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Why do we need to rationalize the fraction, can't the calculator still give an answer? And after that process, isn't the number still involved with an irrational number?

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You are right insofar as numerical computation is concerned. Indeed, $\frac{1}{\sqrt{3}}$ is easier to evaluate with a calculator than $\frac{\sqrt{3}}{3}$ (fewer key presses, particularly if you have a "$1/x$" button. In the old days it was different, since division by an ugly number was harder than multiplication by an ugly number. But the idea is still very useful, though often, for example at the beginnings of calculus, it is the numerator that needs to look nice. –  André Nicolas Mar 16 '12 at 0:00
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@Victor, it's better not to accept answers too quickly since this will discourage other members from replying. –  Bill Dubuque Mar 16 '12 at 0:39

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Rationalizing the denominator (RTD) is useful because it often simplifies problems, e.g. converting division by an algebraic number or function to division by a simpler "rational" one. For example if you search prior posts using that term you'll find examples where is is used when computing limits in order to eliminate "apparent singularities" in the denominator of algebraic functions (e.g. functions involving radical such as square-roots and cube-roots). Also it proves useful when testing divisibility of algebraic integers, since it converts testing the divisibility of two algebraic integers to the simpler problem of testing whether a rational integer divides an algebraic number.

In this prior question is an example where RTD transforms a limit of indeterminate form into a simple determinate limit by way of cancelling apparent singularity at $\rm\ x = a\ $

$$\rm \frac{x^2-a\sqrt{ax}}{\sqrt{ax}-a}\ =\ \frac{x^2-a\sqrt{ax}}{\sqrt{ax}-a} \ \frac{\sqrt{ax}+a}{\sqrt{ax}+a}\ =\ \frac{ax\:(x-a)+\sqrt{ax}\ (x^2-a^2) }{a\:(x-a) }\ =\ x+(x+a)\sqrt{\frac{x}{a}}$$

Here's another example from number theory showing how RTD serves to reduce divisibility of algebraic integers to rational integers. Consider the Gaussian integers $\rm\ \mathbb I = \{ m + n\ i\ : \ m,n\in \mathbb Z \}\ $. As in any ring we define divisibilty by $\rm\ a\ |\ b\ in\ \mathbb I \iff b/a \in \mathbb I\:.\ $ Suppose we wish to know if $\rm\ 2+3\ i\ |\ 91\ in\ \mathbb I\:,\:$ i.e. is $\rm\ w = 91/(2+3\ i)\in \mathbb I\ ?\ $ Now in fact $\rm\:\mathbb I\:$ happens to have a division algorithm which we could apply. But it is simpler to RTD yielding $\rm\ w = 91\ (2-3\ i)/(2^2+3^2) = 7\ (2-3\ i)\ $ so, indeed, $\rm\: w\in \mathbb I\:.\ $ More generally we can often reduce problems about algebraic numbers to problems about rational numbers by taking norms, traces, etc. In fact this is how Kronecker constructed his divisor theory for algebraic integers, see e.g. Harold Edwards: Divisor Theory.

Here's an example of realizing the denominator to prove that the real part is zero:

$$\displaystyle\ z\bar z\: =\: 1\ \:\Rightarrow\:\ \frac{(1-z)\:(1+\bar z)}{(1+z)\:(1+\bar z)}\: =\: \frac{\bar z-z}{|1+z|^2}\: =\: \frac{r\:i}{s},\ \ r,s\in \mathbb R$$

Generally, rationalizing denominators allows one to lift "existence of inverses of elements $\ne 0\:$" from a base field (e.g. $\mathbb R$) to an algebraic extension field (e.g. $\mathbb C$). Namely, since $\mathbb R$ is a field, $\rm\ 0\ne r\in \mathbb R\ \Rightarrow\ r^{-1}\in \mathbb R\:,\:$ so

$$\rm 0\ne\alpha\in\mathbb C\ \ \Rightarrow\ \ 0\ne\alpha\alpha' = r\in \mathbb R\ \ \Rightarrow\ \ \frac{1}\alpha\ =\ \frac{\alpha'}{\alpha\:\alpha'}\ = \frac{\alpha'}r\in\mathbb C $$

Thus $\:$ field $\mathbb R\ \Rightarrow\:\: $ field $\mathbb C\ $ by using the norm $\rm\:\alpha\to\alpha\ \alpha'\:$ to lift existence of inverses from $\mathbb R$ to $\mathbb C\:.$ See this post for more.

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