Exercise 1.4 - Chapter 1 Set Language 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $1.4$
Question 1.
If $\mathrm{P}=\{1,2,5,7,9\}, \mathrm{Q}=\{2,3,5,9,11\}, \mathrm{R}=\{3,4,5,7,9\}$ and $\mathrm{S}=\{2,3,4,5,8\}$, then find
(i) $(P \cup Q) \cup R$
(ii) $(P \cap Q) \cap S$
(iii) $(Q \cap S) \cap R$
Solution:
(i) $(\mathrm{P} \cup \mathrm{Q}) \cup \mathrm{R}$
$(P \cup Q)=\{1,2,5,7,9\} \cup\{2,3,5,9,11\}=\{1,2,3,5,7,9,11\}$
$(\mathrm{P} \cup \mathrm{Q}) \cup \mathrm{R}=\{1,2,3,5,7,9,11\} \cup\{3,4,5,7,9\}=\{1,2,3,4,5,7,9,11\}$
(ii) $(P \cap Q) \cap S$
$(P \cap Q)=\{1, \underline{2}, \underline{5}, 7, \underline{9}\} \cap\{\underline{2}, 3, \underline{5}, \underline{9}, 11\}=\{2,5,9\}$
$(P \cap Q) \cap S=\{2,5,9\} \cap\{2,3,4,5,8\}=\{2,5\}$
(iii) $(\mathrm{Q} \cap \mathrm{S}) \cap \mathrm{R}$
$(\mathrm{Q} \cap \mathrm{S})=\{\underline{2}, \underline{3}, \underline{5}, 9,11\} \cap\{\underline{2}, \underline{3}, 4, \underline{5}, 8\}=\{2,3,5\}$
$(Q \cap S) \cap R=\{2,3,5\} \cap\{3,4,5,7,9\}=\{3,5\}$

Question 2.
Test for the commutative property of union and intersection of the sets
$\mathrm{P}=\{\mathrm{x}: \mathrm{x}$ is a real number between 2 and 7$\}$ and
$\mathrm{Q}=\{\mathrm{x}: \mathrm{x}$ is an irrational number between 2 and 7$\}$
Solution:
Commutative Property of union of sets
$(A \cup B)^{\prime}=(B \cup A)$
Here $P=\{3,4,5,6\}, Q=\{\sqrt{3}, \sqrt{5}, \sqrt{6}\}$
$\mathrm{P} \cup \mathrm{Q}=\{3,4,5,6\} \cup\{\sqrt{3}, \sqrt{5}, \sqrt{6}\}=\{3,4,5,6, \sqrt{3}, \sqrt{5}, \sqrt{6}\}$
$\mathrm{Q} \cup \mathrm{P}=\{\sqrt{3}, \sqrt{5}, \sqrt{6}\} \cup\{3,4,5,6\}=\{\sqrt{3}, \sqrt{5}, \sqrt{6}, 3,4,5,6\}$
(1) $=(2)$
$\therefore \mathrm{P} \cup \mathrm{Q}=\mathrm{Q} \cup \mathrm{P}$
$\therefore$ It is verified that union of sets is commutative.
Commutative Property of intersection of sets $(P \cap Q)=(Q \cap P)$
$P \cap Q=\{3,4,5,6\} \cap\{\sqrt{3}, \sqrt{5}, \sqrt{6}\}=\{\} \ldots \ldots \ldots .$ (1)
$\mathrm{Q} \cap \mathrm{P}=\{\sqrt{3}, \sqrt{5}, \sqrt{6}\} \cap\{3,4,5,6\}=\{\} \ldots \ldots \ldots \ldots$ (2)

From (1) and (2)
$P \cap Q=Q \cap P$
$\therefore$ It is verified that intersection of sets is commutative.
 

Question $3 .$
If $\mathrm{A}=\{\mathrm{p}, \mathrm{q}, \mathrm{r}, \mathrm{s}\}, \mathrm{B}=\{\mathrm{m}, \mathrm{n}, \mathrm{q}, \mathrm{s}, \mathrm{t}\}$ and $\mathrm{C}=\{\mathrm{m}, \mathrm{n}, \mathrm{p}, \mathrm{q}, \mathrm{s}\}$, then verify the associative property of union of sets.
Solution:
Associative Property of union of sets
$A \cup(B \cup C)=(A \cup B) \cup C)$
$B \cup C=\{m, n, q, s, t\} \cup\{m, n, p, q, s\}=\{m, n, p, q, s, t\}$
$A \cup(B \cup C)=\{p, q, r, s\} \cup\{m, n, p, q, s, t\}=\{m, n, p, q, r, s, t\} \ldots \ldots . \ldots . . .$ (1)
$(A \cup B)=\{p, q, r, s\} \cup\{m, n, q, s, L\}-\{p, q, r, s, m, n, t\}$
$(A \cup B) \cup C=\{p, q, r, s, m, n, t\} \cup\{m, n, p, q, s\}=\{p, q, r, s, m, n, t\} \ldots \ldots \ldots \ldots \ldots$ (2)
From (1) \& (2)
It is verified that $A \cup(B \cup C)=(A \cup B) \cup C$
 

Question $4 .$
Verify the associative property of intersection of sets for $\mathrm{A}=\{-11, \sqrt{2}, \sqrt{5}, 7\}, \mathrm{B}=\{$ $\sqrt{3}, \sqrt{5}, 6,13\}$ and $\mathrm{C}=\{\sqrt{2}, \sqrt{3}, \sqrt{5}, 9\}$
Solution:
Associative Property of intersection of sets $A \cap(B \cap C)=(A \cap B) \cap C)$
$\begin{aligned}
&\mathrm{B} \cap \mathrm{C}=\{\sqrt{3}, \sqrt{5}, 6,13\} \cap\{\sqrt{2}, \sqrt{3}, \sqrt{5}, 9\}=\{\sqrt{3}, \sqrt{5}\} \\
&A \cap(B \cap C)=\{-11, \sqrt{2}, \sqrt{5}, 7\} \cap\{\sqrt{3}, \sqrt{5}\}=\{\sqrt{5}\} \\
&A \cap B=\{-11, \sqrt{2}, \sqrt{5}, 7\} \cap\{\sqrt{3}, \sqrt{5}, 6,13\}=\{\sqrt{5}\} \\
&(A \cap B) \cap C=\{\sqrt{5}\} \cap\{\sqrt{2}, \sqrt{3}, \sqrt{5}, 9\}=\{\sqrt{5}\} \\
&\text { From (1) and (2), it is verified that } A \cap(B \cap C)=(A \cap B) \cap C
\end{aligned}$

Question $5 .$
If $\mathrm{A}=\{\mathrm{x}: \mathrm{x}=2, \mathrm{n} \in \mathrm{W}$ and $\mathrm{n}<4\}, \mathrm{B}=\{\mathrm{x}: \mathrm{x}=2 \mathrm{n}, \mathrm{n} \in \mathrm{N}$ and $\mathrm{n} \leq 4\}$ and $\mathrm{C}=\{0,1,2,5$, $6\}$, then verify the associative property of intersection of sets.
Solution:
$\begin{aligned}
&A=\left\{x: x=2^{n}, n \in W, n<4\right\} \\
&\Rightarrow x=2^{\circ}=1 \\
&x=2^{1}=2 \\
&x=2^{2}=4 \\
&x=2^{3}=8 \\
&\therefore A=\{1,2,4,8\}
\end{aligned}$
$\begin{aligned}
&B=\{x: x=2 n, n \in N \text { and } n \leq 4\} \\
&\Rightarrow x=2 \times 1=2 \\
&x=2 \times 2=4 \\
&x=2 \times 3=6 \\
&x=2 \times 4=8 \\
&\therefore B=\{2,4,6,8\} \\
&C=\{0,1,2,5,6\}
\end{aligned}$
Associative property of intersection of sets
$A \cap(B \cap C)=(A \cap B) \cap C$
$B \cap C=\{2,6\}$
$A \cap(B \cap C)=\{1,2,4,8\} \cap\{2,6\}=\{2\}$
$A \cap B=\{1,2,4,8\} \cap\{2,4,6,8\}=\{2,4,8\}$
$(A \cap B) \cap C=\{2,4,8\} \cap\{0,1,2,5,6\}=\{2\} \ldots \ldots \ldots \ldots \ldots \ldots$ (2)
From (1) and (2), It is verified that $A \cap(B \cap C)=(A \cap B) \cap C$