In this page, you can find the complete solutions of the third exercise of Logic, Sets, and Real Number System chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.

In the above-mentioned book, logic, sets, and real number system is the 1st chapter and has four exercises only. Out of which, this is the solution of the third exercise.

Check: Basic Mathematics Grade 11 (Sukunda Publication) Guide:
Grade 11 Basic Mathematics by Sukunda Pustak Vawan Notes and Solutions | Nepal

 

Disclaimer:

Answers mentioned here are not solved by teachers. These are the solutions written by a student of Grade 11. Answers are all correct. However, the language or process of solving the questions might be informal and in examinations, you might have to add little more language and show more calculations than what has been done here. So, we highly encourage you to view these solutions as guide rather than just copying everything mentioned here. Few questions have been typed while most of them have been updated as pictures.

Sets Exercises

$\text{1}. If A \cap B = \phi, \text{prove that}$

$a) A \subseteq \overline{B}$

Solution:
Let x be an element of set A.
$A \implies \{x: x \in A\}$
$\implies \{x: x \notin \overline{A} \}$

$\implies \{x: x \in \overline{B}\}$ $[\because A \cap B = \phi]$


$b) B \cap \overline{A} = B$

Solution:
$B \implies \{x: x \in B \ and \ x \in \overline{A}\}$

$\implies \{x: x \in  B \ and \ x \notin A \}$

$\implies \{x:x \in B \ and \ x \in B\}$ $[\because A \cap B = \phi]$

$\implies \{ x: x \in B \} $

$\implies B$

$c) A \cup \overline{B} = \overline{B}$

Solution:
$A \cup \overline{B} \implies \{x: x \in A or x  \in \overline{B}\}$

$\implies \{x:x \in \overline{B} or x \in \overline{B}\}$ $[\because A \cap B = \phi]$

$\implies \{x: x \in \overline{B}\}$

$\implies \overline{B}$

2. $\text\{If\} A \subseteq B, \text\{prove that\}$

a) $\overline{B\} \subseteq \overline{A\}$

Solution:
$\overline{B\} = \{x: x \in \overline{B\}\}$

$= \{x: x \notin B\}$
$= \{x: x \notin A\}$ $\because A \subseteq B$

$= \{x: x\in \overline{A\}\}$
$= \overline{A\}$
$\implies \overline{B\} \subseteq \overline{A\}$


b) $A \cup B = B$

Solution:

$A \cup B = \{x:x \in A or x \in B\}$

$\implies \{x: x \in B or x \in B\}$ $\because A \subseteq B$

$\implies \{x: x \in B\}$
$\implies B$
Hence, $A \cup B = B$

c) $A \cap B = A$

Solution:
$A \cap B = \{x:x \in A and x \in B\}$

$\implies \{x: x \in A and x \in A\}$ $\because A \subseteq B$

$\implies \{x: x \in A\}$
$\implies A$
Hence, $A \cap B = A$

3. Prove that

a) $B - A = B \cap \overline{A\}$

Solution:

$B - A \implies \{x: x \in B and x \notin A\}$

$\implies \{x:x \in B and x \in \overline{A\}\}$

$\implies B \cap \overline{A\}$


b) $A - \overline{B\} = A \cap B$

Solution:
$A - \overline{B\} \implies \{x:x \in A and x \notin \overline{B\}\}$

$\implies\{x: x \in A and x \in B\}$

$\implies A \cap B$

c) $A - B = \overline{B\} - \overline{A\}$

Solution:
$A - B \implies \{x: x \in A and x \notin B\}$

$\implies \{x: x \notin \overline{A\} and x \in \overline{B\}\}$

$\implies \{x: x \in \overline{B\} and x \notin \overline{A\}\}$

$\implies \overline{B\} - \overline{A\}$


4. If A, B and C are the subsets of a universal set U, prove that

$a) A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$

Solution:

$A \cup (B \cap C) \implies \{x:x \in A or x \in B \cap C\}$

$\implies\{ x:x \in A or (x \in B and x \in C)\}$

$\implies \{x: (x \in A or x \in B) and (x \in A or x \in C)\}$

$\implies\{x: x \in A \cup B and x \in A cup C\}$

$\implies (A \cup B) \cap (A \cup C)$



$b) A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

Solution:
$A \cap (B \cup C) \implies \{x: x \in A and x \in B \cup C\}$

$\implies \{x: x\in A and (x \in B or x \in C)\}$

$\implies \{x: (x\in A and x \in B) or (x \in A and x \in C)\}$

$\implies \{x: x \in A \cap B or x \in A \cap C\}$

$\implies (A \cap B) \cup (A \cap C)$


5. If A, B, and C are three subsets of a universal set U, prove that:

Both $A'$ and $\overline{A\}$ represent the complement of a set A.

$a) (A \cup B)' = A' \cap B'$

Solution:

$(A \cup B)' \implies \{x:x \notin A and x \notin B\}$

$\implies \{x: x \in \overline{A\} and x \in \overline{B\}\}$

$\implies A' \cap B'$


$b) (A \cap B)' = A' \cup B'$

Solution:

$(A \cap B)' \implies \{x:x \notin A \text\{or\} x \notin B\}$

$\implies \{x: x \in A' \text\{or\} \in B'\}$

$\implies A' \cup B'$


$c) A - (B \cup C) = (A-B) \cap (A - c) = (A-B) -C$

Solution:

$A - (B \cup C) \implies \\{ x: x \in A and x \notin B \cup C \\}$

$\implies \\{ x: x \in A \ and \ ( x \notin B \ and \ x \notin C) \\}$

$\implies \\{x: ( x \in A \ and \ x \notin B ) \ and \ ( x \in A \ and \ x \notin C) \\}$

$\implies \\{ x: x \in A - B \ and \ x \in A - C \\}$

$\implies (A-B) \cap (A-C)$


Also,

$A - (B \cup C) \implies \{ x: x \in A and x \notin B \cup C \}$

$\implies \{ x: x \in A \ and \ ( x \notin B \ and \ x \notin C) \}$

$\implies \{x: ( x \in A \ and \ x \notin B ) \ and \ ( x \in A \ and \ x \notin C) \}$

$\implies \{ x: x \in A - B \ and \ x \notin C \}$

$\implies (A-B) - C$


$d) A - (B \cap C) = (A -B) \cup (A -C)$

Solution:

$A - (B \cap C) \implies \{x: x \in A \text\{and\} x \notin B \cap C\}$

$\implies \{x: x \in A \text\{and\} ( x \notin B \text\{or\} x \notin C)\}$

$\implies \{x: (x \in A \text\{and\} x \notin B) \text\{or\} (x \in A \text\{and\} x \notin C)\}$

$\implies \{x: x \in A - B \text\{or\} x \in A - C\}$

$\implies A - B \cup A - C$

About the Textbook:

Name: Basic Mathematics Grade XI
Author(s): D.R. Bajracharya | R.M. Shresththa | M.B. Singh | Y.R. Sthapit | B.C. Bajracharya
Publisher: Sukunda Pustak Bhawan (Bhotahity, Kathmandu)
Telephone: 5320379, 5353537
Price: 695 /- (2078 BS)