) independently: [A] Let a <= b <= n, [B] e <= f <= n for some n. Since a/b < e/f, a < b, e < f, and since a + e < b + f, it is also true that [1] a/b < (a + e)/(b + f) (Why? Hint: look at the inequality a(a + e) < b(b + f), and at (a + e)/(b + f) < b/a). "Manipulate" this inequality around and see what happens.) Therefore, a/b < (a + e)/(b + f). With a similar argument, (a + e)/(b + f) < e/f. Therefore a/b < (a + e)/(b + f) < e/f is true. Now for proving (ii), let GCD(a + e, b + f) = k => 1. then (a + e)/(b + f) = kc/kd = c/d, for some integers c and d, with GCD(c, d) = 1. There is a theorem in number theory that says if GCD(c, d) = 1, then the equation cx + dy = 1 has a solution for (x, y). Similarly, for GCD(a, b) =1, the equation bx + dy = 1 has a solution for (x, y), and the solutions will be (b, -a), and (c, -d) respectively (you might have to prove this to convince yourself). One reference that has a theorem and proof close to this last result is _The Theory of Numbers: An Introduction_, Anthony A. Gioia, Markham Publishing. This is an older publication dating back to the sixties. In {A] and [B] above, the number n has a connection to Farey series. For example, for the Farey Series 0/1, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 2/5, 3/7, 1/2, 4/7, 3/5, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 1, for this Farey series, n = 7. The conditions [A] and [B] above are actually required for a Farey series. - Doctor Nitrogen, The Math Forum http://mathforum.org/dr.math/ |
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