Continued fraction.

• Jan 25th 2011, 07:05 PM
Mathest
Continued fraction.
I edited this because I believe this is a better way to put it, but now I can't change the title

with 1/(1/(c+1/d)+b)) + a = 40/31, I need to find..

a + b + c + d.
However, I have no clue where to start
• Jan 27th 2011, 10:54 AM
Otterified
It's a continued fraction
Continued fraction - Wikipedia, the free encyclopedia

$40/31 = a + \frac{1}{b+\frac{1}{c+\frac{1}{d}}}
$

This is a continued fraction. The wikipedia article I linked will give you a better explanation of all this, but here's how you can create a continued fraction for a number (and thus determine a+b+c+d).

Start with your number written as a mixed fraction, or 1 and 9/31. a is equal to the integer part of this number, so a = 1. Take the fractional part, find its reciprocal, and then write that as a mixed fraction, and repeat the process. 9/31 -> 31/9 = 3 and 4/9, so b = 3. 4/9 -> 9/4 = 2 and 1/4, so c = 2. 1/4 -> 4/1 = 4, so d = 4. Therefore, a + b + c + d = 1 + 2 + 3 + 4 = 10.