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### Rationalizing Roots in More Denominators

```Date: 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.

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

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

-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

```

```
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/
```
Associated Topics:
High School Square & Cube Roots

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