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
@ancientmathemetician:
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.
Thanks
Best Answer
$$(\sqrt a+\sqrt b+\sqrt c)(s+t\sqrt a+u\sqrt b+v\sqrt c+w\sqrt{ab}+x\sqrt{ac}+y\sqrt{bc}+z\sqrt{abc})=1$$ The rational coefficients of the various square-roots give you eight equations in eight unknowns $s$ to $z$.
It looks like it works for any number of square-roots. A $2^N\times2^N$ matrix has to be inverted. I don't know how to tell if it's going to be singular.
The matrix is very sparse - mostly zeros - so this isn't the most efficient method. I think it shows there is an answer.