Question

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 1

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.