# 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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Answer

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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.

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 \[\pi ,\sqrt 5 ,\sqrt[3]{2}\] .

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.

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 \[\pi ,\sqrt 5 ,\sqrt[3]{2}\] .

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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