Rationalization

We know that irrational numbers are those which can’t be expressed in the ‘p/q’ form where ‘p’ and ‘q’ are integers. But these rational numbers can be used in rational fractions as either numerator or denominator. When these numbers are present in numerators of fractions, calculations can be done. But when these exist in denominators of fractions, they make calculations more difficult and complicated. To avoid such complications in the numeric calculations, we use method of rationalization. Hence, rationalization can be defined as the process by which we eliminate radicals present in the denominators of fractions. 


To understand the concept in a better let us have a look at below solved examples based on rationalization:


1. Rationalization by multiplication of both numerator and denominator by a root:

(i) Rationalize \(\frac{1}{\sqrt{2}}\).


Solution: 

Since \(\sqrt{2}\) is a irrational number and is present in denominator of the fraction. So, we first need to rationalize it. This can be done by multiplying both numerator and denominator by \(\sqrt{2}\). So,

\(\frac{1}{\sqrt{2}}\)\(\times\) \(\frac{\sqrt{2}}{\sqrt{2}}\)

⟹ \(\frac{\sqrt{2}}{2}\)


(ii) Rationalize \(\frac{1}{\sqrt{5}}\).

Solution: 

Since \(\sqrt{5}\) is a irrational number and is present in denominator of the fraction. So, we first need to rationalize it. This can be done by multiplying both numerator and denominator by \(\sqrt{5}\). So,

\(\frac{1}{\sqrt{5}}\)\(\times\) \(\frac{\sqrt{5}}{\sqrt{5}}\)

⟹ \(\frac{\sqrt{5}}{5}\)


(iii) Rationalize \(\frac{1}{\sqrt{11}}\).

Solution: 

Since \(\sqrt{11}\) is a irrational number and is present in denominator of the fraction. So, we first need to rationalize it. This can be done by multiplying both numerator and denominator by \(\sqrt{11}\). So,

\(\frac{1}{\sqrt{11}}\)\(\times\)\(\frac{\sqrt{11}}{\sqrt{11}}\)

⟹ \(\frac{\sqrt{11}}{11}\)


2. Rationalization by multiplication with conjugate.

In the fractions that have irrational numbers in the form of addition or subtraction in the denominators of fraction, we use the method of multiplication with conjugate for rationalizing the fraction and making the problem a simplified.

We have, (x + \(\sqrt{y}\))(x - \(\sqrt{y}\)) = x\(^{2}\) - \((\sqrt{y})^{2}\) = (x\(^{2}\) - y) which is a rational number.

Thus, by multiplying the irrational number (x + \(\sqrt{y}\)) by the irrational number (x - \(\sqrt{y}\)) we get a rational numebr. Here, (x - \(\sqrt{y}\))  is the rationalising factor of (x + \(\sqrt{y}\)). Similarly, (x + \(\sqrt{y}\)) is the rationalising factor of (x - \(\sqrt{y}\)).

The irrational number (x - \(\sqrt{y}\)) is also called the conjugate irrational number, or conjugate, of (x + \(\sqrt{y}\)). Similarly, (x + \(\sqrt{y}\)) is the conjugate of (x - \(\sqrt{y}\)).

For example:

The conjugate of (5 + \(\sqrt{7}\)) is (5 - \(\sqrt{7}\))

The conjugate of (5 - \(\sqrt{7}\)) is (5 + \(\sqrt{7}\))

The conjugate of (10 + \(\sqrt{3}\)) is (10 - \(\sqrt{3}\))

The conjugate of (10 - \(\sqrt{3}\)) is (10 + \(\sqrt{3}\))


Below given are the examples on rationalizing the fractions by multiplying with conjugate:

(i) Rationalize \(\frac{1}{4 + \sqrt{2}}\).

Solution: 

Since, the given problem has irrational term in the denominator with addition and subtraction format. So we need to rationalize using the method of multiplication by conjugate. So,

\(\frac{1}{4 + \sqrt{2}}\)  \(\times\) \(\frac{4 - \sqrt{2}}{4 - \sqrt{2}}\)

⟹ \(\frac{4 - \sqrt{2}}{4^{2} - \sqrt{2^{2}}}\), [Since, (a + b)(a - b) = a\(^{2}\) - b\(^{2}\)}]

⟹ \(\frac{4 - \sqrt{2}}{16 - 2}\)

⟹ \(\frac{4 - \sqrt{2}}{14}\)

So, the required rationalized number is: 

\(\frac{4 - \sqrt{2}}{14}\)


(ii) Rationalize\(\frac{1}{3 - \sqrt{5}}\).

Solution:

Since, the given problem has irrational term in the denominator with addition and subtraction format. So we need to rationalize using the method of multiplication by conjugate. So,

\(\frac{1}{3 - \sqrt{5}}\) \(\times\)  \(\frac{3 + \sqrt{5}}{3 + \sqrt{5}}\)

⟹ \(\frac{3 + \sqrt{5}}{3^{2} - \sqrt{5^{2}}}\), [Since, (a + b)(a - b) = a\(^{2}\) - b\(^{2}\)}]

⟹ \(\frac{3 + \sqrt{5}}{9-5}\)

⟹ \(\frac{3 + \sqrt{5}}{4}\)

⟹ So, the required rationalized number is \(\frac{3 + \sqrt{5}}{4}\)


Irrational Numbers

Definition of Irrational Numbers

Representation of Irrational Numbers on The Number Line

Comparison between Two Irrational Numbers

Comparison between Rational and Irrational Numbers

Rationalization

Problems on Irrational Numbers

Problems on Rationalizing the Denominator

Worksheet on Irrational Numbers






9th Grade Math

From Rationalization to HOME PAGE


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