# Rationalizing the denominator having square roots and cube roots

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: $$\frac{1}{2+\sqrt 2+\sqrt{2}}$$

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?

• Rationalize away the square root first, then think at how to rationalize denominators like $\,a+ b \sqrt{n}+c \sqrt{n^2}\,$. – dxiv Aug 1 '17 at 22:56
• 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 1 '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? – Frpzzd Aug 1 '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 1 '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 1 '17 at 23:32

## 3 Answers

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 $5 + \sqrt7$ multiply by $5^2 - 5*\sqrt7 + (\sqrt7)^2$ to get $(5 + \sqrt7)(5^2 - 5*\sqrt7 + (\sqrt7)^2) = 5^3 + 5^2\sqrt7 -5^2\sqrt7 - 5*(\sqrt7)^2 + 5*(\sqrt7)^2 + (\sqrt7)^3 = 125 + 7$.

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

First let's get rid of the $\sqrt2$ term. So we multiply top and bottom by $(2+\sqrt 2)^2 - (2 +\sqrt 2)*\sqrt2 + (\sqrt2)^2$ to get $(2 + \sqrt 2 + \sqrt 2)*[(2+\sqrt 2)^2 - (2 +\sqrt 2)*\sqrt2 + (\sqrt2)^2] = (2 + \sqrt 2)^3 + 2= 8 + 12 \sqrt 2 + 12\sqrt 2 + 2\sqrt 2 + 2 = 10 + 26\sqrt 2$. 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 2} \\&= \frac {(2 + \sqrt 2)^2 - (2+\sqrt2)\sqrt2 + \sqrt2^2}{(2+\sqrt 2)^3 + 2}\\&= \frac {(4 + 4\sqrt 2 + 2) -2\sqrt 2 - \sqrt 2\sqrt2 + \sqrt2^2}{10 + 26\sqrt 2}\\&= \frac {[(4 + 4\sqrt 2 + 2) -2\sqrt 2 - \sqrt 2\sqrt2 + \sqrt2^2](10 - 26\sqrt{2})}{100 - 2*26^2} \end{align}

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

• Beautiful answer! – Frpzzd Aug 1 '17 at 23:27
• It's always possible For certain values of "always" ;-) This won't work out of the box for $1 + 2 \sqrt{2}+3 \sqrt{4}$ for example. – dxiv Aug 1 '17 at 23:28
• Beautiful? Ugly as sin, if you ask me.... but doable. – fleablood Aug 1 '17 at 23:29
• @dxiv $\dfrac{1}{89} \left(\sqrt{4}+16 \sqrt{2}-11\right)\quad$ the method of the minimal polynomial is fantastic... provided you have Mathematica to generate it :) – Raffaele Aug 2 '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 $(1+2 \sqrt2+ 3\sqrt{4})(1+a \sqrt2+ b \sqrt{4})$ and solve for $a,b$ such that the coefficients of $\sqrt{2}, \sqrt{4}$ are both $0$. – dxiv Aug 2 '17 at 20:38

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

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

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

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 $$\sqrt 2 + \sqrt2 + 2$$, 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.

• This is precisely what I described in a comment to the question. – Bill Dubuque Aug 1 '17 at 23:34
• My remark was merely to help readers understand how the replies are related. Nothing was intended about "the answer standing" (whatever that means). – Bill Dubuque Aug 1 '17 at 23:38
• I apologize. What I meant was that I did not want to delete my answer because it coincided with a comment. But I took your comment above in the wrong taste. Forgive me. – астон вілла олоф мэллбэрг Aug 1 '17 at 23:45
• Lovely method, provided you have Mathematica to generate the minimal polynomial :) – Raffaele Aug 2 '17 at 11:45
• Ah, but of course that is true. – астон вілла олоф мэллбэрг Aug 2 '17 at 11:46