Classify the following numbers as rational or irrational:
(i) $ 2-\sqrt{5} $
(ii) $ (3+\sqrt{23})-\sqrt{23} $
(iii) $ \frac{2 \sqrt{7}}{7 \sqrt{7}} $
(iv) $ \frac{1}{\sqrt{2}} $
(v) $ 2 \pi $

AcademicMathematicsNCERTClass 9

To do:

We have to classify the given numbers as rational or irrational.

Solution:

 (i) We know that,

$\sqrt{5}=2.236067..........$

The decimal expansion of \( \sqrt{5} \) is non-terminating and non-recurring.

Therefore, \( 2-\sqrt{5} \) is an irrational number.

(ii) $(3+\sqrt{23})-\sqrt{23}=3+\sqrt{23}-\sqrt{23}$

$=3$

$=\frac{3}{1}$

The number $\frac{3}{1}$ is in $\frac{p}{q}$ form.

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

(iii) \( \frac{2 \sqrt{7}}{7 \sqrt{7}}=\frac{2}{7} \)

The number $\frac{2}{7}$ is in $\frac{p}{q}$ form.

Hence, \( \frac{2 \sqrt{7}}{7 \sqrt{7}} \) is a rational number.

(iv) \( \frac{1}{\sqrt{2}} \)

Rationalising \( \frac{1}{\sqrt{2}} \), we get,

$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\times\frac{\sqrt2}{\sqrt{2}}$

$=\frac{1\times\sqrt2}{\sqrt{2}\sqrt2}$

$=\frac{\sqrt{2}}{2}$

$\sqrt{2}=1.4142135..........$

The decimal expansion of \( \sqrt{2} \) is non-terminating and non-recurring.

Therefore, \( \frac{1}{\sqrt{2}} \) is an irrational number.

(v) \( 2 \pi \)

$\pi=3.1415........$

This implies,

$2\pi=2\times(3.1415........)$

$=6.2830.........$

The number $6.2830.........$ is non-terminating non-repeating (non-recurring).

Therefore, \( 2 \pi \) is an irrational number.

raja
Updated on 10-Oct-2022 13:38:51

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