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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!

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    $\begingroup$ For an algebraic integer $\alpha$ a basic problem is finding its minimal polynomial $p(x)$ so that $p(\alpha) = 0$. Now $p_0\neq 0$, so you will get $$\frac{1}{\alpha} = \frac{-(p_d\alpha^{d-1} + \cdots + p_1)}{p_0}$$ If $P(x),Q(x)$ are minimal polynomials of $\alpha,\beta$, then $\alpha+\beta$ is a root of $res_x(P(x),Q(z-x))$ where $res$ is the resultant function. So minimal poly is a factor of it, usually the whole thing. Something similar exists for products $\alpha\beta$, so you can slowly build up to the final polynomial. $\endgroup$
    – Yong Hao Ng
    Commented Jan 18, 2020 at 7:40

2 Answers 2

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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.

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Grouping will be helpful.

$$\frac{1}{\sqrt2+\sqrt3+\sqrt5+\sqrt7}=\frac{1}{(\sqrt2+\sqrt3)+(\sqrt5+\sqrt7)}$$

For the first rationalizing factor consider $$(\sqrt2+\sqrt3)-(\sqrt5+\sqrt7)$$

The resulting denominator will be $$5+2\sqrt 6 -(12+2\sqrt {35}) = -7+2(\sqrt 6 - \sqrt {35})$$

The next factor will be $$-7-2(\sqrt 6 - \sqrt {35})$$ and the denominator changes to $$49-4(41-2\sqrt {210}) = -115+8\sqrt {210}$$

You can take over from here.

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  • $\begingroup$ That's all good, but what about the general case? $\endgroup$
    – stoic-santiago
    Commented Jan 18, 2020 at 7:57

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