Question

The process of changing irrational denominator into rational denominator is called:
A. Rationalisation
B. Irrationalisation
C. Conjugate
D. Radicands

Answer 1

Hint: In any fraction, if the denominator is irrational, we will directly multiply the denominator and the numerator of the fraction by the conjugate of the irrational denominator. Then we will apply the algebraic property, $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$, to simplify and get the fraction with a rational denominator.

Complete step-by-step solution
Let us consider any fraction with an irrational denominator.
$p = \dfrac{1}{{4 - \sqrt 2 }}$

Here, in this expression for p we see that in the denominator there is $\sqrt 2 $ , which is an irrational number and hence it is making the denominator irrational.

In order to find the value of the above expression where the denominator is rational, we will multiply the denominator and the numerator by the conjugate of the denominator, $4 - \sqrt 2 $ . The conjugate of $4 - \sqrt 2 $ is $4 + \sqrt 2 $.

\[
   \Rightarrow p = \dfrac{{1 \times \left( {4 + \sqrt 2 } \right)}}{{\left( {4 - \sqrt 2 } \right) \times \left( {4 + \sqrt 2 } \right)}} \\
   \Rightarrow p = \dfrac{{4 + \sqrt 2 }}{{\left( {4 - \sqrt 2 } \right)\left( {4 + \sqrt 2 } \right)}} \\
\]

Using the algebraic property, $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$, we get;
\[
  p = \dfrac{{4 + \sqrt 2 }}{{{4^2} - {{\left( {\sqrt 2 } \right)}^2}}} \\
   \Rightarrow p = \dfrac{{4 + \sqrt 2 }}{{16 - 2}} \\
   \Rightarrow p = \dfrac{{4 + \sqrt 2 }}{{14}} \\
\]

Thus, as 14 is a rational number hence an irrational denominator has been converted to a rational one.

This process of changing irrational denominator to a rational denominator is known as rationalization.

Hence, option (A) is the correct option.

Note: While doing the conversion of the denominator from irrational to rational, make sure you properly determine the conjugate of the irrational denominator, as it is the only crucial point in the calculation. Also, this is done just to bring the denominator in the form where the algebraic property, $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ can be applied. If you are not getting this form in the denominator, this means that the conjugate you determined was wrong.