Question

The negation of $$\left( {p \vee \sim q} \right) \wedge q{\text{ is}}$$

A. $$\left( { \sim p \vee q} \right) \wedge \sim q$$
B. $$\left( {p \wedge \sim q} \right) \vee q$$
C. $$\left( { \sim p \wedge q} \right) \vee \sim q$$  
D. $$\left( {p \wedge \sim q} \right) \vee \sim q$$
Answer :   $$\left( { \sim p \wedge q} \right) \vee \sim q$$
Solution :
$$ \sim \left\{ {\left( {p \vee \left( { \sim q} \right)} \right) \wedge q} \right\} = \left( { \sim \left( {p \vee \left( { \sim q} \right)} \right)} \right) \vee \left( { \sim q} \right)$$
By De Morgan's Law,
$$ \sim \left( {A \wedge B} \right) = \left( { \sim A} \right) \vee \left( { \sim B} \right)$$
$$ = \left( {\left( { \sim p} \right) \wedge \left( { \sim \left( { \sim q} \right)} \right)} \right) \vee \left( { \sim q} \right)$$       [Using De Morgan’s law again]
$$ = \left( { \sim p \wedge q} \right) \vee \left( { \sim q} \right)\left[ {\because \,\, \sim \left( { \sim q} \right) = q} \right]$$

Releted MCQ Question on
Algebra >> Mathematical Reasoning

Releted Question 1

Let $$p$$ be the statement “$$x$$ is an irrational number”, $$q$$ be the statement “$$y$$ is a transcendental number”, and $$r$$ be the statement “$$x$$ is a rational number if $$f y$$  is a transcendental number”.
Statement - 1 : $$r$$ is equivalent to either $$q$$ or $$p$$
Statement - 2 : $$r$$ is equivalent to $$ \sim \left( {p \leftrightarrow \sim q} \right).$$

A. Statement - 1 is false, Statement - 2 is true
B. Statement - 1 is true, Statement - 2 is true ; Statement - 2 is a correct explanation for Statement - 1
C. Statement - 1 is true, Statement - 2 is true ; Statement - 2 is not a correct explanation for Statement - 1
D. none of these
Releted Question 2

The statement $$p \to \left( {q \to p} \right)$$   is equivalent to

A. $$p \to \left( {p \to q} \right)$$
B. $$p \to \left( {p \vee q} \right)$$
C. $$p \to \left( {p \wedge q} \right)$$
D. $$p \to \left( {p \leftrightarrow q} \right)$$
Releted Question 3

Statement - 1 : $$ \sim \left( {p \leftrightarrow \sim q} \right)$$   is equivalent to $${p \leftrightarrow q}.$$
Statement - 2 : $$ \sim \left( {p \leftrightarrow \sim q} \right)$$   is a tautology

A. Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1.
B. Statement - 1 is true, Statement - 2 is false.
C. Statement - 1 is false, Statement - 2 is true.
D. Statement - 1 is true, Statement - 2 is true, Statement - 2 is a correct explanation for statement - 1
Releted Question 4

Consider the following statements
$$P$$ : Suman is brilliant
$$Q$$ : Suman is rich
$$R$$ : Suman is honest
The negation of the statement “Suman is brilliant and dishonest if and only if Suman is rich” can be expressed as

A. $$ \sim \left( {Q \leftrightarrow \left( {P \wedge \sim R} \right)} \right)$$
B. $$ \sim Q \leftrightarrow \sim P \wedge R$$
C. $$ \sim \left( {P \wedge \sim R} \right) \leftrightarrow Q$$
D. $$ \sim P \wedge \left( {Q \leftrightarrow \sim R} \right)$$

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Mathematical Reasoning


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