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In middle-school mathematics, the teachers always tell you that if you have radicals on the denominator of a fraction, then it isn't fit to be a final answer - you have to rationalize the denominator, or get rid of all of the radicals in the denominator by moving them to numerator.

Rationalizing the denominator is usually very easy, and can be done quickly using its conjugate. For example, consider $$\frac{1}{2+\sqrt 2}$$ This denominator can be easily rationalized by using its conjugate: $$\frac{2-\sqrt 2}{(2+\sqrt 2)(2-\sqrt 2)}$$ $$\frac{2-\sqrt 2}{4-2}$$ $$\frac{2-\sqrt 2}{2}$$

However, I have stumbled upon a new class of denominator-rationalization problems that I can't figure out how to solve. I was thoroughly stumped when I tried to rationalize the denominator of this fraction: $$\frac{1}{2+\sqrt 2+\sqrt[3]{2}}$$

Can anybody figure out how to rationalize this? Is it even possible?

Or, more interestingly, if anyone suspects that it is not possible, how might one prove something like this?

David G. Stork
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Franklin Pezzuti Dyer
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  • Rationalize away the square root first, then think at how to rationalize denominators like $,a+ b \sqrt[3]{n}+c \sqrt[3]{n^2},$. – dxiv Aug 01 '17 at 22:56
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    Hint: $ $ if $,\alpha,$ is a root of $,c - x f(x),$ then $,\alpha f(\alpha) = c\ $ So you need only find some polynomial having the denominator $,\alpha,$ as root. Wlog we can choose $,c\neq 0,$ over a field (or domain). – Bill Dubuque Aug 01 '17 at 23:02
  • @BillDubuque Have you already found that there is a solution, or are you just proposing that as how you would begin to attack the problem? – Franklin Pezzuti Dyer Aug 01 '17 at 23:11
  • @Nilknarf The method I described works for any denominator that is algebraic, i.e. a root of some polynomial with rational coefficients. Rationals, and roots of rationals are algebraic, and algebraics are closed under sums and products. So they include numbers of the type you exhibited. – Bill Dubuque Aug 01 '17 at 23:15
  • @Nilknarf In case it wasn't clear, here is how to apply what I wrote to rationalize the denominator: $$ 0\neq \alpha f(\alpha) = c\in\Bbb Q\ \Rightarrow\ \dfrac{\beta}{\alpha} = \dfrac{\beta, f(\alpha)}{\alpha, f(\alpha)} = \dfrac{\beta, f(\alpha)}{c}$$ – Bill Dubuque Aug 01 '17 at 23:32
  • From a somewhat more advanced point of view, the reason it's possible is that $\mathbb{Q}[2^{1/6}] = { a + b \cdot 2^{1/6} + c \cdot 2^{1/3} + d \cdot 2^{1/2} + e \cdot 2^{2/3} + f \cdot 2^{5/6} : a, \ldots, f \in \mathbb{Q} }$ is a field and $2 + \sqrt{2} + \sqrt[3]{2}$ is in this field, so its inverse can also be expressed in this way. You probably won't have much luck trying to rationalize $\frac{1}{\pi - 1}$, though. – Daniel Schepler Aug 01 '17 at 23:38
  • @Daniel Indeed $,\alpha f(\alpha) = c,\Rightarrow, \alpha^{-1} = f(\alpha)/c,$ in the notation in my above comment. – Bill Dubuque Aug 01 '17 at 23:42
  • By writing out $(a + \cdots + f \cdot 2^{5/6}) (2 + 2^{1/3} + 2^{1/2})$, setting it equal to 1, and solving the resulting set of 6 linear equations in 6 unknowns (with the help of Maxima!) I got the inverse is $\frac{1}{46} (10 - 14 \cdot 2^{1/6} - 8 \cdot 2^{1/3} + 2 \cdot 2^{1/2} + 11 \cdot 2^{2/3} + 3 \cdot 2^{5/6})$. – Daniel Schepler Aug 02 '17 at 00:07
  • Multiply by all Galois conjugates – Akiva Weinberger Aug 02 '17 at 02:28

4 Answers4

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Hint:  for a shortcut in this particular case, let $a = 2+\sqrt{2}$ then use that:

$$ \frac{1}{a+\sqrt[3]{2}} = \frac{a^2-a\sqrt[3]{2}+\sqrt[3]{4}}{a^3+2} $$

The denominator now contains only integers and terms in $\sqrt{2}\,$ after expansion, which is the case you know how to rationalize.

dxiv
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It's always possible.

You want to multiply top and bottom by $M$ to get that $denominator*M$ has not radical.

As you have figured out: If the denominator is $a + b\sqrt{c}$ you mulitply by the conjugate to get $(a + b\sqrt{c})(a - b\sqrt{c}) = a^2 - b^2*c$.

This will also work with $(\sqrt a + \sqrt b)(\sqrt a - \sqrt b) = a - b$.

So it's the same idea for $a + \sqrt[k] b$. The trick is to realize that $(a + \sqrt[k]b)(a^{k-1} - a^{k-2}\sqrt[k]b + a^{k-3}(\sqrt[k]b)^2-..... \pm a(\sqrt[k]b)^{k-2} \mp (\sqrt[k]b)^{k-1} = a^k \pm b$.

Example: To deradicalize $5 + \sqrt[3]7$ multiply by $5^2 - 5*\sqrt[3]7 + (\sqrt[3]7)^2$ to get $(5 + \sqrt[3]7)(5^2 - 5*\sqrt[3]7 + (\sqrt[3]7)^2) = 5^3 + 5^2\sqrt[3]7 -5^2\sqrt[3]7 - 5*(\sqrt[3]7)^2 + 5*(\sqrt[3]7)^2 + (\sqrt[3]7)^3 = 125 + 7$.

So to deradicalize $(2 + \sqrt 2 + \sqrt[3] 2)$ just deradicalize it term by term.

First let's get rid of the $\sqrt[3]2$ term. So we multiply top and bottom by $(2+\sqrt 2)^2 - (2 +\sqrt 2)*\sqrt[3]2 + (\sqrt[3]2)^2$ to get $(2 + \sqrt 2 + \sqrt[3] 2)*[(2+\sqrt 2)^2 - (2 +\sqrt 2)*\sqrt[3]2 + (\sqrt[3]2)^2] = (2 + \sqrt 2)^3 + 2= 8 + 12 \sqrt 2 + 12\sqrt 2 + 2\sqrt 2 + 2 = 10 + 26\sqrt 2$. Then we multiply that by $10 - 26 \sqrt 2$ to get $(10 + 26\sqrt 2)(10 - 26\sqrt 2) = 100 - 2*26^2$.

So example:

\begin{align} &\frac 1 {2 + \sqrt 2 + \sqrt[3] 2} \\&= \frac {(2 + \sqrt 2)^2 - (2+\sqrt2)\sqrt[3]2 + \sqrt[3]2^2}{(2+\sqrt 2)^3 + 2}\\&= \frac {(4 + 4\sqrt 2 + 2) -2\sqrt[3] 2 - \sqrt 2\sqrt[3]2 + \sqrt[3]2^2}{10 + 26\sqrt 2}\\&= \frac {[(4 + 4\sqrt 2 + 2) -2\sqrt[3] 2 - \sqrt 2\sqrt[3]2 + \sqrt[3]2^2](10 - 26\sqrt{2})}{100 - 2*26^2} \end{align}

Okay... admittedly that is a bear... but it is doable.

Siong Thye Goh
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fleablood
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  • Beautiful answer! – Franklin Pezzuti Dyer Aug 01 '17 at 23:27
  • It's always possible For certain values of "always" ;-) This won't work out of the box for $1 + 2 \sqrt[3]{2}+3 \sqrt[3]{4}$ for example. – dxiv Aug 01 '17 at 23:28
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    Beautiful? Ugly as sin, if you ask me.... but doable. – fleablood Aug 01 '17 at 23:29
  • @dxiv $\dfrac{1}{89} \left(\sqrt[3]{4}+16 \sqrt[3]{2}-11\right)\quad$ the method of the minimal polynomial is fantastic... provided you have Mathematica to generate it :) – Raffaele Aug 02 '17 at 11:41
  • @Raffaele Right about the CAS, in general. However, the simple example from my previous comment can be worked out by hand fairly easily. Just expand $(1+2 \sqrt[3]2+ 3\sqrt[3]{4})(1+a \sqrt[3]2+ b \sqrt[3]{4})$ and solve for $a,b$ such that the coefficients of $\sqrt[3]{2}, \sqrt[3]{4}$ are both $0$. – dxiv Aug 02 '17 at 20:38
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There is a very general procedure for this kind of a question.

You see, in essence, rationalization of a fraction is converting the denominator into a rational, right? But indeed, it's a deeper process : Suppose a fraction of the form $\frac 1b$ can be rationalized, say written in the form $\frac cd$, where $d$ is rational.

Cross multiplying, we get $bc = d$, or that $b$ times something is a rational. This equates to the invertibility of the number $b$ in the field of real numbers, which is true all the time. So every fraction of this kind is indeed rationalizable. But the question then comes down to how to find this inverse.

One way of finding the inverse is finding the minimal polynomial with rational coefficients, which $b$ satisfies. I'll explain why.

Suppose $b$ satisfies the polynomial $\sum_{i=0}^n a_ix^i = 0$. Then, $\sum_{i=0}^n a_ib^i = 0$, so that $\sum_{i=1}^n a_ib^i = -a_0$, from where it follows that $b \left(\sum_{i=1}^n a_i b^{i-1}\right) = a_0$.

Rewriting, $$ \frac{1}{b} = \frac{\left(\sum_{i=1}^n a_i b^{i-1}\right)}{a_0} $$

which is the rationalized form.

So all you need to do, is to find a polynomial which the given surd, in our case $\sqrt 2 + \sqrt[3]2 + 2$, satisfies.

The minimal such polynomial is $x^6 - 12 x^5 + 54 x^4 - 116 x^3 + 132 x^2 - 120 x + 92$, which I found online. There's a better answer above on how to actually find a polynomial, so I will avoid that part, but at least this shows that fractions with "algebraic" denominators can be rationalized using the polynomial they satisfy.

Sarvesh Ravichandran Iyer
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Fleablood gave a method for your particular example but you seem to suggest that you're looking for a general solution. In this case, a formula may be given by $$\frac{1}{\sqrt[6]{a}+\sqrt[6]{b}+\sqrt[6]{c}}=\frac{\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}-\sqrt[6]{ab}-\sqrt[6]{ac}-\sqrt[6]{bc}\right)\left(\left(3\sqrt[6]{abc}+\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^{2}-3\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\sqrt[6]{abc}\right)\left(\left(a+3b+3c\right)\sqrt{a}+\left(3a+b+3c\right)\sqrt{b}-\left(3a+3b+c\right)\sqrt{c}+21\sqrt{abc}\right)\left(2\left(a+3b+3c\right)\left(3a+b+3c\right)\sqrt{ab}+42\left(3a+3b+c\right)\sqrt{abc^{2}}-a\left(a+3b+3c\right)^{2}-b\left(3a+b+3c\right)^{2}+c\left(3a+3b+c\right)^{2}+441abc\right)}{4ab\left(21c\left(3a+3b+c\right)+\left(a+3b+3c\right)\left(3a+b+3c\right)\right)^{2}-\left(a\left(a+3b+3c\right)^{2}+b\left(3a+b+3c\right)^{2}-c\left(3a+3b+c\right)^{2}-441abc\right)^{2}}$$ Which can be used to rationalize any $3$ term denominator with a mix of square and cube roots. It's a beast, but it works.

Baby Hearty Bear
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