# How to rationalize denominator?

Suppose $c$ is not a complete square integer, ${a_0},{a_1} \in \mathbb{Q}$, we have $$\frac{1}{{{a_0} + {a_1}\sqrt c }} = \frac{{{a_0} - {a_1}\sqrt c }}{{a_0^2 - a_1^2c}}.$$

We need to show ${{a_0} + {a_1}\sqrt c } = 0$ iff ${a_0} = {a_1} = 0$. I know it's not hard.

Suppose $c$ is not a complete cube integer, ${a_0},{a_1},{a_2} \in \mathbb{Q}$, how can we deal with $$\frac{1}{{{a_0} + {a_1}\sqrt[3]{c} + {a_2}\sqrt[3]{{{c^2}}}}}$$ similarly?

Any help will be appreciated.

• If we show $a_0,a_1=0$ wont it be like that number is $\infty$ – Archis Welankar Nov 19 '15 at 6:11
• Let $\lambda$ and $\mu$ be the two other cube roots (complex) of $c$. To rationalize, multiply top and bottom by $(a_0+a_1\lambda+a_2\lambda^2)(a_0+a_1\mu+a_2\mu^2)$. There is a lot of simplification, but it is still somewhat messy. – André Nicolas Nov 19 '15 at 6:24
• Hint for the first part: if $a_0+a_1\sqrt c=0$, then $-a_0/a_1 = \sqrt c$. – Greg Martin Nov 19 '15 at 6:42
• I think you mean $\frac1{a_0+a_1\sqrt[3]{c}+a_2\sqrt[3]{c^2}}$. – marty cohen Nov 19 '15 at 7:25

Rationalizing a quadratic surd $a+b\sqrt{d}$ entails multiplying it by its conjugate. In this case, the word "conjugate" is as simple as replacing a binomial sum $A+B$ with a difference $A-B$ and vice versa. However, the word "conjugate" has a much more advanced generalization in the contexts of field theory and Galois theory.

Suppose that $K$ is a field (like $\mathbb{Q},\mathbb{R},\mathbb{C}$), and $f(x)$ is a polynomial in the variable $x$ with coefficients from $K$. If $f(x)$ is irreducible (cannot be factored using other polynomial also with coefficients from $K$) and $\alpha$ is a root of $f(x)$ (so $\alpha$ may not be in $K$, it may be in a bigger field $L$ containing $K$), then the nontrivial conjugates of $\alpha$ are all the other roots of $f(x)$.

If $L$ is a sufficiently nice "extension" of $K$, there are a set of "symmetries" of $L$ (functions $L\to L$) which do nothing to the elements of $K$, preserve the arithmetic operations of addition and multiplication on $L$, and in particular it follows that they permute the roots of $f(x)$. Indeed, they will permute them "transitively," meaning given any two roots $\alpha,\alpha'$ there is a "symmetry" $\sigma:L\to L$ for which $\sigma(\alpha)=\alpha'$.

Let's see this concretely. Say $K=\mathbb{Q}$ and $f(x)=x^2+1$. One root is $\alpha=i$, the other is $\alpha'=-i$, and there is a "symmetry" $\sigma:\mathbb{C}\to\mathbb{C}$ which does nothing to real numbers and $\sigma(i)=-i$. In fact, this is just complex conjugation $\sigma(a+bi)=a-bi$. It "preserves the operations" of addition and subtraction in the sense that $\sigma(z+w)=\sigma(z)+\sigma(w)$ and $\sigma(zw)=\sigma(z)\sigma(w)$ for all $z,w\in\mathbb{C}$.

One thing to notice is that $\sigma(r)=r$ is not only true for $r\in\mathbb{R}$, it is only true for real numbers $r$.

Another example. Let $K=\mathbb{Q}$ and $f(x)=x^2-2$. The roots are $\alpha=\sqrt{2}$ and $\alpha'=-\sqrt{2}$. These exist in the bigger field $L=\mathbb{Q}(\sqrt{2})$. (By definition this means "the smallest field containing $\sqrt{2}$ and $\mathbb{Q}$ within $\mathbb{C}$," but this happens to be the same as $\{a+b\sqrt{2}:a,b\in\mathbb{Q}\}$, which you can prove as an exercise). Once again there is a nontrivial symmetry $\sigma:\mathbb{Q}(\sqrt{2})\to\mathbb{Q}(\sqrt{2})$ which does nothing to rationals, preserves operations, and $\sigma(\sqrt{2})=-\sqrt{2}$. The function is given by the "conjugate," as in $\sigma(a+b\sqrt{2})=a-b\sqrt{2}$; you can check by hand this satisfies the properties $\sigma(x+y)=\sigma(x)+\sigma(y)$ and $\sigma(xy)=\sigma(x)\sigma(y)$ for all $x=a+b\sqrt{d}$, $y=p+q\sqrt{d}$.

Now let $K=\mathbb{Q}$ and $f(x)$ be any irreducible rational-coefficient polynomial, $\alpha$ a root, and $L$ the smallest subfield of $\mathbb{C}$ containing all of the roots of $\mathbb{Q}$. (Note this may be bigger than $\mathbb{Q}(\alpha)$; there may be some other roots $\alpha'$ of $f(x)$ that cannot be expressed in terms of $\alpha$ using rational functions as every element in $\mathbb{Q}(\alpha)$ is.) There may be more than one symmetry $\sigma$, there could be multiple. In any case, if we apply all of the symmetries $\sigma_1,\sigma_2,\cdots$ to $\alpha$ we get all of the roots $\alpha_1,\alpha_2,\cdots$ of the polynomial $f(x)$. Then if we multiply them all together, by Vieta's formulas we find the product of all the zeros $\alpha_1\cdots\alpha_n$ is $\pm$ the constant coefficient of $f(x)$, in particular the product is rational.

That is: the product of an algebraic number and all of its conjugates will be rational.

Let's apply this with $f(x)=x^3-d$, where $d$ is not a perfect cube. Without loss of generality, $d$ is just an integer. The roots are $\sqrt[3]{d},~\omega\sqrt[3]{d}$, and $\omega^2\sqrt[3]{d}$, where $\sqrt[3]{d}$ is the real cube root of $d$ and $1,\omega,\omega^2$ are the three cube roots of unity. In particular, $\omega=\exp(2\pi i/3)$.

The two nontrivial symmetries $\sigma_1$ and $\sigma_2$ satisfy $\sigma_1(\sqrt[3]{d})=\omega\sqrt[3]{d}$ and $\sigma_2(\sqrt[3]{d})=\omega^2\sqrt[3]{d}$, so

$$\begin{array}{ll} \sigma_1(a+b\sqrt[3]{d}+c\sqrt[3]{d^2}) & =a+b\sigma_1(\sqrt[3]{d})+\sigma_1(\sqrt[3]{d})^2 \\ & = a+b\omega\sqrt[3]{d}+b\omega^2\sqrt[3]{d^2} \end{array}$$

$$\begin{array}{ll} \sigma_2(a+b\sqrt[3]{d}+c\sqrt[3]{d^2}) & =a+b\sigma_2(\sqrt[3]{d})+c\sigma_2(\sqrt[3]{d})^2 \\ & = a+b\omega^2\sqrt[3]{d}+\omega\sqrt[3]{d^2} \end{array}$$

because $\sigma_{1,2}$ preserve addition and multiplication and $(\omega^2)^2=\omega$ at the end. Of course the trivial symmetry $\sigma_0$ doesn't do anything, $\sigma_0(a+b\sqrt[3]{d}+c\sqrt[3]{d^2})=a+b\sqrt[3]{d}+c\sqrt[3]{d^2}$.

Thus to rationalize $x=a+b\sqrt[3]{d}+c\sqrt[3]{d^2}$, one multiplies it by its conjugates $\sigma_1(x)$ and $\sigma_2(x)$.

While $\sigma_1(x)$ and $\sigma_2(x)$ may not be real numbers, the product $\sigma_0(x)\sigma_1(x)\sigma_2(x)$ will be rational, and further since $\sigma_0(x)$ is rational, the product $\sigma_1(x)\sigma_2(x)$ will be rational as well, which is what we would multiply the numerator by if $x$ was in the denominator.

(I make no claim this is the easiest or most efficient way of rationalizing the denominator - I'm sure it isn't. I just wanted to showcase some Galois theory.)

You want to invert $a_0+a_1r+a_2r^2$, where $r^3=c$. This means finding $b_0$, $b_1$ and $b_2$ such that $$(a_0+a_1r+a_2r^2)(b_0+b_1r+b_2r^2)=1$$ and this translates into a linear system, because $a_0+a_1r+a_2r^2=0$ if and only if $a_0=a_1=a_2=0$. This follows from the fact that $x^3-c$ is irreducible (it has no rational root by assumption), so no polynomial of degree less than $3$ can have $r$ as root.

The system can be written down and the solution even explicitly expressed via Cramer's rule: just some patience is needed.

I found a way. Let ${p^3} = c$, we write

$\left\{ \begin{gathered} {a_0} + {a_1}p + {a_2}{p^2} = S \hfill \\ {a_2}c + {a_0}p + {a_1}{p^2} = pS \hfill \\ {a_1}c + {a_2}cp + {a_0}{p^2} = {p^2}S \hfill \\ \end{gathered} \right.$

So

$\left| {\begin{array}{*{20}{c}} {{a_0} - S}&{{a_1}}&{{a_2}} \\ {{a_2}c - pS}&{{a_0}}&{{a_1}} \\ {{a_1}c - {p^2}S}&{{a_2}c}&{{a_0}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {{a_0}}&{{a_1}}&{{a_2}} \\ {{a_2}c}&{{a_0}}&{{a_1}} \\ {{a_1}c}&{{a_2}c}&{{a_0}} \end{array}} \right| - S\left| {\begin{array}{*{20}{c}} 1&{{a_1}}&{{a_2}} \\ p&{{a_0}}&{{a_1}} \\ {{p^2}}&{{a_2}c}&{{a_0}} \end{array}} \right| = 0$

And

$\frac{1}{S} = \left| {\begin{array}{*{20}{c}} 1&{{a_1}}&{{a_2}} \\ p&{{a_0}}&{{a_1}} \\ {{p^2}}&{{a_2}c}&{{a_0}} \end{array}} \right|{\left| {\begin{array}{*{20}{c}} {{a_0}}&{{a_1}}&{{a_2}} \\ {{a_2}c}&{{a_0}}&{{a_1}} \\ {{a_1}c}&{{a_2}c}&{{a_0}} \end{array}} \right|^{ - 1}}$

But I cannot see why $\left| {\begin{array}{*{20}{c}} {{a_0}}&{{a_1}}&{{a_2}} \\ {{a_2}c}&{{a_0}}&{{a_1}} \\ {{a_1}c}&{{a_2}c}&{{a_0}} \end{array}} \right| \ne 0$.