# CBSE 10th Mathematics | Irrational Numbers and Solved Examples

## IRRATIONAL NUMBERS

All real numbers which are not rational are called irrational numbers. $\displaystyle \sqrt{2}$, $\displaystyle \sqrt[3]{3}$, $\displaystyle -\sqrt{5}$ are some examples of irrational numbers.

There are decimals which are non-terminating and non-recurring decimal.

Example: 0.303003000300003…

Hence, we can conclude that

An irrational number is a non-terminating and non-recurring decimal and cannot be put in the form $\displaystyle \frac{p}{q}$ where p and q are both co-prime integers and q ¹ 0.

## Solved Examples Based on Irrational Numbers

### Question:

Prove that $\displaystyle \sqrt{2}$ is not a rational number.

### Solution:

Let $\displaystyle \sqrt{2}$ is a rational number

\    $\displaystyle \sqrt{2}=\frac{p}{q}$    [p and q are co-prime and q ¹ 0]

Squaring both sides

$\displaystyle 2=\frac{{{p}^{2}}}{{{q}^{2}}}$

Þ    $\displaystyle 2{{q}^{2}}={{p}^{2}}$

Þ    $\displaystyle {{p}^{2}}$ is even or p is even.

Let p = 2r

Þ    $\displaystyle 2{{q}^{2}}={{(2r)}^{2}}=4{{r}^{2}}$

Þ    $\displaystyle {{q}^{2}}=2{{r}^{2}}$

Þ    $\displaystyle {{q}^{2}}$ is even so q is even.

Hence, p and q have 2 as a common factor or p and q are not co-prime.

So, our supposition is wrong.

\    $\displaystyle \sqrt{2}$ is not a rational number.

### Question:

Prove that $\displaystyle \sqrt{3}+\sqrt{5}$ is an irrational number.

### Solution:

Suppose $\displaystyle \sqrt{3}+\sqrt{5}$ is a rational number and can be taken as $\displaystyle \frac{a}{b}$, b ¹ 0 and a, b are co-prime.

Þ    $\displaystyle \sqrt{3}+\sqrt{5}=\frac{a}{b}$            [Rational]

Squaring both sides

$\displaystyle 3+5+2\sqrt{3}\sqrt{5}=\frac{{{a}^{2}}}{{{b}^{2}}}$

Þ    $\displaystyle 8+2\sqrt{15}=\frac{{{a}^{2}}}{{{b}^{2}}}$

Þ    $\displaystyle \sqrt{15}=\frac{{{a}^{2}}-8{{b}^{2}}}{2{{b}^{2}}}$

LHS is $\displaystyle \sqrt{15}$ which is irrational while RHS is rational.

So, our supposition is wrong.

Hence, $\displaystyle \sqrt{3}+\sqrt{5}$ is not a rational number.

### Question:

Show that there is no positive integer n for which $\displaystyle \sqrt{n-1}+\sqrt{n+1}$ is rational and can be expressed in the form $\displaystyle \frac{a}{b}$, b ¹ 0 and a & b are co-prime.

### Solution:

Let there be a positive integer n for which $\displaystyle \sqrt{n-1}+\sqrt{n+1}$ is rational.

Which means     $\displaystyle \sqrt{n-1}+\sqrt{n+1}=\frac{a}{b}$         … (i)

or    $\displaystyle \frac{1}{\sqrt{n-1}+\sqrt{n+1}}=\frac{b}{a}$

Rationalizing LHS, we get

Þ    $\displaystyle \frac{1}{\sqrt{n-1}+\sqrt{n+1}}\times \frac{\sqrt{n-1}-\sqrt{n+1}}{\sqrt{n-1}-\sqrt{n+1}}=\frac{b}{a}$

Þ    $\displaystyle \frac{\sqrt{n-1}-\sqrt{n+1}}{n-1-n-1}=\frac{b}{a}$

Þ    $\displaystyle \frac{\sqrt{n+1}-\sqrt{n-1}}{2}=\frac{b}{a}$

or    $\displaystyle \sqrt{n+1}-\sqrt{n-1}=\frac{2b}{a}$        …(ii)

Adding and subtracting (i) and (ii) we get

$\displaystyle 2\sqrt{n+1}=\frac{a}{b}+\frac{2b}{a}$ and $\displaystyle 2\sqrt{n-1}=\frac{a}{b}-\frac{2b}{a}$

Þ    $\displaystyle \sqrt{n+1}=\frac{{{a}^{2}}+2{{b}^{2}}}{2ab}$ and $\displaystyle \sqrt{n-1}=\frac{{{a}^{2}}-2{{b}^{2}}}{2ab}$

Þ    RHS of both are rational.

\    $\displaystyle \sqrt{n+1}$ and $\displaystyle \sqrt{n-1}$ are also rational.

Þ    $\displaystyle (n+1)$ and $\displaystyle (n-1)$ are perfect squares of positive integers.

This is impossible as any two perfect squares differ at least by 3.

Hence, there is no positive integer n for which ($\displaystyle \sqrt{n-1}+\sqrt{n+1}$) is rational.

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