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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 $
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.