Exercise 2.1 - Chapter 2 Real Numbers 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.1$ : Chapter 2 - Real Numbers - 9th Maths Guide Samacheer Kalvi Solutions
Question $1 .$
Which arrow best shows the position of $\frac{11}{3}$ on the number line?

Solution:
$\frac{11}{3}=3.666 \ldots . .=3.7 \text { (nearly) }$
$\therefore \mathrm{D}$ arrow best shows the position of $\frac{11}{3}$ on the number line.

Question $2 .$
Find any three rational numbers between $\frac{-7}{11}$. $\frac{2}{11}$.
Solution:
Three rational numbers between $\frac{-7}{11}$ and $\frac{2}{11}$ are $\frac{-6}{11}, \frac{-5}{11}, \frac{-4}{11}, \ldots \ldots, \frac{-1}{11}$
 

Question $3 .$
Find any five rational numbers between
(i) $\frac{1}{4}$ and $\frac{1}{5}$
(ii) $0.1$ and $0.11$
(iii) $-1$ and $-2$
Solution:
(i) $\mathrm{a}=\frac{1}{4}, \mathrm{~b}=\frac{1}{5}$

Let $\mathrm{q}_{1}, \mathrm{q}_{2}, \mathrm{q}_{3}, \mathrm{q}_{4}$ and $\mathrm{q}_{5}$ be five rational numbers, $\mathrm{q}_{1}=\frac{1}{2}(\mathrm{a}+\mathrm{b})$
$\begin{aligned}
&=\frac{1}{2}\left(\frac{1}{4}+\frac{1}{5}\right)=\frac{1}{2}\left(\frac{5+4}{20}\right)=\frac{1}{2}\left(\frac{9}{20}\right)=\frac{9}{40} \\
&q_{2}=\frac{1}{2}\left(a+q_{1}\right)=\frac{1}{4} \text { and } \frac{9}{40}=\frac{1}{2}\left(\frac{1}{4}+\frac{9}{40}\right)=\frac{1}{2}\left(\frac{10+9}{40}\right)=\frac{1}{2}\left(\frac{19}{40}\right)=\frac{19}{80} \\
&q_{3}=\frac{1}{2}\left(a+q_{2}\right)=\frac{1}{2}\left(\frac{1}{4}+\frac{19}{80}\right)=\frac{1}{2}\left(\frac{20+19}{80}\right)=\frac{1}{2}\left(\frac{39}{80}\right)=\frac{39}{160} \\
&q_{4}=\frac{1}{2}\left(a+q_{3}\right)=\frac{1}{4} \text { and } \frac{39}{160}=\frac{1}{2}\left(\frac{1}{4}+\frac{39}{160}\right)=\frac{1}{2}\left(\frac{40+39}{160}\right)=\frac{1}{2}\left(\frac{79}{160}\right)=\frac{79}{320} \\
&q_{5}=\frac{1}{2}\left(a+q_{4}\right)=\frac{1}{4} \text { and } \frac{79}{320}=\frac{1}{2}\left(\frac{1}{4}+\frac{79}{320}\right)=\frac{1}{2}\left(\frac{80+79}{320}\right)=\frac{1}{2}\left(\frac{159}{320}\right)=\frac{159}{640}
\end{aligned}$
Hence five rational numbers between $\frac{1}{4}$ and $\frac{1}{5}$ are $\frac{9}{40}, \frac{19}{80}, \frac{39}{160}, \frac{79}{320}, \frac{159}{640}$
(ii) The rational numbers between $0.1$ and $0.11$ are $0.101,0.102,0.103, \ldots . .0 .109$.
(iii) $-1$ and $-2$

Let $\mathrm{q}_{1}, \mathrm{q}_{2}, \mathrm{q}_{3}, \mathrm{q}_{4}$ and $\mathrm{q}_{5}$ be five rational numbers.
$\begin{aligned}
q_{1}=\frac{1}{2}(\mathrm{a}+\mathrm{b}) &=\frac{1}{2}((-1)+(-2))=\frac{1}{2}(-3)=\frac{-3}{2} \\
q_{2}=\frac{1}{2}\left(a+q_{1}\right) &=\frac{1}{2}\left(-1+\frac{-3}{2}\right)=\frac{1}{2}\left(\frac{-2+(-3)}{2}\right)=\frac{1}{2}\left(\frac{-5}{2}\right)=\frac{-5}{4} \\
q_{3}=\frac{1}{2}\left(a+q_{2}\right) &=-1 \text { and } \frac{-5}{4} \\
&=\frac{1}{2}\left(-1-\frac{5}{4}\right)=\frac{1}{2}\left(\frac{-4+(-5)}{4}\right)=\frac{1}{2}\left(\frac{-9}{4}\right)=\frac{-9}{8} \\
q_{4}=\frac{1}{2}\left(a+q_{3}\right) &=-1 \text { and } \frac{-9}{8} \\
&=\frac{1}{2}\left(-1+\frac{-9}{8}\right)=\frac{1}{2}\left(\frac{(-8)+(-9)}{8}\right)=\frac{1}{2}\left(\frac{-17}{8}\right)=\frac{-17}{16} \\
q_{5}=\frac{1}{2}\left(a+q_{4}\right) &=-1 \text { and } \frac{-17}{16} \\
&=\frac{1}{2}\left(-1+\frac{-17}{16}\right)=\frac{1}{2}\left(\frac{(-16)+(-17)}{16}\right)=\frac{1}{2}\left(\frac{-33}{16}\right)=\frac{-33}{32}
\end{aligned}$
The five rational numbers between $-1$ and $-2$ are $\frac{-3}{2}, \frac{-5}{4}, \frac{-9}{8}, \frac{-17}{16}, \frac{-33}{32}$.