How do you rationalize the denominator of #1/(sqrt(2)+sqrt(3)+sqrt(5))#?

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Jim H Share
Dec 6, 2016

This is a very good question!

If it were simply #1/(sqrta+sqrtb)#, we would use the conjugate,

and we would multiply by #(sqrta-sqrtb)/(sqrta-sqrtb)#.

Let's try something like that and see if it works. (This is what we do with problems of kinds we have not seen before. Try something and see if it works.)

#1/(sqrt(2)+sqrt(3)+sqrt(5)) = 1/(sqrt(2)+(sqrt(3)+sqrt(5)))#

# = 1/([sqrt(2)+(sqrt(3)+sqrt(5))]) ([sqrt(2)-(sqrt(3)+sqrt(5))])/([sqrt(2)-(sqrt(3)+sqrt(5))])#

# = ([sqrt(2)-(sqrt(3)+sqrt(5))]) / ([sqrt(2) + (sqrt(3)+sqrt(5))][sqrt(2)-(sqrt(3)+sqrt(5))])#

# = ([sqrt(2)-(sqrt(3)+sqrt(5))]) / (2 -(sqrt(3)+sqrt(5))^2 )#

# = ([sqrt(2)-(sqrt(3)+sqrt(5))]) / (2 -(3+2sqrt(15)+5) )#

# = ([sqrt(2)-(sqrt(3)+sqrt(5))]) / (-6 - 2sqrt(15)) #

Did that help? (Yes, it did. We now have a more familiar looking problem.

# = ([sqrt(2)-(sqrt(3)+sqrt(5))] [-6 + 2sqrt(15)]) / ([-6 - 2sqrt(15)][-6 + 2sqrt(15)])#

# = ([sqrt(2)-(sqrt(3)+sqrt(5))] [-6 + 2sqrt(15)]) / (36-4(15))#

# = ([sqrt(2)-(sqrt(3)+sqrt(5))] [-6 + 2sqrt(15)]) / -24#

# = ([sqrt(2)-(sqrt(3)+sqrt(5))] [3 - sqrt(15)]) / 12#

Multiply the numerator if you like, to get:

# = ([sqrt(2)-sqrt(3)-sqrt(5)] [3 - sqrt(15)]) / 12#

# =( [sqrt(2)-sqrt(3)-sqrt(5)] [3 - sqrt(15)]) / 12#

# =(3sqrt(2)-3sqrt(3)-3sqrt(5)- sqrt(30)+sqrt45+sqrt75) / 12#

# =(3sqrt(2)-3sqrt(3)-3sqrt(5)- sqrt(30)+3sqrt5+5sqrt3) / 12#

# =(3sqrt(2)+2sqrt(3)- sqrt(30)) / 12#

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Alan P. Share
May 8, 2015

I've decided to add my version to Jim's as a demonstration or perhaps a warning about the variety of forms the outcome could take:
#1/(sqrt(2)+sqrt(3)+sqrt(5))#

Consider a simpler problem:
If we were asked to rationalize the denominator of
#1/(x+sqrt(5))#

we would simply multiply both the numerator and denominator by the conjugate of the denominator

#1/(x+sqrt(5))xx(x-sqrt(5))/(x-sqrt(5))#

#= (x-sqrt(5))/(x^2-5)#

In this case #x=(sqrt(2)+sqrt(3))#
and the result would be

#=(sqrt(2)+sqrt(3)-sqrt(5))/((sqrt(2)+sqrt(3))^2-5#

#=(sqrt(2)+sqrt(3)-sqrt(5))/(2+2sqrt(2)sqrt(3)+3 -5)#

#=(sqrt(2)+sqrt(3)-sqrt(5))/(2sqrt(2*3))#

In order to complete the rationalization of the denominator we would need to multiply both the numerator and denominator by #sqrt(6)=sqrt(2*3)#

#(sqrt(2)+sqrt(3)-sqrt(5))/(2sqrt(2*3)) xx sqrt(6)/sqrt(2*3)#

#= (sqrt(12)+sqrt(18)-sqrt(30))/(2(2*3))#

#=(2sqrt(3)+3sqrt(2)-sqrt(2*3*5))/12#

I avoided posting this because I am bothered by the result seems to imply that the sequence of the denominator terms is reflected in the result. This should not be true but I have not (yet) been able to come up with a final result where the terms are interchangeable.

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Max G. Share
Mar 23, 2017

Answer:

A more general denominator

Explanation:

If we employ the more general denominator of
#sqrt(a) + sqrt(b)+sqrt(c)#

then we arrive at a numerator of

#(a-b-c)sqrt(a)+(b-a-c)sqrt(b)+(c-a-b)sqrt(c)+2sqrt(abc)#

and a more general (rational) denominator of

#a^2 + b^2 + c^2-2(ab+ac+bc)#

The same work as we see for 2, 3, and 5 yields this result.

In the example given, c = a + b, which eliminates the term that would have contained #sqrt(5)#. Furthermore any time c = a + b, the more general solution simplifies to a numerator of

#bsqrt(a)+asqrt(b)-sqrt(ab(a+b))#
and a denominator of 2ab.

In the specific instance here we have
#(3sqrt(2)+2sqrt(3)-sqrt(30))/12#

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