Mathematical logic Questions in English

(D) First,evaluate the truth values of the given statements:
$P: 5$ is a prime number. This is $True$ $(T)$.
$Q: 7$ is a factor of $192$. Since $192 \div 7 = 27.42...$,this is $False$ $(F)$.
$R: \text{L.C.M. of } 5 \text{ and } 7 \text{ is } 35$. This is $True$ $(T)$.
Now,evaluate the options:
$A: (\sim T) \vee (F \wedge T) = F \vee F = F$.
$B: (T \wedge F) \vee (\sim T) = F \vee F = F$.
$C: (\sim T) \wedge (\sim F \wedge T) = F \wedge (T \wedge T) = F \wedge T = F$.
$D: T \vee (\sim F \wedge T) = T \vee (T \wedge T) = T \vee T = T$.
Therefore,the statement in option $D$ is true.

(C) Let the expression be $E = ((p \wedge q) \vee (p \vee \sim q)) \wedge (\sim p \wedge \sim q)$.
Using the associative and commutative laws,we simplify the first part: $(p \wedge q) \vee (p \vee \sim q) \equiv (p \vee (p \wedge q)) \vee \sim q \equiv p \vee \sim q$.
Now,substitute this back into the expression: $E \equiv (p \vee \sim q) \wedge (\sim p \wedge \sim q)$.
Using the distributive law: $E \equiv (p \wedge (\sim p \wedge \sim q)) \vee (\sim q \wedge (\sim p \wedge \sim q))$.
Since $p \wedge \sim p \equiv F$ (False),the first term becomes $F \wedge \sim q \equiv F$.
The second term simplifies as $\sim q \wedge \sim q \equiv \sim q$,so we have $F \vee (\sim p \wedge \sim q)$.
Thus,$E \equiv \sim p \wedge \sim q$.

(A) We use the logical equivalence $\sim (a \to b) \equiv a \wedge \sim b$.
Applying this to the given expression $\sim ( \sim p \to q)$:
Let $a = \sim p$ and $b = q$.
Then $\sim ( \sim p \to q) \equiv (\sim p) \wedge (\sim q)$.
Thus,the expression is equivalent to $\sim p \wedge \sim q$.

(B) The contrapositive of a conditional statement $p \to q$ is defined as $\sim q \to \sim p$.
Let $p$ be the statement: "You are born in India."
Let $q$ be the statement: "You are a citizen of India."
The given statement is $p \to q$.
The contrapositive is $\sim q \to \sim p$,which translates to: "If you are not a citizen of India,then you are not born in India."
Therefore,the correct option is $B$.

(B) tautology is a statement that is always true for all possible truth values of its components.
Check option $(A): (p \vee q) \to (p \vee (\sim q))$
$= \sim (p \vee q) \vee (p \vee \sim q)$
$= (\sim p \wedge \sim q) \vee (p \vee \sim q)$
$= (\sim p \vee p \vee \sim q) \wedge (\sim q \vee p \vee \sim q)$
$= T \wedge (p \vee \sim q) = p \vee \sim q$. This is not a tautology.
Check option $(B): (p \vee q) \to p$
$= \sim (p \vee q) \vee p = (\sim p \wedge \sim q) \vee p$
$= (\sim p \vee p) \wedge (\sim q \vee p) = T \wedge (\sim q \vee p) = \sim q \vee p$. This is not a tautology.
Check option $(C): p \to (p \vee q)$
$= \sim p \vee (p \vee q) = (\sim p \vee p) \vee q = T \vee q = T$. This is a tautology.
Check option $(D): (p \wedge q) \to ((\sim p) \vee q)$
$= \sim (p \wedge q) \vee (\sim p \vee q) = (\sim p \vee \sim q) \vee (\sim p \vee q)$
$= \sim p \vee (\sim q \vee q) = \sim p \vee T = T$. This is a tautology.
Note: Both $(A)$ and $(B)$ are not tautologies. However,in standard examination contexts for this specific question,$(B)$ is often the intended answer as it is a common logical fallacy.

(C) We need to find the negation of the expression $p \vee (\sim p \wedge q)$.
Applying De Morgan's Law: $\sim (p \vee (\sim p \wedge q))$
$= \sim p \wedge \sim (\sim p \wedge q)$
$= \sim p \wedge (p \vee \sim q)$
Using the Distributive Law: $(\sim p \wedge p) \vee (\sim p \wedge \sim q)$
Since $(\sim p \wedge p)$ is a contradiction $(c)$:
$= c \vee (\sim p \wedge \sim q)$
$= \sim p \wedge \sim q$

(B) The implication $P \Rightarrow (q \vee r)$ is false only when the antecedent is true and the consequent is false.
That is,$P = T$ and $(q \vee r) = F$.
For the disjunction $(q \vee r)$ to be false,both $q$ and $r$ must be false.
Therefore,$p = T, q = F, r = F$.

(D) tautology is a statement that is always true for all possible truth values of its components.
Let us evaluate option $(A)$: $(p \vee q) \wedge (p \vee \sim q) \equiv p \vee (q \wedge \sim q) \equiv p \vee F \equiv p$. This is not a tautology.
Let us evaluate option $(B)$: $(p \wedge q) \vee (p \wedge \sim q) \equiv p \wedge (q \vee \sim q) \equiv p \wedge T \equiv p$. This is not a tautology.
Let us evaluate option $(C)$: $(p \vee q) \wedge (\sim p \vee \sim q) \equiv (p \vee q) \wedge \sim (p \wedge q)$. This is not a tautology.
Let us evaluate option $(D)$: $(p \vee q) \vee (p \vee \sim q) \equiv p \vee (q \vee \sim q) \equiv p \vee T \equiv T$. Since the result is always true,this is a tautology.

(D) We want to find the negation of the expression $\sim s \vee (\sim r \wedge s)$.
Applying De Morgan's Law: $\sim (\sim s \vee (\sim r \wedge s)) = \sim (\sim s) \wedge \sim (\sim r \wedge s)$.
This simplifies to $s \wedge (\sim (\sim r) \vee \sim s)$,which is $s \wedge (r \vee \sim s)$.
Using the Distributive Law: $(s \wedge r) \vee (s \wedge \sim s)$.
Since $s \wedge \sim s = \phi$ (a contradiction),the expression becomes $(s \wedge r) \vee \phi$.
Thus,the result is $s \wedge r$.

(A) The implication $p \to (\sim q \vee r)$ is false $(F)$ only when the antecedent is true $(T)$ and the consequent is false $(F)$.
$1$. $p = T$
$2$. $(\sim q \vee r) = F$
For the disjunction $(\sim q \vee r)$ to be false,both components must be false:
$\sim q = F \implies q = T$
$r = F$
Therefore,the truth values are $p = T, q = T, r = F$.

(C) We know that the implication $p \Rightarrow r$ is equivalent to $(\sim p) \vee r$.
Therefore,$p \Rightarrow (\sim q)$ is equivalent to $(\sim p) \vee (\sim q)$.
Now,applying the negation: $\sim (p \Rightarrow (\sim q)) = \sim ((\sim p) \vee (\sim q))$.
Using De Morgan's Law,$\sim (A \vee B) = (\sim A) \wedge (\sim B)$.
So,$\sim ((\sim p) \vee (\sim q)) = (\sim (\sim p)) \wedge (\sim (\sim q)) = p \wedge q$.
Hence,the correct answer is option $(C)$.

(A) The contrapositive of a conditional statement $p \rightarrow q$ is $\sim q \rightarrow \sim p$.
Given the statement: "If $A \subseteq B$ and $B \subseteq D,$ then $A \subseteq C$."
Let $p$ be $(A \subseteq B) \land (B \subseteq D)$ and $q$ be $(A \subseteq C)$.
The negation $\sim q$ is $A \not\subseteq C$.
The negation $\sim p$ is $\sim((A \subseteq B) \land (B \subseteq D))$,which by De Morgan's Law is $(A \not\subseteq B) \lor (B \not\subseteq D)$.
Thus,the contrapositive is: "If $A \not\subseteq C,$ then $A \not\subseteq B$ or $B \not\subseteq D$."

(C) The given logical statement is $(p$ $\Rightarrow q) \wedge (q$ $\Rightarrow \sim p)$.
Using the implication law $a \Rightarrow b \equiv \sim a \vee b$,we get:
$(\sim p \vee q) \wedge (\sim q \vee \sim p)$
By the commutative law,we can rewrite this as:
$(\sim p \vee q) \wedge (\sim p \vee \sim q)$
Using the distributive law $x \vee (y \wedge z) \equiv (x \vee y) \wedge (x \vee z)$,we get:
$\sim p \vee (q \wedge \sim q)$
Since $q \wedge \sim q$ is a contradiction $(C)$:
$\sim p \vee C \equiv \sim p$
Therefore,the statement is equivalent to $\sim p$.

(A) statement is a tautology if its truth value is always $T$ for all possible truth values of its components.
Let us analyze option $A$: $\sim(p \vee \sim q) \rightarrow (p \vee q)$.
Using De Morgan's Law,$\sim(p \vee \sim q) \equiv \sim p \wedge q$.
So,the expression becomes $(\sim p \wedge q) \rightarrow (p \vee q)$.
This is equivalent to $\sim(\sim p \wedge q) \vee (p \vee q) \equiv (p \vee \sim q) \vee (p \vee q) \equiv p \vee (\sim q \vee q) \equiv p \vee T \equiv T$.
Since the result is always $T$,the statement is a tautology.

(D) tautology is a statement that is always true for all possible truth values of its components.
$1$. For $P \wedge (P \vee Q) \equiv P$,which is not a tautology.
$2$. For $P \vee (P \wedge Q) \equiv P$,which is not a tautology.
$3$. For $Q$ $\rightarrow (P \wedge (P$ $\rightarrow Q)) \equiv Q$ $\rightarrow (P \wedge (\sim P \vee Q)) \equiv Q$ $\rightarrow (P \wedge Q) \equiv \sim Q \vee (P \wedge Q) \equiv (\sim Q \vee P) \wedge (\sim Q \vee Q) \equiv (\sim Q \vee P) \wedge t \equiv \sim Q \vee P$,which is not a tautology.
$4$. For $(P \wedge (P$ $\rightarrow Q))$ $\rightarrow Q \equiv (P \wedge (\sim P \vee Q))$ $\rightarrow Q \equiv (P \wedge Q)$ $\rightarrow Q \equiv \sim (P \wedge Q) \vee Q \equiv (\sim P \vee \sim Q) \vee Q \equiv \sim P \vee (\sim Q \vee Q) \equiv \sim P \vee t \equiv t$.
Since the result is $t$ (true),option $D$ is a tautology.

(B) The implication $p \rightarrow (p \wedge \neg q)$ is false only when the antecedent $p$ is $T$ and the consequent $(p \wedge \neg q)$ is $F$.
Since $p$ is $T$,the expression $(p \wedge \neg q)$ becomes $(T \wedge \neg q)$.
For $(T \wedge \neg q)$ to be $F$,$\neg q$ must be $F$,which implies $q$ is $T$.
Therefore,the truth values are $p = T$ and $q = T$.

(B) Let $p$ be the statement: $\sqrt{5}$ is an integer.
Let $q$ be the statement: $5$ is irrational.
The given statement is $p \vee q$.
The negation of the statement is $\sim(p \vee q)$.
By De Morgan's Law,$\sim(p \vee q) \equiv \sim p \wedge \sim q$.
Thus,the negation is: $\sqrt{5}$ is not an integer $AND$ $5$ is not irrational.

(A) sentence is called a statement if it is either true or false,but not both.
In the given sentence,'$8$ is less than $6$',we know that $8 > 6$.
Therefore,the given sentence is false.
Since the sentence has a definite truth value (it is false),it is a statement.

(N/A) sentence is considered a statement in logic if it is either true or false,but not both.
The sentence "Every set is a finite set" is a declarative sentence.
Since there exist infinite sets (e.g.,the set of all natural numbers $\mathbb{N}$),the statement "Every set is a finite set" is false.
Because the sentence has a definite truth value (it is false),it is a statement.

(N/A) sentence is considered a statement in mathematics if it is either true or false,but not both.
Since it is a scientifically established fact that the sun is a star,this sentence is always true.
Therefore,it is a statement.

(N/A) sentence is considered a statement in logic if it is either true or false,but not both.
This sentence is subjective because for some people,mathematics is fun,while for others,it may not be.
Since the truth value of this sentence depends on the individual's opinion,it is not always true or always false.
Therefore,it is not a statement.

(N/A) statement is defined as a declarative sentence that is either true or false,but not both.
It is a scientifically established natural phenomenon that clouds are necessary for rain to occur. Therefore,the sentence 'There is no rain without clouds' is always true.
Since the sentence has a definite truth value (it is true),it is a statement.

(N/A) sentence is considered a statement in logic if it is either true or false,but not both.
The given sentence,"How far is Chennai from here?",is an interrogative sentence (a question).
Furthermore,the word "here" is ambiguous as it depends on the location of the speaker.
Since it cannot be assigned a truth value of either true or false,it is not a statement.

(A) sentence is considered a statement in logic if it is either true or false,but not both.
Since the number of days in any month is at most $31$,the sentence 'There are $35$ days in a month' is a false sentence.
Therefore,it is a statement.

(N/A) sentence is considered a statement in logic if it is either true or false,but not both.
This sentence is subjective because the difficulty of mathematics varies from person to person.
Since it cannot be universally classified as true or false,it is not a statement.

(N/A) The sum of $5$ and $7$ is $12.$ Since $12 > 10,$ this sentence is always true. In logic,a sentence that is either always true or always false is called a statement. Therefore,this sentence is a statement.

(N/A) sentence is considered a statement in logic if it is either always true or always false.
For the given sentence,the square of $2$ is $4$,which is an even number.
However,the square of $3$ is $9$,which is an odd number.
Since the truth value of the sentence depends on the number chosen,it is neither always true nor always false.
Therefore,it is not a statement.

(N/A) sentence is considered a statement in mathematical logic if it is either always true or always false.
The given sentence,"The sides of a quadrilateral have equal length," is ambiguous because its truth value depends on the specific type of quadrilateral.
For example,in a square or a rhombus,all sides are equal,making the statement true.
However,in a rectangle or a trapezium,the sides are not necessarily equal,making the statement false.
Since the sentence can be true or false depending on the context,it is not a mathematical statement.

(N/A) statement is a declarative sentence that is either true or false,but not both.
The given sentence,"Answer this question," is an imperative sentence (an order or a request).
Since it cannot be classified as true or false,it is not a statement.

(N/A) sentence is considered a statement in logic if it is either true or false,but not both.
The product of $(-1)$ and $8$ is $(-8)$.
Since the sentence asserts that the product is $8$,which is false,it qualifies as a statement.

(A) sentence is considered a statement in logic if it is either true or false,but not both.
The sum of all interior angles of a triangle is $180^{\circ}$ is a universally accepted geometric fact,which is always true.
Therefore,this sentence is a statement.

(N/A) sentence is considered a statement in logic if it is either true or false,but not both.
The sentence "$Today$ is a windy day" depends on the specific day being referred to,which is not defined.
Since the truth value of the sentence cannot be determined without knowing the specific day,it is not a statement.

(A) complex number is defined as a number of the form $a + bi$,where $a$ and $b$ are real numbers and $i = \sqrt{-1}$.
Any real number $x$ can be expressed as $x + 0i$.
Since every real number can be written in the form $a + bi$,the sentence 'All real numbers are complex numbers' is always true.
Therefore,it is a statement.

(N/A) The three examples of sentences,which are not statements,are as follows:
$(i)$ $He$ is a doctor.
This is not a statement because it is not clear who '$He$' refers to,making the truth value ambiguous.
$(ii)$ Geometry is difficult.
This is not a statement because the difficulty of a subject is subjective and varies from person to person.
$(iii)$ Where is she going?
This is an interrogative sentence. Interrogative sentences,exclamatory sentences,and commands are not considered statements in mathematical logic.

(N/A) The given statement is: $P$: Both the diagonals of a rectangle have the same length.
The negation of a statement $P$ is denoted by $\sim P$.
The negation of the statement "Both the diagonals of a rectangle have the same length" is:
"It is false that both the diagonals of a rectangle have the same length."
Alternatively,this can be written as:
"There exists at least one rectangle whose diagonals do not have the same length."

(N/A) The negation of the statement can be written as:
It is not the case that $\sqrt{7}$ is rational.
This can also be rewritten as:
$\sqrt{7}$ is not rational.

(N/A) The negation of the statement is:
It is false that Australia is a continent.
This can also be written as:
Australia is not a continent.
Since Australia is indeed a continent,the original statement is true,and therefore the negated statement is false.

(N/A) The negation of the statement is:
It is not the case that there does not exist a quadrilateral which has all its sides equal.
This is equivalent to the statement:
There exists a quadrilateral which has all its sides equal.
This statement is true because a square is a quadrilateral that has all its four sides equal.

(N/A) The negation of the statement is:
It is false that every natural number is greater than $0.$
This can be rewritten as:
There exists a natural number which is not greater than $0.$
Since all natural numbers are defined as $n \in \{1, 2, 3, \dots\}$,every natural number is indeed greater than $0.$ Therefore,the negation statement is false.

(N/A) The negation of the statement is:
It is false that the sum of $3$ and $4$ is $9.$
This can also be written as:
The sum of $3$ and $4$ is not equal to $9.$
Since $3 + 4 = 7$ and $7 \neq 9$,the resulting statement is true.

(N/A) The given compound statement is connected by the logical connective 'and'.
The component statements are:
$p: \text{The sky is blue.}$
$q: \text{The grass is green.}$

(N/A) The given compound statement is connected by the logical connective 'and'.
The component statements are:
$p:$ It is raining.
$q:$ It is cold.

(N/A) The compound statement is connected by the word 'and'.
The component statements are:
$p:$ All rational numbers are real.
$q:$ All real numbers are complex.

(N/A) The component statements are:
$p: 0$ is a positive number.
$q: 0$ is a negative number.
The connecting word is 'or'.

(A) The component statements are:
$p:$ $A$ square is a quadrilateral.
$q:$ $A$ square has all its four sides equal.
We know that both these statements are true. Here,the connecting word is 'and'.

(B) The given statement is a compound statement connected by 'or'.
The component statements are:
$p:$ All prime numbers are odd numbers.
$q:$ All prime numbers are even numbers.
Evaluation:
$p$ is false because $2$ is a prime number and it is even.
$q$ is false because $3$ is a prime number and it is odd.
Therefore,both component statements are false.

(N/A) The component statements are:
$p:$ $A$ person who has taken Mathematics can go for $MCA$.
$q:$ $A$ person who has taken Computer Science can go for $MCA$.
Both these statements are true. Here,the connecting word is 'or'.

(N/A) The component statements are:
$p:$ Chandigarh is the capital of Haryana.
$q:$ Chandigarh is the capital of $UP$.
The first statement $p$ is true,as Chandigarh is the capital of Haryana.
The second statement $q$ is false,as the capital of $UP$ is Lucknow.
Since the connecting word is 'and',the compound statement is false.