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Why do all school algebra texts define simplest form for expressions with radicals to not allow a radical in the denominator. For the classic example, $1/\sqrt{3}$ needs to be "simplified" to $\sqrt{3}/3$.

Is there a mathematical or other reason?

And does the same apply to exponential notation -- are students expected to "simplify" $3^{-1/2}$ to $3^{1/2}/3$ ?

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    $\begingroup$ Where is this? Pretty silly to enforce such rules, IMO. $\endgroup$
    – Aryabhata
    Mar 9, 2011 at 23:44
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    $\begingroup$ My schooling up until university required the same; and I never really understood it. The best explanation I could think of is that it is intuitively simpler to divide by an integer than an irrational; and they wanted us to think of things that way. But they had no problem dividing by π..... $\endgroup$
    – Steve
    Mar 9, 2011 at 23:53
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    $\begingroup$ My Theory: There is no longer any good reason to do so. $\endgroup$
    – Eric Naslund
    Mar 10, 2011 at 0:06
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    $\begingroup$ This is a duplicate, but I can't figure out how to find the original. $\endgroup$
    – Qiaochu Yuan
    Mar 10, 2011 at 0:07
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    $\begingroup$ I've added a new answer, with a little bit on why canonical forms generally are used. When you learned how to reduce fractions to lowest terms, you were learning a canonical form. This is another. $\endgroup$
    – Michael Hardy
    Jun 13, 2012 at 21:41

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The usual reason I've heard is that dividing by integers is computationally easier -- it's easier to find, say, $(5\sqrt{3})/3$ by computing $5 \times \sqrt{3} \approx 8.66 $ and then dividing by $3$ to get $2.89$ then to find $5/\sqrt{3}$ by dividing $5/1.73$ directly.

The slightly cynical teacher in me wants to say that the reason for demanding no radicals in the denominator is so there is only one right answer to each question, which simplifies grading.

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    $\begingroup$ Simplifying grading is an admirable goal! Especially for the sake of the teaching assistants :-) $\endgroup$
    – Aryabhata
    Mar 10, 2011 at 0:22
  • $\begingroup$ The "one right answer" theory actually has a mathematical basis. Another way of putting it is that disallowing radials in the denominator brings us pretty close to a canonical form. That can probably be worked out formally in terms of abstract algebra -- is there a standard or easy result on that? $\endgroup$
    – David Lewis
    Mar 10, 2011 at 1:39
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    $\begingroup$ @Moron: Under the subliminal influence of all the recent questions using "grade" for "degree", I momentarily misinterpreted your above comment as one about simplifying degrees - a method which, alas, some high-school teachers possess too much competence. $\endgroup$
    – Bill Dubuque
    Mar 10, 2011 at 2:39
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I have had students seem surprised when I told them that I have no preference for $\tan\left(\frac{\pi}{6}\right)$ to be "simplified" to $\frac{\sqrt 3}{3}$, for example. A high school teacher whom I told this was even more surprised, but no reason was given in defense of requiring the practice. Nonetheless, to some extent I am glad that high school teachers often tell students that they have to rationalize denominators, because it means that many of them will at least be familiar with the idea when the time comes that "rationalizing" (both denominators and numerators) does serve a purpose in simplifying algebraic expressions, in particular when finding limits in calculus.

So maybe $\frac{1}{4+\sqrt 3}$ is no less simple than $\frac{4-\sqrt 3}{13}$, but $\frac{h}{\sqrt{x+h}-\sqrt{x}}$ is harder to deal with than $\sqrt{x+h}+\sqrt{x}$ when $h$ goes to $0$, and this comes more easily if you already know how to rationalize denominators with numbers.

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  • $\begingroup$ Personally I'd argue that $1/(4+\sqrt{3})$ is in some sense ``simpler'' than $(4-\sqrt{3})/13$, in that I can pretty quickly say $1/(4+\sqrt{3})$ is around $0.17$ but $(4-\sqrt{3})/13$ is a bit harder to work with. This is because I have a good sense of which numbers are which reciprocals of which other numbers, so $1/x$ generally seems simple to me from a mental-computation point of view. $\endgroup$
    – Michael Lugo
    Mar 10, 2011 at 0:09
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I heard from a high school math teacher once that this was done back in the day (before calculators) so that people could look up the values of these expressions in tables. If this is indeed the case, there is no good reason for this practice now.

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    $\begingroup$ This is what I have been told too. I don't require it in my class. $\endgroup$
    – Brian
    Mar 10, 2011 at 0:14
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    $\begingroup$ @Brian: I don't buy that argument. One can look up and substitute an approximate value of a radical whether or not it is in the numerator or denominator. Perhaps what they meant to say is that subsequent computations may be simpler if the radicals are all in the numerator since generally it is easier to (manually) multiply by reals than to divide by them. Is that what your teacher actually said? $\endgroup$
    – Bill Dubuque
    Mar 10, 2011 at 19:03
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This is known as rationalizing the denominator (RTD). $ $ As the name suggests, it simplifies by transforming an irrational divisor into a rational divisor. As explained here, this is a prototypical instance of the method of simpler multiples. This can lead to all sorts of simplifications. Below are a couple prototypical examples.

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

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

Here's a number-theoretic example showing how RTD reduces divisibility of algebraic integers to rational integers. Consider the Gaussian integers $\ \mathbb I = \{ m + n\ i\ : \ m,n\in \mathbb Z \},\,$ where $\,i^2 = -1.\,$ As usual we define divisibilty by $\ a\mid b\,\ {\rm in}\,\ \mathbb I \!\iff\! b/a \in \mathbb I,\,$ for $\,b\neq 0.\,$ Suppose we wish to know if $\ 2\!+\!3\, i\mid 91\ \,{\rm in}\,\ \mathbb I,\,$ i.e. is $\ w = 91/(2\!+\!3\, i)\in \mathbb I\ ?\ $ $\,\mathbb I\,$ happens to have a division algorithm which we could apply. But it is simpler to RTD: $\, w = 91\ (2\!-\!3\, i)/(2^2\!+\!3^2) = 7 (2\!-\!3\, i)\, $ so, indeed, $\, w\in \mathbb I$.

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.

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  • $\begingroup$ No question that RTD is a good operation, especially to eliminate binomials from the denominator, and can lead to very useful results. But why require it for every expression just because it's useful in a few circumstances. Historically and mathematically, why did it get to be that way? $\endgroup$
    – David Lewis
    Mar 10, 2011 at 1:40
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    $\begingroup$ @David: It's not a "few" circumstances but, rather, "many" circumstances". As I attempted to illustrate above, RTD can serve to eliminate obfuscating terms and. hence, serve to make more transparent innate structure - e.g. the cancellation of the apparent singularity above. One applies RTD for the same reasons that one applies any normal form transformation e.g. reducing fractions to lowest terms, or expanding them into partial fractions for integration, or putting them into square-free form, etc. $\endgroup$
    – Bill Dubuque
    Mar 10, 2011 at 1:52
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    $\begingroup$ @David: Alas, probably to make it so that underqualified high school math teachers wouldn't have to think about it and could simply "lay down the law". $\endgroup$
    – Arturo Magidin
    Mar 10, 2011 at 2:10
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    $\begingroup$ @Arturo -- both points well-taken. I myself tell students, somewhat tongue-in-cheek, that it's "math teacher form" designed to torture students, but sometimes with a point for real problems. So I make sure they get plenty of practice on the "skill", but don't insist on it for every answer. The conjugate method is obviously very crucial here. $\endgroup$
    – David Lewis
    Mar 10, 2011 at 2:46
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The form with neither denominators in radicals nor radicals in denominators and with only squarefree expressions under square-root signs, etc., is a canonical form, and two expressions are equal precisely if they're the same when put into canonical form.

When are two fractions equal? How do you know that $\dfrac{51}{68}$ is the same as $\dfrac{39}{52}$? They're both the same when reduced to lowest terms.

How do you know that $\dfrac{1}{3+\sqrt{5}}$ is the same as $\dfrac{3\sqrt{5}-5}{4\sqrt{5}}$? Again, they're the same when put into canonical form.

How do you know that $\dfrac{13+i}{1+2i}$ is the same as $\dfrac{61}{6+5i}$? Same thing.

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Don't blame your high school teacher

They are right

$\frac{1}{\sqrt{3}}$ is correctly "simplified" to $\frac {\sqrt{3}}{3}$.

Its difficult to find value $\frac{1}{\sqrt{3}}$ as $\sqrt{3}$ is irrational number

Its easy to find value of $\frac {\sqrt{3}}{3}$ as here we are dividing it with 3

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    $\begingroup$ I don't really see what value this answer adds. The question is over two years old and already has a number of better answers. $\endgroup$
    – Antonio Vargas
    Oct 9, 2013 at 13:06
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    $\begingroup$ I just want to convey my views.Its two years old question but not closed.Also its is not accepted yet $\endgroup$
    – rst
    Oct 9, 2013 at 13:12
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    $\begingroup$ Why is it difficult to find the value of $\frac{1}{\sqrt{3}}$? $\sqrt{3}$ has the continued fraction representation $[1;1,2,1,2,\ldots]$, so the reciprocal is $[0;1,1,2,1,2,\ldots]$. Terminate at any point for best approximations. How else are you computing it without a calculator (which doesn't care how it's written)? $\endgroup$
    – Zack Wolske
    Jul 9, 2014 at 0:13

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