Mathematical Reasoning MCQ Questions & Answers in Algebra | Maths

Whatever the truth value of $$p$$ may be, $$p \vee \left( { \sim p} \right)$$   is always true. Hence, $$p \vee \left( { \sim p} \right)$$   is a tautology.

Since, $$ \sim \left( {p \vee q} \right) \equiv \left( { \sim p \, \wedge \sim q} \right){\text{and }} \sim \left( {p \wedge q} \right) \equiv \left( { \sim p \wedge q} \right)$$
So option $$\left( B \right)$$ and $$\left( D \right)$$ are not true.
$$\left( {p \to q} \right) \equiv p \wedge \sim q),$$     so option $$\left( C \right)$$ are not true.
$$\eqalign{ & {\text{Now, }}p \to q \sim p \vee q \cr & \sim q \to \sim p \equiv \left[ { \sim \left( { \sim q} \right) \vee \sim p} \right] \equiv q \, \vee \sim p \equiv \, \sim p \vee q \cr & p \to q \equiv \sim q \to \sim p \cr} $$

'Two plus two is four', is a statement.

$$p : 3 + 3 = 33 , q : 1 + 2 = 12$$
Truth value of both $$p$$ and $$q$$ is $$F.$$
$$\therefore \,\, \sim \left( {F \vee F} \right) \equiv \,\, \sim F \equiv T$$

The inverse of the proposition $$\left( {p \, \wedge \sim q} \right) \to r{\text{ is }} \sim \left( {p \, \wedge \sim q} \right) \to \, \sim r$$
$$\eqalign{ & \equiv \,\, \sim p \, \vee \sim \left( { \sim q} \right) \to \, \sim r \cr & \equiv \,\, \sim p \vee q \to \, \sim r \cr} $$

$$p \Rightarrow \left( { \sim p \vee q} \right)$$    is false means $$p$$ is true and $${ \sim p \vee q}$$   is false.
$$ \Rightarrow p$$  is true and both $$\sim p$$  and $$q$$ are false
$$ \Rightarrow p$$  is true and $$q$$ is false.

$$\eqalign{ & \left( {p \wedge \sim q} \right) \vee q \vee \left( { \sim p \wedge q} \right) \cr & \Rightarrow \,\,\left\{ {\left( {p \vee q} \right) \wedge \left( { \sim q \vee q} \right)} \right\} \vee \left( { \sim p \wedge q} \right) \cr & \Rightarrow \,\,\left\{ {\left( {p \vee q} \right) \wedge T} \right\} \vee \left( { \sim p \wedge q} \right) \cr & \Rightarrow \,\,\left( {p \vee q} \right) \vee \left( { \sim p \wedge q} \right) \cr & \Rightarrow \,\,\left\{ {\left( {p \vee q} \right) \vee \sim p} \right\} \wedge \left( {p \vee q \vee q} \right) \cr & \Rightarrow \,\,T \wedge \left( {p \vee q} \right) \cr & \Rightarrow \,\,p \vee q \cr} $$

The inverse of $$p \Rightarrow \, \sim q{\text{ is }} \sim p \Rightarrow q.$$
The contrapositive of $$ \sim p \Rightarrow q{\text{ is }} \sim q \Rightarrow p.$$
[ $$\because $$ contrapositive of $$p \Rightarrow q{\text{ is }} \sim q \Rightarrow \, \sim p$$    ].

$$\eqalign{ & \sim \left( {p \Rightarrow q} \right) \equiv p \wedge \sim q \cr & \therefore \,\, \sim \left( { \sim p \Rightarrow \, \sim q} \right) \equiv \,\, \sim p \wedge \sim \left( { \sim q} \right) \equiv \,\, \sim p \wedge q \cr & {\text{Thus, }} \sim \left( { \sim p \Rightarrow \, \sim q} \right) \equiv \,\, \sim p \wedge q \cr} $$

Given : $$P :$$ He is intelligent.
$$Q =$$  He is strong.
Symbolic form of
“It is not true that he is either intelligent or strong” is $$ \sim \left( {P \vee Q} \right)$$