Square Root of 3 minus 1Date: 09/24/97 at 04:12:40 From: Brian Lam Subject: Very Challenging Integer Problem Dr. Maths, This is a very hard and challenging problem for a student aged 14 to work out without some long and hard thinking. World you please help me with it? Express sqrt 3 minus 1 as a continued fraction. Thank you! Brian Lam Date: 09/24/97 at 08:59:05 From: Doctor Rob Subject: Re: Very Challenging Integer Problem Brian, To solve this problem, you need to know the general way to expand a quadratic surd as a continued fraction. In general, let t(0) be any number you want to expand. Then for each n >= 0, let a(n) = [t(n)] (greatest integer less than or equal to t(n)), and let t(n+1) = 1/(t(n)-a(n)) Continue this until either the denominator is zero (and the process ends), or until t(k) = t(i) for some k and i (and the process is obviously periodic). Then the continued fraction of t(0) is given by 1 t(0) = a(0) + ------------------------ = [a(0); a(1), a(2), a(3), ...] 1 a(1) + ----------------- 1 a(2) + ---------- 1 a(3) + --- ... When t(0) is a quadratic surd, it can be put into the form (P + R*Sqrt[D])/Q, where P, Q, R, and D are integers, and D > 0 and not divisible by a square, and Q is a divisor of P^2 - R^2*D. When you subtract the integer part a(0), you get a quantity of the same form (but different P), and satisfying the same conditions. When you take the reciprocal, and then rationalize the denominator, you get another quantity of the same form (but different P and Q, and possibly R), and satisfying the same conditions. In the case at hand, P = -1, Q = 1, R = 1, and D = 3, when you begin. The values of t(n) and a(n) can be tabulated: n t(n) a(n) t(n)-a(n) 0 (Sqrt[3]-1)/1 0 (Sqrt[3]-1)/1 1 (Sqrt[3]+1)/2 1 (Sqrt[3]-1)/2 2 ... ... ... You continue until t(n) = t(i) for some i < n. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 09/24/97 at 09:59:41 From: Doctor Anthony Subject: Re: Very Challenging Integers Problem When dealing with surd quantities like [sqrt(3)-1] in the numerator of an expression we shall be multiplying top and bottom by the conjugate to get [sqrt(3)-1][sqrt(3+1)] 3-1 2 ---------------------- = ----------- = ------------ sqrt(3)+1 sqrt(3)+1 sqrt(3)+1 Sqrt(3) - 1 = 0 + sqrt(3)-1 = [sqrt(3)-1][sqrt(3+1)] --------- ---------------------- 1 sqrt(3)+1 = 2 1 --------- = -------------- ..............(1) sqrt(3)+1 [sqrt(3)+1]/2 We now work on the next fraction [sqrt(3)+1]/2 sqrt(3)+1 sqrt(3)+1 sqrt(3)-1 multiply top and --------- = 1 + --------- - 1 = 1 + --------- bottom by conjugate 2 2 2 1 = 1 + ---------- sqrt(3)+1 And the next fraction is [sqrt(3)+1]/1 sqrt(3)+1 sqrt(3)+1 sqrt(3)-1 ---------- = 2 + ----------- - 2 = 2 + ---------- 1 1 1 2 = 2 + ---------- sqrt(3)+1 1 = 2 + ------------ [sqrt(3)+1]/2 But if you look back at expression (1) you will see that we have already met this continued fraction, so from here on we have a recurring continued fraction. The whole number parts of the continued fractions will then go 1, 2, 1, 2 for evermore. Writing this in the shorthand for continued fractions we get: 1 1 1 1 sqrt(3)-1 = 0 + ---- ---- ----- ----- ............ 1+ 2+ 1+ 2+ -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/