How to Rationalize the Denominator

Co-authored by Hannah Madden Reviewed by Joseph Meyer

Last Updated: December 13, 2022 Fact Checked

Traditionally, a radical or irrational number cannot be left in the denominator (the bottom) of a fraction. When a radical does appear in the denominator, you need to multiply the fraction by a term or set of terms that can remove that radical expression. While the use of calculators make rationalizing fractions a bit dated, this technique may still be tested in class.

Part 1

Part 1 of 4:

Rationalizing a Monomial Denominator

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$Step 1 Examine the fraction.$

1
Examine the fraction. A fraction is written correctly when there is no radical in the denominator. If the denominator contains a square root or other radical, you must multiply both the top and bottom by a number that can get rid of that radical. Note that the numerator can contain a radical. Don't worry about the numerator.^{[1]
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Research source}
- ${\frac {7{\sqrt {3}}}{2{\sqrt {7}}}}$
- We can see that there is a ${\sqrt {7}}$ in the denominator.
2
Multiply the numerator and denominator by the radical in the denominator. A fraction with a monomial term in the denominator is the easiest to rationalize. Both the top and bottom of the fraction must be multiplied by the same term, because what you are really doing is multiplying by 1.
- ${\frac {7{\sqrt {3}}}{2{\sqrt {7}}}}\cdot {\frac {\sqrt {7}}{\sqrt {7}}}$
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3
Simplify as needed. The fraction has now been rationalized.^{[2]
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Research source}
- ${\frac {7{\sqrt {3}}}{2{\sqrt {7}}}}\cdot {\frac {\sqrt {7}}{\sqrt {7}}}={\frac {7{\sqrt {21}}}{14}}={\frac {\sqrt {21}}{2}}$

Part 2

Part 2 of 4:

Rationalizing a Binomial Denominator

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1
Examine the fraction. If your fraction contains a sum of two terms in the denominator, at least one of which is irrational, then you cannot multiply the fraction by it in the numerator and denominator.^{[3]
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Research source}
- ${\frac {4}{2+{\sqrt {2}}}}$
- To see why this is the case, write an arbitrary fraction ${\frac {1}{a+b}},$ where $a$ and $b$ are irrational. Then the expression $(a+b)(a+b)=a^{2}+2ab+b^{2}$ contains a cross-term $2ab.$ If at least one of $a$ and $b$ is irrational, then the cross-term will contain a radical.
- Let's see how this works with our example.
  - ${\frac {4}{2+{\sqrt {2}}}}\cdot {\frac {2+{\sqrt {2}}}{2+{\sqrt {2}}}}={\frac {4(2+{\sqrt {2}})}{4+4{\sqrt {2}}+2}}$
- As you can see, there's no way we can get rid of the $4{\sqrt {2}}$ in the denominator after doing this.
$Step 2 Multiply the fraction by the conjugate of the denominator.$

2
Multiply the fraction by the conjugate of the denominator. The conjugate of an expression is the same expression with the sign reversed.^{[4]
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Research source} For example, the conjugate of $2+{\sqrt {2}}$ $2+{\sqrt {2}}$ is $2-{\sqrt {2}}.$ $2-{\sqrt {2}}.$
- ${\frac {4}{2+{\sqrt {2}}}}\cdot {\frac {2-{\sqrt {2}}}{2-{\sqrt {2}}}}$
- Why does the conjugate work? Going back to our arbitrary fraction ${\frac {1}{a+b}},$ multiplying by the conjugate in the numerator and denominator results in the denominator being $(a+b)(a-b)=a^{2}-b^{2}.$ The key here is that there are no cross-terms. Since both of these terms are being squared, any square roots will be eliminated.
3
Simplify as needed.^{[5]
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Research source}
- ${\frac {4}{2+{\sqrt {2}}}}\cdot {\frac {2-{\sqrt {2}}}{2-{\sqrt {2}}}}={\frac {4(2-{\sqrt {2}})}{4-2}}=4-2{\sqrt {2}}$

Part 3

Part 3 of 4:

Working with Reciprocals

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1
Examine the problem. If you are asked to write the reciprocal of a set of terms containing a radical, you will need to rationalize before simplifying. Use the method for monomial or binomial denominators, depending on whichever applies to the problem.^{[6]
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Research source}
- $2-{\sqrt {3}}$
2
Write the reciprocal as it would usually appear. A reciprocal is created when you invert the fraction.^{[7]
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Research source} Our expression $2-{\sqrt {3}}$ $2-{\sqrt {3}}$ is actually a fraction. It's just being divided by 1.
- ${\frac {1}{2-{\sqrt {3}}}}$
3
Multiply by something that can get rid of the radical on the bottom. Remember, you're actually multiplying by 1, so you have to multiply both the numerator and denominator. Our example is a binomial, so multiply the top and bottom by the conjugate.^{[8]
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Research source}
- ${\frac {1}{2-{\sqrt {3}}}}\cdot {\frac {2+{\sqrt {3}}}{2+{\sqrt {3}}}}$
4
Simplify as needed.
- ${\frac {1}{2-{\sqrt {3}}}}\cdot {\frac {2+{\sqrt {3}}}{2+{\sqrt {3}}}}={\frac {2+{\sqrt {3}}}{4-3}}=2+{\sqrt {3}}$
- Do not be thrown off by the fact that the reciprocal is the conjugate. This is just a coincidence.

Part 4

Part 4 of 4:

Rationalizing Denominators with a Cube Root

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$Step 1 Examine the fraction.$

1
Examine the fraction. You can also expect to face cube roots in the denominator at some point, though they are rarer. This method also generalizes to roots of any index.^{[9]
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Research source}
- ${\frac {3}{\sqrt[{3}]{3}}}$
2
Rewrite the denominator in terms of exponents. Finding an expression that will rationalize the denominator here will be a bit different because we cannot simply multiply by the radical.^{[10]
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Research source}
- ${\frac {3}{3^{1/3}}}$
3
Multiply the top and bottom by something that makes the exponent in the denominator 1. In our case, we are dealing with a cube root, so multiply by ${\frac {3^{2/3}}{3^{2/3}}}.$ ${\frac {3^{{2/3}}}{3^{{2/3}}}}.$ Remember that exponents turn a multiplication problem into an addition problem by the property $a^{b}a^{c}=a^{b+c}.$ $a^{{b}}a^{{c}}=a^{{b+c}}.$ ^{[11]
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Research source}
- ${\frac {3}{3^{1/3}}}\cdot {\frac {3^{2/3}}{3^{2/3}}}$
- This can generalize to nth roots in the denominator. If we have ${\frac {1}{a^{1/n}}},$ we multiply the top and bottom by $a^{1-{\frac {1}{n}}}.$ This will make the exponent in the denominator 1.
4
Simplify as needed.^{[12]
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Research source}
- ${\frac {3}{3^{1/3}}}\cdot {\frac {3^{2/3}}{3^{2/3}}}=3^{2/3}$
- If you need to write it in radical form, factor out the $1/3.$ $1/3.$
  - $3^{2/3}=(3^{2})^{1/3}={\sqrt[{3}]{9}}$

Community Q&A

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Question

How do I rationalize with three terms?

Community Answer

Something like 1/(1+root2 + root3)? If so, group as 1+(root2 + root3) and multiply through by the "difference of squares conjugate" 1-(root2 + root3). That makes the denominator -4 - root6, which is still irrational, but did improve from two irrational terms to only one. So repeat the same trick by multiplying through by -4+root6 and the denominator is rationalized.

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In your pictures, what does the point mean?

Donagan

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If you're asking about the dots that are placed between various fractions, those are multiplication signs. For example, in the article's second image we see (7√3) / (2√7), then a dot, then (√7 / √7). That means we multiply the first fraction by the second fraction (numerator times numerator, and denominator times denominator), giving us (7√21) / 14, which simplifies to √21 / 2. (Incidentally, the article shows some other dots that are not between fractions. Those are merely "bullet points.")

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How can I rationalize the denominator with a cube root that has a variable?

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If it is a binomial expression, follow the steps outlined in method 2.

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