Exercise 12.2 - Chapter 12 Discrete Mathematics 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $12.2$
Question 1.
Let $\mathrm{p}$ : Jupiter is a planet and $\mathrm{q}$ : India is an island be any two simple statements. Give verbal sentence describing each of the following statements.
(i) $\neg p$
(ii) $p \wedge \neg q$
(iii) $\neg p \vee q$
(iv) $p \rightarrow \neg q$
(v) $p \leftrightarrow q$
Solution:
(i) $\neg p:$ Jupiter is not a planet
(ii) $p \wedge \neg q$ : Jupiter is not a planet and India is not an island
(iii) $\neg p \vee q$ : Jupiter is not a planet or India is an island.
(iv) $p \rightarrow \neg q$ : If Jupiter is a planet then India is not an island
(v) $p \leftrightarrow q$ : If Jupiter is a planet if and only if India is an island
Question 2.
Write each of the following sentences in symbolic form using statement variables $\mathrm{p}$ and $q$.
(i) 19 is not a prime number and all the angles of a triangle are equal.
(ii) 19 is a prime number or all the angles of a triangle are not equal
(iii) 19 is a prime number and all the angles of a triangle are equal
(iv) 19 is not a prime number
Solution:
$\mathrm{p}: 19$ is a prime number
$\mathrm{q}$ : All the angles of a triangle are equal
(i) $\neg p \wedge q$
(ii) $p \vee \neg q$
(iii) $p \wedge q$
(iv) $\neg p$
Question 3.
Determine the truth value of each of the following statements
(i) If $6+2=5$, then the milk is white.
(ii) China is in Europe or $\sqrt{3}$ is an integer
(iii) It is not true that $5+5=9$ or Earth is a planet
(iv) 11 is a prime number and all the sides of a rectangle are equal
Solution:
(i) $p, \rightarrow q$ is True : T
(ii) $p \vee q$ is False: F
(iii) $\neg p \vee q$ is True : $\mathrm{T}$
(iv) $p \wedge q$ is False: $\mathrm{F}$
Question 4.
Which one of the following sentences is a proposition?
(i) $4+7=12$
(ii) What are you doing?
(iii) $3^{\mathrm{n}} \leq 81, \mathrm{n} \in \mathrm{N}$

(iv) Peacock is our national bird
(v) How tall this mountain is!
Solution:
(i) is a proposition
(ii) not a proposition
(iii) is a proposition
(iv) is a proposition
(v) not a proposition
Question 5.
Write the converse, inverse, and contrapositive of each of the following implication.
(i) If $x$ and $y$ are numbers such that $x=y$, then $x^2=y^2$
(ii) If a quadrilateral is a square then it is a rectangle
Solution:
(i) Converse: If $x$ and $y$ are numbers such that $x^2=y^2$ then $x=y$.
Inverse: If $x$ and $y$ are numbers such that $x \neq y$ then $x^2 \neq y^2$.
Contrapositive : If $x$ and $v$ are numbers such that $x^2 \neq y^2$ then $x \neq y$.
(ii) Converse: If a quadrilateral is a rectangle then it is a square.
Inverse: If a quadrilateral is not a square then it is not a rectangle.
Contrapositive : If a quadrilateral is not a rectangle then it is not a square.
Question 6.
Construct the truth table for the following statements.
(i) $\neg p \wedge \neg q$
(ii) $\neg(p \wedge \neg q)$
(iii) $(p \vee q) \vee \neg q$
(iv) $(\neg p \rightarrow r) \wedge(p \leftrightarrow q)$
Solution:
(i) Truth table for $\neg p \wedge \neg q$

$\text { (ii) Truth table for } \neg(p \wedge \neg q)$

$\text { (iii) Truth table for }(p \vee q) \vee \neg q$

$\text { (iv) Truth table for }(\neg p \rightarrow r) \wedge(p \leftrightarrow q)$

Question 7.
Verify whether the following compound propositions are tautologies or contradictions or contingency
(i) $(p \wedge q) \wedge \neg(p \vee q)$
(ii) $((p \vee q) \wedge \neg p) \rightarrow q$
(iii) $(p \rightarrow q) \leftrightarrow(\neg p \rightarrow q)$
(iv) $((p \rightarrow q) \wedge(q \rightarrow r)) \rightarrow(p \rightarrow r)$
Solution:
(i) Truth table for $(p \wedge q) \wedge \neg(p \vee q)$

In the above Truth table the last column entries are ' $F$ '. So the given propositions is a contradiction.
(ii) Truth table for $((p \vee q) \wedge \neg p) \rightarrow q$

In the above truth table the last column entries are ' $\mathrm{T}$ '. So the given propositions is a tautology.
(iii) Truth table for $(p \rightarrow q) \leftrightarrow(\neg p \rightarrow \neg q)$

In the above truth table the entries in the last column are a combination of' $\mathrm{T}$ ' and ' $\mathrm{F}$ '. So the given statement is neither propositions is neither tautology nor a contradiction. It is a contingency.
(iv) Truth table for $((p \rightarrow q) \wedge(q \rightarrow r)) \rightarrow(p \rightarrow r)$

The last column entires are ' $\mathrm{T}$ '. So the given proposition is a tautology.
Question 8.
Show that $(i) \neg(p \wedge q) \equiv \neg p \vee \neg q$
(ii) $\neg(p \rightarrow q) \equiv p \wedge \neg q$
Solution:
(i) Truth table for $\neg(p \wedge q)$

$\text { Truth table for } \neg p \vee \neg q$

The entries in the column $\neg(p \wedge q)$ and $\neg p \vee \neg q$ are identical. So, $\neg(p \wedge q) \equiv \neg p \vee \neg q$
(ii) To prove $\neg(p \rightarrow q) \equiv p \wedge \neg q$.
Truth table for $\neg(p \rightarrow q)$

$\text { Truth table for } p \wedge \neg q$

The entries in the column $\neg(p \rightarrow q)$ and $p \wedge \neg q$ are identical and hence they are equivalent.
Question 9.
Prove that $q \rightarrow p \equiv \neg p \rightarrow \neg q$
Solution:
Truth table for $q \rightarrow p$

$\text { Truth table for } \neg p \rightarrow \neg q$

The entries in the column corresponding to $\mathrm{q} \rightarrow \mathrm{p}$ and $\neg p \rightarrow \neg q$ are identical and hence they are equivalent.
Question 10.
Show that $\mathrm{p} \rightarrow \mathrm{q}$ and $\mathrm{q} \rightarrow \mathrm{p}$ are not equivalent
Solution:
Truth table for $p \rightarrow q$

$\text { Truth table for } q \rightarrow p$

The entries in the column corresponding to $\mathrm{p} \rightarrow \mathrm{q}$ and $\mathrm{q} \rightarrow \mathrm{p}$ are not identical, hence they are not equivalent.
Question 11.
Show that $\neg(p \leftrightarrow q) \equiv p \leftrightarrow \neg q$
Solution:
Truth table for $\neg(p \leftrightarrow q)$

$\text { Trûth table for } p \leftrightarrow \neg q$

Question 12.
Check whether the statement $\mathrm{p} \rightarrow(\mathrm{q} \rightarrow \mathrm{p})$ is a tautology or a contradiction without using the truth table. Solution:
$
\begin{aligned}
p \rightarrow(q \rightarrow p) & \equiv \neg p \vee(q \rightarrow p) & & \text { (Conditional equivalence) } \\
& \equiv(q \rightarrow p) \vee \neg p & & \text { (Commutative law) } \\
& \equiv(\neg q \vee p) \vee \neg p & & \text { (Conditional equivalence) } \\
& \equiv \neg q \vee(p \vee \neg p) & & \text { (Associative law) } \\
& \equiv \neg q \vee \mathrm{T} & & \text { (Complement law) } \\
& \equiv \mathrm{T} & & \text { (Identity law) }
\end{aligned}
$
$\therefore p \rightarrow(q \rightarrow p)$ is a tautology.
Question 13.
Using truth table check whether the statements $\neg(p \vee q) \vee(\neg p \wedge q)$ and $\neg p$ are logically equivalent. Solution:
Truth table for $\neg(p \vee q) \vee(\neg p \wedge q)$

In the above table $\mathrm{C}_7$ and $\mathrm{C}_5$ are identical $\Rightarrow \neg(p \vee q) \vee(\neg p \wedge q) \equiv \neg p$ is verified.
Question 14.
Prove $p \rightarrow(q \rightarrow r) \equiv(p \wedge q) \rightarrow r$ without using truth table
Solution:
$
\begin{aligned}
& p \rightarrow(q \rightarrow r) \equiv \neg p \vee(q \rightarrow r) \quad \text { (Conditional equivalence) } \\
& \equiv(q \rightarrow r) \vee(\neg p) \quad \text { (Commutative law) } \\
& \equiv(\neg q \vee r) \vee(\neg p) \text { (Conditional equivalence) } \\
& \equiv(\neg p) \vee(\neg q \vee r) \text { (Commutative law) } \\
& \equiv(\neg p \vee \neg q) \vee r \text { (Associative law) } \\
& \equiv \neg(p \wedge q) \vee r \quad \text { (Demorgan's law) } \\
& \equiv(p \wedge q) \rightarrow r \quad \text { (Conditional equivalence) } \\
& \therefore p \rightarrow(q \rightarrow r) \equiv(p \wedge q) \rightarrow r \\
&
\end{aligned}
$
Question 15.
Prove that $p \rightarrow(\neg q \vee r) \equiv \neg p \vee(\neg q \vee r)$ using truth table.
Solution:
Truth table for $p \rightarrow(\neg q \vee r)$