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Rationalizing Denominators

Date: 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 

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


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


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)


  (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.


-Doctor Rob,  The Math Forum
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Associated Topics:
High School Basic Algebra

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