When working with fractions, we often want to simplify as much as possible to make our work easier. Denominators are important when adding or subtracting fractions – especially when square roots and other radicals are involved.

So, can radicals be in the denominator? **Radicals can appear in the denominator, and this is a perfectly valid mathematical expression. However, we usually eliminate radicals from the denominator of a fraction in order to simplify it. After we rationalize the denominator, it is a rational number (often an integer) with no radical symbol.**

Of course, some denominators with radicals are easier to rationalize than others. For example, denominators with two radical terms require a little extra work to rationalize.

In this article, we’ll talk about radicals in the denominator and how to rationalize a denominator. We’ll also look at several examples to make the concept clear.

Let’s get started.

## Can Radicals Be In The Denominator?

Radicals can be in the denominator of a fraction. For example, the following are perfectly valid fractions:

**1 / √2****√3 / √5****(9 – √7) / √6****4 / (1 + √8)**

However, it is standard procedure to remove radicals from the denominator. By convention, we say that a fraction is simplified only after we rationalize the denominator.

### How To Rationalize The Denominator

To rationalize the denominator of a fraction, we multiply by a fraction that is equal to 1 (such as √3/√3). Since multiplying by 1 does not change a number, we get the same value after rationalizing the denominator (just in a different form).

To decide what to multiply by in the numerator and denominator, we look at the index of the radical. The goal is to make the index 1 (that is, eliminate the radical).

As a reminder: the *index* of a radical is the value to the upper-left of the radical symbol. If no value is written, the index is assumed to be 2.

For example:

**√5 has an index of 2 (no value is written to the top left of the radical symbol, so assume 2)**^{3}√7 has an index of 3^{4}√8 has an index of 4^{5}√10 has an index of 5**etc.**

So, if the index of the radical is n, then the power is 1/n (the reciprocal of the index). For example:

**√5 = 5**^{1/2}^{3}√6 = 6^{1/3}^{4}√3 = 3^{1/4}**etc.**

We calculate 1 – P, where P is the power. Then, we multiply by the radicand raised to the power 1 – P.

As a reminder: the *radicand* of a radical is the value under the radical symbol. For example:

**√5 has a radicand of 5**^{3}√7 has a radicand of 7^{4}√8 has a radicand of 8^{5}√10 has a radicand of 10**etc.**

To summarize, here are the steps to rationalize a denominator with one term that contains a radical:

**First, express the radical from the denominator in the form R**^{P}(radicand raised to a power). For the radical^{N}√R, we would get R^{1/N}(the base is the radicand R, and the power is the reciprocal of the index N).**Next, calculate 1 – P.****Then, find R**^{1-P}(radicand raised to the power of 1 – P).**Now, rewrite R**^{1-P}as a radical. For R^{1-P}, we would get^{1/(1-P)}√R.**Then, multiply both the numerator and denominator of the original fraction by**^{1/(1-P)}√R (the radical from step 4).**Finally, simplify the numerator and denominator as much as possible. The denominator should now be rational.**

The following table shows some values of N and corresponding powers P and 1-P:

Index (N) | Power (P=1/N) | Power (1-P) |
---|---|---|

2 | 1/2 | 1/2 |

3 | 1/3 | 2/3 |

4 | 1/4 | 3/4 |

5 | 1/5 | 4/5 |

6 | 1/6 | 5/6 |

N | 1/N | (N-1)/N |

the power for the corresponding rational

exponent, and the value of 1 – power.

Now let’s take a look at some examples of how to rationalize a denominator that contains one or more radicals.

#### Example 1: Rationalizing A Denominator (Monomial)

Let’s say we have the fraction 1 / √3.

Step 1 is to express the radical from the denominator in the form R^{P}. Since the radical is √3, we have R = 3 (radicand) and N = 2 (index). So, we get 3^{1/2} (with a base of 3 and a power of ½).

Step 2 is to calculate 1 – P = 1 – (1/2) = 1/2.

Step 3 is to find R^{1-P} = 3^{1/2.}

Step 4 is to rewrite R^{1-P }as a radical. For 3^{1/2}, we get √3.

Step 5 is to multiply both the numerator and denominator of the original fraction by √3 (see below).

Step 6 is to simplify the numerator and denominator as much as possible.

So, we want to multiply by √3 in the numerator and denominator of the original fraction. Here are the steps:

**1 / √3**[the original fraction]**=1*√3 / √3*√3**[multiplied by √3 in the numerator and denominator]**=√3 / 3**[√3*√3 = 3]

Now the denominator is rationalized, since the radicals are gone (3 is an integer, which is rational). This fraction is as simple as we can make it.

#### Example 2: Rationalizing A Denominator (Monomial)

Let’s say we have the fraction 4 / 5√6.

Step 1 is to express the radical from the denominator in the form R^{P}. Since the radical is √6, we have R = 6 (radicand) and N = 2 (index). So, we get 6^{1/2} (with a base of 6 and a power of 1/2).

Step 2 is to calculate 1 – P = 1 – (1/2) = 1/2.

Step 3 is to find R^{1-P} = 6^{1/2}.

Step 4 is to rewrite R^{1-P }as a radical. For 6^{1/2}, we get √6.

Step 5 is to multiply both the numerator and denominator of the original fraction by √6 (see below).

Step 6 is to simplify the numerator and denominator as much as possible.

So, we want to multiply by √6 in the numerator and denominator of the original fraction. Here are the steps:

**4 / 5√6**[the original fraction]**=4*√6****/ 5√6√6**[multiplied by √6 in the numerator and denominator]**=4√6 / 5*6**[√6*√6 = 6]**=4√6 / 30****=2√6 / 15**

Now the denominator is rationalized, since the radicals are gone (15 is an integer, which is rational). This fraction is as simple as we can make it.

#### Example 3: Rationalizing A Denominator (Monomial)

Let’s say we have the fraction 1 / ^{3}√5.

Step 1 is to express the radical from the denominator in the form R^{P}. Since the radical is ^{3}√5, we have R = 5 (radicand) and N = 3 (index). So, we get 5^{1/3} (with a base of 5 and a power of 1/3).

Step 2 is to calculate 1 – P = 1 – (1/3) = 2/3.

Step 3 is to find R^{1-P} = 5^{2/3}.

Step 4 is to rewrite R^{1-P }as a radical. For 5^{2/3}, we get ^{3}√(5^{2}).

Step 5 is to multiply both the numerator and denominator of the original fraction by ^{3}√(5^{2}) (see below).

Step 6 is to simplify the numerator and denominator as much as possible.

So, we want to multiply by ^{3}√(5^{2}) in the numerator and denominator of the original fraction. Here are the steps:

**1 /**[the original fraction]^{3}√5**=1***[multiplied by^{3}√(5^{2}) /^{3}√5*^{3}√(5^{2})^{3}√(5^{2}) in the numerator and denominator]**=**^{3}√(5^{2}) /^{3}√(5*5^{2})**=**^{3}√(5^{2}) /^{3}√(5^{3})**=**^{3}√(5^{2}) / 5

Now the denominator is rationalized, since the radicals are gone (5 is an integer, which is rational). This fraction is as simple as we can make it.

(Alternatively, we can write the final fraction as 5^{2/3} / 5, using a rational exponent instead of a radical for the numerator).

#### Example 4: Rationalizing A Denominator (Binomial)

Let’s say we have the fraction 3 / (2 + √5).

Since the denominator is a binomial, we need to modify our approach a bit.

Instead of multiplying the numerator and denominator by √5, we instead multiply by 2 – √5 (this is called the *conjugate *of 2 + √5).

Let’s see how this works:

**3 / (2 + √5)**[the original fraction]**=3*(2 – √5) / (2 + √5)*(2 – √5)**[multiplied by (2 – √5) in the numerator and denominator]**=(6 – 3√5) / (2 + √5)*(2 – √5)**[distribute the 3 through the parentheses in the numerator]**=(6 – 3√5) / (4 – 2√5 + 2√5 – 5)**[used FOIL on the product of binomials in the denominator]**=(6 – 3√5) / (-1)**[cancel and combine like terms]**=-6 + 3√5**

Now the denominator is rationalized, since the radicals are gone (1 is an integer, which is rational). This fraction is as simple as we can make it.

#### Example 5: Rationalizing A Denominator (Binomial)

Let’s say we have the fraction 1 / (√2 + ^{3}√2).

Since the denominator is a binomial with two different radicals with a different index for each one (2 and 3), we need to modify our approach again.

Instead of multiplying the numerator and denominator by √2 or ^{3}√2, we instead multiply by (√2 – ^{3}√2) (this is called the *conjugate *of (√2 + ^{3}√2).

Let’s see how this works:

**1 / (√2 +**[the original fraction]^{3}√2)**=1*(√2 –**[multiplied by (√2 –^{3}√2) / (√2 +^{3}√2)*(√2 –^{3}√2)^{3}√2) in the numerator and denominator]**=1*(√2 –**[used FOIL on the product of binomials in the denominator]^{3}√2) / (2 – √2^{3}√2 + √2^{3}√2 –^{3}√(2^{2}))**=1*(√2 –**[cancel and combine like terms]^{3}√2) / (2 –^{3}√(2^{2}))

Now we only have one radical in the denominator, instead of two. The radical is ^{3}√(2^{2}). Instead of factoring a square, we factor a cube.

The formula for a difference of cubes is:

**a**^{3}– b^{3}= (a – b)(a^{2}+ ab + b^{2})

If we use a = 2 and b = ^{3}√(2^{2}) in the formula above, we get:

**2**^{3}– (^{3}√(2^{2}))^{3}= (2 –^{3}√(2^{2}))(2^{2}+ 2(^{3}√(2^{2})) + (^{3}√(2^{2}))^{2})**8 – (2**^{2}) = (2 –^{3}√(2^{2}))(4 + 2(^{3}√(2^{2})) + (^{3}√(2^{4}))**8 – 4 = (2 –**^{3}√(2^{2}))(4 + 2(^{3}√(2^{2})) + (^{3}√(2^{4}))**4 = (2 –**^{3}√(2^{2}))(4 + 2(^{3}√(2^{2})) + (^{3}√(2^{4}))

So, if we multiply the numerator and denominator by (4 + 2(^{3}√(2^{2})) + (^{3}√(2^{4})), we will get a result of 4 in the denominator (a rational number).

## Conclusion

Now you know how to rationalize a denominator if there are radicals present (whether monomials or binomials). This will help when simplifying fractions to find a least common denominator for addition or subtraction.

You can learn more about square roots in these articles:

- What is a radical equation?
- Radicals & rational exponents
- Can a square root be negative?
- Common questions about square roots
- Adding, multiplying, and dividing square roots
- How to do square roots by hand
- What is a square root used for? (real life applications)

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