Simplifying Clear the root

How to Rationalize a Denominator

How to rationalize the denominator: multiply the top and bottom to clear the radical, so 1/√2 becomes √2/2, and use the conjugate on a two-term bottom.

By SimplifyCalc 8 min read

To rationalize a denominator, you multiply the top and bottom by whatever clears the radical, so 1/√2 1/sqrt(2) becomes √2/2 sqrt(2)/2 . The bottom √2 · √2 sqrt(2)*sqrt(2) turns into 2, a plain whole number, and the value has not changed because you multiplied by √2/√2 sqrt(2)/sqrt(2) , which is just one. That is the whole idea, and you can watch it happen on any example by pasting it into the simplify calculator and reading the steps it shows.

The move is always the same shape: find a factor that removes the root, multiply top and bottom by it, then tidy up. What changes is which factor does the job. This guide walks the three cases you will meet, from a single square root up to a two-term denominator, and each one leans on the same trick you just saw.

Why bother clearing the radical

The habit is partly historical. Before calculators, people divided by hand, and dividing by a whole number is far kinder than dividing by a decimal that runs on forever. Working out 1/√2 1/sqrt(2) meant dividing 1 by 1.41421356…, which is slow and error-prone. Rewriting it as √2/2 sqrt(2)/2 turns the job into 1.41421356… divided by 2, which anyone can do in a moment. The radical still exists, but it sits where the arithmetic is easy.

There is a second reason that still matters. A rationalized form is easier to compare and combine. Two fractions with whole-number denominators can be added straight away, and it is easier to see at a glance whether √2/2 sqrt(2)/2 and another value are equal when the roots sit on top. Standard form makes matching, adding, and checking faster.

Now the honest part. Many modern courses no longer require you to rationalize, because a calculator divides by a decimal just as happily as by a whole number. Some teachers still mark an un-rationalized denominator as not fully simplified; others do not care at all. Conventions differ by curriculum, so follow your own course and teacher. This guide shows you how to do it well when it is asked for, not a claim that it is always required.

The three cases, easiest first

A single square root

This is the base case, the one from the opening. When the denominator is a single square root, multiply top and bottom by that same root. A square root times itself gives the number inside, so √a · √a = a sqrt(a)*sqrt(a) , and the radical is gone from the bottom. For 1/√7 1/sqrt(7) you multiply by √7/√7 sqrt(7)/sqrt(7) and land on √7/7 sqrt(7)/7 .

One thing to watch: simplify the radical first if you can. A denominator like 2/√8 2/sqrt(8) hides an easier problem, because √8 = 2√2 sqrt(8) . Reducing before you rationalize saves you from clearing a bigger root than you need to.

A higher root

Cube roots and beyond need a little more care, because one copy no longer finishes the job. A cube root clears only when the inside reaches a perfect cube. For 1/∛3 1/nthroot(3,3) , multiplying by a single ∛3 nthroot(3,3) gives ∛9 nthroot(9,3) on the bottom, which is still a radical. You need the inside to reach ∛27 = 3 nthroot(27,3) , so you multiply by ∛9 nthroot(9,3) instead, supplying the two missing copies of 3 in one step. The rule generalizes: for an nth root, supply whatever brings the inside up to a perfect nth power.

If the variable version of this interests you, the same counting logic drives radicals with variables, where you track how many copies of each factor sit under the root.

A two-term denominator: the conjugate

When the denominator has two terms, such as 2 + √5 2+sqrt(5) , multiplying by a single root does not help, because the root only touches one of the two terms and a new radical appears from the other. The fix is the conjugate: the same two terms with the middle sign flipped, so the conjugate of 2 + √5 2+sqrt(5) is 2 − √5 2-sqrt(5) .

Multiply a two-term expression by its conjugate and you get a difference of squares, (a + √b)(a − √b) = a² − b (a+sqrt(b))(a-sqrt(b)) . The cross terms cancel and the square removes the radical. That difference-of-squares pattern is the same one the homepage already names in its rules reference, so rather than re-derive it here, borrow it: the conjugate works precisely because squaring a square root gives back a whole number, and the middle terms wipe out.

Conjugates also show up when you clean up a fraction stacked on another fraction, which is why the same skill carries over to complex fractions.

Three worked examples, easy to exam-level

Watch the single method hold as the denominator gets harder. Each step names the rule it uses.

Rationalizing, three levels

Warm-up a single square root
1/√2 1/sqrt(2)
  1. Multiply top and bottom by the same root, √2/√2 sqrt(2)/sqrt(2) , which equals one so the value is unchanged.
  2. Simplify the denominator using √a · √a = a sqrt(2)*sqrt(2) : here √2 · √2 = 2 sqrt(2)*sqrt(2) .
  3. Simplify the numerator: 1 · √2 = √2 1*sqrt(2) .
  4. Write the result: √2/2 sqrt(2)/2 .
Answer √2/2 sqrt(2)/2
Typical a higher root, more than one copy
5/∛3 5/nthroot(3,3)
  1. Decide what clears the cube root. One copy is not enough: to clear ∛3 nthroot(3,3) you need two more copies, which is ∛(3²) = ∛9 nthroot(9,3) , so the denominator can reach ∛(3³) = 3 nthroot(27,3) .
  2. Multiply top and bottom by ∛9 nthroot(9,3) , a form of one.
  3. Simplify the numerator: 5 · ∛9 = 5∛9 5*nthroot(9,3) .
  4. Simplify the denominator using the completed cube: ∛27 = 3 nthroot(27,3) . Result: 5∛9/3 5*nthroot(9,3)/3 .
Answer 5∛9/3 5*nthroot(9,3)/3
Exam-level a binomial, use the conjugate
3/(2 + √5) 3/(2+sqrt(5))
  1. Multiply top and bottom by the conjugate, (2 − √5)/(2 − √5) (2-sqrt(5))/(2-sqrt(5)) .
  2. Simplify the denominator with the difference of squares: (2 + √5)(2 − √5) = 2² − 5 = 4 − 5 = −1 (2+sqrt(5))(2-sqrt(5)) .
  3. Simplify the numerator: 3(2 − √5) = 6 − 3√5 3*(2-sqrt(5)) .
  4. Divide by −1: (6 − 3√5)/(−1) = 3√5 − 6 (6-3*sqrt(5))/(-1) . Result: 3√5 − 6 3*sqrt(5)-6 .
Answer 3√5 − 6 3*sqrt(5)-6

Where people slip

Almost every lost mark on this topic comes from one of four places. None are hard once you have seen them named.

One copy is not enough for a higher root

With a square root, a single copy clears it. With a cube root or higher, it does not. Multiplying 1/∛3 1/nthroot(3,3) by a single ∛3 nthroot(3,3) leaves ∛9 nthroot(9,3) on the bottom, still a radical. Count what the inside needs to reach a full power first, then supply exactly that. For a cube root, the inside has to become a perfect cube; for a fourth root, a perfect fourth power.

Flipping only one sign in the conjugate

The conjugate flips the middle sign and nothing else. The conjugate of 2 + √5 2+sqrt(5) is 2 − √5 2-sqrt(5) , not −2 − √5 -2-sqrt(5) and not −2 + √5 -2+sqrt(5) . Change the one sign between the terms, leave both terms themselves alone, and the difference of squares does the rest.

Forgetting to reduce at the end

Clearing the radical is not always the last step. The denominator often comes out as a number that shares a factor with the numerator, or as a negative you can absorb. In the exam-level example the bottom landed on −1, and dividing through was the real finish. Always glance back for a common factor once the root is gone.

Rationalizing before simplifying the radical

A denominator like 2/√8 2/sqrt(8) tempts you to multiply by √8 sqrt(8) straight away. Simplify first: √8 = 2√2 sqrt(8) , and the problem shrinks to 1/√2 1/sqrt(2) . Reducing the radical before you rationalize keeps the numbers small and the arithmetic clean.

How do you know you are finished?

Rationalizing has a short, exact finish line. You are done when both of these are true:

  • No radical remains in any denominator. Not on the whole fraction, not on a piece of it. If a root is still underneath, there is one more multiplication to do.
  • The fraction is reduced. After you clear the radical, the numerator and denominator often share a common factor. In the exam-level example, dividing by −1 was the reduction step; in others you might cancel a 2 or a 3. Simplest form means lowest terms.

A quick self-check: read the denominator out loud. If you say a root anywhere in it, you are not finished. If it is a whole number and nothing cancels, you are.

Practice: rationalizing denominators

Practice: rationalizing denominators

Try each one on paper first, then open the answers to check your working.

  1. Rationalize 1/√3 1/sqrt(3) .
  2. Rationalize 5/√5 5/sqrt(5) .
  3. Rationalize 2/√8 2/sqrt(8) .
  4. Rationalize 1/√x 1/sqrt(x) for x > 0 x>0 .
  5. Rationalize 1/∛4 1/nthroot(4,3) .
  6. Rationalize 4/(3 − √7) 4/(3-sqrt(7)) .
  7. Rationalize 1/(√x + √y) 1/(sqrt(x)+sqrt(y)) for x, y > 0 x>0, y>0 and x ≠ y x != y .
Show answers
  1. Multiply by √3/√3 sqrt(3)/sqrt(3) : 1/√3 = √3/3 sqrt(3)/3 .
  2. Multiply by √5/√5 sqrt(5)/sqrt(5) : the top becomes 5√5 5*sqrt(5) and the bottom 5, which reduces to √5 sqrt(5) .
  3. Simplify first: √8 = 2√2 sqrt(8) , so 2/(2√2) = 1/√2 = √2/2 sqrt(2)/2 .
  4. Multiply by √x/√x sqrt(x)/sqrt(x) : 1/√x = √x/x sqrt(x)/x .
  5. To reach ∛8 = 2 nthroot(8,3) , multiply by ∛2 nthroot(2,3) : 1/∛4 = ∛2/2 nthroot(2,3)/2 .
  6. Multiply by the conjugate (3 + √7) 3+sqrt(7) : the bottom is 9 − 7 = 2 9-7 , the top 4(3 + √7) = 12 + 4√7 4*(3+sqrt(7)) , which reduces to 6 + 2√7 6+2*sqrt(7) .
  7. Multiply by the conjugate (√x − √y) sqrt(x)-sqrt(y) : the bottom is x − y x-y , giving (√x − √y)/(x − y) (sqrt(x)-sqrt(y))/(x-y) .

Check what your course expects

Rationalizing is a formatting choice more than a mathematical necessity. The value of 1/√2 1/sqrt(2) and √2/2 sqrt(2)/2 is identical; only the writing differs, and a calculator handles either form without complaint. What varies is whether your marker wants the radical off the bottom. Some curricula insist on it as part of simplest form, some treat it as optional, and the reasons are more historical and stylistic than deep. The formal statement, if you want the reference, lives on Wikipedia’s rationalisation page.

So learn the three moves well, keep the simplify calculator handy to check a tricky conjugate, and let your own course tell you when the job is actually required.

The next time a root sits in a denominator, ask first whether your course wants it gone, then reach for the matching root, the extra copies, or the conjugate.

Frequently asked questions

How do you rationalize a denominator with one square root?

Multiply the top and bottom by that same square root. For 1/√2, multiply by √2/√2. The bottom becomes √2 · √2 = 2, and the answer is √2/2. The value is unchanged because √2/√2 equals one.

What is a conjugate and when do I use it?

A conjugate is the same two-term expression with the middle sign flipped, so the conjugate of 2 + √5 is 2 − √5. You use it when the denominator has two terms and one of them is a radical. Multiplying a pair of conjugates gives a difference of squares, (a + √b)(a − √b) = a² − b, which has no radical in it.

How do you rationalize a cube root in the denominator?

One copy is not enough. To clear ∛3 you need the inside to reach a full cube, so you multiply by ∛9, since 3 · 9 = 27 and ∛27 = 3. In general, for an nth root you supply whatever copies bring the inside up to a perfect nth power.

Do I still have to rationalize the denominator?

It depends on your course. The habit comes from a time before calculators, when dividing by a whole number was easier than dividing by a long decimal. Many modern classes no longer require it, while others still mark it as simplest form. Follow your own teacher or curriculum.

How do I know the denominator is fully rationalized?

No radical is left in any denominator, and the fraction is reduced to lowest terms. If a root still sits underneath, or the numerator and denominator share a common factor, there is one more step to do.

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