Associated Topics || Dr. Math Home || Search Dr. Math

### 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/
```
Associated Topics:
High School Sequences, Series

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search