Rationalizing DenominatorsDate: 07/12/97 at 15:11:49 From: Jim Summerlin Subject: Rationalizing denominators Dear Dr. Math, HELP!! I just can't seen to do this. How can I rationalize a denominator with odd power radicals? This one for example 1/[ (64^(1/5))+(36^(1/5))+(24^(1/5))+(16^(1/5))] It looks better with radicals but I don't know how to enter symbols in this text box. Can you rationalize that denominator? Thanks for your help. Jim Summerlin Date: 07/16/97 at 16:19:06 From: Doctor Rob Subject: Re: Rationalizing denominators Yes, you can, but it will end up with a ghastly numerator. In order to do this, you need to drag in a 5-th root of unity, call it z. Then it satisfies (z^5 - 1)/(z - 1) = z^4 + z^3 + z^2 + z + 1 = 0. You then multiply both numerator and denominator by the product of all z^a*(64^(1/5))+z^b*(36^(1/5))+z^c*(24^(1/5))+z^d*(16^(1/5)) for all 4-tuples (a, b, c, d) with 0 <= a, b, c, d <= 4, except for the 4-tuple (0,0,0,0), which is already there in the denominator. This is a list of 255 additional factors. Now expand everything in sight, and simplify using first z^5 = 1 and then the above irreducible quartic polynomial equation satisfied by z. It may be hard to believe, but all the z's will miraculously vanish from the expression, and all the radicals will vanish from the denominator, and you will have your answer. Of course not all the radicals vanish from the numerator! The denominator turns out to be 2551318400000^25, or about 5.831779 * 10^303 ! In this particular case, we didn't need b or d at all, because (64^(1/5))^4 = 16*16^(1/5) and (24^(1/5))^2/(16^(1/5)) = 36^(1/5). If we hadn't used the b and d, we would have gotten a denominator of only 2551318400000. After reducing to lowest terms, the denominator is only 510263680. Computing the numerator I leave as an exercise ( :-) !). Actually it only consumed 7 lines of Mathematica output. Convinced? -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/http://mathforum.org/dr.math/ |
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