$$p :$$ we control population, $$q :$$ we prosper
$$\therefore $$ we have $${p \Rightarrow q}$$
Its negation is $$ \sim \left( {p \Rightarrow q} \right){\text{ i}}{\text{.e}}{\text{., }}p \wedge \sim q$$
i.e., we control population but we do not prosper.

$$\left( B \right)\frac{8}{4} = 2,\frac{{64}}{4} = 16;$$    but 4 is not prime.
Hence $$P \wedge Q \to R,{\text{false}}$$
$$\left( C \right)\frac{{{{\left( 6 \right)}^2}}}{{12}} = \frac{{36}}{{12}} = 3;$$    but 12 is not prime
Hence $$Q \to R,{\text{false}}$$
$$\left( D \right)\frac{{{{\left( 4 \right)}^2}}}{8} = \frac{{16}}{8} = 2;\frac{4}{8}$$     is not an integer
Hence $$Q \to P,{\text{false}}$$

Let $$p$$ and $$q$$ be two proposition given by $$p : 2^2 = 5, q : 1$$    get first class
Here give statement is $$p \to q$$
So contrapositive of $$p \to q{\text{ is }} \sim q \to \, \sim p$$
i.e., if I do not get first class then $${2^2} \ne 5.$$

"Please do me a favour" is not a statement.

Given that
$$p : 4$$  is an even prime number.
$$q : 6$$  is a divisor of 12.
and $$r :\,$$ the HCF of 4 and 6 is 2.
$$\therefore \,\, \sim p \vee \left( {q \wedge r} \right){\text{ is true}}{\text{.}}$$

$${p \wedge q}$$  means Mathematics is interesting and Mathematics is difficult.

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

All the statements in $$\left( A \right),\left( B \right)$$  and $$\left( C \right)$$ are equivalent and each is the negation of $$p.$$

$$p$$	$$ \sim p$$	$$p \Rightarrow \sim p$$	$$ \sim p \Rightarrow p$$	$$\left( {p \Rightarrow \sim p} \right) \wedge \left( { \sim p \Rightarrow p} \right)$$
T	F	F	T	F
F	T	T	F	F

Clearly, $$\left( {p \Rightarrow \, \sim p} \right) \wedge \left( { \sim p \Rightarrow p} \right)$$     is a contradiction.

We consider following truth table.

$$p$$	$$q$$	$$ \sim p$$	$$ \sim q$$	$$p \wedge q$$	$$p \vee q$$	$$\left( { \sim \left( {p \vee q} \right)} \right)$$	$$\left( {p \wedge q} \right) \wedge \left( { \sim \left( {p \vee q} \right)} \right)$$
T	T	F	F	T	T	F	F
T	F	F	T	F	T	F	F
F	T	T	F	F	T	F	F
F	F	T	T	F	F	T	F

Clearly last column of the above truth table contains only $$F.$$ Hence $$\left( {p \wedge q} \right) \wedge \left( { \sim \left( {p \vee q} \right.} \right)$$    is a contradiction.