Exercise 1.5 Number Systems NCERT Solutions Class 9

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Chapter 1 Ex.1.5 Question 1

Classify the following numbers as rational or irrational:

(i) \(\begin{align}  \quad 2 - \sqrt 5 \end{align}\)

(ii) \(\begin{align}(3 + \sqrt {23} ) - \sqrt {23} \end{align}\)

(iii) \(\begin{align}\frac{{2\sqrt 7 }}{{7\sqrt 7 }} \end{align}\)

(iv) \(\begin{align}\frac{1}{{\sqrt 2 }} \end{align}\)

(v) \(\begin{align}2\pi \end{align}\)

 

Solution

Video Solution

\(\begin{align}({\rm{i}})\,\,\,\,2 - \sqrt 5 \end{align}\)

Steps:

The sum or difference of a rational number and an irrational number is always irrational.

Here \(2\) is a rational number and \(\begin{align} \sqrt{5} \end{align}\) is an irrational number. Hence \(\begin{align} 2 -\sqrt{5} \end{align}\) is an irrational number.

\(\begin{align}\rm{(ii)}\quad(3 + \sqrt {23} ) - \sqrt {23} \end{align}\)

Steps:

By simplifying we get only \(3\).

\(\begin{align}3 = \frac{3}{1},\end{align}\) which is in the form of \(\begin{align}\frac{p}{q},\end{align}\)

Hence is \(\begin{align}(3 + \sqrt {23} ) - \sqrt {23} \end{align}\) a rational number.

\(\begin{align}\rm{(iii)}\quad \frac{{2\sqrt 7 }}{{7\sqrt 7 }}\end{align}\)

Steps:

\(\begin{align}\frac{{2\sqrt 7 }}{{7\sqrt 7 }} = \frac{2}{7},\end{align}\) which is in the form of \(\begin{align}\frac{p}{q},\end{align}\)

Hence \(\begin{align}\frac{{2\sqrt 7 }}{{7\sqrt 7 }}\end{align}\) is a rational number.

\(\begin{align}\rm{(iv)}\quad \frac{1}{{\sqrt 2 }}\end{align}\)

Steps:

\(\begin{align} \frac{1}{\sqrt{2}} &=\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\ &=\frac{\sqrt{2}}{2} \\ &=\frac{1.414 \ldots}{2} \end{align}\)

\(\begin{align}= 0.702……\end{align}\) is a non - terminating, non-recurring decimal  and therefore it is irrational. Hence \(\begin{align}\,\frac{1}{{\sqrt 2 }}\,\end{align}\) is an irrational number.

\(\begin{align}\rm{(v)}\quad 2{\rm{\pi }}\end{align}\)

Steps:

\(\begin{align}~2\text{ }\!\!\pi\!\!\text{ }=2\times 3.1415...\end{align}\)

\(\pi\) is an irrational number whose value is non-terminating and non-recurring.

\(2\) is a rational number.

Product of a non-zero rational number and irrational number is an irrational number.

Hence \(\begin{align}2\pi\end{align}\) is irrational.

Chapter 1 Ex.1.5 Question 2

Simplify each of the following expressions

(i) \( (3 + \sqrt 3 )\,(2 + \sqrt 2 )\)    

(ii) \((3 + \sqrt 3 )\,(3 - \sqrt 3 )\)

(iii)\({(\sqrt 5 + \sqrt 2 )^2}\)    

(iv) \((5 - \sqrt 2 )\,(5 + \sqrt 2 )\)

 

Solution

Video Solution

Steps:

(i) \(\begin{align}(3 + \sqrt 3 )\,(2 + \sqrt 2 )\end{align}\)

By Distributive property

\[\begin{align}=6+3 \sqrt{2}+2 \sqrt{3}+\sqrt{6} \end{align}\]

(ii) \(\begin{align}(3 + \sqrt 3 )\,(3 - \sqrt 3 )\end{align}\)

Using the identity:

\[\begin{align}(a+b)(a-b) &=a^{2}-b^{2} \\(3+\sqrt{3}) 3-\sqrt{3} &=9-3 \\ &=6 \end{align}\]

(iii) \(\begin{align}{(\sqrt 5 + \sqrt 2 )^2}\,\end{align}\)

Steps:

Using the identity: \((a+b)^{2} =a^{2}+2 a b+b^{2}\)

\[\begin{align}&=(\sqrt{5})^{2}+2 \times \sqrt{5} \times \sqrt{2}+(\sqrt{2})^{2} \\ &=(5+2 \sqrt{10}+2) \\ &=7+2 \sqrt{10} \end{align}\]

(iv) \(\begin{align}(5 - \sqrt 2 )\,(5 + \sqrt 2 )\end{align}\)

Steps:

Using the identity: \((a+b)(a-b) =a^{2}-b^{2}\)

\[\begin{align}&(\sqrt{5})^{2}-(\sqrt{2})^{2} \\ &=5-2 \\ &=3 \end{align}\]

Chapter 1 Ex.1.5 Question 3

Recall, \(\begin{align}\pi \end{align}\) is defined as the ratio of circumference (say \(c\)) to its diameter (say \(d\)). That is \(\begin{align}{\rm{\pi = }}\frac{c}{d}.\end{align}\). This seems to contradict the fact that \(\begin{align}\pi \end{align}\) is irrational. How will you resolve this contradiction?

 

Solution

Video Solution

Steps:

Writing \(\pi\) as \(\begin{align}\frac{22}{7}\end{align}\) is only an approximate value and so we can’t conclude that it is in the form of a rational. In fact, the value of \(\pi\) is calculating as non-terminating, non-recurring decimal as \(\pi=3.14159\dots\) Whereas

 If we calculate the value of \(\begin{align}\frac{22}{7}\end{align}\) it gives \(3.142857\) and hence\(\begin{align}\pi\ne\frac{22}{7}\end{align}\)

In conclusion \(\pi\) is an irrational number.

Chapter 1 Ex.1.5 Question 4

Represent \(\begin{align}\sqrt {9.3} \end{align}\) on the number line.

 

Solution

Video Solution

Steps:

Draw a line and take \(AB = 9.3\) units on it.

From \(B\) measure a distance of \(1\) unit and mark \(C\) on the number line. Make the midpoint of \(AC\) as \(O.\)

With \(‘O’\) as center and \(OC\) as radius, draw a semicircle.

At \(B\), draw a perpendicular to cut the semicircle at \(D\), with \(B\) as center and \(BD\) as radius draw an arc to cut the number line at \(E\). Taking \(B\) as the origin the distance \({BE}=\sqrt{9.3}\) and hence \(E\) represents \(\sqrt{9.3}\) .

Proof:

\[\begin{align}AB &=9.3, BC=1 \\ AC &=A B+BC=10.3 \\ OC &=\frac{AC}{2}=\frac{10.3}{2}=5.15 \\ OC&=OD=5.15\\O B &=OC-BC\\&=5.15-1=4.15 \end{align}\]

In right angled\(\begin{align}\Delta OBD,\end{align}\)

\[\begin{align} {BD}^{2} &={OD}^{2}-{OB}^{2}\\ &=(5.15)^{2}-(4.15)^{2} \\ &=(5.15+4.15)(5.15-4.15)\\&\rm{Using} \quad a^{2}-b^{2}=(a+b)(a-b) \\  &=9.3 \times 1\\  &=9.3 \\ BD &=\sqrt{9.3}=BE \end{align}\]

Chapter 1 Ex.1.5 Question 5

Rationalize the denominators of the following:

(i) \(\begin{align}\frac{1}{{\sqrt 7 }} \end{align}\)

(ii)\(\begin{align}\frac{1}{{\sqrt 7 - \sqrt 6 }}\end{align}\)

(iii) \(\begin{align}\frac{1}{{\sqrt 5 + \sqrt 2 }}\end{align}\)

(iv) \(\begin{align}\frac{1}{{\sqrt 7 - 2}}\end{align}\)

 

Solution

Video Solution

Steps:

(i) \(\begin{align}\frac{1}{{\sqrt 7 }} = \frac{1}{{\sqrt 7 }} \times \frac{{\sqrt 7 }}{{\sqrt 7 }}\end{align}\)

(Dividing and multiplying by \(\sqrt{7}\) )

\[=\frac{\sqrt{7}}{7}\]

(ii) \(\begin{align}\frac{1}{{\sqrt 7 - \sqrt 6 }}\end{align}\)

Steps:

Dividing and multiplying by \(\begin{align}\sqrt 7 + \sqrt 6 \end{align}\), we get

\[\begin{align} &=\frac{1}{\sqrt{7}-\sqrt{6}} \times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}-\sqrt{6}} \\ &=\frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}\\&\left(\text { Using }(a+b)(a-b)=a^{2}-b^{2}\right) \\ &=\frac{\sqrt{7}+\sqrt{6}}{7-6} \\ &=\sqrt{7}+\sqrt{6} \end{align}\]

(iii) \(\begin{align}\frac{1}{{\sqrt 5 + \sqrt 2 }}\end{align}\)

Steps:

Dividing and multiplying by \(\begin{align}\sqrt 5 - \sqrt 2 \end{align}\) , we get

\[\begin{align} &=\frac{1}{\sqrt{5}+\sqrt{2}} \times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\ &=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^{2}-(\sqrt{2})^{2}} \\&\text{Using }(a+b)(a-b)=\left(a^{2}-b^{2}\right)\\ &=\frac{\sqrt{5}-\sqrt{2}}{5-2} \\ &=\frac{\sqrt{5}-\sqrt{2}}{3} \end{align}\]

(iv) \(\begin{align}\frac{1}{{\sqrt 7 - 2}} & \end{align}\)

Steps:

Dividing and multiplying by \(\begin{align}\sqrt 7 + 2,\end{align}\), we get

\[\begin{align} &=\frac{1}{\sqrt{7}-2} \times \frac{\sqrt{7}+2}{\sqrt{7}+2} \\ &=\frac{\sqrt{7}+2}{(\sqrt{7})^{2}-(2)^{2}}\\&\left(\text{Using }(a+b)(a-b)={a}^{2}-{b}^{2}\right)\\ &=\frac{\sqrt{7}+2}{7-4} \\ &=\frac{\sqrt{7}+2}{3} \end{align}\]

  
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