Just something that I've been thinking about lately, and can't figure how to generalize efficiently:
Say we have a surd of the form $\frac{1}{\sqrt2+\sqrt3+\sqrt5+\sqrt7}$, or something along those lines. How can I find the rationalizing factor, without brute force? Does there exist one?
We could add more numbers of the form $\sqrt p$ in the denominator, where $p$ is prime, and even multiply by certain coefficients for the very general case –
$$\frac{1}{(a_1\sqrt2+a_2\sqrt3+a_3\sqrt5+a_4\sqrt7+…a_i\sqrt p +…)}$$, where $a_i \in Z$.
Any ideas? This seems like an interesting problem!
Best Answer
You have for example
$$(a+b\sqrt 2+c\sqrt 3 +d\sqrt 6 )(a-b\sqrt 2 +c\sqrt 3 -d\sqrt 6)(a+b\sqrt 2-c\sqrt 3 -d\sqrt 6 )(a-b\sqrt 2-c\sqrt 3+d\sqrt 6)=a^4 - 4 a^2 b^2 - 6 a^2 c^2 - 12 a^2 d^2 + 48 a b c d + 4 b^4 - 12 b^2 c^2 - 24 b^2 d^2 + 9 c^4 - 36 c^2 d^2 + 36 d^4$$
Note that the second term has the signs on the terms with $\sqrt 2$ flipped, the third on the ones with $\sqrt 3$ and the fourth in both simultaneously. So you can easily rationalize this:
$$\frac 1{a+b\sqrt 2+c\sqrt 3 +d\sqrt 6}=\frac{(a-b\sqrt 2 +c\sqrt 3 -d\sqrt 6)(a+b\sqrt 2-c\sqrt 3 -d\sqrt 6 )(a-b\sqrt 2-c\sqrt 3+d\sqrt 6)}{a^4 - 4 a^2 b^2 - 6 a^2 c^2 - 12 a^2 d^2 + 48 a b c d + 4 b^4 - 12 b^2 c^2 - 24 b^2 d^2 + 9 c^4 - 36 c^2 d^2 + 36 d^4}$$
You can always do this, no matter how many square roots you have. But in general if you have the square roots of n primes you have to multiply $n^2$ terms, which is a lot of effort. So it's better to choose one of the primes to remove from the expression and repeat, for example:
$$ \begin{split}\frac 1{3\sqrt 2-\sqrt 3+2\sqrt 5+\sqrt 7}&=\frac{3\sqrt 2-\sqrt 3+2\sqrt 5-\sqrt 7}{34 - 6\sqrt 6 + 12 \sqrt{10} - 4 \sqrt{15}}\\ &=\frac{(3\sqrt 2-\sqrt 3+2\sqrt 5-\sqrt 7)(34-6\sqrt 6 -12\sqrt{10}+4\sqrt{15})}{72\sqrt 6 -308}\end{split}$$
Note that in the second line I flipped the signs on the $\sqrt{10}$ and $\sqrt{15}$ terms, which are the multiples of 5. Lots of questions like this can be solved with abstract algebra.