Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A teaching question. What is or might be simplest form for rational exponents? My search turns up little for the concept other than a translation of the simplest form for n-th root radicals. But that leads to ugly things such as...

$\large{5^\frac{7}{3}\rightarrow 5^2\centerdot 5^\frac{1}{3}}$

$\large{x^{-\frac{2}{3}}\rightarrow x^{-1}\centerdot x^\frac{1}{3}}$

...when the whole point of rational exponents is beauty and elegance.

share|improve this question
1  
what do you mean? whats wrong with $x^{m/n}$ with $m/n$ in lowest terms? –  yoyo Mar 30 '11 at 0:17
    
I mean, is there a concept that makes sense. Lowest terms for m/n works, as long as you stick to non-negative bases. Seems you should also factor into primes and combine factors with the same base. But beyond that, I can't see the point of mimicking the radical rules. Actually, that is enough to serve as a nice normal form for monomials with monomial factors. –  David Lewis Mar 30 '11 at 0:24

2 Answers 2

What's "simplest" is highly context dependent - context which you have not supplied. However, one important aspect of simplicity is computational effectivity. In order to have any hope of effectively computing with such radicals it is essential to normalize them so that any algebraic relationships are made explicit. For example, suppose that you are computing with expressions involving $\:5\:$ raised to the powers $\ 1/6,\ 1/10,\:$ and $\:1/15\:.\:$ Then such expressions can all be expressed as polynomials in the single radical $\:5^{1/30}\:,\:$ e.g. $\rm\:5^{1/6} = (5^{1/30})^5\:.\:$ Said in the language of field theory $\rm\ \mathbb Q(5^{1/6},\:5^{1/10},\:5^{1/15}) \subset \mathbb Q(5^{1/30})\:.\:$ Therefore arithmetic operations on such radical expressions reduce to simple efficient polynomial operations, viz. operations in the ring $\rm\ Q[x]\ (mod\ x^{30}-5)\:.\:$

Such normalization is even more crucial when working with multiple radicands. For example $\ \sqrt{6}\ \sqrt{10}\ \sqrt{15}\ =\ 30\:,\:$ so this algebraic dependence needs to be eliminated by choosing two of the three radicands as a basis, say $\:\sqrt{6},\ \sqrt{10}\:,\ $ then eliminating $\:\sqrt{15}\ $ using $\rm\ \sqrt{15}\ =\ \sqrt{6}\ \sqrt{10}/2\:.\:$ Developing effective algorithms for computing with such radicands (e.g. denesting) is a highly nontrivial task, based on the Galois theory of radical extensions (Kummer theory). This is the theory at the basis of such effective algorithms implemented in computer algebra systems.

share|improve this answer
    
Well, I did kinda state a context -- it's a teaching question, by which I meant high school algebra 2. In that context, simplest form rules seem kind of arbitrary, motivated as much by driving algebraic mastery as any mathematical basis. For example, the old rule about no radicals in the denominator sees, as far as I can determine, to come from avoiding long decimals in the divisor when doing long division in the pre-calculator era. So while I appreciate hearing about the math behind normal forms, etc, I have a more modest goal. –  David Lewis Mar 30 '11 at 12:39
    
One of the advantages to simplifying a radical expression is that two expressions are known to be or not be equal once they are both simplified completely. For high school education, this reason should suffice since multiple choice questions are popular in the US. The choice between writing your expressions with positive or negative exponents does seem to be a matter of taste though. –  Joshua Shane Liberman Mar 30 '11 at 14:52
    
@David: It's not clear what "rules" you refer to (besides the laws of exponents). As I said, the driving force here for rule-based computation are certain normal forms that facilitate effective computation in radical extension fields. Just as in your prior question about rationalizing denominators, these normal forms reduce complexity - here by eliminating algebraic dependences among radical monomials. I suspect that any "rules" you may find in high-school textbooks are ad-hoc and probably should be ignored. To properly appreciate these matters requires knowledge of abstract algebra. –  Bill Dubuque Mar 30 '11 at 16:28
    
@David: As for your prior question on rationalizing denominators, I hoped that the examples I gave their might serve to convince you that the motivations for such are much deeper than trivial reasons such as "avoiding long decimals in the divisor". I'm disappointed to learn that you were apparently not swayed by these arguments. Please feel free to continue the discussion in comments there if further clarification may prove helpful. –  Bill Dubuque Mar 30 '11 at 16:28
    
@Joshua: Of course deciding equality or, equivalently, zero-equivalence, is one component of what I refer to above as effective calculation in radical extensions. This is one of the primary reasons for reducing expressions to tractable normal forms. –  Bill Dubuque Mar 30 '11 at 16:37

"Lowest terms" for a rational is usually understood to mean $a/b$, with $a$ and $b$ integers, $b\gt 0$, and $\gcd(a,b)=1$. (That is, the "sign" goes with the numerator).

For $p/q$ written in lowest terms, the definition of $a^{p/q}$ is that $a^{p/q} = r$ where $r$ is the unique nonnegative real number such that $r^q = a^p$; that is, $\displaystyle a^{p/q} = \sqrt[q]{a^p}$. This includes the cases with $p\gt q$ and with $-p\gt q$. If you want only fractional exponents of the form $p/q$ with $0\lt p\lt q$, then of course this can be done fairly easily, but it does not seem to me to lead to much simplification for algebraic manipulations (though it may for actual computations).

Rational exponents in lowest terms with even denominator are only, a priori, defined for positive bases when working in the reals.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.