Consider the following expression: $1/(\sqrt a_1 ± \sqrt a_2 ± ... \sqrt a_n)$ where $a_1 ... a_n$ are positive integers
I would like a general algorithm by which I can convert expressions of this form into expression of the following form: exp/k
where k is some positive integer and exp is an expression consisting of additions, subtractions, multiplications and square roots but no divisions.
Here is an example of a simple case:
$1/(\sqrt 2 + \sqrt 3) = (\sqrt 3 - \sqrt 2)/(3 - 2) = \sqrt 3 - \sqrt 2$
Here I multiplied the top and bottom by the conjugate. This technique can be used in the general problem for n≤4 by breaking the sum up into pairs of roots. I will illustrate this below in the case where all square root terms are added rather than having subtraction mixed in.
consider this denominator: $(\sqrt a + \sqrt b) + (\sqrt c + \sqrt d)$
multiplying by its conjugate: $(\sqrt a + \sqrt b) - (\sqrt c + \sqrt d)$ yields:
$(\sqrt a + \sqrt b)^2 - (\sqrt c + \sqrt d)^2$
$a+b+2\sqrt (ab) - c - d - 2\sqrt (cd) = $
$ A + (\sqrt B - \sqrt C)$ where $A= a+b-c-d$ and $B=4ab$ and $C=4cd$
multiplying by its conjugate $ A - (\sqrt B - \sqrt C)$ yields:
$ A^2 - (\sqrt B - \sqrt C)^2 =$
$ A^2 - B - C + 2\sqrt (BC) =$
$ D + \sqrt E =$ where $D=A^2-B-C$ and $E=4BC$
now multiply by its conjugate $ D - \sqrt E$ yields:
$ D^2 - E$ which is an integer
I would like to know how to generalize this to n=5 and beyond, or if this is not possible, why that is the case. Thanks
I am not sure how to do this for the case n=8. Consider the following expression: $(\sqrt a_1 + \sqrt a_2 + \sqrt a_3 + \sqrt a_4) + (\sqrt a_5 + \sqrt a_6 + \sqrt a_7 + \sqrt a_8)$
I could multiply by its conjugate which would yield the following
$(\sqrt a_1 + \sqrt a_2 + \sqrt a_3 + \sqrt a_4)^2 - (\sqrt a_5 + \sqrt a_6 + \sqrt a_7 + \sqrt a_8)^2$
However once each squared term is distributed out there will be 6 new square root terms introduced rather than just the original 4. Below I will do the first term:
$(\sqrt a_1 + \sqrt a_2 + \sqrt a_3 + \sqrt a_4)^2 =$
$a_1 + a_2 + a_3 + a_4 + 2\sqrt (a_1a_2) + 2\sqrt (a_1a_3) + 2\sqrt (a_1a_4) + 2\sqrt (a_2a_3) + 2\sqrt (a_2a_4) + 2\sqrt (a_3a_4)$
as a result, I go from 8 roots in the denominator to 12 and a single non root term
If instead I broke up the expression into a group of 5 square roots and a group of 3 square roots I would end up with 10 roots from the group of 5 and 3 from the group of 3 totalling 13.
in either case the number of roots does not go down, which is the trick I have been exploiting. It seems to me a new trick of sorts is required to find the general solution.