CBSE 10th Mathematics | Irrational Numbers and Solved Examples

 

IRRATIONAL NUMBERS    

 

All real numbers which are not rational are called irrational numbers. \displaystyle \sqrt{2}, \displaystyle \sqrt[3]{3}, \displaystyle -\sqrt{5} are some examples of irrational numbers.

    There are decimals which are non-terminating and non-recurring decimal.

    Example: 0.303003000300003…

    Hence, we can conclude that

    An irrational number is a non-terminating and non-recurring decimal and cannot be put in the form \displaystyle \frac{p}{q} where p and q are both co-prime integers and q ¹ 0.

 

Solved Examples Based on Irrational Numbers

 

Question:

 

Prove that \displaystyle \sqrt{2} is not a rational number.

 

Solution:

 

Let \displaystyle \sqrt{2} is a rational number

        \    \displaystyle \sqrt{2}=\frac{p}{q}    [p and q are co-prime and q ¹ 0]    

        Squaring both sides

            \displaystyle 2=\frac{{{p}^{2}}}{{{q}^{2}}}

        Þ    \displaystyle 2{{q}^{2}}={{p}^{2}}

        Þ    \displaystyle {{p}^{2}} is even or p is even.

        Let p = 2r

        Þ    \displaystyle 2{{q}^{2}}={{(2r)}^{2}}=4{{r}^{2}}

        Þ    \displaystyle {{q}^{2}}=2{{r}^{2}}

        Þ    \displaystyle {{q}^{2}} is even so q is even.

        Hence, p and q have 2 as a common factor or p and q are not co-prime.

        So, our supposition is wrong.

        \    \displaystyle \sqrt{2} is not a rational number.

 

Question:

 

Prove that \displaystyle \sqrt{3}+\sqrt{5} is an irrational number.

 

Solution:

 

Suppose \displaystyle \sqrt{3}+\sqrt{5} is a rational number and can be taken as \displaystyle \frac{a}{b}, b ¹ 0 and a, b are co-prime.

        Þ    \displaystyle \sqrt{3}+\sqrt{5}=\frac{a}{b}            [Rational]             

        Squaring both sides

            \displaystyle 3+5+2\sqrt{3}\sqrt{5}=\frac{{{a}^{2}}}{{{b}^{2}}}

        Þ    \displaystyle 8+2\sqrt{15}=\frac{{{a}^{2}}}{{{b}^{2}}}

        Þ    \displaystyle \sqrt{15}=\frac{{{a}^{2}}-8{{b}^{2}}}{2{{b}^{2}}}

        LHS is \displaystyle \sqrt{15} which is irrational while RHS is rational.

        So, our supposition is wrong.

        Hence, \displaystyle \sqrt{3}+\sqrt{5} is not a rational number.

 

Question:

 

Show that there is no positive integer n for which \displaystyle \sqrt{n-1}+\sqrt{n+1} is rational and can be expressed in the form \displaystyle \frac{a}{b}, b ¹ 0 and a & b are co-prime.

 

Solution:

    

Let there be a positive integer n for which \displaystyle \sqrt{n-1}+\sqrt{n+1} is rational.

        Which means     \displaystyle \sqrt{n-1}+\sqrt{n+1}=\frac{a}{b}         … (i)

        or    \displaystyle \frac{1}{\sqrt{n-1}+\sqrt{n+1}}=\frac{b}{a}         

        Rationalizing LHS, we get

        Þ    \displaystyle \frac{1}{\sqrt{n-1}+\sqrt{n+1}}\times \frac{\sqrt{n-1}-\sqrt{n+1}}{\sqrt{n-1}-\sqrt{n+1}}=\frac{b}{a}

        Þ    \displaystyle \frac{\sqrt{n-1}-\sqrt{n+1}}{n-1-n-1}=\frac{b}{a}

        Þ    \displaystyle \frac{\sqrt{n+1}-\sqrt{n-1}}{2}=\frac{b}{a}

        or    \displaystyle \sqrt{n+1}-\sqrt{n-1}=\frac{2b}{a}        …(ii)

        Adding and subtracting (i) and (ii) we get

            \displaystyle 2\sqrt{n+1}=\frac{a}{b}+\frac{2b}{a} and \displaystyle 2\sqrt{n-1}=\frac{a}{b}-\frac{2b}{a}

        Þ    \displaystyle \sqrt{n+1}=\frac{{{a}^{2}}+2{{b}^{2}}}{2ab} and \displaystyle \sqrt{n-1}=\frac{{{a}^{2}}-2{{b}^{2}}}{2ab}

        Þ    RHS of both are rational.

        \    \displaystyle \sqrt{n+1} and \displaystyle \sqrt{n-1} are also rational.

        Þ    \displaystyle (n+1) and \displaystyle (n-1) are perfect squares of positive integers.

        This is impossible as any two perfect squares differ at least by 3.

        Hence, there is no positive integer n for which (\displaystyle \sqrt{n-1}+\sqrt{n+1}) is rational.

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