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# 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). $2\pi$  Answer Verified
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.

Complete step-by-step answer:
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.
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Rational Numbers  Rational Numbers on a Number line  Rational Numbers Between Two Rational Numbers  Rational Numbers and Their Properties  Rational and Irrational Numbers  Operations on Rational Numbers  Decimal Expansion of Rational Numbers  Difference Between Rational and Irrational Numbers  CBSE Class 7 Maths Chapter 9 - Rational Numbers Formulas  CBSE Class 8 Maths Chapter 1 - Rational Numbers Formulas  