Classify the following numbers as rational or irrational with justification:
$(i)$ $\sqrt{\frac{9}{27}}$
$(ii)$ $\frac{\sqrt{28}}{\sqrt{343}}$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) $(i)$ $\sqrt{\frac{9}{27}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$. Since $\sqrt{3}$ is an irrational number,the quotient of a rational number $(1)$ and an irrational number $(\sqrt{3})$ is an irrational number.
$(ii)$ $\frac{\sqrt{28}}{\sqrt{343}} = \sqrt{\frac{28}{343}} = \sqrt{\frac{4 \times 7}{49 \times 7}} = \sqrt{\frac{4}{49}} = \frac{2}{7}$. Since $\frac{2}{7}$ is in the form $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$,it is a rational number.

Explore More

Similar Questions

If $(\frac{2}{5})^{5} \times (\frac{25}{4})^{3} = (\frac{5}{2})^{3x-2}$,then find $x$.

Find three rational numbers between $0.1$ and $0.11$.

Simplify the following:
$\frac{\sqrt{24}}{8} + \frac{\sqrt{54}}{9}$

For each question,select the proper option from four options given,to make the statement true: The decimal expansion of $-\frac{11}{4}$ is..........

If $x = 3 + 2\sqrt{2}$,then find the value of $x^{2} + \frac{1}{x^{2}}$ and $x^{3} + \frac{1}{x^{3}}$.

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo