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

Updated On May 15, 2024

By SaraNextGen

Basic Algebra
EX 2.1
Question 1.
Classify each element of $\left\{\sqrt{7}, \frac{-1}{4}, 0,3.14,4, \frac{22}{7}\right\}$ as a member of $\mathbb{N}, \mathbb{Q}, \mathbb{R}, \mathbb{Z}$
Solution:
$\frac{-1}{4}$ is a rational number (i.e.) $\frac{-1}{4} \in \mathrm{Q}$,
$3.14 \in \mathrm{Q}$
0,4 are integers and $0 \in Z, 4 \in \mathrm{N}, Z, Q$
$\frac{22}{7} \in \mathrm{Q}$
Question 2.
Prove that $\sqrt{3}$ is an irrational number.
(Hint: Follow the method that we have used to prove $\sqrt{2} \notin \mathrm{Q}$.
Solution:
Suppose that $\sqrt{3}$ is rational $\mathrm{P}$

Then $\sqrt{3}=\frac{p}{q}$ (where $p$ and $q$ are integers which are co-prime)
$
\begin{aligned}
& \Rightarrow \quad p=\sqrt{3} q \Rightarrow p^2=3 q^2 \\
& \frac{p^2}{3}=q^2 \Rightarrow 3 \text { is a factor of } p \\
& \text { So let } p=3 c \\
& \text { substituting } p=3 c \text { in (1) we get } \\
& (3 c)^2=3 q^2 \Rightarrow 9 c^2=3 q^2 \\
& \Rightarrow c^2=\frac{3 q^2}{9}=\frac{q^2}{3} \\
&
\end{aligned}
$
Question 3.
Are there two distinct irrational numbers such that their difference is a rational number? Justify. Solution:
Taking two irrational numbers as $3+\sqrt{2}$ and $1+\sqrt{2}$
Their difference is a rational number. But if we take two irrational numbers as $2-\sqrt{3}$ and $4+$ $\sqrt{7}$.
Their difference is again an irrational number. So unless we know the two irrational numbers we cannot say that their difference is a rational number or irrational number.
Question 4.
Find two irrational numbers such that their sum is a rational number. Can you find two irrational numbers whose product is a rational number.
Solution:
(i) Let the two irrational numbers as $2+\sqrt{3}$ and $3-\sqrt{3}$
Their sum is $2+\sqrt{3}+3-3 \sqrt{3}$ which is a rational number.
But the sum of $3+\sqrt{5}$ and $4-\sqrt{7}$ is not a rational number. So the sum of two irrational numbers is either rational or irrational.
(ii) Again taking two irrational numbers as $\pi$ and $\frac{3}{\pi}$ their product is $\sqrt{3}$ and $\sqrt{2}=\sqrt{3} \times \sqrt{2}$ which is irrational, So the product of two irrational numbers is either rational or irrational.

Question 5.
Find a positive number smaller than $\frac{1}{2^{1000}}$. Justify.
Solution:
$
\text { We know } \begin{aligned}
\frac{1}{2} & =0.5 \\
\frac{1}{2^2} & =(0.5)^2=0.25 \\
\frac{1}{2^3} & =(0.5)^3=0.125 \\
\frac{1}{2^4} & =(0.5)^4=0.0625 \\
\vdots & =(0.5)^{1000} \simeq 0 \\
\frac{1}{2^{1000}} & =0
\end{aligned}
$
There will not be a positive number smaller than 0 . So there will not be a +ve number smaller than $\frac{1}{2^{1000}}$