Exercise 2.1-Additional Problems - Chapter 2 Basic Algebra 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions
Question 1.
Prove that $\sqrt{5}$ is an irrational number.
Solution:
Suppose that $\sqrt{5}$ is rational
$\begin{aligned} \text {. Then } \sqrt{5} & =\frac{p}{q} \text { (where } p \text { and } q \text { are integers which are co-prime) } \\ p & =\sqrt{5} q \Rightarrow p^2=5 q^2 \\ \frac{p^2}{5} & =q^2 \Rightarrow 5 \text { is a factor of } p\end{aligned}$
So let $\mathrm{p}=5 \mathrm{c}$
substitute $\mathrm{p}=5 \mathrm{c}$ in (1) we get
$(5 c)^2=5 q^2 \Rightarrow 25 c^2=5 q^2$
$\Rightarrow c^2 \quad=\frac{5 q^2}{25}=\frac{q^2}{5}$
$\Rightarrow 5$ is a factor of $\mathrm{q}$ also
So 5 is a factor of $\mathrm{p}$ and $\mathrm{q}$ which is a contradiction.
$\Rightarrow \sqrt{5}$ is not a rational number
$\Rightarrow \sqrt{5}$ is an irrational number

Question 2.
Prove that $0.33333=\frac{1}{3}$
Solution:
Let $x=0.33333 \ldots$
$
\begin{aligned}
& 10 \mathrm{x}=3.3333 \\
& 10 \mathrm{x}-\mathrm{x}=9 \mathrm{x}=3 \\
& \Rightarrow \quad x=\frac{3}{9}=\frac{1}{3}
\end{aligned}
$