Rationalizing a Denominator with Multiple Cube RootsDate: 04/22/2011 at 05:08:13 From: Steve (but you may call me stevie) Subject: Rationalize a denominator with multiple cube roots Hi Dr. Math, Or may I call you 'doc'? I have the following problem: Given this fraction of rational numbers a, b, c, and q in Q: 1/(a + b*CBRT(q) + c*CBRT(q)^2) I was wondering if it is possible to express this general formula without non-rational numbers in the bottom of the fraction -- in other words, get rid of the cube roots in the denominator. I know this should be possible, as we know from basic algebra that this is isomorphic with Q(CBRT(q)): Q[x]/(x^3 - q) Moreover, all elements of the first one are of the form q1 + q2*X + q3*X^2 With this knowledge, it should be clear that what I am trying to achieve should be possible. Alas, I still fail to get the right form of that general case. I've tried to use a formula involving (a + b)(a^2 + ab + b^2), but this does not seem to work on this case. Could you help me out, doc? Date: 04/22/2011 at 19:42:19 From: Doctor Vogler Subject: Re: Rationalize a denominator with multiple cube roots Hi Stevie, Thanks for writing to Dr. Math. Before we take up your question, let's start with a simpler root. To rationalize a denominator with a square root, like ... a + b*SQRT(q), ... you multiply the numerator and the denominator by the *conjugate* of your number. The conjugate is what you get by replacing SQRT(q) by the other square root of q, namely -SQRT(q): a - b*SQRT(q). Well, in the case of cube roots, there are two other conjugates of your number. One replaces CBRT(q) by w*CBRT(q). The other replaces CBRT(q) by w^2*CBRT(q), where ... w^2 + w + 1 = 0 ... is the complex number w = (-1 + SQRT(-3))/2, ... which also has w^3 = 1. (Notice that those are the other two complex cube roots of q. These conjugates are determined by the three complex embeddings of the number field you mentioned, Q(CBRT(q)), into the complex numbers.) In other words, the other two conjugates of ... a + b*CBRT(q) + c*CBRT(q)^2 ... are the two complex numbers ... a + b*w*CBRT(q) + c*w^2*CBRT(q)^2 ... and ... a + b*w^2*CBRT(q) + c*w*CBRT(q)^2. Assuming that a, b, c, and CBRT(q) are real numbers, as you have, the product of those two numbers is also a real number, so the w's go away: (a + b*w*CBRT(q) + c*w^2*CBRT(q)^2) * (a + b*w^2*CBRT(q) + c*w*CBRT(q)^2) = (a^2 - c*b*q) + (c^2*q - b*a)*CBRT(q) + (b^2 - c*a)*CBRT(q)^2 Well, if you multiply the numerator and the denominator of your fraction by this number, then the denominator will be a polynomial in a, b, c, and q, namely: a^3 + (b^3 - 3*c*b*a)*q + c^3*q^2. That is also the norm of your denominator as an element of Q(CBRT(q)). If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/