Rationalize (i.e., get rid of radicals in the denominator) in $\dfrac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$, $\dfrac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}}$, and, in general $\dfrac{1}{\sum_{i=1}^n \sqrt{a_i}}$.

I have submitted my work so far as an answer.

• It might be easier to solve $$\dfrac{1}{x_1+\dots x_n} = \dfrac{F_n(x_1,\dots,x_n)}{G_n(x_1^2,x_2^2,\dots,x_n^2)}$$ Certainly easier to write out in LaTeX. :) – Thomas Andrews Jul 19 '13 at 17:52
• There is a smoother way. but a proof along your lines will work. Induction assumption: we can make $\frac{1}{\sqrt{a_1}+\cdots +\sqrt{a_n}+s}$ radical-free. Now replace $s$ by $\sqrt{a_{n+1}}+t$. The expression now has some even powers of $\sqrt{a_{n+1}}$, which are OK, and some terms in $\sqrt{a_{n+1}}$, which can be deradicalized by the base case. – André Nicolas Jul 19 '13 at 17:57
• A slightly different approach is to consider the terms of the form $(\sqrt{a_1}\pm\sqrt{a_2}\pm\ldots\pm\sqrt{a_n})$ and look at what happens if they are multiplied together. – Peter Košinár Jul 19 '13 at 18:08
• Maybe the answers they kindly gave me in (math.stackexchange.com/questions/421179/…) will help – chubakueno Jul 20 '13 at 0:18

A smooth way to do it, that involves some theory, is to look at $$\frac{1}{\sqrt{t_1}+ \sqrt{t_2}+\cdots+\sqrt{t_n}},$$ where the $t_i$ are indeterminates. Multiply top and bottom by $\pm\sqrt{t_1}\pm\sqrt{t_2}\pm \cdots\pm\sqrt{t_n}$, where the $\pm$ range independently, avoiding only the current denominator. So we use $2^n-1$ terms.

The denominator we get is invariant under the transformation that takes one of the $\sqrt{t_i}$ to $-\sqrt{t_i}$, so it is radical-free. We can cut the terms by a factor of $2$ by not using the $-\sqrt{t_1}$ terms, at the cost of making the proof a little longer.

An induction proof roughly along the lines of yours will also work. It is useful for the induction to prove the stronger result that $\frac{1}{\sqrt{t_1}+ \sqrt{t_2}+\cdots+\sqrt{t_n}+s}$ can be deradicalized, where $s$ is an indeterminate.

The induction assumption is that the denominator of $\frac{1}{\sqrt{t_1}+ \sqrt{t_2}+\cdots+\sqrt{t_k}+s}$ can be deradicalized. Replace $s$ by $\sqrt{t_{k+1}}+s$ in this deradicalization, and expand.

The denominator then has shape $P(s)+\sqrt{t_{k+1}}Q(s)$, where $P$ and $Q$ are radical-free polynomials. Multiply by $P(s)-\sqrt{t_{k+1}}Q(s)$.

• This reminds me of the proof that the algebraic numbers form a field. – marty cohen Jul 19 '13 at 20:34

I'll work out the first case, and, after that, show my initial work on doing the general case by induction.

\begin{align} \dfrac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}} &= \dfrac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}} \dfrac{\sqrt{a}+\sqrt{b}-\sqrt{c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}} \\ &=\dfrac{\sqrt{a}+\sqrt{b}-\sqrt{c}}{(\sqrt{a}+\sqrt{b})^2-c}\\ &=\dfrac{\sqrt{a}+\sqrt{b}-\sqrt{c}}{a+b-c+2\sqrt{ab}}\\ &=\dfrac{\sqrt{a}+\sqrt{b}-\sqrt{c}}{a+b-c+2\sqrt{ab}} \dfrac{a+b-c-2\sqrt{ab}}{a+b-c-2\sqrt{ab}}\\ &=\dfrac{(\sqrt{a}+\sqrt{b}-\sqrt{c})(a+b-c-2\sqrt{ab})}{(a+b-c)^2-4ab}\\ \end{align}

Here is an attempt to do the general case by induction.

If $\dfrac{1}{\sum_{i=1}^n \sqrt{a_i}} =\dfrac{F_n}{G_n}$, where $G_n$ is radical-free,

\begin{align} \dfrac{1}{\sum_{i=1}^{n+1} \sqrt{a_i}} &= \dfrac{1}{\sum_{i=1}^n \sqrt{a_i}+\sqrt{a_{n+1}}}\\ &= \dfrac{1}{\sum_{i=1}^n \sqrt{a_i}+\sqrt{a_{n+1}}} \dfrac{\sum_{i=1}^n \sqrt{a_i}-\sqrt{a_{n+1}}}{\sum_{i=1}^n \sqrt{a_i}-\sqrt{a_{n+1}}}\\ &= \dfrac{\sum_{i=1}^n \sqrt{a_i}-\sqrt{a_{n+1}}} {(\sum_{i=1}^n \sqrt{a_i})^2-a_{n+1}}\\ &= \dfrac{G_n/F_n-\sqrt{a_{n+1}}} {(G_n/F_n)^2-a_{n+1}}\\ &= \dfrac{F_nG_n-F_n^2\sqrt{a_{n+1}}} {G_n^2-F_n^2a_{n+1}}\\ \end{align}

I would like to be able to say at this point that $F_{n+1} = F_nG_n-F_n^2\sqrt{a_{n+1}}$ and $G_{n+1} = G_n^2-F_n^2a_{n+1}$. The problem is that the denominator is $not$ radical-free, since it contains $F_n^2$.

I'll leave it at that for now.

• Why not include the work in the question? – Thomas Andrews Jul 19 '13 at 17:21
• Unfortunately, this approach only works up to three terms. If you try to apply this reasoning to four, you'll see that the the number of radicals in denominator will actually increase (from four to six) due to the squaring. It only gets worse for larger numbers. – Peter Košinár Jul 19 '13 at 17:59
• As to why I didn't include the work in the question, that is due to my unfamiliarity with the protocol here. Since I am being asked this question, I assume that I am supposed to put my work in the question, not as a separate answer. I was guided by the fact that, when entering a question, I am given the opportunity to also enter an answer. – marty cohen Jul 19 '13 at 20:32

As you know, the identity $$(a+b)(a-b) \;\; = \;\; a^2 - b^2$$ shows that if you start with $a+b$ and multiply by $a-b,$ you'll get an expression that contains only even powers of $a$ and $b.$ Also, if you start with $a-b$ and multiply by $a+b,$ you'll get an expression that contains only even powers of $a$ and $b.$

One way to generalize this to a trinomial is to expand $$(a+b+c)(a+b-c)(a-b+c)(a-b-c)$$ By using a little strategy, this isn't as difficult as you might think: $$[(a+b)+c][(a+b)-c]\cdot[(a-b)+c][(a-b)-c] \;\; = \;\; \left[(a+b)^2 - c^2 \right] \cdot \left[(a-b)^2 - c^2 \right]$$ $$= \;\; (a+b)^{2}(a-b)^2 - (a+b)^{2}c^2 - c^{2}(a-b)^2 + c^4$$ $$= \;\; \left[(a+b)(a-b)\right]^2 \; + \; \left[-a^2c^2 - 2abc^2 - b^2c^2 - a^2c^2 + 2abc^2 - b^2c^2 \right] + c^4$$ $$= \;\; \left(a^2 - b^2\right)^2 \; - \; 2c^2\left(a^2 + b^2\right) \; + \; c^4$$ Notice that the result is an expression that contains only even powers of $a,$ $b,$ and $c.$ Also, notice that I didn't have to completely expand it to see that only even powers will remain.

This product of trinomials tells you that if, for example, you start with $a-b+c$ and multiply by $(a+b+c)(a+b-c)(a-b-c),$ you'll get an expression that contains only even powers of $a,$ $b,$ and $c.$ A shorthand way of saying this is that an appropriate "conjugate" of a $(-,+)$ pattern is the product of the remaining $3$ patterns. That is, an appropriate "conjugate" for a $(-,+)$ pattern is the product $(+,+)(+,-)(-,-).$

Without going into why this works (while I was writing this answer it appears that André Nicolas gave a brief, but nice, indication of how this can be done), if you start with $(a-b+c-d),$ which we might call a $(-,+,-)$ pattern, and multiply it by all of the $7$ corresponding patterns $(-,+,+),$ $(-,-,+),$ $(-,-,-),$ $(+,+,+),$ $(+,+,-),$ $(+,-,+),$ and $(+,-,-),$ you'll get an expression that contains only even powers of $a,$ $b,$ $c,$ and $d.$

The same method continues to work for additive combinations of $5$ terms, additive combinations of $6$ terms, etc. However, doing this by hand will get very tedious very soon since, for an additive combination of $5$ terms, you'll be multiplying it by $2^{4} - 1 = 15$ different expressions each consisting of $5$ terms. For an additive combination of $6$ terms, you'll be multiplying it by $2^{5} - 1 = 31$ different expressions each consisting of $6$ terms.

This method can be generalized to take care of sums in which various roots (square, cube, etc.) appear, but things quickly get even more complicated because you wind up multiplying by "$n$th roots of unity" combinations instead of sign combinations. (The signs $+$ and $-$ can be interpreted as multiplying by $1$ and by $-1,$ the two square roots of unity.) For example, radicals can be cleared from $\sqrt{a} + \sqrt[3]{b} + \sqrt{c},$ which has "sign pattern" $(+,+),$ by multiplying it by all $5$ of the "sign patterns" $(+,-),$ $(\omega,+),$ $(\omega,-),$ $({\omega}^2,+),$ and $({\omega}^2,-),$ where $\omega$ is the "first non-real cube root of $1$" as you travel counterclockwise along the unit circle in $\mathbb C$ beginning at $1$ on the positive real axis.