Exercise 1.4 (Revised) - Chapter 1 - Number Systems - Ncert Solutions class 9 - Maths

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Chapter 1 - Number Systems | NCERT Solutions Class 9 Maths

Ex 1.4 Question 1.

Classify the following numbers as rational or irrational:
(i) $2-\sqrt{5}$
(ii) $(3+\sqrt{23})-\sqrt{23}$
(iii) $\frac{2 \sqrt{7}}{7 \sqrt{7}}$
(iv) $\frac{1}{\sqrt{2}}$
(v) $2 \pi$
Answer

(i) $2-\sqrt{5}$

We know that $\sqrt{5}=2.236 \ldots$, which is an irrational number.
$
\begin{aligned}
& 2-\sqrt{5}=2-2.236 \ldots \\
& =-0.236 \ldots,
\end{aligned}
$
which is also an irrational number.

Therefore, we conclude that $2-\sqrt{5}$ is an irrational number.
$
\begin{aligned}
& \text { (ii) }(3+\sqrt{23})-\sqrt{23} \\
& (3+\sqrt{23})-\sqrt{23}=3+\sqrt{23}-\sqrt{23} \\
& =3
\end{aligned}
$

Therefore, we conclude that $(3+\sqrt{23})-\sqrt{23}$ is a rational number.
(iii) $\frac{2 \sqrt{7}}{7 \sqrt{7}}$

We can cancel $\sqrt{7}$ in the numerator and denominator, as $\sqrt{7}$ is the common number in numerator as well as denominator, to get
$
\frac{2 \sqrt{7}}{7 \sqrt{7}}=\frac{2}{7}
$

Therefore, we conclude that $\frac{2 \sqrt{7}}{7 \sqrt{7}}$ is a rational number.
(iv) $\frac{1}{\sqrt{2}}$

We know that $\sqrt{2}=1.414 \ldots$, which is an irrational number.

We can conclude that, when 1 is divided by $\sqrt{2}$, we will get an irrational number.

Therefore, we conclude that $\frac{1}{\sqrt{2}}$ is an irrational number.
(v) $2 \pi$

We know that $\pi=3.1415 \ldots$, which is an irrational number.

We can conclude that $2 \pi_{\text {will }}$ also be an irrational number.

Therefore, we conclude that $2 \pi$ is an irrational number.

Ex 1.4 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) $(5-\sqrt{2})(5+\sqrt{2})$

Answer.

(i) $(3+\sqrt{3})(2+\sqrt{2})$
We need to apply distributive law to find value of $(3+\sqrt{3})(2+\sqrt{2})$.
$
\begin{aligned}
& (3+\sqrt{3})(2+\sqrt{2})=3(2+\sqrt{2})+\sqrt{3}(2+\sqrt{2}) \\
& =6+3 \sqrt{2}+2 \sqrt{3}+\sqrt{6}
\end{aligned}
$

Therefore, on simplifying $(3+\sqrt{3})(2+\sqrt{2})$, we get $6+3 \sqrt{2}+2 \sqrt{3}+\sqrt{6}$.
(ii) $(3+\sqrt{3})(3-\sqrt{3})$

We need to apply distributive law to find value of $(3+\sqrt{3})(3-\sqrt{3})$.
$
\begin{aligned}
& (3+\sqrt{3})(3-\sqrt{3})=3(3-\sqrt{3})+\sqrt{3}(3-\sqrt{3}) \\
& =9-3 \sqrt{3}+3 \sqrt{3}-3
\end{aligned}
$

Therefore, on simplifying $(3+\sqrt{3})(3-\sqrt{3})$, we get 6 .
(iii) $(\sqrt{5}+\sqrt{2})^2$

We need to apply the formula $(a+b)^2=a^2+2 a b+b^2$ to find value of $(\sqrt{5}+\sqrt{2})^2$.
$
\begin{aligned}
& (\sqrt{5}+\sqrt{2})^2=(\sqrt{5})^2+2 \times \sqrt{5} \times \sqrt{2}+(\sqrt{2})^2 \\
& =5+2 \sqrt{10}+2 \\
& =7+2 \sqrt{10}
\end{aligned}
$

$
=6
$

Therefore, on simplifying $(\sqrt{5}+\sqrt{2})^2$, we get $7+2 \sqrt{10}$.
(iv) $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$

We need to apply the formula $(a-b)(a+b)=a^2-b^2$ to find value of $(\sqrt{5}+\sqrt{2})^2$.
$
\begin{aligned}
& (\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=(\sqrt{5})^2-(\sqrt{2})^2 \\
& =5-2 \\
& =3
\end{aligned}
$

Therefore, on simplifying $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$, we get 3 .

Ex 1.4 Question 3.

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

Answer.

We know that when we measure the length of a line or a figure by using a scale or any device, we do not get an exact measurement. In fact, we get an approximate rational value. So, we are not able to realize that either circumference (c) or diameter(d) of a circle is irrational.

Therefore, we can conclude that as such there is not any contradiction regarding the value of $\pi$ and we realize that the value of $\pi$ is irrational.

Ex 1.4 Question 4.

Represent 9.3 on the number line.

Answer.

Mark the distance 9.3 units from a fixed point $A$ on a given line to obtain a point $B$ such that $A B=$ 9.3 units. From B mark a distance of 1 unit and call the new point as $C$. Find the mid-point of $A C$ and call that point as $\mathrm{O}$. Draw a semi-circle with centre $\mathrm{O}$ and radius $\mathrm{OC}=5.15$ units. Draw a line perpendicular to $A C$ passing through $B$ cutting the semi-circle at $D$. Then $B D=\sqrt{9.3}$.

Ex 1.4 Question 5.

Rationalize the denominators of the following:
(i) $\frac{1}{\sqrt{7}}$
(ii) $\frac{1}{\sqrt{7}-\sqrt{6}}$
(iii) $\frac{1}{\sqrt{5}+\sqrt{2}}$
(iv) $\frac{1}{\sqrt{7}-2}$

Answer.

(i) $\frac{1}{\sqrt{7}}$

We need to multiply the numerator and denominator of $\frac{1}{\sqrt{7}}$ by $\sqrt{7}$, to get
$
\frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}
$

Therefore, we conclude that on rationalizing the denominator of $\frac{1}{\sqrt{7}}$, we get $\frac{\sqrt{7}}{7}$.

(ii) $\frac{1}{\sqrt{7}-\sqrt{6}}$

We need to multiply the numerator and denominator of $\frac{1}{\sqrt{7}-\sqrt{6}}$ by $\sqrt{7}+\sqrt{6}$, to get
$
\frac{1}{\sqrt{7}-\sqrt{6}} \times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}=\frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})}
$

We need to apply the formula $(a-b)(a+b)=a^2-b^2$ in the denominator to get
$
\begin{aligned}
& \frac{1}{\sqrt{7}-\sqrt{6}}=\frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^2-(\sqrt{6})^2} \\
& =\frac{\sqrt{7}+\sqrt{6}}{7-6} \\
& =\sqrt{7}+\sqrt{6}
\end{aligned}
$

Therefore, we conclude that on rationalizing the denominator of $\frac{1}{\sqrt{7}-\sqrt{6}}$, we get $\sqrt{7}+\sqrt{6}$.

(iii) $\frac{1}{\sqrt{5}+\sqrt{2}}$

We need to multiply the numerator and denominator of $\frac{1}{\sqrt{5}+\sqrt{2}}$ by $\sqrt{5}-\sqrt{2}$, to get
$
\frac{1}{\sqrt{5}+\sqrt{2}} \times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}
$

We need to apply the formula $(a-b)(a+b)=a^2-b^2$ in the denominator to get
$
\begin{aligned}
& \frac{1}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^2-(\sqrt{2})^2} \\
& =\frac{\sqrt{5}-\sqrt{2}}{5-2} \\
& =\frac{\sqrt{5}-\sqrt{2}}{3} .
\end{aligned}
$

Therefore, we conclude that on rationalizing the denominator of $\frac{1}{\sqrt{5}+\sqrt{2}}$, we get $\frac{\sqrt{5}-\sqrt{2}}{3}$.

(iv) $\frac{1}{\sqrt{7}-2}$

We need to multiply the numerator and denominator of $\frac{1}{\sqrt{7}-2}$ by $\sqrt{7}+2$, to get
$
\frac{1}{\sqrt{7}-2} \times \frac{\sqrt{7}+2}{\sqrt{7}+2}=\frac{\sqrt{7}+2}{(\sqrt{7}-2)(\sqrt{7}+2)}
$

We need to apply the formula $(a-b)(a+b)=a^2-b^2$ in the denominator to get
$
\begin{aligned}
& \frac{1}{\sqrt{7}-2}=\frac{\sqrt{7}+2}{(\sqrt{7})^2-(2)^2} \\
& =\frac{\sqrt{7}+2}{7-4} \\
& =\frac{\sqrt{7}+2}{3} .
\end{aligned}
$

Therefore, we conclude that on rationalizing the denominator of $\frac{1}{\sqrt{7}-2}$, we get $\frac{\sqrt{7}+2}{3}$.