Rationalisation (mathematics) Facts for Kids

Root rationalisation is a way to make fractions look simpler by getting rid of square roots (or other roots) from the bottom part of the fraction. The bottom part of a fraction is called the denominator.

Imagine you have a fraction like $\frac{1}{\sqrt{2}}$ . It's often easier to work with this fraction if the square root is not in the denominator. Rationalisation helps us move that root to the top part (the numerator) or remove it completely from the fraction.

When the denominator has a single square root, like $\sqrt{5}$ , you multiply both the top and bottom of the fraction by that same square root. This works because multiplying a square root by itself makes the root disappear (for example, $\sqrt{5} \cdot \sqrt{5} = 5$ ).

If the denominator has a sum or difference involving a square root, like $2 + \sqrt{3}$ , you use a special trick. You multiply both the top and bottom by something called the "conjugate." The conjugate is the same expression but with the opposite sign in the middle (so for $2 + \sqrt{3}$ , the conjugate is $2 - \sqrt{3}$ ). This trick uses a math rule called the "difference of two squares" which helps the square roots disappear from the denominator.

Rationalising a Single Square Root or Cube Root
- Example 1: Square Root
- Example 2: Cube Root
Dealing with More Square Roots
- Example 1: Two Square Roots
- Example 2: Square Root with an Imaginary Number
See also

Rationalising a Single Square Root or Cube Root

To rationalise a fraction, you always multiply the numerator and the denominator by the same number or expression. This is like multiplying by 1, so it doesn't change the value of the fraction.

Example 1: Square Root

Let's look at this fraction:

\frac{10}{\sqrt{5}}

To get rid of the $\sqrt{5}$ in the denominator, we multiply both the top and bottom by $\sqrt{5}$ :

\frac{10}{\sqrt{5}} = \frac{10}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}

Now, we multiply the numerators together and the denominators together:

\frac{10 \cdot \sqrt{5}}{\sqrt{5} \cdot \sqrt{5}} = \frac{10\sqrt{5}}{(\sqrt{5})^2}

Since $(\sqrt{5})^2$ means $\sqrt{5}$ multiplied by itself, it equals 5.

\frac{10\sqrt{5}}{5}

Finally, we can simplify the numbers:

\frac{10\sqrt{5}}{5} = 2\sqrt{5}

The square root is now gone from the denominator!

Example 2: Cube Root

What if we have a cube root, like $\sqrt[3]{a}$ ? Let's try this fraction:

\frac{10}{\sqrt[3]{a}}

To make a cube root disappear, you need to multiply it by itself two more times. So, we multiply by $\sqrt[3]{a}^2$ (which is $\sqrt[3]{a} \cdot \sqrt[3]{a}$ ):

\frac{10}{\sqrt[3]{a}} = \frac{10}{\sqrt[3]{a}} \cdot \frac{\sqrt[3]{a}^2}{\sqrt[3]{a}^2}

Multiply the top and bottom:

\frac{10 \cdot \sqrt[3]{a}^2}{\sqrt[3]{a} \cdot \sqrt[3]{a}^2} = \frac{10\sqrt[3]{a}^2}{\sqrt[3]{a}^3}

Since $\sqrt[3]{a}^3$ means $\sqrt[3]{a}$ multiplied by itself three times, it equals a.

\frac{10\sqrt[3]{a}^2}{a}

The cube root is now gone from the denominator!

Dealing with More Square Roots

When the denominator has two terms, one or both of which are square roots, we use a special trick with something called a "conjugate."

If you have an expression like $\sqrt{2} + \sqrt{3}$ , its conjugate is $\sqrt{2} - \sqrt{3}$ . The only difference is the sign in the middle.

When you multiply an expression by its conjugate, you use the "difference of two squares" rule: $(x+y)(x-y) = x^2 - y^2$ . This rule is very helpful because it makes the square roots disappear!

Example 1: Two Square Roots

Let's rationalise this fraction:

\frac{3}{\sqrt{3}+\sqrt{5}}

The denominator is $\sqrt{3}+\sqrt{5}$ . Its conjugate is $\sqrt{3}-\sqrt{5}$ . We multiply both the top and bottom by the conjugate:

\frac{3}{\sqrt{3}+\sqrt{5}} \cdot \frac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}}

Now, multiply the numerators and the denominators:

\frac{3(\sqrt{3}-\sqrt{5})}{(\sqrt{3}+\sqrt{5})(\sqrt{3}-\sqrt{5})}

For the denominator, we use the difference of two squares rule: $(\sqrt{3})^2 - (\sqrt{5})^2$ .

\frac{3(\sqrt{3}-\sqrt{5})}{3 - 5}

Simplify the denominator:

\frac{3(\sqrt{3}-\sqrt{5})}{-2}

This is the rationalised form of the fraction.

Example 2: Square Root with an Imaginary Number

This process also works with complex numbers that involve $i$ , where $i = \sqrt{-1}$ . Let's rationalise this fraction:

\frac{7}{1+\sqrt{-5}}

Remember that $\sqrt{-5}$ can be written as $\sqrt{5}i$ . So the denominator is $1+\sqrt{5}i$ . The conjugate of $1+\sqrt{-5}$ is $1-\sqrt{-5}$ . We multiply both the top and bottom by the conjugate: