Square Root of 3 minus 1

Date: 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/