In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP!

# Ex.1.5 Q1 Number System Solution - NCERT Maths Class 9

Go back to  'Ex.1.5'

## Question

Classify the following numbers as rational or irrational:

(i) \begin{align} \quad 2 - \sqrt 5 \end{align}

(ii) \begin{align}(3 + \sqrt {23} ) - \sqrt {23} \end{align}

(iii) \begin{align}\frac{{2\sqrt 7 }}{{7\sqrt 7 }} \end{align}

(iv) \begin{align}\frac{1}{{\sqrt 2 }} \end{align}

(v) \begin{align}2\pi \end{align}

Video Solution
Number Systems
Ex 1.5 | Question 1

## Text Solution

\begin{align}({\rm{i}})\,\,\,\,2 - \sqrt 5 \end{align}

Steps:

The sum or difference of a rational number and an irrational number is always irrational.

Here $$2$$ is a rational number and \begin{align} \sqrt{5} \end{align} is an irrational number. Hence \begin{align} 2 -\sqrt{5} \end{align} is an irrational number.

\begin{align}\rm{(ii)}\quad(3 + \sqrt {23} ) - \sqrt {23} \end{align}

Steps:

By simplifying we get only $$3$$.

\begin{align}3 = \frac{3}{1},\end{align} which is in the form of \begin{align}\frac{p}{q},\end{align}

Hence is \begin{align}(3 + \sqrt {23} ) - \sqrt {23} \end{align} a rational number.

\begin{align}\rm{(iii)}\quad \frac{{2\sqrt 7 }}{{7\sqrt 7 }}\end{align}

Steps:

\begin{align}\frac{{2\sqrt 7 }}{{7\sqrt 7 }} = \frac{2}{7},\end{align} which is in the form of \begin{align}\frac{p}{q},\end{align}

Hence \begin{align}\frac{{2\sqrt 7 }}{{7\sqrt 7 }}\end{align} is a rational number.

\begin{align}\rm{(iv)}\quad \frac{1}{{\sqrt 2 }}\end{align}

Steps:

\begin{align} \frac{1}{\sqrt{2}} &=\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\ &=\frac{\sqrt{2}}{2} \\ &=\frac{1.414 \ldots}{2} \end{align}

\begin{align}= 0.702……\end{align} is a non - terminating, non-recurring decimal  and therefore it is irrational. Hence \begin{align}\,\frac{1}{{\sqrt 2 }}\,\end{align} is an irrational number.

\begin{align}\rm{(v)}\quad 2{\rm{\pi }}\end{align}

Steps:

\begin{align}~2\text{ }\!\!\pi\!\!\text{ }=2\times 3.1415...\end{align}

$$\pi$$ is an irrational number whose value is non-terminating and non-recurring.

$$2$$ is a rational number.

Product of a non-zero rational number and irrational number is an irrational number.

Hence \begin{align}2\pi\end{align} is irrational.

Video Solution
Number Systems
Ex 1.5 | Question 1

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school