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

Last updated date: 25th Mar 2023
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Hint: Simplify the numbers and check if they can be represented in $\dfrac{p}{q}$ form, where p and q are integers and $q \ne 0$ . If they can be represented, then they belong to rational numbers, if not, then they are irrational numbers.

A number that is in the form $\dfrac{p}{q}$ or can be simplified to the form $\dfrac{p}{q}$ where p and q are integers and $q \ne 0$ is called a rational number. They can also be represented in the decimal form as terminating decimals or non-terminating recurring decimals. Examples include $22,\dfrac{5}{7},\dfrac{1}{2}$ .
A real number that can’t be represented in the form $\dfrac{p}{q}$ where both p and q are integers and $q \ne 0$ is called an irrational number. These numbers are non-terminating and non-recurring decimals.
Examples include $\pi ,\sqrt 5 ,\sqrt{2}$ .
Now, having the knowledge of rational and irrational numbers, we can classify the numbers into these categories.
(i). The number $2 - \sqrt 5$ , is the difference between 2 and $\sqrt 5$ .
2 is a rational number because it can be represented as $\dfrac{2}{1}$ where 2 and 1 are integers.
$\sqrt 5$ is an irrational number because it can’t be represented in the form of rational numbers.
We know that the sum or difference of a rational and an irrational number is an irrational number.
Hence, $2 - \sqrt 5$ is irrational.
(ii). The number $(3 + \sqrt {23} ) - \sqrt {23}$ can be simplified as follows:
$(3 + \sqrt {23} ) - \sqrt {23} = 3 + \sqrt {23} - \sqrt {23}$
Cancelling $\sqrt {23}$ , we have:
$(3 + \sqrt {23} ) - \sqrt {23} = 3$
We know that 3 is a rational number since it can be represented as $\dfrac{3}{1}$ where 3 and 1 are integers.
Hence, $(3 + \sqrt {23} ) - \sqrt {23}$ is rational.
(iii). The number $\dfrac{{2\sqrt 7 }}{{7\sqrt 7 }}$ can be simplified by cancelling $\sqrt 7$ as follows:
$\dfrac{{2\sqrt 7 }}{{7\sqrt 7 }} = \dfrac{2}{7}$
We can see that $\dfrac{2}{7}$ is in the rational form since 2 and 7 are integers.
Hence, $\dfrac{{2\sqrt 7 }}{{7\sqrt 7 }}$ is rational.
(iv). We can simplify the number $\dfrac{1}{{\sqrt 2 }}$ by multiplying numerator and denominator by $\sqrt 2$ .
$\dfrac{1}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}$
$\dfrac{1}{{\sqrt 2 }} = \dfrac{{\sqrt 2 }}{2}$
We know that $\sqrt 2$ is irrational and 2 is rational.
Division of an irrational number by a rational number, results in an irrational number.
Hence, $\dfrac{1}{{\sqrt 2 }}$ is irrational.
(v). In the number $2\pi$ , 2 is rational and $\pi$ is irrational.
Multiplication of a rational and an irrational number is an irrational number.
Hence, $2\pi$ is irrational.

Note: You might conclude that options (ii) and (iii) are irrational numbers because they contain the square root terms $\sqrt {23}$ and $\sqrt 7$ respectively, but it is wrong. Simplify the number completely and then check for the $\dfrac{p}{q}$ form.