Mathematical Reasoning Test - 18

The contrapositive of $$(\sim p\wedge q)\rightarrow \sim r$$ is,
$$\equiv     \sim (\sim r) \rightarrow \sim  (\sim p\wedge q)$$
$$\equiv     r \rightarrow  ( p   \vee \sim q)$$

(i) The component statements of " Mumbai is the capital of Rajasthan or Maharashtra" are
$$p :$$ Mumbai is the capital of Rajasthan.
$$q :$$ Mumbai is the capital of Maharashtra.
We note that $$p$$ is false and $$q$$ is true, so the compound statement is true

(ii) The component statements of $$\displaystyle \sqrt{3}$$ is a rational or an irrational are
$$p :$$ $$\displaystyle \sqrt{3}$$ is a rational number.
$$q:$$ $$\displaystyle \sqrt{3}$$ is an irrational number.
We note that $$p$$ is false and $$q$$ is true, so the compound statement is true.

(iii) The component statements of $$125$$ is a multiple of $$7$$ or $$8$$ are
$$p:125$$ is a multiple of $$7.$$
$$q:125$$ is a multiple of $$8.$$
We note that $$p$$ and $$q$$ both are false statements, so compound statement is false.

(iv) The component statements of "A rectangle is a quadrilateral or a regular hexagon." are
$$p: A$$ rectangle is a quadrilateral.
$$q: A$$ rectangle is a regular hexagon.
We note that $$p$$ is true and $$q$$ is false, so the compound statement is true.

The converse of $$p \rightarrow (q \rightarrow r)$$ is,
$$\equiv  (q \to r) \to p \equiv  (\sim q \vee r)  \to p \equiv (q\wedge \sim r) \to p$$

The contrapositive of $$\sim p \rightarrow ( q \rightarrow \sim r)$$ is

Let $$p, q, r$$ be three statements defined as
$$p$$ : a number $$N$$ is divisible by $$15$$
$$q$$ : number $$N$$ is divisible by $$5$$
$$r$$ : number $$N$$ is divisible by $$3$$
Here given statement is $$p \rightarrow (q \vee r)$$ 
Here negative of above statement is
$$\sim (p \rightarrow (q \vee  r))\equiv p \wedge (\sim (q \vee r)$$
$$\equiv p \wedge ( \sim q \wedge \sim r)$$
i.e. A number is divisible by $$15$$ and it is not divisible by $$5$$ and $$3$$.

This follows the following pattern
$$\alpha |\beta \beta |\alpha \alpha \alpha |\beta \beta \beta \beta |\alpha \alpha \alpha \alpha \alpha |\beta \beta \beta \beta \beta \beta $$
Therefore $$\beta\alpha\beta\alpha$$ is the missing part.
Hence, option 'B' is correct..

The converse of $$p \rightarrow (q \rightarrow r)$$ is,

The negation of the statement $$q \vee (p \wedge \sim r) $$ is
$$\equiv \sim (q \vee (p \wedge \sim r) ) \equiv  \sim q \wedge (\sim p \vee \sim(\sim r))\equiv \sim q \wedge (\sim p \vee r)\equiv \sim q \wedge (p\to r) $$

A. $$(p \rightarrow q) \wedge r \equiv (T \rightarrow F) \wedge T \equiv F \wedge T \equiv F$$
B. $$(p \rightarrow q) \vee \sim r\equiv (T \rightarrow F) \vee F \equiv F \vee F \equiv F$$
C. $$(p \wedge q) \vee (q \wedge r)\equiv(T \wedge F) \vee (F \wedge T)\equiv F\vee F \equiv F $$
D. $$(p \rightarrow q) \rightarrow r \equiv (T \rightarrow F) \rightarrow T\equiv F\rightarrow T \equiv T$$