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

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Question

Classify the following numbers as rational or irrational:

\(\begin{align}&{\rm{(i}}) \quad 2 - \sqrt 5 \qquad \rm{(ii)}\quad (3 + \sqrt {23} ) - \sqrt {23} \qquad \rm{(iii)}\quad \frac{{2\sqrt 7 }}{{7\sqrt 7 }}\\&\rm{(iv)}\quad \frac{1}{{\sqrt 2 }} \qquad \;\;\rm{(v)}\quad 2\pi \end{align}\)

 

 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.

  
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