Rationalizing Roots in More DenominatorsDate: 02/16/2011 at 16:57:56 From: Steven Subject: Rationalize the Denominator: 1/[2^(1/2)+3^(1/3)+...n^(1/n)] It's said that all denominator expressions should be rationalized for simplicity, but I was wondering if there is a systematic approach for rationalizing the bottom of a fraction that contains multiple and different roots, such as 1/[2^(1/2) + 3^(1/3) + 4^(1/4)]? Is rationalization of such denominators even possible? I've been trying to look this up and have only come up with answers solving either the same roots in the denominator, or just two different roots, or monomials and binomials, which doesn't apply here. I've looked into roots of unity, but that becomes pretty complex. Maybe you can express the different roots as a common root? For example, 2^(1/2)+3^(1/3) = 2^(3/6)+3^(2/6) But I don't think I can just come up with a new math law. Date: 02/17/2011 at 21:11:43 From: Doctor Vogler Subject: Re: Rationalize the Denominator: 1/[2^(1/2)+3^(1/3)+...n^(1/n)] Hi Steven, Thanks for writing to Dr. Math. You can rationalize a denominator like this one, but it becomes very, very messy. The idea is to multiply the numerator and the denominator by every conjugate of the denominator. The simplest case is when there is only one conjugate, which is what you get when the denominator is a quadratic surd like 2 + 3*sqrt(7). And the conjugate is, of course, 2 - 3*sqrt(7). This task becomes more complicated when there are more conjugates, such as when you have a cube root, as in 1 + 3^(1/3). Then the two conjugates are ... 1 + w*3^(1/3) 1 + w^2*3^(1/3), ... where w = (-1 + sqrt(-3))/2 is a primitive cube root of 1 (so w^3 = 1; but w is not 1, so w^2 + w + 1 = 0). Each of those two conjugates is a complex number, but their product is real-valued: (1 + w*3^(1/3))(1 + w^2*3^(1/3)) = 1 + (w + w^2)*3^(1/3) + w^3*3^(2/3) = 1 - 3^(1/3) + 3^(2/3) So you can multiply the numerator and denominator by ... 1 - 3^(1/3) + 3^(2/3), ... and in the denominator, you get [1 + 3^(1/3)][1 - 3^(1/3) + 3^(2/3)] = 4. When you have a sum of different types of algebraic numbers, then you have to run through all of the conjugates. So if your denominator is ... 2^(1/2) + 3^(1/3), ... then the conjugates of 2^(1/2) are itself and -2^(1/2), while the conjugates of 3^(1/3) are itself, w*3^(1/3), and w^2*3^(1/3). This means that the other conjugates of their sum are the five numbers 2^(1/2) + w*3^(1/3), 2^(1/2) + w^2*3^(1/3), -2^(1/2) + 3^(1/3), -2^(1/2) + w*3^(1/3), and -2^(1/2) + w^2*3^(1/3). The product of those five numbers is a real number; and when you multiply by the denominator 2^(1/2) + 3^(1/3), you get a rational number (indeed, an integer). Similarly, if there are more terms in the sum, you have even more conjugates. 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/ Date: 03/02/2011 at 20:23:19 From: Steven Subject: Rationalize the Denominator: 1/[2^(1/2)+3^(1/3)+...n^(1/n)] Thanks for the great (and fast) answer. Sorry I couldn't comment back recently; school has been bogging me down. But, on to business. So you don't actually have to know the roots of unity? just the laws? You have enlightened me! If w is a root of unity of n, then ... 1 + w + w^2 + ... w^(n-1) = 0 ... and ... w^n = 1. I used this and did the last example you gave me. I multiplied all five conjugates of 2^(1/2) + 3^(1/3): 2^(1/2) + w*3^(1/3), 2^(1/2) + w^2*3^(1/3), -2^(1/2) + 3^(1/3), -2^(1/2) + w*3^(1/3), and -2^(1/2) + w^2*3^(1/3). You're right, the output gets pretty ugly ;) but here it is: 2 + 2^(1/2)*w^2*3^(1/3) - 2 + 2^(1/2)*3^(1/3) - 2 + 2^(1/2)*w*3^(1/3) - 2 + 2^(1/2)*w^2*3^(1/3) + 2^(1/2)*w*3^(1/3) + w^3*3^(2/3) - 2^(1/2)*w*3^(1/3) + w*3^(2/3) - 2^(1/2)*w*3^(1/3) + w^2*3^(2/3) - 2^(1/2)*w*3^(1/3) + w^3*3^(2/3) - 2 + 2^(1/2)*3^(1/3) - 2 + 2^(1/2)*w*3^(1/3) - 2 + 2^(1/2)*w^2*3^(1/3) - 2^(1/2)*w^2*3^(1/3) + w^2*3^(2/3) - 2^(1/2)*w^2*3^(1/3) + w^3*3^(2/3) - 2^(1/2)*w^2*3^(1/3) + w^4*3^(2/3) + 2 - 2^(1/2)*w*3^(1/3) + 2 - 2^(1/2)*w^2*3^(1/3) - 2^(1/2)*3^(1/3) + w*3^(2/3) - 2^(1/2)*3^(1/3) + w^2*3^(2/3) + 2 - 2^(1/2)*w^2*3^(1/3) - 2^(1/2)*w*3^(1/3) + w^3*3^(2/3) It looks a lot better after combining like terms: - 2(2) - 2(2^(1/2)*w*3^(1/3)) - 2(2^(1/2)*w^2*3^(1/3)) + 2(w*3^(2/3)) + 3(w^2*3^(2/3)) + 4(w^3*3^(2/3)) + w^4*3^(2/3) I then factored out a -2 from the first group: -4 - 2(2^(1/2))(w*3^(1/3) + w^2*3^(1/3)) + (w*3^(2/3) + w^2*3^(2/3)) + (w*3^(2/3) + w^2*3^(2/3)) + w^2*3^(2/3) + 4(w^3*3^(2/3)) + w^4*3^(2/3) Next, I factored out all the cube roots: -4 - 2(2^(1/2))(3^(1/3))(w + w^2) + (3^(2/3))(w + w^2) + (3^(2/3))(w + w^2) + w^2*3^(2/3) + 4(w^3*3^(2/3)) + w^4*3^(2/3) I used my newly conceived laws to get rid of some of the w's: -4 + 2(2^(1/2))(3^(1/3)) - 3^(2/3) - 3^(2/3) + w^2*3^(2/3) + 4(3^(2/3)) + w^4*3^(2/3) Combining like terms again, -4 + 2(2^(1/2))(3^(1/3)) + 2(3^(2/3)) + w^2*3^(2/3) + w^4*3^(2/3) Here, I got a little tricky, factoring out a 3^(2/3) from the last terms: -4 + 2(2^(1/2))(3^(1/3)) + 2(3^(2/3)) + 3^(2/3)(w^2 + w^4) Then a w^2: -4 + 2(2^(1/2))(3^(1/3)) + 2(3^(2/3)) + 3^(2/3)(w^2)(w^2 + 1) Now, if 1 + w + w^2 = 0 then w^2 + 1 = -w, right? So, -4 + 2(2^(1/2))(3^(1/3)) + 2(3^(2/3)) + 3^(2/3)(w^2)(-w) Multiplying the last of the w's: -4 + 2(2^(1/2))(3^(1/3)) + 2(3^(2/3)) + 3^(2/3)(-w^3) And it's gone: -4 + 2(2^(1/2))(3^(1/3)) + 2(3^(2/3)) - 3^(2/3) Combining the last like terms, I got my final conjugate: -4 + 2(2^(1/2))(3^(1/3)) + 3^(2/3) I multiplied this with the original denominator -- but still get a radical: -4(2^(1/2)) + 3(2^(1/2))(3^(2/3)) + 3 Am I missing something? Is this right? -2^(1/2) = -(2^(1/2)), not 2^(1/2)i Thanks again for your help! Date: 03/03/2011 at 16:11:13 From: Doctor Vogler Subject: Re: Rationalize the Denominator: 1/[2^(1/2)+3^(1/3)+...n^(1/n)] Hi Steven, Your first expression is equal to your last expression, so I believe the algebra in between, but neither one is equal to the product of those five conjugates, so you must have made a mistake in computing the first expression. But you are correct that you will always be able to eliminate all of the roots of unity. When I multiply together those conjugates, I get (-2*3^(2/3) - 3^(1/3) - 4)*2^(1/2) + (3*3^(2/3) + 4*3^(1/3) + 6) And the product of this number with ... 2^(1/2) + 3^(1/3) ... is 1. (I admit, I did the multiplication in a math program, rather than by hand.) - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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