In this chapter we will learn to rationalize irrational number which is present in denominator of the fraction.

## Irrational number in denominator

We know that irrational number cannot be represented in the form of fraction p / q.

But** there are irrational numbers which are present in the denominator of the fraction**.

Since the value of irrational number are non repeating and non terminating in nature, it gets very difficult to find the exact value of fraction.**For example,**

Consider the number \mathtt{\ \frac{1}{\sqrt{2}}}

Here \mathtt{\sqrt{2}} is an irrational number whose value is given as;

\mathtt{\sqrt{2} \ =\ 1.41428\ .\ .\ .\ }

So the fraction \mathtt{\ \frac{1}{\sqrt{2}}} can be represented as;

⟹ \mathtt{\frac{1}{\mathtt{\ 1.41428\ .\ .}}}

Imagine how difficult is to find the exact value of above fraction.

So, **in order to simplify the fraction with irrational number**, **we will use the rationalization technique**.

In simple words, using rationalization, we are trying to bring the irrational number in the numerator for better calculation.

## Rationalization of Irrational number

Consider the fraction with irrational number \mathtt{\frac{1}{\sqrt{2}}}

To simplify the fraction, simply multiply numerator and denominator by same irrational number and do the further calculation.

\mathtt{\Longrightarrow \ \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{\sqrt{2}}{2}}

Hence, the fraction \mathtt{\frac{1}{\sqrt{2}}} is rationalized to \mathtt{\frac{\sqrt{2}}{2}}

I hope you understood the above process. Given below are some examples for better clarity

**Example 01**

Simplify the fraction \mathtt{\frac{1}{\sqrt{13}}} **Solution**

Multiply numerator and denominator by \mathtt{\sqrt{13}}

\mathtt{\Longrightarrow \ \frac{1}{\sqrt{13}} \times \frac{\sqrt{13}}{\sqrt{13}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{\sqrt{13}}{\left(\sqrt{13}\right)^{2}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{\sqrt{13}}{13}}

Hence, the fraction is simplified to \mathtt{\frac{\sqrt{13}}{13}}

**Example 02**

Simplify the fraction \mathtt{\frac{1}{\sqrt{55}}}

**Solution**

Multiply numerator and denominator by \mathtt{\sqrt{55}}

\mathtt{\Longrightarrow \ \frac{1}{\sqrt{55}} \times \frac{\sqrt{55}}{\sqrt{55}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{\sqrt{55}}{\left(\sqrt{55}\right)^{2}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{\sqrt{55}}{55}}

Hence, \mathtt{\frac{\sqrt{55}}{55}} is the required fraction.

## Rationalizing irrational number in subtraction / addition form

In some type of fractions, **denominator contains irrational number added / subtracted with other number**.

Common examples of such fractions are \mathtt{\frac{1}{a+\sqrt{x}} \ \ or\ \frac{1}{a-\sqrt{x}}} .

To simplify such fractions, **we have to multiply the given number with its conjugate**.

\mathtt{a+\sqrt{x}} conjugate is \mathtt{a-\sqrt{x}} .

\mathtt{a-\sqrt{x}} conjugate is \mathtt{a+\sqrt{x}} .

Given below are some examples for further clarity.**Example 01**

Simplify \mathtt{\frac{1}{2+\sqrt{5}}}

**Solution**

Conjugate of \mathtt{2+\sqrt{5}} is \mathtt{2-\sqrt{5}} .

Multiply numerator and denominator by conjugate.

\mathtt{\Longrightarrow \ \frac{1}{2+\sqrt{5}} \times \frac{2-\sqrt{5}}{2-\sqrt{5}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{2-\sqrt{5}}{\left( 2+\sqrt{5}\right)\left( 2-\sqrt{5}\right)}}

Referring the formula;

\mathtt{( a+b)( a-b) =a^{2} -b^{2}}

Using the formula, we get;

\mathtt{\Longrightarrow \ \frac{2-\sqrt{5}}{2^{2} -\left(\sqrt{5}\right)^{2}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{2-\sqrt{5}}{4-5}}\\\ \\ \mathtt{\Longrightarrow \ \frac{2-\sqrt{5}}{-1}}\\\ \\ \mathtt{\Longrightarrow \ -2+\sqrt{5}}

Hence, the irrational number is transferred from denominator to numerator.

**Example 02**

Rationalize \mathtt{\frac{1}{6-\sqrt{11}}}

**Solution**

The conjugate of \mathtt{6-\sqrt{11}} is \mathtt{6+\sqrt{11}} .

Multiply numerator and denominator by conjugate.

\mathtt{\Longrightarrow \ \frac{1}{6-\sqrt{11}} \times \frac{6+\sqrt{11}}{6+\sqrt{11}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{6+\sqrt{11}}{\left( 6-\sqrt{11}\right)\left( 6+\sqrt{11}\right)}}

Referring the formula;

\mathtt{( a+b)( a-b) =a^{2} -b^{2}}

Using the formula;

\mathtt{\Longrightarrow \ \frac{6+\sqrt{11}}{6^{2} -\left(\sqrt{11}\right)^{2}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{6+\sqrt{11}}{36-11}}\\\ \\ \mathtt{\Longrightarrow \ \frac{6+\sqrt{11}}{25}}

Hence, the fraction is simplified to \mathtt{\frac{6+\sqrt{11}}{25}}

**Example 03**

Rationalize \mathtt{\frac{1}{\sqrt{2} +\sqrt{3}}}

**Solution**

Conjugate of \mathtt{\sqrt{2} +\sqrt{3}} is \mathtt{\sqrt{2} -\sqrt{3}}

Multiplying numerator and denominator by conjugate.

\mathtt{\Longrightarrow \ \frac{1}{\sqrt{2} +\sqrt{3}} \times \frac{\sqrt{2} -\sqrt{3}}{\sqrt{2} -\sqrt{3}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{\sqrt{2} -\sqrt{3}}{\left(\sqrt{2} +\sqrt{3}\right)\left(\sqrt{2} -\sqrt{3}\right)}}

Referring the formula;

\mathtt{( a+b)( a-b) =a^{2} -b^{2}}

Using the formula, we get;

\mathtt{\Longrightarrow \ \frac{\sqrt{2} -\sqrt{3}}{\left(\sqrt{2}\right)^{2} -\left(\sqrt{3}\right)^{2}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{\sqrt{2} -\sqrt{3}}{2-3}}\\\ \\ \mathtt{\Longrightarrow \ \frac{\sqrt{2} -\sqrt{3}}{-1}}\\\ \\ \mathtt{\Longrightarrow \ \sqrt{3} -\sqrt{2}}

Hence, the fraction is rationalized to \mathtt{\sqrt{3} -\sqrt{2}}

**Next chapter** : **Questions on irrational numbers**