# 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) $2 \pi$

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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.

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