Chapter 14 Mathematical Reasoning (Additional Questions)

A proposition is a declarative sentence that is either true or false, but not both. It must have a definite truth value that is universally accepted at a given time.

Let's analyze each option based on the definition:

(A) "What is your name?" - This is an interrogative sentence (a question). It asks for information and does not make a statement that can be judged as true or false. Therefore, it is not a proposition.

(B) "Delhi is the capital of India." - This is a declarative sentence. It makes a statement about the relationship between Delhi and India. This statement is objectively true. Since it is a declarative sentence with a definite truth value (True), it is a proposition.

(C) "Close the door." - This is an imperative sentence (a command). It gives an instruction and does not make a statement that can be judged as true or false. Therefore, it is not a proposition.

(D) "$x + 1 = 5$." - This is an open sentence or a predicate. Its truth value depends on the specific value assigned to the variable $x$. For instance, if $x=4$, the statement "$4+1=5$" is true. If $x=2$, the statement "$2+1=5$" (which is $3=5$) is false. Since its truth value is not fixed and depends on the variable, it is not a proposition unless $x$ is quantified (e.g., "For all $x$, $x+1=5$") or assigned a specific value.

Comparing the options with the definition of a proposition, only option (B) is a declarative sentence with a definite truth value.

Thus, the correct option is (B) Delhi is the capital of India.

The definition provided states that a proposition is a sentence that is either true or false, but not both. We need to identify which type of sentence fits this description.

Let's examine the given options:

(A) A question (interrogative sentence) asks for information. It does not make a statement that can be assigned a truth value (true or false). For example, "Where is the library?" is a question, not a proposition.

(B) A command (imperative sentence) gives an order or instruction. It also does not make a statement that can be assigned a truth value. For example, "Sit down." is a command, not a proposition.

(C) A declarative sentence makes a statement or assertion. Such a statement can be objectively determined to be either true or false (but not both) in a specific context. For example, "The sun rises in the east." is a declarative sentence that is true, making it a proposition. "Cats bark." is a declarative sentence that is false, also making it a proposition.

(D) An exclamation (exclamatory sentence) expresses strong emotion. It does not make a statement that can be assigned a truth value. For example, "What a beautiful day!" is an exclamation, not a proposition.

Based on the analysis, the definition of a proposition precisely describes the nature of a declarative sentence that has a definite truth value.

Thus, the definition refers to a declarative sentence.

The correct option is (C) A declarative sentence.

The negation of a statement $P$, denoted by $\neg P$, is a statement that is true whenever $P$ is false, and false whenever $P$ is true. Essentially, it is the statement that denies the truth of the original statement.

Let the given statement be $P$: "The sun is shining".

We are looking for the negation of this statement, $\neg P$.

Let's evaluate the options:

(A) "The sun is not shining." - This statement is true if and only if the original statement "The sun is shining" is false. This directly fits the definition of negation.

(B) "Is the sun shining?" - This is an interrogative sentence (a question) and not a statement that can be true or false. Therefore, it cannot be the negation of a proposition.

(C) "The sun is setting." - This is a different statement. The truth of "The sun is setting" does not necessarily mean "The sun is shining" is false (e.g., the sun can be shining while setting). This is not the logical negation.

(D) "The moon is shining." - This statement is unrelated to whether the sun is shining or not. It is a separate proposition and not the negation of the given statement.

The sentence that directly denies the original statement "The sun is shining" is "The sun is not shining".

Thus, the negation of the statement "The sun is shining" is "The sun is not shining".

The correct option is (A) The sun is not shining.

We are given two simple statements:

$p$: "It is raining"

$q$: "The ground is wet"

The compound statement is "It is raining and the ground is wet".

The word "and" is a logical connective used to combine two statements. In propositional logic, the connective corresponding to "and" is called conjunction.

The symbol for conjunction is $\land$.

Therefore, the compound statement "It is raining and the ground is wet" which combines statement $p$ and statement $q$ using "and", is symbolically represented as $p \land q$.

Let's look at the given options and their corresponding meanings:

(A) $p \lor q$ represents the disjunction "It is raining or the ground is wet".

(B) $p \land q$ represents the conjunction "It is raining and the ground is wet".

(C) $p \to q$ represents the implication "If it is raining, then the ground is wet".

(D) $\neg p$ represents the negation "It is not raining".

Comparing the given compound statement with the meanings of the options, we find that $p \land q$ correctly represents "It is raining and the ground is wet".

The correct option is (B) $p \land q$.

In propositional logic, different symbols are used to represent logical connectives which combine simple statements to form compound statements. The connective 'or' is used to represent the disjunction of two statements.

Let's list the common logical connectives and their standard symbols:

- Conjunction (and): represented by $\land$

- Disjunction (or): represented by $\lor$

- Negation (not): represented by $\neg$

- Implication (if... then...): represented by $\to$ (or $\Rightarrow$)

- Biconditional (...if and only if...): represented by $\leftrightarrow$ (or $\Leftrightarrow$)

The question asks for the symbol that represents the logical connective 'or'. From the list above, the symbol for disjunction ('or') is $\lor$.

Let's check the given options:

(A) $\land$ represents 'and'.

(B) $\lor$ represents 'or'.

(C) $\neg$ represents 'not'.

(D) $\to$ represents 'if... then...'.

The symbol corresponding to 'or' is $\lor$.

The correct option is (B) $\lor$.

The original statement is a universal statement: "All birds can fly". This statement asserts that every single bird has the property of being able to fly.

The negation of a statement is a statement that is true when the original statement is false, and false when the original statement is true.

If the statement "All birds can fly" is false, it means that it is not true that every single bird can fly. This implies that there must be at least one bird that cannot fly.

Conversely, if there is at least one bird that cannot fly, then the statement "All birds can fly" is false.

Let's analyze the options:

(A) "All birds cannot fly." - This means every single bird cannot fly. This is a very strong statement. If some birds can fly and some cannot, this statement is false, but so is the original statement. This is not the correct negation. This is the negation of "Some birds can fly".

(B) "No birds can fly." - This is equivalent to "All birds cannot fly". This means for every bird, it cannot fly. Same as (A), not the correct negation. This is the negation of "Some birds can fly".

(C) "Some birds cannot fly." - This means there exists at least one bird that cannot fly. If this is true, then it is not true that all birds can fly, so the original statement is false. If this is false (meaning no birds cannot fly, i.e., all birds *can* fly), then the original statement is true. This matches the definition of negation.

(D) "Some birds can fly." - This means there exists at least one bird that can fly. This is not the negation of "All birds can fly". For example, if only some birds can fly (and some cannot), the original statement "All birds can fly" is false, but this statement "Some birds can fly" is true. The negation should be true when the original is false and vice versa.

The negation of a universal statement "All P are Q" is an existential statement "Some P are not Q".

In this case, P is "birds" and Q is "can fly". The negation of "All birds can fly" is "Some birds cannot fly".

The correct option is (C) Some birds cannot fly.

This is an Assertion-Reason type question. We need to evaluate the truthfulness of both the Assertion (A) and the Reason (R) and then determine if the Reason correctly explains the Assertion.

Let's analyze the Assertion (A): "The statement 'If a number is divisible by 6, then it is divisible by 3' is a true statement."

This is a conditional statement of the form "If P, then Q", where P is "a number is divisible by 6" and Q is "it is divisible by 3". To verify if this statement is true, we need to check if whenever P is true, Q is also true.

Consider any number $n$ that is divisible by 6. By the definition of divisibility, this means there exists an integer $k$ such that:

$n = 6k$

We can rewrite the number 6 as the product of 3 and 2:

$6 = 3 \times 2$

Substituting this into the equation for $n$:

$n = (3 \times 2)k$

Using the associative property of multiplication, we can group the terms differently:

$n = 3 \times (2k)$

Let $m = 2k$. Since $k$ is an integer, $2k$ is also an integer. So, $m$ is an integer.

Thus, we have $n = 3m$, where $m$ is an integer. By the definition of divisibility, this means $n$ is divisible by 3.

This shows that any number divisible by 6 is indeed divisible by 3. Therefore, the Assertion (A) is a true statement.

Now let's analyze the Reason (R): "If $n$ is divisible by 6, $n = 6k$ for some integer $k$. $n = 3(2k)$, which means $n$ is divisible by 3."

This statement provides a sequence of steps starting from the premise "If $n$ is divisible by 6" and concluding that "$n$ is divisible by 3".

- "If $n$ is divisible by 6, $n = 6k$ for some integer $k$." - This is the correct definition of divisibility by 6.

- "$n = 3(2k)$" - This is a correct algebraic manipulation, rewriting $6k$ as $3 \times (2k)$.

- "which means $n$ is divisible by 3." - This is the correct conclusion based on the form $n = 3 \times (\text{an integer})$.

The steps provided in Reason (R) are logically sound and mathematically correct. Therefore, the Reason (R) is a true statement.

Finally, we examine the relationship between the Assertion and the Reason. The Reason (R) provides the exact mathematical derivation and explanation for why a number divisible by 6 is necessarily divisible by 3. It explains the underlying principle behind the truth of the Assertion (A).

Therefore, Reason (R) is the correct explanation for Assertion (A).

Based on our analysis:

- Assertion (A) is true.

- Reason (R) is true.

- Reason (R) is the correct explanation for Assertion (A).

This corresponds to option (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

In propositional logic, for a given conditional statement (implication) of the form "$p \to q$" (read as "If $p$, then $q$"), there are related conditional statements: the converse, the inverse, and the contrapositive.

The original implication is $p \to q$.

The converse of $p \to q$ is formed by interchanging the hypothesis (antecedent) $p$ and the conclusion (consequent) $q$. This results in the statement $q \to p$ (read as "If $q$, then $p$").

The inverse of $p \to q$ is formed by negating both the hypothesis and the conclusion: $\neg p \to \neg q$ (read as "If not $p$, then not $q$").

The contrapositive of $p \to q$ is formed by interchanging and negating both the hypothesis and the conclusion: $\neg q \to \neg p$ (read as "If not $q$, then not $p$").

The question asks for the converse of $p \to q$. According to the definition, the converse is $q \to p$.

Let's look at the options:

(A) $q \to p$ - This matches the definition of the converse.

(B) $\neg p \to \neg q$ - This is the inverse.

(C) $\neg q \to \neg p$ - This is the contrapositive.

(D) $p \land q$ - This is the conjunction (p and q).

The correct option is (A) $q \to p$.

We need to match each type of statement with its correct example from the provided lists.

Let's analyze each type and find its corresponding example:

(i) Simple Statement: A simple statement is a declarative sentence that does not contain any other sentence as a component.

Example (b) "The sky is blue." is a single, straightforward declarative sentence without any logical connectives combining other statements. This is a simple statement.

Match: (i) - (b)

(ii) Compound Statement with 'and': A compound statement formed by joining two statements with the connective 'and'.

Example (d) "It is sunny and warm." joins the simple statements "It is sunny" and "It is warm" using the connective 'and'. This is a compound statement with 'and'.

Match: (ii) - (d)

(iii) Compound Statement with 'or': A compound statement formed by joining two statements with the connective 'or'.

Example (c) "The sun is hot or the moon is cold." joins the simple statements "The sun is hot" and "The moon is cold" using the connective 'or'. This is a compound statement with 'or'.

Match: (iii) - (c)

(iv) Implication: A conditional statement of the form "If P, then Q".

Example (a) "If it is cold, then it is snowing." has the structure "If P, then Q", where P is "it is cold" and Q is "it is snowing". This is an implication.

Match: (iv) - (a)

Combining the matches, we have:

(i) - (b)

(ii) - (d)

(iii) - (c)

(iv) - (a)

Now let's check the options to find the one that matches our determined pairs.

Option (A) is (i)-(b), (ii)-(d), (iii)-(c), (iv)-(a). This matches our results.

The correct option is (A) (i)-(b), (ii)-(d), (iii)-(c), (iv)-(a).

In mathematical reasoning, a statement (or proposition) is defined as a declarative sentence that is either true or false, but not both. Sentences that are questions, commands, or exclamations are generally not considered statements.

Let's examine each option:

(A) "$1 + 1 = 2$." - This is a declarative sentence making a claim about a mathematical equality. This claim is objectively true. Since it is a declarative sentence with a definite truth value, it is a statement.

(B) "The square of a negative number is positive." - This is a declarative sentence making a mathematical assertion. This assertion is objectively true (for any real number $x < 0$, $x^2 > 0$). Since it is a declarative sentence with a definite truth value, it is a statement.

(C) "All triangles are equilateral." - This is a declarative sentence making a claim about triangles. This claim is objectively false (e.g., a right triangle or an isosceles triangle is not equilateral). Since it is a declarative sentence with a definite truth value, it is a statement.

(D) "What a beautiful day!" - This is an exclamatory sentence expressing feeling or emotion. It does not make a claim that can be judged as true or false. Therefore, it is not a statement in the context of mathematical reasoning.

The question asks which of the given options is NOT a statement in mathematical reasoning. Based on the analysis, option (D) does not fit the definition of a statement.

The correct option is (D) What a beautiful day!

The given statement is an existential statement. It asserts that there is at least one number $x$ that satisfies the condition $x^2 < x$.

The structure of the statement is "There exists $x$ such that $P(x)$", where $P(x)$ is the predicate "$x^2 < x$". Symbolically, this is $\exists x, P(x)$.

The negation of an existential statement "There exists $x$ such that $P(x)$" is a universal statement that asserts that for all $x$, the property $P(x)$ is false.

The negation of $\exists x, P(x)$ is $\neg (\exists x, P(x))$.

Using the rules of predicate logic, the negation of an existential quantifier ($\exists$) is a universal quantifier ($\forall$) applied to the negation of the predicate.

So, $\neg (\exists x, P(x)) \equiv \forall x, \neg P(x)$.

In our case, $P(x)$ is "$x^2 < x$".

The negation of $P(x)$, $\neg P(x)$, is the negation of "$x^2 < x$". The negation of "less than" ($<$) is "greater than or equal to" ($\geq$).

So, $\neg P(x)$ is "$x^2 \geq x$".

Applying the rule $\forall x, \neg P(x)$, the negation of the original statement is "For all numbers $x$, $x^2 \geq x$".

Let's examine the given options:

(A) "There exists a number $x$ such that $x^2 \geq x$." - This is an existential statement with the negated predicate. It is not the negation of the original statement.

(B) "For all numbers $x$, $x^2 \geq x$." - This is a universal statement with the negated predicate. This matches our derivation.

(C) "For all numbers $x$, $x^2 < x$." - This is a universal statement with the original predicate. It is not the negation.

(D) "There exists no number $x$ such that $x^2 < x$." - This statement means "It is false that there exists a number $x$ such that $x^2 < x$". This is logically equivalent to "For all numbers $x$, it is not the case that $x^2 < x$", which is "For all numbers $x$, $x^2 \geq x$". While logically equivalent to (B), option (B) is the standard form obtained by directly applying the negation rules for quantifiers and predicates ($\neg \exists x P(x) \equiv \forall x \neg P(x)$). Option (B) is the most direct representation of the negation.

The standard negation of "There exists $x$ such that $P(x)$" is "For all $x$, not $P(x)$".

Therefore, the negation of "There exists a number $x$ such that $x^2 < x$" is "For all numbers $x$, $x^2 \geq x$".

The correct option is (B) For all numbers $x$, $x^2 \geq x$.

The given statement is "If today is Monday, then tomorrow is Tuesday".

This statement has the structure "If P, then Q", where:

P is the statement "today is Monday".

Q is the statement "tomorrow is Tuesday".

Let's consider the types of compound statements based on logical connectives:

- Conjunction uses the connective 'and' to combine two statements (P and Q, written as $P \land Q$). Example: "Today is Monday and tomorrow is Tuesday."

- Disjunction uses the connective 'or' to combine two statements (P or Q, written as $P \lor Q$). Example: "Today is Monday or tomorrow is Tuesday."

- Implication (or Conditional) uses the structure 'If P, then Q' to combine two statements (written as $P \to Q$). Example: "If today is Monday, then tomorrow is Tuesday."

- Negation is formed by denying a single statement ($\neg P$). Example: "Today is not Monday."

The given statement "If today is Monday, then tomorrow is Tuesday" directly matches the structure and definition of an implication ($P \to Q$).

Comparing this with the given options:

(A) Conjunction - does not match.

(B) Disjunction - does not match.

(C) Implication - matches.

(D) Negation - does not match.

Therefore, the statement is an example of an implication.

The correct option is (C) Implication.

For a given conditional statement (implication) of the form "$p \to q$" (read as "If $p$, then $q$"), the contrapositive is a related conditional statement.

The contrapositive of $p \to q$ is formed by two steps:

1. Negate both the hypothesis ($p$) and the conclusion ($q$), resulting in $\neg p$ and $\neg q$.

2. Interchange the negated hypothesis and the negated conclusion. The negated conclusion becomes the new hypothesis, and the negated hypothesis becomes the new conclusion.

Applying these steps to $p \to q$, we negate $p$ to get $\neg p$, and we negate $q$ to get $\neg q$. Then, we interchange them to form the implication $\neg q \to \neg p$ (read as "If not $q$, then not $p$").

The standard related conditional statements derived from $p \to q$ are:

- Original Implication: $p \to q$

- Converse: $q \to p$

- Inverse: $\neg p \to \neg q$

- Contrapositive: $\neg q \to \neg p$

The question asks for the contrapositive of $p \to q$. According to the definition, the contrapositive is $\neg q \to \neg p$.

Let's examine the given options:

(A) $q \to p$ - This is the converse.

(B) $\neg p \to \neg q$ - This is the inverse.

(C) $\neg q \to \neg p$ - This matches the definition of the contrapositive.

(D) $p \land \neg q$ - This represents "p and not q". This is the negation of the implication $p \to q$.

The symbol for the contrapositive of $p \to q$ is $\neg q \to \neg p$.

The correct option is (C) $\neg q \to \neg p$.

We are given two simple statements related to an integer $n$:

$p$: "An integer $n$ is even"

$q$: "Its square $n^2$ is even"

The original statement provided in the case study is "If an integer is even, then its square is even."

Let's rephrase this statement using the defined variables $p$ and $q$:

"If an integer is even, then its square is even."

"If $p$, then $q$."

A statement with the structure "If $P$, then $Q$" is a conditional statement or implication, and it is symbolically represented as $P \to Q$.

In this case, $P$ is the statement $p$ and $Q$ is the statement $q$.

Therefore, the statement "If $p$, then $q$" is represented by $p \to q$.

Let's look at the given options and their meanings:

(A) $p \land q$ represents the conjunction "p and q", which means "An integer $n$ is even and its square $n^2$ is even."

(B) $p \lor q$ represents the disjunction "p or q", which means "An integer $n$ is even or its square $n^2$ is even."

(C) $p \to q$ represents the implication "If p, then q", which means "If an integer $n$ is even, then its square $n^2$ is even." This matches the original statement.

(D) $q \to p$ represents the implication "If q, then p", which means "If the square of an integer $n^2$ is even, then the integer $n$ is even." This is the converse of the original statement.

The original statement is correctly represented by the implication $p \to q$.

The correct option is (C) $p \to q$.

In the previous question (Question 14), we defined the statements:

$p$: "An integer $n$ is even"

$q$: "Its square $n^2$ is even"

The original statement is "If an integer is even, then its square is even", which is represented as the implication $p \to q$.

The converse of an implication $p \to q$ is obtained by interchanging the hypothesis ($p$) and the conclusion ($q$). The converse is $q \to p$.

Now, let's translate the converse $q \to p$ back into English using the definitions of $p$ and $q$:

$q \to p$ means "If $q$, then $p$".

Substituting the statements for $p$ and $q$:

"If its square $n^2$ is even, then an integer $n$ is even."

This can be phrased as "If the square of an integer is even, then the integer is even."

Let's compare this with the given options:

(A) If an integer is even, then its square is not even. (This is $p \to \neg q$)

(B) If an integer's square is even, then the integer is even. (This is $q \to p$)

(C) If an integer is not even, then its square is not even. (This is $\neg p \to \neg q$, the inverse)

(D) If an integer's square is not even, then the integer is not even. (This is $\neg q \to \neg p$, the contrapositive)

The statement corresponding to the converse $q \to p$ is "If an integer's square is even, then the integer is even."

The correct option is (B) If an integer's square is even, then the integer is even.

From the Case Study in Question 14, we have the statements:

$p$: "An integer $n$ is even"

$q$: "Its square $n^2$ is even"

The original statement is the implication $p \to q$: "If an integer is even, then its square is even."

The contrapositive of an implication $p \to q$ is obtained by negating both the hypothesis ($p$) and the conclusion ($q$) and then interchanging them. The contrapositive is $\neg q \to \neg p$.

First, let's find the negations of $p$ and $q$:

$\neg p$: The negation of "An integer $n$ is even" is "An integer $n$ is not even". An integer that is not even is odd. So, $\neg p$ is "An integer $n$ is odd".

$\neg q$: The negation of "Its square $n^2$ is even" is "Its square $n^2$ is not even". An integer square that is not even is odd. So, $\neg q$ is "Its square $n^2$ is odd".

Now, we form the contrapositive $\neg q \to \neg p$:

"If $\neg q$, then $\neg p$."

Substituting the English phrases for $\neg q$ and $\neg p$:

"If an integer's square is not even, then the integer is not even."

This can also be phrased using the term 'odd': "If an integer's square is odd, then the integer is odd."

Let's compare this with the given options:

(A) If an integer is even, then its square is not even. ($p \to \neg q$)

(B) If an integer's square is even, then the integer is even. ($q \to p$, the converse)

(C) If an integer is not even, then its square is not even. ($\neg p \to \neg q$, the inverse)

(D) If an integer's square is not even, then the integer is not even. ($\neg q \to \neg p$, the contrapositive)

The statement corresponding to the contrapositive $\neg q \to \neg p$ is "If an integer's square is not even, then the integer is not even."

The correct option is (D) If an integer's square is not even, then the integer is not even.

A conjunction of two statements $p$ and $q$, denoted by $p \land q$, is a compound statement formed by connecting $p$ and $q$ with the word "and".

The truth value of a compound statement depends on the truth values of its simple components and the logical connective used. For a conjunction $p \land q$, the rule for determining its truth value is as follows:

The conjunction $p \land q$ is true only if both $p$ and $q$ are true. In all other cases (i.e., if $p$ is false, or $q$ is false, or both are false), the conjunction $p \land q$ is false.

This can be summarized in a truth table:

Looking at the truth table, the statement $p \land q$ has a truth value of 'True' in only one case: when $p$ is True and $q$ is True.

Now let's examine the given options in light of this rule:

(A) "$p$ is true or $q$ is true." - This describes the condition for a disjunction ($p \lor q$) to be true.

(B) "$p$ is true and $q$ is true." - This exactly matches the condition for a conjunction ($p \land q$) to be true.

(C) "$p$ is false and $q$ is false." - If this condition is met, the conjunction $p \land q$ is false.

(D) "$p$ is false or $q$ is false." - This condition means that at least one of $p$ or $q$ is false. If this is met, the conjunction $p \land q$ is false.

The condition under which the truth value of a conjunction ($p \land q$) is true is when both $p$ is true and $q$ is true.

The correct option is (B) $p$ is true and $q$ is true.

A disjunction of two statements $p$ and $q$, denoted by $p \lor q$, is a compound statement formed by connecting $p$ and $q$ with the word "or".

The truth value of a disjunction $p \lor q$ is determined as follows:

The disjunction $p \lor q$ is true if at least one of the statements $p$ or $q$ is true (this includes the case where both are true). This corresponds to the inclusive 'or'.

Consequently, the disjunction $p \lor q$ is false only when both $p$ and $q$ are false.

This can be summarized in a truth table:

Looking at the truth table, the statement $p \lor q$ has a truth value of 'False' in only one case: when $p$ is False and $q$ is False.

Now let's examine the given options in light of this rule:

(A) "$p$ is true or $q$ is true." - If this condition is met, the disjunction $p \lor q$ is true.

(B) "$p$ is true and $q$ is true." - If this condition is met, the disjunction $p \lor q$ is true.

(C) "$p$ is false and $q$ is false." - This exactly matches the condition for a disjunction ($p \lor q$) to be false.

(D) "$p$ is false or $q$ is false." - If this condition is met, the disjunction $p \lor q$ is true (unless both are false, in which case it's false, but this option doesn't specify *only* when it's false).

The condition under which the truth value of a disjunction ($p \lor q$) is false is when both $p$ is false and $q$ is false.

The correct option is (C) $p$ is false and $q$ is false.

An implication or conditional statement "$p \to q$" (read as "If $p$, then $q$") is a compound statement where $p$ is the hypothesis (antecedent) and $q$ is the conclusion (consequent).

The truth value of an implication $p \to q$ is defined such that it is false in only one specific case. It is considered true in all other cases.

The rule for the truth value of $p \to q$ is as follows:

The implication $p \to q$ is false if and only if the hypothesis $p$ is true and the conclusion $q$ is false.

In all other possible combinations of truth values for $p$ and $q$ (True-True, False-True, False-False), the implication $p \to q$ is true.

This can be illustrated with a truth table:

Looking at the truth table, the truth value of $p \to q$ is 'False' exclusively in the second row, where $p$ is True and $q$ is False.

Now let's examine the given options:

(A) "$p$ is true and $q$ is true." - If this is the case, $p \to q$ is True.

(B) "$p$ is false and $q$ is true." - If this is the case, $p \to q$ is True.

(C) "$p$ is true and $q$ is false." - If this is the case, $p \to q$ is False. This matches the condition we identified.

(D) "$p$ is false and $q$ is false." - If this is the case, $p \to q$ is True.

The condition under which the truth value of an implication ($p \to q$) is false is when the hypothesis ($p$) is true and the conclusion ($q$) is false.

The correct option is (C) $p$ is true and $q$ is false.

The given statement is an existential statement. It claims that there is at least one element (a student) in a set (all students) that satisfies a certain property (did not pass the exam).

The structure is "There exists $x$ such that $P(x)$", where $x$ refers to a student, and $P(x)$ is the predicate "$x$ did not pass the exam". Symbolically, this is $\exists x, P(x)$.

The negation of an existential statement "There exists $x$ such that $P(x)$" is a universal statement that claims that for every element $x$ in the set, the property $P(x)$ is false.

The negation of $\exists x, P(x)$ is $\neg (\exists x, P(x))$, which is logically equivalent to $\forall x, \neg P(x)$. This translates to "For all $x$, it is not the case that $P(x)$".

In our case, the predicate $P(x)$ is "$x$ did not pass the exam".

The negation of $P(x)$, $\neg P(x)$, is "It is not the case that $x$ did not pass the exam". This is equivalent to "$x$ passed the exam".

Applying the negation rule $\forall x, \neg P(x)$, the negation of the original statement "There exists a student who did not pass the exam" is "For all students, it is not the case that the student did not pass the exam".

This simplifies to "For all students, the student passed the exam".

Let's check the options:

(A) "There exists a student who passed the exam." - This is $\exists x, \neg P(x)$.

(B) "For all students, the student did not pass the exam." - This is $\forall x, P(x)$.

(C) "For all students, the student passed the exam." - This is $\forall x, \neg P(x)$. This matches our derived negation.

(D) "There is no student who passed the exam." - This is $\neg (\exists x, \neg P(x))$.

The statement that is logically equivalent to the negation of "There exists a student who did not pass the exam" is "For all students, the student passed the exam".

The correct option is (C) For all students, the student passed the exam.

In mathematical logic, quantifiers are symbols that indicate the number of elements in a domain of discourse that satisfy a given property or relation. The two most common types of quantifiers are the universal quantifier and the existential quantifier.

The phrase "For all", "For every", or "For any" is used to express that a statement applies to every member of a set or domain. This is known as the universal quantification.

The symbol used for the universal quantifier is $\forall$. For example, the statement "For all numbers $x$, $x^2 \geq 0$" is written as $\forall x, x^2 \geq 0$.

The phrase "There exists", "There is at least one", or "For some" is used to express that there is at least one member of a set or domain that satisfies a statement. This is known as the existential quantification.

The symbol used for the existential quantifier is $\exists$. For example, the statement "There exists a number $x$ such that $x > 5$" is written as $\exists x, x > 5$.

Based on these definitions, the quantifier "For all" is also known as the Universal Quantifier.

Let's look at the options:

(A) Existential Quantifier: Refers to "There exists".

(B) Universal Quantifier: Refers to "For all". This matches.

(C) Uniqueness Quantifier: Refers to "There exists exactly one" (symbol $\exists!$).

(D) Negation Quantifier: This is not a standard term in logic for a type of quantifier. Negation ($\neg$) is a logical operator.

The correct option is (B) Universal Quantifier.

For a given conditional statement (implication) of the form "$p \to q$" (read as "If $p$, then $q$"), the inverse is a related conditional statement.

The inverse of $p \to q$ is formed by negating both the hypothesis ($p$) and the conclusion ($q$), while keeping the direction of the implication.

Applying this to $p \to q$, we negate $p$ to get $\neg p$, and we negate $q$ to get $\neg q$. The inverse is the implication with the negated hypothesis and the negated conclusion: $\neg p \to \neg q$ (read as "If not $p$, then not $q$").

The standard related conditional statements derived from $p \to q$ are:

- Original Implication: $p \to q$

- Converse: $q \to p$

- Inverse: $\neg p \to \neg q$

- Contrapositive: $\neg q \to \neg p$

The question asks for the inverse of $p \to q$. According to the definition, the inverse is $\neg p \to \neg q$.

Let's examine the given options:

(A) $q \to p$ - This is the converse.

(B) $\neg p \to \neg q$ - This matches the definition of the inverse.

(C) $\neg q \to \neg p$ - This is the contrapositive.

(D) $p \land \neg q$ - This represents "p and not q". This is the negation of the implication $p \to q$.

The symbol for the inverse of $p \to q$ is $\neg p \to \neg q$.

The correct option is (B) $\neg p \to \neg q$.

Two statements are said to be logically equivalent if they have the same truth value for all possible truth values of their constituent simple statements. In other words, they have identical truth tables.

Let's examine the truth tables for the statements involved in the options.

Now let's look at the pairs in the options and compare their truth value columns:

(A) $p \to q$ (Column 3) and $q \to p$ (Column 4). Their truth values are different in rows 2 and 3. They are not logically equivalent.

(B) $p \to q$ (Column 3) and $\neg p \to \neg q$ (Column 7). Their truth values are different in rows 2 and 3. They are not logically equivalent. (Note that $\neg p \to \neg q$ has the same truth values as $q \to p$, which is the converse).

(C) $p \to q$ (Column 3) and $\neg q \to \neg p$ (Column 8). Their truth values are identical in all rows (T, F, T, T). They are logically equivalent.

(D) $p \land q$ (Column 9) and $p \lor q$ (Column 10). Their truth values are different in rows 2, 3, and 4. They are not logically equivalent.

The pair of statements that are logically equivalent is the original implication and its contrapositive.

The correct option is (C) $p \to q$ and $\neg q \to \neg p$.

Let the given statement be an implication of the form $p \to q$.

Let $p$: "A number is an integer".

Let $q$: "A number is a rational number".

The original statement is "If a number is an integer, then it is a rational number", which can be written symbolically as $p \to q$.

The converse of an implication $p \to q$ is obtained by interchanging the hypothesis ($p$) and the conclusion ($q$). The converse is the statement $q \to p$.

Now, let's translate the converse $q \to p$ back into English using the definitions of $p$ and $q$:

$q \to p$ means "If $q$, then $p$".

Substituting the English phrases: "If a number is a rational number, then it is an integer."

Let's compare this with the given options:

(A) "If a number is not an integer, then it is not a rational number." This is $\neg p \to \neg q$ (the inverse).

(B) "If a number is a rational number, then it is an integer." This is $q \to p$ (the converse). This matches our result.

(C) "If a number is not a rational number, then it is not an integer." This is $\neg q \to \neg p$ (the contrapositive).

(D) "A number is an integer and it is a rational number." This is $p \land q$ (a conjunction).

The statement that represents the converse of the original implication is "If a number is a rational number, then it is an integer."

The correct option is (B) If a number is a rational number, then it is an integer.

Note that the original statement ($p \to q$) is true (every integer is rational). However, its converse ($q \to p$) is false (e.g., $1/2$ is a rational number, but it is not an integer). This illustrates that an implication and its converse are not logically equivalent.

The given statement is "Some students like mathematics". This is an existential statement, meaning it asserts the existence of at least one element (student) satisfying a property (liking mathematics).

Let $S$ be the set of students, and let $P(x)$ be the predicate "$x$ likes mathematics", where $x$ is a student in $S$. The original statement can be written symbolically as $\exists x \in S, P(x)$.

The negation of an existential statement $\exists x, P(x)$ is a universal statement $\forall x, \neg P(x)$. This translates to "For all $x$, it is not the case that $P(x)$".

In our case, $P(x)$ is "$x$ likes mathematics". The negation $\neg P(x)$ is "$x$ does not like mathematics".

So, the negation of the original statement $\exists x \in S, P(x)$ is $\forall x \in S, \neg P(x)$.

Translating $\forall x \in S, \neg P(x)$ back into English, we get "For all students $x$, $x$ does not like mathematics". This means that for every single student, it is true that they do not like mathematics. This is equivalent to stating that "No student likes mathematics".

We need to complete the sentence "________ students like mathematics" to make it equivalent to "No student likes mathematics".

Let's consider the options:

(A) "All students like mathematics" ($\forall x, P(x)$). This is the opposite meaning of "No students like mathematics".

(B) "No students like mathematics" ($\forall x, \neg P(x)$). This matches our derived negation.

(C) "Some students like mathematics" ($\exists x, P(x)$). This is the original statement.

(D) "Exactly one student likes mathematics" ($\exists! x, P(x)$). This is a different type of statement.

To make the completion sentence the negation of the original statement, the blank should be filled with "No".

The correct option is (B) No.

We need to evaluate the truthfulness of both the Assertion (A) and the Reason (R) and then determine if the Reason correctly explains the Assertion.

Let's analyze the Assertion (A): "The statement '2 + 2 = 5' is a proposition."

Recall the definition of a proposition: A proposition is a declarative sentence that is either true or false, but not both.

The statement "$2 + 2 = 5$" is a declarative sentence. It makes a specific claim.

The claim "$2 + 2 = 5$" is mathematically false.

Since it is a declarative sentence and has a definite truth value (False), it fits the definition of a proposition. The definition does not require the statement to be true, only that it *can* be determined to be true or false.

Therefore, Assertion (A) is a true statement.

Now let's analyze the Reason (R): "A proposition must be a true statement."

This statement claims that a proposition must necessarily be true.

However, as per the definition of a proposition, it can be either true or false. For example, the statement "The Earth is flat" is a proposition, and it is false.

The Reason (R) contradicts the definition of a proposition by stating that it *must* be true.

Therefore, Reason (R) is a false statement.

Based on the analysis:

- Assertion (A) is true.

- Reason (R) is false.

When the Reason is false, it cannot be the correct explanation for the Assertion, regardless of the Assertion's truth value.

This matches option (C).

The correct option is (C) A is true but R is false.

The statement "For all $x$, $P(x)$ is true" is a universal statement, often written symbolically as $\forall x, P(x)$. It asserts that the property $P(x)$ holds for every element $x$ in the specified domain of discourse.

We need to find a suitable method for proving such a statement. Let's evaluate the given options:

(A) "Prove $P(a)$ is true for a specific value $a$." - Proving that a statement $P(x)$ is true for just one specific instance ($x=a$) does not guarantee that it is true for all possible values of $x$ in the domain. This method is insufficient to prove a universal statement.

(B) "Find a counterexample where $P(x)$ is false." - A counterexample is an element $x$ in the domain for which $P(x)$ is false. Finding a single counterexample is the standard method used to disprove a universal statement, not to prove it. If a counterexample exists, the statement "For all $x$, $P(x)$ is true" is false.

(C) "Prove that $P(x)$ is true for an arbitrary element $x$ from the domain." - This is a fundamental method for proving universal statements. The strategy is to select an element $x$ from the domain, without making any assumptions about $x$ other than that it belongs to the domain, and then logically demonstrate that the property $P(x)$ must hold for this $x$. Since $x$ represents any arbitrary element from the domain, the conclusion $P(x)$ is true generalizes to all elements in the domain. This is a suitable method for proving a universal statement.

(D) "Use mathematical induction (if the domain is integers)." - Mathematical induction is a powerful and suitable proof technique specifically used to prove universal statements about the set of natural numbers or integers (starting from a base case). However, this method is only applicable to specific types of domains and properties that can be related to the structure of integers. Option (C) describes a more general method applicable to universal statements over any domain, not limited to integers or properties suitable for induction. While induction is suitable *in specific cases*, proving for an arbitrary element is a more general and widely applicable proof strategy for universal quantification.

Among the given options, proving the property for an arbitrary element from the domain is the most general and directly suitable method for establishing the truth of a universal statement "For all $x$, $P(x)$ is true".

The correct option is (C) Prove that $P(x)$ is true for an arbitrary element $x$ from the domain.

Let the given statement be an implication of the form $p \to q$.

Let $p$: "A triangle is equilateral".

Let $q$: "It is isosceles".

The original statement is "If a triangle is equilateral, then it is isosceles", which can be written symbolically as $p \to q$. The problem states that this is a true statement, which is correct as an equilateral triangle has three equal sides, and thus at least two equal sides, satisfying the definition of an isosceles triangle.

The inverse of an implication $p \to q$ is obtained by negating both the hypothesis ($p$) and the conclusion ($q$) and forming a new implication with these negations, keeping the original direction. The inverse is the statement $\neg p \to \neg q$.

First, let's find the negations of $p$ and $q$:

$\neg p$: The negation of "A triangle is equilateral" is "A triangle is not equilateral".

$\neg q$: The negation of "It is isosceles" is "It is not isosceles".

Now, we form the inverse $\neg p \to \neg q$:

"If $\neg p$, then $\neg q$."

Substituting the English phrases: "If a triangle is not equilateral, then it is not isosceles."

Let's compare this with the given options:

(A) "If a triangle is isosceles, then it is equilateral." This is $q \to p$ (the converse).

(B) "If a triangle is not equilateral, then it is not isosceles." This is $\neg p \to \neg q$ (the inverse). This matches our result.

(C) "If a triangle is not isosceles, then it is not equilateral." This is $\neg q \to \neg p$ (the contrapositive).

(D) "If a triangle is isosceles, then it is not equilateral." This is $q \to \neg p$.

The statement that represents the inverse of the original implication is "If a triangle is not equilateral, then it is not isosceles."

The correct option is (B) If a triangle is not equilateral, then it is not isosceles.

Note that the inverse of a statement is logically equivalent to its converse, but generally not logically equivalent to the original statement or its contrapositive. In this case, the original statement ($p \to q$) is true, but its inverse ($\neg p \to \neg q$) is false (e.g., an isosceles triangle with angles $50^\circ, 50^\circ, 80^\circ$ is not equilateral, but it is isosceles).

The given statement is a conditional statement of the form "If P, then Q".

Let P be the statement "The roads are wet".

Let Q be the statement "It has rained".

The given statement is "If the roads are wet, then it has rained", which is represented symbolically as P $\to$ Q.

In problems involving causality or common relationships (like rain causing wet roads), the implication representing the usual cause-effect relationship is often considered the "original" statement from which others are derived. In this context, the usual cause is rain and the effect is wet roads.

Let's define the simple statements according to the common causality:

Let $p$: "It has rained".

Let $q$: "The roads are wet".

The implication representing the cause-effect is $p \to q$: "If it has rained, then the roads are wet". Let's consider this as the implicit original statement.

Now let's see how the given statement "If the roads are wet, then it has rained" relates to this implicit original statement $p \to q$.

The given statement is "If $q$, then $p$", which is written symbolically as $q \to p$.

The relationship between an implication $p \to q$ and the implication $q \to p$ is that $q \to p$ is the converse of $p \to q$.

Let's check the definitions of the related conditional statements derived from $p \to q$:

- Original statement: $p \to q$ ("If it has rained, then the roads are wet")

- Converse: $q \to p$ ("If the roads are wet, then it has rained")

- Inverse: $\neg p \to \neg q$ ("If it has not rained, then the roads are not wet")

- Contrapositive: $\neg q \to \neg p$ ("If the roads are not wet, then it has not rained")

The given statement "If the roads are wet, then it has rained" matches the structure of the converse ($q \to p$).

Thus, the statement is the converse of the implied original statement.

The correct option is (B) Converse.

In mathematical logic, quantifiers are used to specify the quantity of elements in the domain of discourse that satisfy a given predicate. The two primary quantifiers are the universal quantifier and the existential quantifier.

The universal quantifier represents the phrase "For all" or "For every". Its symbol is $\forall$. For example, $\forall x, P(x)$ means "For all $x$, $P(x)$ is true".

The existential quantifier represents the phrase "There exists" or "There is at least one". Its symbol is $\exists$. For example, $\exists x, P(x)$ means "There exists an $x$ such that $P(x)$ is true".

Let's look at the given options:

(A) $\forall$: This is the symbol for the universal quantifier ("For all").

(B) $\exists$: This is the symbol for the existential quantifier ("There exists"). This matches the question.

(C) $\neg$: This is the symbol for negation ("not").

(D) $\to$: This is the symbol for implication ("if... then...").

The symbol that represents the existential quantifier is $\exists$.

The correct option is (B) $\exists$.

We are given three statements about the attendance in a class:

$p$: All students attended the class.

$q$: Some students did not attend the class.

$r$: No student attended the class.

Let $S$ be the set of students and $A(x)$ be the predicate "$x$ attended the class". We can represent these statements symbolically:

$p$: $\forall x \in S, A(x)$ (For all students $x$, $x$ attended)

$q$: $\exists x \in S, \neg A(x)$ (There exists a student $x$ such that $x$ did not attend)

$r$: $\forall x \in S, \neg A(x)$ (For all students $x$, $x$ did not attend, which is equivalent to "No student attended")

We are given that statement $p$ is false.

If $p$ is false, then $\neg p$ is true.

$\neg p$ is the negation of the statement "All students attended the class" ($\forall x, A(x)$).

The negation of a universal statement $\forall x, P(x)$ is the existential statement $\exists x, \neg P(x)$.

Therefore, $\neg (\forall x \in S, A(x)) \equiv \exists x \in S, \neg A(x)$.

The statement $\exists x \in S, \neg A(x)$ translates to "There exists a student $x$ such that $x$ did not attend the class".

Looking at the given statements, this is precisely the meaning of statement $q$: "Some students did not attend the class".

So, if $p$ is false, then $q$ must be true.

Let's check if $r$ must be true when $p$ is false.

If $r$ is true ("No student attended"), this means every student did not attend. If every student did not attend, then it is definitely true that at least one student did not attend (assuming there is at least one student in the class), so $q$ would be true. Also, if no student attended, it is false that all students attended, so $p$ would be false. So, $r$ being true implies $p$ is false and $q$ is true.

However, consider a scenario where some students attended and some did not. In this case, "All students attended" ($p$) is false, and "Some students did not attend" ($q$) is true. But "No student attended" ($r$) is also false.

Thus, if $p$ is false, $q$ must be true, but $r$ is not necessarily true. $r$ is true only in the specific case where *all* students did not attend, which is one possibility when $p$ is false, but not the only one.

Therefore, if statement $p$ ("All students attended the class") is false, it necessarily means that there is at least one student who did not attend the class, which is statement $q$ ("Some students did not attend the class").

The correct option is (A) $q$ is true.

We recall the statements from the previous question:

$p$: All students attended the class. ($\forall x, A(x)$)

$q$: Some students did not attend the class. ($\exists x, \neg A(x)$)

$r$: No student attended the class. ($\forall x, \neg A(x)$)

We are given that statement $q$ is false.

If $q$ is false, then its negation, $\neg q$, must be true.

The negation of $q$ ($\exists x, \neg A(x)$) is $\neg (\exists x, \neg A(x))$.

Using the rules of predicate logic, the negation of an existential statement $\exists x, P(x)$ is $\forall x, \neg P(x)$. Applying this rule, the negation of $\exists x, \neg A(x)$ is $\forall x, \neg (\neg A(x))$.

The double negation $\neg (\neg A(x))$ is equivalent to $A(x)$.

So, $\neg q \equiv \forall x \in S, A(x)$.

The statement $\forall x \in S, A(x)$ translates to "For all students $x$, $x$ attended the class".

Looking at the given statements, this is precisely the meaning of statement $p$: "All students attended the class".

So, if $q$ is false, then $p$ must be true, and the statement "All students attended the class" (which is $p$) must be true.

Let's examine the given options:

(A) $p$ is true. - This matches our derivation.

(B) $r$ is true. - If $p$ is true ("All students attended"), then it's impossible for $r$ to be true ("No student attended"), unless the class has no students, which is usually not implied in such problems. If there is at least one student, $p$ being true implies $r$ is false. So $r$ is not necessarily true, and is usually false in this case.

(C) The negation of $q$ is true, i.e., "All students attended the class". - This is also true, as shown above. The negation of $q$ is indeed $p$.

(D) Both (A) and (C). - Since (A) states that $p$ is true, and (C) states that the negation of $q$ is true and is equivalent to "All students attended the class" (which is $p$), options (A) and (C) essentially say the same thing. If $q$ is false, $p$ is true, which means "All students attended the class" is true. Thus, both conditions mentioned in (D) are met.

If statement $q$ ("Some students did not attend the class") is false, it means there is no student who did not attend the class. This is equivalent to saying that every student attended the class. This is precisely what statement $p$ ("All students attended the class") asserts. Statement (C) restates this fact explicitly, stating that the negation of $q$ is true and giving its meaning, which is $p$. Since (A) simply states that $p$ is true, and (C) provides the logical equivalence that makes (A) true, both (A) and (C) describe the consequence of $q$ being false. Option (D) correctly captures that both assertions made in (A) and (C) are true.

The correct option is (D) Both (A) and (C).

An implication or conditional statement "$p \to q$" (read as "If $p$, then $q$") is a compound statement where $p$ is the hypothesis and $q$ is the conclusion.

The truth value of an implication $p \to q$ is defined to be true in all cases except one. The specific case where the implication is false is when the hypothesis ($p$) is true and the conclusion ($q$) is false. In all other combinations of truth values for $p$ and $q$, the implication $p \to q$ is true.

We can summarize the truth values of $p \to q$ in a truth table:

The question asks for the case where the truth value of $p \to q$ is NOT true (i.e., it is false).

Looking at the truth table, the implication $p \to q$ is false only in the second row, which corresponds to the case where $p$ is True and $q$ is False.

Let's examine the given options:

(A) $p$ is true and $q$ is true. - In this case, $p \to q$ is True.

(B) $p$ is false and $q$ is true. - In this case, $p \to q$ is True.

(C) $p$ is true and $q$ is false. - In this case, $p \to q$ is False. This is the exception.

(D) $p$ is false and $q$ is false. - In this case, $p \to q$ is True.

The implication $p \to q$ is true in all cases except when $p$ is true and $q$ is false.

The correct option is (C) $p$ is true and $q$ is false.

We are asked to identify the true statement regarding the logical equivalence between an implication and its related conditional statements (converse, inverse, contrapositive).

Let the original implication be $p \to q$.

The related statements are:

- Original Implication: $p \to q$

- Converse: $q \to p$

- Inverse: $\neg p \to \neg q$

- Contrapositive: $\neg q \to \neg p$

Recall that two statements are logically equivalent if they have the same truth values for all possible truth assignments of their variables. We can refer to the truth table constructed in Question 23, which includes columns for all these statements:

Now let's evaluate the truthfulness of each option based on the truth table columns:

(A) The converse ($q \to p$) and inverse ($\neg p \to \neg q$) of an implication are logically equivalent.

Compare column 4 ($q \to p$) and column 5 ($\neg p \to \neg q$). Their truth values are identical in all rows (T, T, F, T). This statement is true.

(B) The converse ($q \to p$) and contrapositive ($\neg q \to \neg p$) of an implication are logically equivalent.

Compare column 4 ($q \to p$) and column 6 ($\neg q \to \neg p$). Their truth values are different in rows 2 and 3. This statement is false.

(C) The inverse ($\neg p \to \neg q$) and contrapositive ($\neg q \to \neg p$) of an implication are logically equivalent.

Compare column 5 ($\neg p \to \neg q$) and column 6 ($\neg q \to \neg p$). Their truth values are different in rows 2 and 3. This statement is false.

(D) An implication ($p \to q$) and its inverse ($\neg p \to \neg q$) are logically equivalent.

Compare column 3 ($p \to q$) and column 5 ($\neg p \to \neg q$). Their truth values are different in rows 2 and 3. This statement is false.

From the analysis, the only true statement is that the converse and the inverse of an implication are logically equivalent. Additionally, the original implication and its contrapositive are logically equivalent.

The correct option is (A) The converse and inverse of an implication are logically equivalent.

Solution:

Let the given statement be $S$. The statement $S$ is "For all $x \in A$, $P(x)$ is true".

In logical notation, this statement can be written as:

$\forall x \in A, P(x)$

The negation of a statement involving a universal quantifier ($\forall$) is a statement involving an existential quantifier ($\exists$).

The general rule for negation is:

The negation of $\forall x, Q(x)$ is $\exists x, \neg Q(x)$.

(The negation of "For all $x$, $Q(x)$ is true" is "There exists an $x$ such that $Q(x)$ is false".)

Applying this rule to our statement $\forall x \in A, P(x)$, where $Q(x)$ is $P(x)$ is true, the negation is $\exists x \in A, \neg (P(x) \text{ is true})$.

Negating "$P(x)$ is true" gives "$P(x)$ is false".

So, the negation of the statement is "There exists $x \in A$ such that $P(x)$ is false".

Let's examine the given options:

(A) For all $x \in A$, $P(x)$ is false. (Logical form: $\forall x \in A, \neg P(x)$) - This is different from the negation.

(B) There exists $x \in A$ such that $P(x)$ is false. (Logical form: $\exists x \in A, \neg P(x)$) - This matches our derived negation.

(C) There exists $x \in A$ such that $P(x)$ is true. (Logical form: $\exists x \in A, P(x)$) - This is not the negation.

(D) For all $x \in A$, $P(x)$ is not false. (This is equivalent to "For all $x \in A$, $P(x)$ is true", which is the original statement itself).

Therefore, the correct negation is "There exists $x \in A$ such that $P(x)$ is false".

The correct option is (B).

Solution:

Let $p$ be the statement "The number is rational".

Let $q$ be the statement "The number is irrational".

The given statement is "$p$ or $q$". In logical symbols, this is $p \lor q$.

We need to find the negation of this statement, which is $\neg (p \lor q)$.

According to De Morgan's Laws, the negation of a disjunction ($p \lor q$) is the conjunction of the negations of the individual statements ($\neg p \land \neg q$).

So, $\neg (p \lor q) \equiv \neg p \land \neg q$.

Now let's find the negations of $p$ and $q$:

$\neg p$ is "The number is not rational".

$\neg q$ is "The number is not irrational".

Combining these with the conjunction ($\land$), the negation $\neg p \land \neg q$ translates to "The number is not rational and the number is not irrational".

So, the negation of the statement "The number is rational or it is irrational" is "The number is not rational and it is not irrational".

Let's examine the given options:

(A) The number is not rational or it is not irrational. (This is $\neg p \lor \neg q$, which is different from $\neg p \land \neg q$)

(B) The number is not rational and it is not irrational. (This is $\neg p \land \neg q$, which matches our result)

(C) The number is rational and it is irrational. (This is $p \land q$, which is not the negation)

(D) The number is not rational or it is irrational. (This is $\neg p \lor q$, which is not the negation)

The correct option is (B).

Solution:

Let the statement be "If $p$ then $q$". Here, $p$ is "a number is odd" and $q$ is "it is prime". The statement is $p \implies q$.

Let's evaluate Assertion (A):

The statement is "If a number is odd, then it is prime".

The assertion claims this statement is false and provides an example: 9 is odd, but 9 is not prime (since $9 = 3 \times 3$).

In the conditional statement $p \implies q$, if $p$ is true (9 is odd) and $q$ is false (9 is not prime), then the statement $p \implies q$ is false.

The example given (9) is a valid counterexample because the hypothesis ($p$) is true, but the conclusion ($q$) is false.

Thus, the statement "If a number is odd, then it is prime" is indeed false.

Therefore, Assertion (A) is true.

Let's evaluate Reason (R):

The reason describes the standard method for determining the truth value of a conditional statement $p \implies q$.

To validate (prove true) "If $p$ then $q$", we assume the hypothesis $p$ is true and logically deduce that the conclusion $q$ must also be true.

To invalidate (prove false) "If $p$ then $q$", we only need to find one specific case where the hypothesis $p$ is true, but the conclusion $q$ is false. This specific case is called a counterexample.

This description in Reason (R) is a fundamental principle in logic regarding conditional statements.

Therefore, Reason (R) is true.

Now, let's determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that the conditional statement is false because of the counterexample 9.

Reason (R) explains that finding a case where $p$ is true and $q$ is false is the method used to invalidate (show false) a conditional statement $p \implies q$.

The counterexample 9 given in (A) (9 is odd, 9 is not prime) fits the description in (R) (a case where $p$ is true and $q$ is false).

Reason (R) provides the logical justification for why the example given in Assertion (A) proves the statement to be false.

Thus, Reason (R) is the correct explanation for Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

The correct option is (A).

Solution:

The statement "$p$ if and only if $q$" is a biconditional statement, denoted by $p \iff q$. This statement is true precisely when $p$ and $q$ have the same truth value (i.e., both are true or both are false), and false otherwise.

Let's analyze the given options:

(A) $p \to q$

This is a conditional statement ("If $p$ then $q$"). Its truth table is different from that of $p \iff q$. For instance, $p \to q$ is true when $p$ is false and $q$ is true, but $p \iff q$ is false in this case.

(B) $(p \to q) \land (q \to p)$

This statement represents the conjunction of two conditional statements: "If $p$ then $q$" and "If $q$ then $p$".

The statement $p \to q$ means that if $p$ is true, then $q$ must be true.

The statement $q \to p$ means that if $q$ is true, then $p$ must be true.

The conjunction $(p \to q) \land (q \to p)$ is true only when both $p \to q$ and $q \to p$ are true.

This combination implies that the truth of $p$ is necessary and sufficient for the truth of $q$. This is exactly what "$p$ if and only if $q$" means.

Thus, $(p \to q) \land (q \to p)$ is logically equivalent to $p \iff q$.

(C) $(p \land q) \lor (\neg p \land \neg q)$

This statement represents the disjunction of two conjunctions:

$(p \land q)$ means "$p$ is true and $q$ is true".

$(\neg p \land \neg q)$ means "$p$ is false and $q$ is false".

The disjunction $(p \land q) \lor (\neg p \land \neg q)$ is true if either ($p$ is true AND $q$ is true) OR ($p$ is false AND $q$ is false) is true.

This statement is true precisely when $p$ and $q$ have the same truth value, which is the definition of $p \iff q$.

Thus, $(p \land q) \lor (\neg p \land \neg q)$ is also logically equivalent to $p \iff q$.

Since both option (B) and option (C) are logically equivalent to the statement "$p$ if and only if $q$", the correct answer includes both.

The correct option is (D) Both (B) and (C).

Solution:

Let the statement containing "and" be represented as "$p$ and $q$", which can be written in logic as $p \land q$.

A conjunction statement ($p \land q$) is true if and only if both of its component statements ($p$ and $q$) are true.

The truth table for $p \land q$ is:

From the truth table, we see that $p \land q$ is true only in the case where $p$ is true and $q$ is true.

To validate (prove true) the statement "$p$ and $q$", you must demonstrate that the truth condition is met. This means you must show that $p$ is true and you must show that $q$ is true.

In other words, you must validate both component statements.

Let's consider the options:

(A) Validate at least one of the component statements. - This is insufficient. If only one is true, the conjunction $p \land q$ is false.

(B) Validate both component statements. - This is correct. If both $p$ and $q$ are validated (shown to be true), then $p \land q$ is true.

(C) Find a case where both are false. - This describes a way to show that the statement $p \land q$ is false, but it is not the method for validating it.

(D) Find a case where at least one is false. - This also describes a way to show that the statement $p \land q$ is false (a counterexample for the conjunction), not for validating it.

Therefore, to validate a statement containing "and", you must validate both component statements.

The correct option is (B).

Solution:

Let the given statement be $S$. The statement $S$ is "Some students scored above 90%".

This statement is an existential statement, meaning "There exists at least one student such that they scored above 90%".

Let $x$ represent a student. Let $P(x)$ be the predicate "$x$ scored above 90%".

The statement can be written in logical notation as:

$\exists x, P(x)$

We need to find the negation of this statement, which is $\neg (\exists x, P(x))$.

According to the rules of logical negation for quantifiers, the negation of an existential statement ($\exists x, P(x)$) is a universal statement with the negated predicate ($\forall x, \neg P(x)$).

So, $\neg (\exists x, P(x)) \equiv \forall x, \neg P(x)$.

Now, let's find the negation of the predicate $P(x)$:

$P(x)$ is "$x$ scored above 90%".

$\neg P(x)$ is "$x$ did not score above 90%".

In the context of scores, "did not score above 90%" is equivalent to "scored below or equal to 90%".

So, $\neg P(x)$ means "$x$ scored below or equal to 90%".

Now, let's translate the negated statement $\forall x, \neg P(x)$ back into English:

"For all $x$, $x$ scored below or equal to 90%".

This can be rephrased as "All students scored below or equal to 90%".

Let's examine the given options:

(A) Some students scored below 90%. ($\exists x, \neg P'(x)$ where $P'(x)$ is "$x$ scored below 90%") - Incorrect.

(B) No student scored above 90%. (This means $\neg (\exists x, P(x))$, which is the definition of the negation) - This is a correct negation.

(C) All students scored above 90%. ($\forall x, P(x)$) - Incorrect.

(D) All students scored below or equal to 90%. ($\forall x, \neg P(x)$) - This matches our derived negation based on the rule $\neg \exists P \equiv \forall \neg P$.

Both options (B) and (D) are logically equivalent and represent the negation of the original statement. However, option (D) directly shows the result of applying the standard logical equivalence $\neg \exists x P(x) \equiv \forall x \neg P(x)$ by explicitly negating the predicate and applying the universal quantifier.

The correct option that represents the negation in the form $\forall x, \neg P(x)$ is (D).

Solution:

In logic, simple statements can be combined to form compound statements using logical connectives.

The word "and" is a fundamental logical connective.

Let $p$ and $q$ be two simple statements.

Combining $p$ and $q$ with "and" forms the compound statement "$p$ and $q$". This is denoted by $p \land q$.

This connective is called a conjunction.

Let's review the terms in the options:

• Disjunction: Connective "or" ($p \lor q$).
• Conjunction: Connective "and" ($p \land q$).
• Implication: Connective "if... then..." ($p \to q$).
• Negation: Connective "not" ($\neg p$).

The word "and" is used to form a conjunction.

The correct option is (B).

The logical connective used to form a compound statement by joining two simple statements with the word "and" is called a Conjunction.

Solution:

Let the given statement be $S$. The statement $S$ is "All even numbers are divisible by 4".

This is a universal statement. Let $x$ represent an even number. Let $P(x)$ be the predicate "$x$ is divisible by 4".

The statement can be written in logical notation as:

$\forall x \in \text{Even Numbers}, P(x)$

We need to find the negation of this statement, which is $\neg (\forall x \in \text{Even Numbers}, P(x))$.

According to the rules of logical negation for quantifiers, the negation of a universal statement ($\forall x, P(x)$) is an existential statement with the negated predicate ($\exists x, \neg P(x)$).

So, $\neg (\forall x \in \text{Even Numbers}, P(x)) \equiv \exists x \in \text{Even Numbers}, \neg P(x)$.

Now, let's find the negation of the predicate $P(x)$:

$P(x)$ is "$x$ is divisible by 4".

$\neg P(x)$ is "$x$ is not divisible by 4".

Now, let's translate the negated statement $\exists x \in \text{Even Numbers}, \neg P(x)$ back into English:

"There exists an even number $x$ such that $x$ is not divisible by 4".

This can be rephrased as "Some even numbers are not divisible by 4".

Let's examine the given options:

(A) All even numbers are not divisible by 4. ($\forall x \in \text{Even Numbers}, \neg P(x)$) - This would mean no even number is divisible by 4, which is incorrect and not the negation.

(B) Some even numbers are not divisible by 4. ($\exists x \in \text{Even Numbers}, \neg P(x)$) - This matches our derived negation.

(C) No even numbers are divisible by 4. (Equivalent to $\forall x \in \text{Even Numbers}, \neg P(x)$) - Incorrect.

(D) Some even numbers are divisible by 4. ($\exists x \in \text{Even Numbers}, P(x)$) - This is an existential statement, but not the negation of the original universal statement.

The correct option is (B).

An example of such an even number is 2 or 6 or 10, etc. 2 is an even number, but it is not divisible by 4. This single counterexample proves the original statement false and confirms the truth of the negation.

Solution:

The statement "$p \to q$" is a conditional statement (an implication), read as "If $p$ then $q$".

The truth value of a conditional statement is determined by the truth values of its components $p$ (the hypothesis or antecedent) and $q$ (the conclusion or consequent).

The truth table for $p \to q$ is as follows:

We are given that the statement "$p \to q$" is false.

Looking at the truth table, the statement $p \to q$ is false in only one specific case:

This occurs in the second row of the table, where $p$ is True and $q$ is False.

Thus, if the statement "$p \to q$" is false, it must be the case that $p$ is true and $q$ is false.

Let's examine the given options:

(A) $p$ is true and $q$ is true. (In this case, $p \to q$ is True)

(B) $p$ is false and $q$ is true. (In this case, $p \to q$ is True)

(C) $p$ is true and $q$ is false. (In this case, $p \to q$ is False) - This matches our result.

(D) $p$ is false and $q$ is false. (In this case, $p \to q$ is True)

The correct option is (C).

Solution:

Let the statement containing "or" be represented as "$p$ or $q$", which can be written in logic as $p \lor q$.

A disjunction statement ($p \lor q$) is true if and only if at least one of its component statements ($p$ or $q$) is true. It is false only when both $p$ and $q$ are false.

The truth table for $p \lor q$ is:

From the truth table, we see that $p \lor q$ is true in any case where $p$ is true, or $q$ is true, or both are true. This means that for the statement "$p$ or $q$" to be true, it is sufficient for at least one of the component statements to be true.

To validate (prove true) the statement "$p$ or $q$", you must demonstrate that the truth condition is met. This means you must show that $p$ is true, or show that $q$ is true, or show that both are true. In other words, you must validate at least one of the component statements.

Let's consider the options:

(A) Validate at least one of the component statements. - This is correct. If you can show $p$ is true, $p \lor q$ is true. If you can show $q$ is true, $p \lor q$ is true. Showing either one is sufficient to validate the disjunction.

(B) Validate both component statements. - While validating both makes $p \lor q$ true, it is not necessary. The disjunction is true even if only one component is true.

(C) Find a case where both are false. - This describes the condition under which the statement $p \lor q$ is false. It's a method for invalidating the statement, not validating it.

(D) Find a case where at least one is false. - If $p$ is false but $q$ is true, $p \lor q$ is true. If $q$ is false but $p$ is true, $p \lor q$ is true. Finding a case where at least one is false does not validate the disjunction.

Therefore, to validate a statement containing "or", you must validate at least one of the component statements.

The correct option is (A).

Solution:

We are asked to find a statement that is logically equivalent to the conjunction $p \land q$.

We can use De Morgan's Laws and the property of double negation to evaluate the given options.

De Morgan's Laws are:

$\neg (p \land q) \equiv \neg p \lor \neg q$

$\neg (p \lor q) \equiv \neg p \land \neg q$

The double negation property is: $\neg (\neg A) \equiv A$

Let's evaluate each option:

(A) $\neg (\neg p \lor \neg q)$

Applying De Morgan's Law for disjunction, $\neg (A \lor B) \equiv \neg A \land \neg B$. Let $A = \neg p$ and $B = \neg q$.

So, $\neg (\neg p \lor \neg q) \equiv \neg (\neg p) \land \neg (\neg q)$.

Using the double negation property, $\neg (\neg p) \equiv p$ and $\neg (\neg q) \equiv q$.

Therefore, $\neg (\neg p \lor \neg q) \equiv p \land q$.

This statement is logically equivalent to $p \land q$.

(B) $\neg p \lor \neg q$

According to De Morgan's first law, $\neg (p \land q) \equiv \neg p \lor \neg q$. So this option is equivalent to the negation of $p \land q$, not $p \land q$ itself.

(C) $\neg (\neg p \land \neg q)$

Applying De Morgan's Law for conjunction, $\neg (A \land B) \equiv \neg A \lor \neg B$. Let $A = \neg p$ and $B = \neg q$.

So, $\neg (\neg p \land \neg q) \equiv \neg (\neg p) \lor \neg (\neg q)$.

Using the double negation property, $\neg (\neg p) \equiv p$ and $\neg (\neg q) \equiv q$.

Therefore, $\neg (\neg p \land \neg q) \equiv p \lor q$. This statement is generally not logically equivalent to $p \land q$.

(D) $\neg p \land \neg q$

According to De Morgan's second law, $\neg (p \lor q) \equiv \neg p \land \neg q$. So this option is equivalent to the negation of $p \lor q$, which is not generally equivalent to $p \land q$.

Based on the analysis, the statement logically equivalent to $p \land q$ is $\neg (\neg p \lor \neg q)$.

The correct option is (A).

Solution:

The given statement is a universal statement: "For every prime number $p$, $p$ is odd".

In logical terms, this can be written as $\forall p \in \{\text{prime numbers}\}, \text{"}p\text{ is odd"}$.

A counterexample to a universal statement $\forall x, P(x)$ is a specific value of $x$ for which the property $P(x)$ is false.

In this case, we are looking for a prime number $p$ such that the statement "$p$ is odd" is false. The negation of "$p$ is odd" is "$p$ is not odd", which means "$p$ is even".

So, a counterexample must be a prime number that is also an even number.

Let's examine the given options to see which one fits this description:

• (A) 3: Is 3 prime? Yes (divisors are 1 and 3). Is 3 even? No, 3 is odd. This is not a counterexample.
• (B) 5: Is 5 prime? Yes (divisors are 1 and 5). Is 5 even? No, 5 is odd. This is not a counterexample.
• (C) 2: Is 2 prime? Yes (divisors are 1 and 2). Is 2 even? Yes, 2 is an even number ($2 = 2 \times 1$). This is a prime number that is even.
• (D) 7: Is 7 prime? Yes (divisors are 1 and 7). Is 7 even? No, 7 is odd. This is not a counterexample.

The only even prime number is 2.

Since 2 is a prime number and it is not odd (it is even), it serves as a counterexample to the statement "For every prime number $p$, $p$ is odd".

The correct option is (C) 2.

Solution:

Let's evaluate Assertion (A):

The original statement is "Every student passed the exam". This is a universal statement: $\forall x \in \text{Students}, P(x)$, where $P(x)$ is "$x$ passed the exam".

The assertion claims its negation is "No student passed the exam".

"No student passed the exam" means "For every student $x$, $x$ did not pass the exam". This is $\forall x \in \text{Students}, \neg P(x)$.

The negation of $\forall x, P(x)$ is $\exists x, \neg P(x)$ ("There exists at least one student who did not pass the exam").

The statement "No student passed the exam" ($\forall x, \neg P(x)$) is *not* logically equivalent to "There exists at least one student who did not pass the exam" ($\exists x, \neg P(x)$).

For example, if out of 10 students, 9 passed and 1 failed, then "Every student passed the exam" is false. The negation "There exists a student who did not pass" is true. However, "No student passed the exam" is false in this scenario.

Thus, the negation of "Every student passed the exam" is not "No student passed the exam". The correct negation is "At least one student did not pass the exam" or "Some student did not pass the exam".

Therefore, Assertion (A) is false.

Let's evaluate Reason (R):

Reason (R) states: "The negation of a universal statement is an existential statement with a negated predicate."

Let the universal statement be $\forall x, P(x)$.

The negation of this statement is $\neg (\forall x, P(x))$.

The rule for negating universal quantifiers is $\neg (\forall x, P(x)) \equiv \exists x, \neg P(x)$.

This translates to "There exists an $x$ such that not $P(x)$". This is indeed an existential statement ($\exists$) with a negated predicate ($\neg P(x)$).

This rule is a fundamental principle of predicate logic.

Therefore, Reason (R) is true.

Since Assertion (A) is false and Reason (R) is true, we look for the option that reflects this.

The correct option is (D) A is false but R is true.

Solution:

In mathematical logic, the statement "$p$ implies $q$" is a conditional statement. It means that if statement $p$ is true, then statement $q$ must also be true.

This type of statement is commonly read as "If $p$, then $q$".

The standard logical connective used to represent implication is the arrow symbol, pointing from the hypothesis ($p$) to the conclusion ($q$).

The notation for "$p$ implies $q$" is $p \to q$ (or sometimes $p \implies q$).

Let's examine the given options and their meanings:

• (A) $p \land q$: This represents the conjunction "$p$ and $q$". It is true only when both $p$ and $q$ are true.
• (B) $p \lor q$: This represents the disjunction "$p$ or $q$". It is true when $p$ is true, or $q$ is true, or both are true.
• (C) $p \to q$: This represents the implication "$p$ implies $q$" or "If $p$, then $q$". This matches the required notation.
• (D) $q \to p$: This represents the implication "$q$ implies $p$" or "If $q$, then $p$". This is the converse of "$p$ implies $q$".

Therefore, the statement "$p$ implies $q$" is written as $p \to q$.

The correct option is (C).

Solution:

The statement $p \to q$ (read as "If $p$ then $q$") is a conditional statement. It is defined to be false only when the hypothesis $p$ is true and the conclusion $q$ is false. In all other cases, it is true.

We can find logical equivalents using truth tables or logical equivalences.

Let's analyze the truth table for $p \to q$:

Now let's evaluate the options:

(A) $\neg p \lor q$

The truth table for $\neg p \lor q$ is:

Comparing the truth table for $\neg p \lor q$ with the truth table for $p \to q$, we see they are identical. Thus, $\neg p \lor q$ is logically equivalent to $p \to q$.

$\div$ $p \to q \equiv \neg p \lor q$

(B) $p \land \neg q$

The conjunction $p \land \neg q$ is true only when $p$ is true and $\neg q$ is true (which means $q$ is false). This is exactly the case where $p \to q$ is false.

Thus, $p \land \neg q$ is the negation of $p \to q$.

$\div \neg (p \to q) \equiv p \land \neg q$.

So, $p \land \neg q$ is not equivalent to $p \to q$ (unless $p \to q$ is a contradiction, which is not generally true).

(C) $\neg (p \land \neg q)$

This is the negation of the statement in option (B).

Since $\neg (p \to q) \equiv p \land \neg q$, then $\neg (\neg (p \to q)) \equiv \neg (p \land \neg q)$.

Using the double negation law, $\neg (\neg A) \equiv A$, we have $\neg (\neg (p \to q)) \equiv p \to q$.

Therefore, $\neg (p \land \neg q) \equiv p \to q$.

Alternatively, using De Morgan's Law and double negation on option (C):

$\neg (p \land \neg q) \equiv \neg p \lor \neg (\neg q)$ $\quad$(De Morgan's Law)

$\equiv \neg p \lor q$ $\quad$(Double Negation)

As shown in option (A), $\neg p \lor q$ is logically equivalent to $p \to q$.

Thus, $\neg (p \land \neg q)$ is logically equivalent to $p \to q$.

(D) Both (A) and (C)

Since both option (A) ($\neg p \lor q$) and option (C) ($\neg (p \land \neg q)$) are logically equivalent to $p \to q$, this option is correct.

The correct option is (D).

Solution:

We are looking for a number that satisfies two properties simultaneously: it must be an even number, and it must be a prime number.

Let's define the properties:

An integer is even if it is divisible by $2$. Examples: $2, 4, 6, 8, ...$

A natural number greater than $1$ is prime if its only positive divisors are $1$ and itself. Examples: $2, 3, 5, 7, 11, ...$

The statement "The number is even and prime" is a conjunction ($p \land q$). For this statement to be true for a given number, both the statement "$p$: The number is even" and the statement "$q$: The number is prime" must be true for that number.

Let's test each option:

• (A) 4: Is 4 even? Yes, because $4 = 2 \times 2$. Is 4 prime? No, because 4 has divisors 1, 2, and 4. Since it's not prime, the statement "4 is even and prime" is false.
• (B) 6: Is 6 even? Yes, because $6 = 2 \times 3$. Is 6 prime? No, because 6 has divisors 1, 2, 3, and 6. Since it's not prime, the statement "6 is even and prime" is false.
• (C) 2: Is 2 even? Yes, because $2 = 2 \times 1$. Is 2 prime? Yes, because its only positive divisors are 1 and 2. Since it is both even and prime, the statement "2 is even and prime" is true.
• (D) 9: Is 9 even? No, 9 is odd. Is 9 prime? No, because 9 has divisors 1, 3, and 9. Since it is neither even nor prime, the statement "9 is even and prime" is false.

The only number among the options that is both even and prime is $2$.

The correct option is (C) 2.

Solution:

The given statement is "If it is raining, the ground is wet".

This statement connects two simpler statements with the phrase "If... then...".

Let $p$ be the statement "it is raining".

Let $q$ be the statement "the ground is wet".

The compound statement is of the form "If $p$, then $q$".

Different types of compound statements are formed using different logical connectives:

• Conjunction: Uses the connective "and" ($p \land q$, read as "$p$ and $q$").
• Disjunction: Uses the connective "or" ($p \lor q$, read as "$p$ or $q$").
• Conditional: Uses the connective "If... then..." ($p \to q$, read as "If $p$, then $q$" or "$p$ implies $q$").
• Biconditional: Uses the connective "if and only if" ($p \iff q$, read as "$p$ if and only if $q$").

The statement "If it is raining, the ground is wet" precisely matches the structure of a conditional statement ("If $p$, then $q$").

Therefore, this type of statement is called a conditional statement.

The correct option is (C).

Solution:

Let the given statement be $p \to q$, where:

$p$: $x > 0$

$q$: $x^2 > 0$ (assuming $x$ is a real number)

The original statement is "If $x > 0$, then $x^2 > 0$".

The contrapositive of a conditional statement $p \to q$ is the statement $\neg q \to \neg p$.

To find the contrapositive, we need to negate both the conclusion ($q$) and the hypothesis ($p$) and then swap their positions, making the negated conclusion the new hypothesis and the negated hypothesis the new conclusion.

First, let's find the negations of $p$ and $q$:

$\neg p$: The negation of "$x > 0$" is "$x$ is not greater than 0", which means "$x \leq 0$".

$\neg q$: The negation of "$x^2 > 0$" is "$x^2$ is not greater than 0", which means "$x^2 \leq 0$".

Now, form the contrapositive $\neg q \to \neg p$:

"If $\neg q$, then $\neg p$".

Substituting the negated statements, we get:

"If $x^2 \leq 0$, then $x \leq 0$".

Let's examine the given options:

(A) If $x^2 > 0$, then $x > 0$. (This is the converse, $q \to p$)

(B) If $x^2 \leq 0$, then $x \leq 0$. (This matches our derived contrapositive)

(C) If $x \leq 0$, then $x^2 \leq 0$. (This is the inverse, $\neg p \to \neg q$)

(D) If $x^2 \leq 0$, then $x < 0$. (This is a variation of the contrapositive, but the conclusion $\neg p$ is $x \leq 0$, not $x < 0$)

The contrapositive of "If $x > 0$, then $x^2 > 0$" is "If $x^2 \leq 0$, then $x \leq 0$".

The correct option is (B).

Note: A conditional statement and its contrapositive are logically equivalent. The original statement "If $x > 0$, then $x^2 > 0$" is indeed true for real numbers (the square of any positive real number is positive). The contrapositive "If $x^2 \leq 0$, then $x \leq 0$" is also true. For real numbers, $x^2 \leq 0$ only happens when $x^2 = 0$, which means $x=0$. If $x=0$, then $x \leq 0$ is true. So the contrapositive is true.

Solution:

The statement "It is cold ________ it is snowing" consists of two simple statements: "It is cold" and "it is snowing". A compound statement is formed by joining simple statements using logical connectives.

Let's examine how each option, when placed in the blank, forms a compound statement using a logical connective:

• (A) If the blank is filled with "implies", the statement becomes "It is cold implies it is snowing". This is a conditional statement, using the logical connective of implication.
• (B) If the blank is filled with "if and only if", the statement becomes "It is cold if and only if it is snowing". This is a biconditional statement, using the logical connective of biconditional.
• (C) If the blank is filled with "or", the statement becomes "It is cold or it is snowing". This is a disjunction, using the logical connective of disjunction.
• (D) If the blank is filled with "and", the statement becomes "It is cold and it is snowing". This is a conjunction, using the logical connective of conjunction.

All of the options (A), (B), (C), and (D) are indeed logical connectives (or phrases representing logical connectives) that can be used to form a compound statement by joining two simple statements.

Given the structure of the multiple-choice question implying a single correct answer, and considering the preceding questions which have focused significantly on conditional statements and the word "implies" (as seen in Question 48), it is highly probable that the question intends to highlight "implies" as one such connective.

Therefore, filling the blank with "implies" results in the compound statement "It is cold implies it is snowing", which uses a logical connective.

The correct option, based on common logical connectives and potential emphasis from surrounding questions, is (A).

Solution:

The statement we want to prove is an existential statement: "There exists an $x$ such that $P(x)$ is true". In logical notation, this is written as $\exists x, P(x)$.

The truth condition for an existential statement $\exists x, P(x)$ is that there is at least one element $x$ in the domain for which the predicate $P(x)$ holds (is true).

Therefore, to prove an existential statement $\exists x, P(x)$, the standard method is to find or construct a specific instance or value of $x$ from the domain and then show that the property $P(x)$ is true for that specific value.

This specific value of $x$ serves as a witness to the existence claimed by the statement.

Let's examine the given options in light of this understanding:

• (A) Prove $P(x)$ is true for all $x$. This is the method for proving a universal statement ($\forall x, P(x)$). If $\forall x, P(x)$ is true, then $\exists x, P(x)$ is also true (assuming the domain is non-empty), but proving the universal statement is much stronger than needed and is not the direct method for proving an existential statement.
• (B) Find at least one value of $x$ for which $P(x)$ is true. This is precisely the method described above for proving an existential statement. You just need one such value.
• (C) Find a value of $x$ for which $P(x)$ is false. This is a method for disproving a universal statement ($\forall x, P(x)$), where $x$ serves as a counterexample. It does not prove an existential statement.
• (D) Use mathematical induction. Mathematical induction is a proof technique primarily used to prove that a property holds for all natural numbers (a type of universal statement), not for proving the existence of a single element with a property.

Therefore, to prove the statement "There exists an $x$ such that $P(x)$ is true", you must find at least one value of $x$ for which $P(x)$ is true.

The correct option is (B).

Solution:

Let the given statement be $S$. The statement $S$ is "Some days are rainy".

This statement is an existential statement, meaning "There exists at least one day such that it is rainy".

Let the domain be "days". Let $P(d)$ be the predicate "day $d$ is rainy".

The statement can be written in logical notation as:

$\exists d, P(d)$

We need to find the negation of this statement, which is $\neg (\exists d, P(d))$.

According to the rules of logical negation for quantifiers, the negation of an existential statement ($\exists d, P(d)$) is a universal statement with the negated predicate ($\forall d, \neg P(d)$).

So, $\neg (\exists d, P(d)) \equiv \forall d, \neg P(d)$.

Now, let's find the negation of the predicate $P(d)$:

$P(d)$ is "day $d$ is rainy".

$\neg P(d)$ is "day $d$ is not rainy".

Now, let's translate the negated statement $\forall d, \neg P(d)$ back into English:

"For all days $d$, day $d$ is not rainy".

This can be rephrased as "All days are not rainy" or "No day is rainy".

Let's examine the given options:

(A) Some days are not rainy. ($\exists d, \neg P(d)$) - This is the negation of "All days are rainy".

(B) All days are rainy. ($\forall d, P(d)$) - This is the universal counterpart, not the negation.

(C) All days are not rainy (or No day is rainy). ($\forall d, \neg P(d)$) - This matches our derived negation.

(D) No days are sunny. - This statement is about being sunny, not being rainy.

The correct option is (C).

Solution:

We are asked to find a statement that is logically equivalent to the disjunction $p \lor q$.

We can use De Morgan's Laws and the property of double negation to evaluate the given options.

De Morgan's Laws are:

$\neg (p \land q) \equiv \neg p \lor \neg q$

$\neg (p \lor q) \equiv \neg p \land \neg q$

The double negation property is: $\neg (\neg A) \equiv A$

Let's evaluate each option:

(A) $\neg (\neg p \land \neg q)$

This statement is the negation of a conjunction $(\neg p \land \neg q)$. Applying De Morgan's Law for conjunction, $\neg (A \land B) \equiv \neg A \lor \neg B$. Let $A = \neg p$ and $B = \neg q$.

So, $\neg (\neg p \land \neg q) \equiv \neg (\neg p) \lor \neg (\neg q)$.

Using the double negation property, $\neg (\neg p) \equiv p$ and $\neg (\neg q) \equiv q$.

Therefore, $\neg (\neg p \land \neg q) \equiv p \lor q$.

This statement is logically equivalent to $p \lor q$.

(B) $\neg p \land \neg q$

According to De Morgan's second law, $\neg (p \lor q) \equiv \neg p \land \neg q$. So this option is equivalent to the negation of $p \lor q$, not $p \lor q$ itself.

(C) $\neg (\neg p \lor \neg q)$

This statement is the negation of a disjunction $(\neg p \lor \neg q)$. Applying De Morgan's Law for disjunction, $\neg (A \lor B) \equiv \neg A \land \neg B$. Let $A = \neg p$ and $B = \neg q$.

So, $\neg (\neg p \lor \neg q) \equiv \neg (\neg p) \land \neg (\neg q)$.

Using the double negation property, $\neg (\neg p) \equiv p$ and $\neg (\neg q) \equiv q$.

Therefore, $\neg (\neg p \lor \neg q) \equiv p \land q$. This statement is generally not logically equivalent to $p \lor q$.

(D) $\neg p \lor \neg q$

According to De Morgan's first law, $\neg (p \land q) \equiv \neg p \lor \neg q$. So this option is equivalent to the negation of $p \land q$, which is not generally equivalent to $p \lor q$.

Based on the analysis, the statement logically equivalent to $p \lor q$ is $\neg (\neg p \land \neg q)$.

The correct option is (A).

Solution:

Let the original statement be $p \to q$, where:

$p$: "A number ends with 0"

$q$: "The number is divisible by 5"

The original statement is "If a number ends with 0, then it is divisible by 5". This statement is true, as any number ending in 0 is a multiple of 10, and any multiple of 10 is also a multiple of 5.

The converse of a conditional statement $p \to q$ is the statement $q \to p$.

The converse of the given statement is "If a number is divisible by 5, then it ends with 0".

To determine if the converse is true, we need to check if every number divisible by 5 necessarily ends with 0.

We can attempt to find a counterexample: a number that is divisible by 5 but does not end with 0.

Consider the number 15.

• Is 15 divisible by 5? Yes, $15 = 5 \times 3$. So, the hypothesis ($q$: "The number is divisible by 5") is true for 15.
• Does 15 end with 0? No, it ends with 5. So, the conclusion ($p$: "A number ends with 0") is false for 15.

Since we found a number (15) for which the hypothesis of the converse is true and the conclusion is false, the converse statement is false.

Therefore, the converse of the statement "If a number ends with 0, then it is divisible by 5" is not true.

Let's examine the given options:

(A) Yes (If a number is divisible by 5, it ends with 0 - False, e.g., 15). The reasoning in parentheses correctly identifies the converse as false using a counterexample, but the "Yes" answer contradicts this correct reasoning.

(B) No. This correctly states that the converse is not true.

(C) Only if the number is positive. The counterexample 15 is positive, and the converse is still false. This condition doesn't make the converse true.

(D) Only if the number is an integer. The concept of ending digits and divisibility typically applies to integers. The counterexample 15 is an integer, and the converse is still false. This condition doesn't make the converse true.

The correct option is (B).

Solution:

The given statement is a compound statement formed by joining two simple statements with the logical connective "or". This is a disjunction.

Let $p$ be the statement "2 is an even number".

Let $q$ be the statement "3 is an odd number".

The compound statement is "$p$ or $q$", which is $p \lor q$ in logical notation.

First, let's determine the truth value of each simple statement:

Statement $p$: "2 is an even number".

An even number is an integer that is divisible by $2$. Since $2 \div 2 = 1$ with no remainder, $2$ is divisible by $2$.

Thus, the statement "2 is an even number" is True.

Statement $q$: "3 is an odd number".

An odd number is an integer that is not divisible by $2$. When $3$ is divided by $2$, the remainder is $1$.

Thus, the statement "3 is an odd number" is True.

Now, we need to find the truth value of the disjunction $p \lor q$, where $p$ is True and $q$ is True.

The truth table for a disjunction ($p \lor q$) states that the statement is true if at least one of the component statements ($p$ or $q$) is true. It is false only when both $p$ and $q$ are false.

In our case, both $p$ is True and $q$ is True.

Referring to the truth table for $p \lor q$:

When $p$ is True and $q$ is True, the disjunction $p \lor q$ is True.

Therefore, the truth value of the statement "2 is an even number or 3 is an odd number" is True.

Comparing this with the given options:

(A) True (True or True) - This matches our finding that both components are true and the resulting disjunction is true.

(B) False - Incorrect.

(C) Depends on context - The truth values of "2 is even" and "3 is odd" are universally accepted in mathematics and do not depend on context.

(D) Cannot be determined - The truth value can be determined based on the truth values of the components and the rule for disjunction.

The correct option is (A).

Solution:

Let the given statement be $S$. The statement $S$ is "The car is red and the bike is blue".

This statement is a conjunction, formed by joining two simple statements with "and".

Let $p$ be the statement "The car is red".

Let $q$ be the statement "The bike is blue".

The given statement is "$p$ and $q$", which is $p \land q$ in logical notation.

We need to find the negation of this statement, which is $\neg (p \land q)$.

According to De Morgan's first law, the negation of a conjunction ($p \land q$) is the disjunction of the negations of the individual statements ($\neg p \lor \neg q$).

So, $\neg (p \land q) \equiv \neg p \lor \neg q$.

Now let's find the negations of $p$ and $q$:

$\neg p$ is "The car is not red".

$\neg q$ is "The bike is not blue".

Combining these with the disjunction ($\lor$), the negation $\neg p \lor \neg q$ translates to "The car is not red or the bike is not blue".

Let's examine the given options:

(A) The car is not red and the bike is not blue. (This is $\neg p \land \neg q$, which is the negation of $p \lor q$)

(B) The car is not red or the bike is not blue. (This is $\neg p \lor \neg q$, which matches our result)

(C) The car is red or the bike is blue. (This is $p \lor q$, which is different from the negation)

(D) The car is not red and the bike is blue. (This is $\neg p \land q$, which is different from the negation)

Therefore, the negation of the statement "The car is red and the bike is blue" is "The car is not red or the bike is not blue".

The correct option is (B).

Solution:

The given statement is a conditional statement of the form "If $p$, then $q$".

Let $p$ be the statement "a number is divisible by 4".

Let $q$ be the statement "the number is divisible by 8".

The statement is $p \to q$: "If a number is divisible by 4, then it is divisible by 8".

A conditional statement $p \to q$ is false if and only if the hypothesis $p$ is true and the conclusion $q$ is false.

To determine if the statement is false, we need to find a number that is divisible by 4 (hypothesis $p$ is true) but is NOT divisible by 8 (conclusion $q$ is false).

Let's consider the number 4.

• Is 4 divisible by 4? Yes, $4 \div 4 = 1$. So, the hypothesis "4 is divisible by 4" is true.
• Is 4 divisible by 8? No, $4 \div 8$ is not an integer. So, the conclusion "4 is divisible by 8" is false.

Since we found a number (4) for which the hypothesis is true and the conclusion is false, the conditional statement "If a number is divisible by 4, then it is divisible by 8" is false.

The number 4 serves as a counterexample.

Let's examine the given options:

(A) True (e.g., 16 -> 8) - The example 16 does not prove the universal statement true; it only shows one case where it holds. The statement is false due to the counterexample.

(B) False (Counterexample: 4 is divisible by 4 but not by 8) - This correctly identifies the statement as false and provides a valid counterexample.

(C) True if the number is even - Any number divisible by 4 is already even ($4n = 2(2n)$). This condition does not change the truth value of the statement.

(D) True if the number is a multiple of 8 - If a number is a multiple of 8, the conclusion ($q$) is true, which makes the conditional statement $p \to q$ true regardless of whether $p$ is true. However, this does not mean the original general statement is true for all numbers divisible by 4.

The existence of the counterexample 4 proves that the statement is false.

The correct option is (B).

Solution:

There are different types of logical reasoning used to draw conclusions.

• Inductive Reasoning: This type of reasoning moves from specific observations or examples to formulate a general conclusion or hypothesis. The conclusion is probable but not guaranteed to be true.
• Deductive Reasoning: This type of reasoning starts with one or more general premises (statements assumed to be true) and uses logical rules to reach a specific conclusion that must necessarily be true if the premises are true.
• Analogical Reasoning: This type of reasoning draws conclusions by comparing similarities between two different things or situations.
• Abductive Reasoning: This type of reasoning starts with an observation and then seeks the simplest and most likely explanation for that observation (often used in diagnosis or hypothesis formation).

The question asks for the type of reasoning that involves drawing a specific conclusion from a general statement.

This direction, from general premises to a specific conclusion that logically follows, is the hallmark of Deductive Reasoning.

For example, a general statement (premise) is "All men are mortal". A specific statement (premise) is "Socrates is a man". Using deductive reasoning, we can conclude the specific statement "Socrates is mortal".

Let's examine the options based on this definition:

(A) Inductive Reasoning - This goes from specific to general.

(B) Deductive Reasoning - This goes from general to specific, where the conclusion is guaranteed if the premises are true.

(C) Analogical Reasoning - This is based on comparison, not general-to-specific rules.

(D) Abductive Reasoning - This is about finding the best explanation, not drawing a certain conclusion from a general statement.

The correct option is (B).

Solution:

Let the given statement be $S$. The statement $S$ is "All numbers are positive".

This statement is a universal statement. Let $x$ represent a number. Let $P(x)$ be the predicate "$x$ is positive".

The statement can be written in logical notation as:

$\forall x, P(x)$

Here, the domain is implicitly all numbers being considered.

We need to find the negation of this statement, which is $\neg (\forall x, P(x))$.

According to the rules of logical negation for quantifiers, the negation of a universal statement ($\forall x, P(x)$) is an existential statement with the negated predicate ($\exists x, \neg P(x)$).

So, $\neg (\forall x, P(x)) \equiv \exists x, \neg P(x)$.

Now, let's find the negation of the predicate $P(x)$:

$P(x)$ is "$x$ is positive".

$\neg P(x)$ is "$x$ is not positive".

For a real number, "not positive" means "less than or equal to zero". So, $\neg P(x)$ means "$x \leq 0$".

Now, let's translate the negated statement $\exists x, \neg P(x)$ back into English:

"There exists an $x$ such that $x$ is not positive".

This can be rephrased as "Some numbers are not positive" or "At least one number is not positive".

Let's examine the given options:

(A) All numbers are negative. ($\forall x, \text{"x is negative"}$) - This is a universal statement with a different predicate.

(B) Some numbers are not positive. ($\exists x, \text{"x is not positive"}$) - This matches our derived negation.

(C) No numbers are positive. (Equivalent to "All numbers are not positive", $\forall x, \neg P(x)$) - This is the negation of "Some numbers are positive" ($\exists x, P(x)$), not the negation of the original statement.

(D) Some numbers are negative. ($\exists x, \text{"x is negative"}$) - This is an existential statement with a different predicate.

Therefore, the negation of "All numbers are positive" is "Some numbers are not positive".

The correct option is (B).

Solution:

The statement "$p \leftrightarrow q$" is a biconditional statement, read as "$p$ if and only if $q$".

A biconditional statement $p \leftrightarrow q$ is true when $p$ and $q$ have the same truth value (both true or both false).

A biconditional statement $p \leftrightarrow q$ is false when $p$ and $q$ have different truth values (one is true and the other is false).

Let's look at the truth table for $p \leftrightarrow q$:

From the truth table, the statement $p \leftrightarrow q$ is false in the following cases:

• $p$ is True and $q$ is False.
• $p$ is False and $q$ is True.

In both these cases, $p$ and $q$ have different truth values.

Let's examine the given options:

(A) $p$ and $q$ have the same truth value. (This corresponds to the cases where $p \leftrightarrow q$ is True)

(B) $p$ and $q$ have different truth values. (This corresponds to the cases where $p \leftrightarrow q$ is False) - This matches our finding.

(C) $p$ is true and $q$ is true. (This is a case where $p \leftrightarrow q$ is True)

(D) $p$ is false and $q$ is false. (This is a case where $p \leftrightarrow q$ is True)

Therefore, the statement "$p \leftrightarrow q$" is false if and only if $p$ and $q$ have different truth values.

The correct option is (B).

Solution:

A compound statement is formed by combining two or more simple statements using logical connectives such as "and", "or", "if... then...", "if and only if", or by negating a simple statement ("not").

We are looking for a compound statement that specifically uses the word "or", which forms a disjunction.

Let's examine each option:

(A) The sum of two even numbers is even. This is a single simple statement about a property of numbers. It does not combine two simpler statements using "or".

(B) The number is an integer and it is positive. This is a compound statement using the word "and". It is a conjunction.

(C) The colour is red or the fruit is an apple. This statement combines two simple statements ("The colour is red" and "the fruit is an apple") using the word "or". This is a disjunction.

(D) If it is cold, I wear a jacket. This is a compound statement using the phrase "If... then...". It is a conditional statement.

The statement that is a compound statement using the word "or" is "The colour is red or the fruit is an apple".

The correct option is (C).

Solution:

Let the given statement be a conditional statement $p \to q$, where:

$p$: "it snows"

$q$: "the roads are slippery"

The original statement is "If it snows, then the roads are slippery".

The contrapositive of a conditional statement $p \to q$ is the statement $\neg q \to \neg p$.

To find the contrapositive, we need to negate both the conclusion ($q$) and the hypothesis ($p$) and then swap their positions, making the negated conclusion the new hypothesis and the negated hypothesis the new conclusion.

First, let's find the negations of $p$ and $q$:

$\neg p$: The negation of "it snows" is "it does not snow".

$\neg q$: The negation of "the roads are slippery" is "the roads are not slippery".

Now, form the contrapositive $\neg q \to \neg p$:

"If $\neg q$, then $\neg p$".

Substituting the negated statements, we get:

"If the roads are not slippery, then it does not snow".

Let's examine the given options:

(A) If the roads are slippery, then it snows. (This is the converse, $q \to p$)

(B) If it does not snow, then the roads are not slippery. (This is the inverse, $\neg p \to \neg q$)

(C) If the roads are not slippery, then it does not snow. (This matches our derived contrapositive)

(D) If the roads are not slippery, then it snows. (This is $\neg q \to p$)

The contrapositive of "If it snows, then the roads are slippery" is "If the roads are not slippery, then it does not snow".

The correct option is (C).

Note: A conditional statement and its contrapositive are logically equivalent. If the original statement is true (which is generally assumed in this context), its contrapositive is also true.

Solution:

Let the given statement be $S$. The statement $S$ is "No cat can bark".

This statement is equivalent to the universal statement "For all cats, the cat cannot bark".

Let the domain be cats. Let $P(x)$ be the predicate "$x$ can bark".

The statement $S$ can be written in logical notation as:

$\forall x, \neg P(x)$

We need to find the negation of this statement, which is $\neg (\forall x, \neg P(x))$.

According to the rules of logical negation for quantifiers, the negation of a universal statement ($\forall x, Q(x)$) is an existential statement with the negated predicate ($\exists x, \neg Q(x)$).

Here, $Q(x)$ is $\neg P(x)$. So, the negation is:

$\neg (\forall x, \neg P(x)) \equiv \exists x, \neg (\neg P(x))$

Using the double negation property, $\neg (\neg P(x)) \equiv P(x)$.

So, the negation is:

$\exists x, P(x)$

Now, let's translate the negated statement $\exists x, P(x)$ back into English:

"There exists a cat $x$ such that $x$ can bark".

This can be rephrased as "Some cat can bark" or "At least one cat can bark".

We need to choose the word that fits in the blank in the phrase "________ cat can bark".

Let's examine the options:

(A) All: "All cat can bark" ($\forall x, P(x)$). This is not the negation.

(B) Some: "Some cat can bark" ($\exists x, P(x)$). This matches our derived negation.

(C) No: "No cat can bark" ($\forall x, \neg P(x)$). This is the original statement.

(D) Exactly one: "Exactly one cat can bark" ($\exists ! x, P(x)$). This is a stronger statement than the standard negation. The standard negation of "None" or "No" is "Some" (meaning "at least one").

The phrase "Some cat can bark" is the standard negation of "No cat can bark".

The correct option is (B).

The negation of the statement "No cat can bark" is "Some cat can bark".

Solution:

Validating a statement means proving that the statement is true.

Let's examine each of the listed methods:

(A) Proof by contradiction:

This method is used to prove a statement $P$ by assuming the negation of the statement, $\neg P$, is true, and then showing that this assumption leads to a logical contradiction (a statement that is always false, such as $Q \land \neg Q$). Since the assumption $\neg P$ leads to a contradiction, it must be false. If $\neg P$ is false, then $P$ must be true.

This is a valid method for proving (validating) a wide range of statements.

(B) Proving the contrapositive:

This method is specifically used to prove conditional statements of the form $p \to q$. The contrapositive of $p \to q$ is $\neg q \to \neg p$. A fundamental logical equivalence states that a conditional statement is logically equivalent to its contrapositive ($p \to q \equiv \neg q \to \neg p$). Therefore, if you can prove the contrapositive $\neg q \to \neg p$ is true, the original statement $p \to q$ is also true.

This is a valid method for proving (validating) conditional statements.

(C) Direct proof:

This is perhaps the most straightforward proof method. To directly prove a statement (often a conditional statement $p \to q$), you start by assuming the hypothesis $p$ is true and use definitions, axioms, and previously proven theorems to logically deduce that the conclusion $q$ must also be true.

This is a valid method for proving (validating) statements.

Since all three methods listed (Proof by contradiction, Proving the contrapositive, and Direct proof) are standard and valid techniques used in mathematics and logic to prove or validate statements, the correct answer is that all of them are methods of validating a statement.

The correct option is (D) All of the above.

We need to determine which of the given sentences are mathematical statements. A sentence is a statement if it is either true or false, but not both.

(i) Every square is a rectangle.

This is a declarative sentence.

It asserts a fact that is verifiable. Every square satisfies the properties of a rectangle (four sides, four right angles, opposite sides parallel and equal).

The sentence can be assigned a truth value, which is True.

Therefore, this sentence is a statement.

(ii) Close the door.

This is an imperative sentence (a command).

It does not assert a fact that can be true or false.

It cannot be assigned a truth value.

Therefore, this sentence is not a statement.

(iii) What is your name?

This is an interrogative sentence (a question).

It asks for information and does not assert a fact that can be true or false.

It cannot be assigned a truth value.

Therefore, this sentence is not a statement.

(iv) The sum of two odd integers is an even integer.

This is a declarative sentence.

It asserts a mathematical property. Let $a$ and $b$ be two odd integers. Then $a = 2m+1$ and $b = 2n+1$ for some integers $m, n$. Their sum is $a+b = (2m+1) + (2n+1) = 2m+2n+2 = 2(m+n+1)$. Since $m+n+1$ is an integer, $a+b$ is an even integer.

The sentence can be assigned a truth value, which is True.

Therefore, this sentence is a statement.

The negation of a statement is a statement that is true when the original statement is false, and false when the original statement is true.

(i) All cats scratch.

This statement claims that the property "scratch" applies to every element in the set "cats". The negation is that this is not the case, meaning there is at least one cat that does not scratch.

The negation is: Not all cats scratch.

Alternatively, the negation can be stated as: There exists a cat which does not scratch.

(ii) India is a country.

This is a simple declarative statement. To negate it, we state the opposite.

The negation is: India is not a country.

(iii) For every real number x, $x > 0$.

This statement uses a universal quantifier ("For every"). The negation of a universally quantified statement is an existentially quantified statement.

The statement says that for all real numbers $x$, the property $x > 0$ holds. The negation is that there is at least one real number $x$ for which the property $x > 0$ does not hold.

The condition "$x > 0$" does not hold if $x \leq 0$.

The negation is: There exists a real number x such that $x \leq 0$.

A compound statement is formed by combining two or more simple statements using logical connectives like "and", "or", etc.

(i) 2 is an even number and 3 is a prime number.

The compound statement is "2 is an even number and 3 is a prime number".

The connective is "and".

The component statements are:

Component 1 (P): 2 is an even number.

Component 2 (Q): 3 is a prime number.

Checking the truth value of each component:

Component 1 (P): 2 is divisible by 2, so it is an even number. This statement is True.

Component 2 (Q): A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. 3 fits this definition. This statement is True.

The compound statement is of the form P and Q. A compound statement connected by "and" is true if and only if both component statements are true.

Since both P and Q are true, the compound statement "2 is an even number and 3 is a prime number" is True.

(ii) All integers are positive or negative.

The compound statement is "All integers are positive or negative".

The connective is "or".

The component statements are:

Component 1 (P): All integers are positive.

Component 2 (Q): All integers are negative.

Checking the truth value of each component:

Component 1 (P): Integers include 0 and negative numbers (e.g., $-1$). So, not all integers are positive. This statement is False.

Component 2 (Q): Integers include 0 and positive numbers (e.g., $1$). So, not all integers are negative. This statement is False.

The compound statement is of the form P or Q. A compound statement connected by "or" (inclusive or) is true if at least one of the component statements is true. It is false only if both component statements are false.

Since both P and Q are false, the compound statement "All integers are positive or negative" is False.

Note: The integer 0 is neither positive nor negative, which makes the statement false.

A statement of the form "P is a sufficient condition for Q" means that if P is true, then Q must be true. This can be written in the "if-then" form as "If P, then Q".

In the given statement:

P = "Being wealthy"

Q = "Being happy"

Using the "if-then" structure, the statement becomes:

If a person is wealthy, then that person is happy.

For a statement of the form "If P, then Q", the contrapositive is "If not Q, then not P".

In the given statement: "If a number is divisible by 9, then it is divisible by 3".

Let P be the statement: "a number is divisible by 9".

Let Q be the statement: "it is divisible by 3".

The negation of P (not P) is: "a number is not divisible by 9".

The negation of Q (not Q) is: "it is not divisible by 3".

The contrapositive statement ("If not Q, then not P") is formed by negating both the conclusion (Q) and the hypothesis (P) and swapping them.

Thus, the contrapositive is: If a number is not divisible by 3, then it is not divisible by 9.

For a conditional statement of the form "If P, then Q", the converse is formed by interchanging the hypothesis (P) and the conclusion (Q). The converse is "If Q, then P".

In the given statement: "If it is raining, then the roads are wet".

Let P be the statement: "it is raining".

Let Q be the statement: "the roads are wet".

The converse ("If Q, then P") is formed by swapping P and Q.

Thus, the converse is: If the roads are wet, then it is raining.

The given statement is: "There exists a number which is equal to its square".

Identifying the Quantifier:

The statement asserts that there is at least one number that satisfies the property of being equal to its square.

The phrase "There exists a number" is the part of the statement that indicates existence.

The quantifier used in this statement is the Existential Quantifier.

Writing the Negation:

Let the original statement be $S$. $S$ is of the form "There exists $x$ such that $P(x)$", where $P(x)$ is the property "$x$ is equal to its square". Mathematically, $P(x)$ is $x = x^2$.

The negation of an existential statement "There exists $x$ such that $P(x)$" is a universal statement "For every $x$, not $P(x)$".

The negation of the property $P(x)$ ($x = x^2$) is "not ($x = x^2$)", which is $x \neq x^2$.

So, the negation of the statement is "For every number, it is not equal to its square".

The negation is: For every number, it is not equal to its square.

The given statement is a universal statement about students. It says that for every student, one of two conditions is true: either they study seriously, or they pass the exam (or both).

Let P be the property "a student studies seriously".

Let Q be the property "a student passes the exam".

The statement about any given student is "P or Q".

The original statement is of the form "All students satisfy (P or Q)".

To negate a universal statement ("All X have property Y"), we state that "There exists an X that does not have property Y".

So, the negation will be "There exists a student who does not satisfy (P or Q)".

Now, we need to find the negation of "(P or Q)". According to De Morgan's laws, the negation of "(P or Q)" is "(not P) and (not Q)".

So, "(not P)" is "a student does not study seriously".

And "(not Q)" is "a student does not pass the exam".

The negation of "(P or Q)" is "a student does not study seriously and a student does not pass the exam".

Combining the negated quantifier and the negated condition, the negation of the original statement is:

There exists a student who does not study seriously and does not pass the exam.

We want to check the validity of the statement "If x is a real number such that $x^2 = 0$, then $x = 0$" using the method of contradiction.

Let the statement be $S$. The statement is of the form "If P, then Q", where:

P: $x$ is a real number such that $x^2 = 0$.

Q: $x = 0$.

The method of contradiction involves assuming that the statement "If P, then Q" is false. A conditional statement "If P, then Q" is false if and only if the hypothesis P is true and the conclusion Q is false.

So, we assume that:

1. $x$ is a real number such that $x^2 = 0$ (P is true).

2. $x \neq 0$ (Q is false).

From our assumption 1, we have $x^2 = 0$.

Since $x$ is a real number, $x^2$ means $x$ multiplied by itself, i.e., $x \times x$.

So, we have the equation:

$x \times x = 0$

In the set of real numbers, if the product of two numbers is zero, then at least one of the numbers must be zero.

Therefore, from $x \times x = 0$, it must be true that $x = 0$ or $x = 0$.

This implies that $x$ must be equal to 0.

So, based on the hypothesis ($x$ is a real number such that $x^2 = 0$), we have deduced that $x = 0$.

However, our assumption for contradiction included that $x \neq 0$.

We have reached a situation where we conclude $x = 0$ and simultaneously assumed $x \neq 0$. This is a contradiction.

Since assuming the statement is false leads to a contradiction, the original statement must be true (valid).

Conclusion:

The statement "If x is a real number such that $x^2 = 0$, then $x = 0$" is valid.

The given statement is "A triangle is equilateral if and only if it is equiangular".

This statement is already written in the "if and only if" form.

The phrase "if and only if" (often abbreviated as "iff") connects two statements, say P and Q, and means that "If P, then Q" and "If Q, then P" are both true.

In this case:

P: A triangle is equilateral.

Q: It is equiangular.

The statement means: (If a triangle is equilateral, then it is equiangular) AND (If a triangle is equiangular, then it is equilateral).

Since the statement is already in the requested format, rewriting it simply results in the same statement.

The rewritten statement is: A triangle is equilateral if and only if it is equiangular.

A tautology is a statement that is always true, regardless of the truth values of its individual components.

Example of a tautology:

"It is raining or it is not raining."

Justification:

Let P be the statement "It is raining".

The given compound statement is of the form "P or not P".

We can analyze the truth value of this compound statement based on the possible truth values of the component statement P.

Case 1: Assume P is True. This means "It is raining" is True.

Then "not P" ("It is not raining") is False.

The compound statement is "True or False", which is always True.

Case 2: Assume P is False. This means "It is raining" is False.

Then "not P" ("It is not raining") is True.

The compound statement is "False or True", which is always True.

In both possible scenarios for the truth value of P, the compound statement "P or not P" is always true.

Therefore, the statement "It is raining or it is not raining" is a tautology.

The given statement is: "For all positive integers $x$ and $y$, $x \cdot y > 0$".

This is a universal statement involving two variables $x$ and $y$. It uses the universal quantifier "For all".

Let the property be $P(x, y)$: $x \cdot y > 0$.

The original statement is "For every positive integer $x$, and for every positive integer $y$, $P(x, y)$ is true".

To negate a universal statement ("For all X, Property Y holds"), we state that "There exists an X such that Property Y does not hold".

When there are multiple universal quantifiers, we negate each one by changing "For all" to "There exists", and negate the condition.

The negation of "For all positive integers $x$, for all positive integers $y$, $x \cdot y > 0$" is "There exists a positive integer $x$, and there exists a positive integer $y$, such that the negation of $x \cdot y > 0$ is true".

The negation of the property $x \cdot y > 0$ is $x \cdot y \leq 0$.

Combining the negation of the quantifiers and the property, the negation of the statement is:

There exist positive integers $x$ and $y$ such that $x \cdot y \leq 0$.

The given statement is a conditional statement of the form "If P, then Q".

Let P be the statement: "the sun is shining".

Let Q be the statement: "there are no clouds".

The original statement is: If P, then Q.

Contrapositive:

The contrapositive of "If P, then Q" is "If not Q, then not P".

First, we find the negations of P and Q:

not P: "the sun is not shining".

not Q: "there are clouds" (the negation of "there are no clouds").

Now, we form the conditional statement "If not Q, then not P".

The contrapositive is: If there are clouds, then the sun is not shining.

Converse:

The converse of "If P, then Q" is "If Q, then P".

We interchange the hypothesis (P) and the conclusion (Q) of the original statement.

The converse is: If there are no clouds, then the sun is shining.

The given compound statement is "A square is a parallelogram or a rhombus".

The connective used is "or".

The component statements are:

Component 1 (P): A square is a parallelogram.

Component 2 (Q): A square is a rhombus.

Checking the truth value of each component statement:

Component 1 (P): A parallelogram is a quadrilateral with opposite sides parallel. A square has opposite sides parallel by definition. Thus, every square is a parallelogram. This statement is True.

Component 2 (Q): A rhombus is a parallelogram with all four sides equal in length. A square is a quadrilateral with four right angles and all four sides equal in length. Since a square has all sides equal, it is a rhombus. This statement is True.

The compound statement is of the form "P or Q". For a compound statement connected by "or" (inclusive or) to be true, at least one of the component statements must be true. It is false only if both components are false.

Since both Component 1 (P) and Component 2 (Q) are true, the compound statement "A square is a parallelogram or a rhombus" is True.

The given statement is a conditional statement of the form "If P, then Q".

Let P be the hypothesis: $3 > 5$.

Let Q be the conclusion: $4 < 6$.

We need to determine the truth values of P and Q.

For P: "$3 > 5$" is a mathematical inequality. Since 3 is not greater than 5, this statement is False.

For Q: "$4 < 6$" is a mathematical inequality. Since 4 is indeed less than 6, this statement is True.

The given statement is "If False, then True".

In logic, a conditional statement "If P, then Q" ($P \to Q$) is considered true in all cases except when the hypothesis (P) is true and the conclusion (Q) is false.

The truth table for $P \to Q$ is as follows:

In our case, P is False and Q is True.

Looking at the third row of the truth table, where P is False and Q is True, the truth value of the conditional statement $P \to Q$ is True.

Therefore, the truth value of the statement "If $3 > 5$, then $4 < 6$" is True.

Justification: The hypothesis "$3 > 5$" is false, and a conditional statement with a false hypothesis is always true, regardless of the truth value of the conclusion.

The given statement is a universal statement: "All students like mathematics or physics".

Let P be the property "a student likes mathematics".

Let Q be the property "a student likes physics".

The statement about any given student is "P or Q".

The original statement is of the form "For all students, (P or Q) is true".

To negate a universal statement ("For all X, Property Y holds"), we use an existential statement ("There exists an X such that Property Y does not hold").

So, the negation will start with "There exists a student...".

Next, we need to negate the property "(P or Q)". According to De Morgan's laws, the negation of "(P or Q)" is "(not P) and (not Q)".

"not P" means "a student does not like mathematics".

"not Q" means "a student does not like physics".

So, "(not P) and (not Q)" means "a student does not like mathematics and does not like physics".

Combining the negated quantifier and the negated condition, the negation of the original statement is:

There exists a student who does not like mathematics and does not like physics.

The given statement is: "The school will be closed if it is a holiday".

This statement is a conditional statement. The word "if" indicates the condition or hypothesis.

The structure of the statement is "Q if P", which is equivalent to "If P, then Q".

In this statement:

P (Hypothesis): It is a holiday.

Q (Conclusion): The school will be closed.

The type of implication is a standard conditional implication (often called a material conditional).

To rewrite the statement using 'if...then...', we follow the structure "If P, then Q".

Rewritten statement: If it is a holiday, then the school will be closed.

The given statement is: "There exists a rectangle which is a square".

This statement uses the existential quantifier "There exists".

Let the set be "rectangles".

Let the property be P(x): "x is a square", where x is an element of the set of rectangles.

The statement is of the form "There exists x such that P(x)".

To negate an existential statement ("There exists X such that Property Y holds"), we use a universal statement ("For all X, Property Y does not hold").

So, the negation will start with "For every rectangle...".

Next, we need to negate the property P(x). The negation of "x is a square" is "x is not a square".

Combining the negated quantifier and the negated property, the negation of the original statement is:

For every rectangle, it is not a square.

Alternatively, using "All": All rectangles are not squares.

The given statement is a conditional statement of the form "If P, then Q".

Let P be the statement: "a number is odd".

Let Q be the statement: "its square is odd".

The original statement is: If P, then Q.

The contrapositive of "If P, then Q" is "If not Q, then not P".

First, we find the negations of P and Q:

not P: "a number is not odd". (A number that is not odd is an even number).

not Q: "its square is not odd". (A number that is not odd is an even number, so "its square is not odd" means "its square is even").

Now, we form the conditional statement "If not Q, then not P".

The contrapositive is: If the square of a number is not odd (i.e., is even), then the number is not odd (i.e., is even).

Rewriting it more smoothly: If the square of a number is even, then the number is even.

The given statement is a conditional statement of the form "If P, then Q".

Let P be the statement: "it snows".

Let Q be the statement: "the temperature is below $0^\circ \text{C}$".

The original statement is: If P, then Q.

The converse of "If P, then Q" is "If Q, then P".

We interchange the hypothesis (P) and the conclusion (Q) of the original statement.

The converse is: If the temperature is below $0^\circ \text{C}$, then it snows.

The given statement is: "For every prime number $p$, $\sqrt{p}$ is an irrational number".

We need to determine if this statement holds true for all prime numbers $p$.

Understanding the terms:

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself (e.g., 2, 3, 5, 7, 11, ...).

An irrational number is a real number that cannot be expressed as a simple fraction $\frac{a}{b}$, where $a$ and $b$ are integers and $b \neq 0$. Numbers like $\sqrt{2}$, $\pi$, $e$ are examples of irrational numbers.

Checking the statement:

Let's consider a few prime numbers:

If $p = 2$ (which is a prime number), $\sqrt{p} = \sqrt{2}$. It is a well-known fact that $\sqrt{2}$ is irrational.

If $p = 3$ (which is a prime number), $\sqrt{p} = \sqrt{3}$. $\sqrt{3}$ is irrational.

If $p = 5$ (which is a prime number), $\sqrt{p} = \sqrt{5}$. $\sqrt{5}$ is irrational.

In fact, it is a fundamental theorem in number theory that the square root of any prime number is irrational.

Justification (Proof by Contradiction):

Assume, for the sake of contradiction, that $\sqrt{p}$ is a rational number for some prime number $p$.

If $\sqrt{p}$ is rational, it can be written in the form $\frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and the fraction is in its simplest form (i.e., $a$ and $b$ have no common factors other than 1, or $\text{gcd}(a, b) = 1$).

So, $\sqrt{p} = \frac{a}{b}$

Squaring both sides, we get $p = \frac{a^2}{b^2}$.

Multiplying by $b^2$, we get $pb^2 = a^2$.

This equation $pb^2 = a^2$ means that $a^2$ is divisible by $p$.

Since $p$ is a prime number, if $p$ divides $a^2$, then $p$ must also divide $a$.

So, we can write $a = pk$ for some integer $k$.

Substitute this into the equation $pb^2 = a^2$:

$pb^2 = (pk)^2$

$pb^2 = p^2k^2$

Since $p$ is prime, $p \neq 0$. We can divide both sides by $p$:

$b^2 = pk^2$

This equation $b^2 = pk^2$ means that $b^2$ is divisible by $p$.

Since $p$ is a prime number, if $p$ divides $b^2$, then $p$ must also divide $b$.

So, $p$ divides both $a$ and $b$.

However, this contradicts our initial assumption that the fraction $\frac{a}{b}$ was in its simplest form, meaning $a$ and $b$ have no common factors other than 1. If $p$ divides both $a$ and $b$, then $p$ is a common factor greater than 1 (since $p$ is prime, $p > 1$).

Since our assumption ($\sqrt{p}$ is rational) leads to a contradiction, the assumption must be false.

Therefore, $\sqrt{p}$ must be irrational for any prime number $p$.

Conclusion:

The statement "For every prime number $p$, $\sqrt{p}$ is an irrational number" is True.

The given statement is a conditional statement of the form "If P, then Q".

Let P be the statement: "it rains".

Let Q be the statement: "the match is cancelled".

The original statement is: If P, then Q.

The negation of a conditional statement "If P, then Q" is equivalent to "P and not Q".

First, we find the negation of Q:

not Q: "the match is not cancelled".

Now, we form the statement "P and not Q".

The negation is: It rains and the match is not cancelled.

To find the negation of a compound statement, we apply the rules of negation for the specific logical connectives involved.

(i) Rahul is tall and intelligent.

This statement is of the form "P and Q", where P is "Rahul is tall" and Q is "Rahul is intelligent".

The negation of "P and Q" is "not P or not Q" (by De Morgan's Law).

Negation of P: Rahul is not tall.

Negation of Q: Rahul is not intelligent.

The negation of the statement is: Rahul is not tall or Rahul is not intelligent.

(ii) All students like mathematics or study physics.

This statement involves a universal quantifier ("All students") and a disjunction ("like mathematics or study physics").

Let R be the property for a student: "likes mathematics or studies physics". The original statement is "For all students, R is true".

The negation of a universal statement "For all X, Property Y holds" is "There exists an X such that Property Y does not hold".

So, the negation starts with "There exists a student...".

Now, we need to negate the property R, which is "likes mathematics or studies physics". Let P' be "likes mathematics" and Q' be "studies physics". R is "P' or Q'".

The negation of "P' or Q'" is "not P' and not Q'" (by De Morgan's Law).

Negation of P': The student does not like mathematics.

Negation of Q': The student does not study physics.

The negation of R is: The student does not like mathematics and does not study physics.

Combining the parts, the negation is: There exists a student who does not like mathematics and does not study physics.

(iii) If it is cold, then it is raining.

This statement is a conditional statement of the form "If P, then Q", where P is "it is cold" and Q is "it is raining".

The negation of "If P, then Q" is "P and not Q".

Negation of Q: It is not raining.

The negation of the statement is: It is cold and it is not raining.

(iv) The sun shines if and only if there are no clouds.

This statement is a biconditional statement of the form "P if and only if Q", where P is "The sun shines" and Q is "there are no clouds".

The negation of "P if and only if Q" means that P and Q have different truth values. This occurs when (P is true and Q is false) or (P is false and Q is true).

Negation of P: The sun does not shine.

Negation of Q: there are clouds (the negation of "there are no clouds").

Case 1: P is true and Q is false. This means "The sun shines" is true AND "there are no clouds" is false (i.e., "there are clouds" is true).

Case 2: P is false and Q is true. This means "The sun shines" is false (i.e., "The sun does not shine" is true) AND "there are no clouds" is true.

The negation of the statement is: (The sun shines and there are clouds) or (The sun does not shine and there are no clouds).

For a conditional statement "If P, then Q":

The converse is "If Q, then P".

The contrapositive is "If not Q, then not P".

(i) If a number is a multiple of 10, then it is a multiple of 5.

Let P: A number is a multiple of 10.

Let Q: It is a multiple of 5.

Negation of P (not P): A number is not a multiple of 10.

Negation of Q (not Q): It is not a multiple of 5.

Converse: If it is a multiple of 5, then a number is a multiple of 10.

Rewritten: If a number is a multiple of 5, then it is a multiple of 10.

Contrapositive: If it is not a multiple of 5, then a number is not a multiple of 10.

Rewritten: If a number is not a multiple of 5, then it is not a multiple of 10.

(ii) If a triangle is equilateral, then it is isosceles.

Let P: A triangle is equilateral.

Let Q: It is isosceles.

Negation of P (not P): A triangle is not equilateral.

Negation of Q (not Q): It is not isosceles.

Converse: If it is isosceles, then a triangle is equilateral.

Rewritten: If a triangle is isosceles, then it is equilateral.

Contrapositive: If it is not isosceles, then a triangle is not equilateral.

Rewritten: If a triangle is not isosceles, then it is not equilateral.

(iii) If you study hard, then you will pass the exam.

Let P: You study hard.

Let Q: You will pass the exam.

Negation of P (not P): You do not study hard.

Negation of Q (not Q): You will not pass the exam.

Converse: If you will pass the exam, then you study hard.

Rewritten: If you will pass the exam, then you study hard.

Contrapositive: If you will not pass the exam, then you do not study hard.

Rewritten: If you will not pass the exam, then you do not study hard.

We want to check the validity of the statement "The sum of an irrational number and a rational number is irrational" using the method of contradiction.

Let the statement be $S$. The statement can be formulated as: If a number is the sum of an irrational number and a rational number, then the number is irrational.

Let P be the hypothesis: A number is the sum of an irrational number and a rational number.

Let Q be the conclusion: The number is irrational.

The method of contradiction involves assuming that the statement "If P, then Q" is false. A conditional statement "If P, then Q" is false if and only if the hypothesis P is true and the conclusion Q is false.

So, we assume that:

1. The hypothesis P is true: There exists a number which is the sum of an irrational number and a rational number.

2. The conclusion Q is false: The number is not irrational (i.e., the number is rational).

Let the irrational number be $I$ and the rational number be $R$. Let their sum be $S$.

From our assumption 1, we have $S = I + R$.

From our assumption 2, we assume $S$ is a rational number.

So, we have the equation $I + R = S$, where $I$ is irrational, $R$ is rational, and $S$ is rational (by assumption).

We can rearrange this equation to isolate $I$:

$I = S - R$

Now, let's consider the properties of rational and irrational numbers.

A rational number can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are integers and $b \neq 0$.

The difference of two rational numbers is always a rational number. Let $S = \frac{p}{q}$ and $R = \frac{r}{s}$, where $p, q, r, s$ are integers and $q, s \neq 0$.

Then $S - R = \frac{p}{q} - \frac{r}{s} = \frac{ps - qr}{qs}$.

Since $p, q, r, s$ are integers, $ps - qr$ is an integer and $qs$ is a non-zero integer (because $q \neq 0$ and $s \neq 0$).

Thus, $\frac{ps - qr}{qs}$ is a rational number.

From the equation $I = S - R$, and the fact that the difference of two rational numbers ($S$ and $R$) is rational, we conclude that $I$ must be a rational number.

However, our initial premise stated that $I$ is an irrational number.

We have reached a situation where we conclude that $I$ is rational, but we started with the fact that $I$ is irrational. This is a contradiction.

Since assuming the negation of the original statement leads to a contradiction, the original statement must be true (valid).

Conclusion:

The statement "The sum of an irrational number and a rational number is irrational" is valid (True).

We want to verify the truth of the statement "If $n$ is an integer and $n^2$ is odd, then $n$ is odd" using the method of contrapositive.

Let the statement be $S$. $S$ is of the form "If P, then Q", where:

P: $n$ is an integer and $n^2$ is odd.

Q: $n$ is odd.

The domain for $n$ is the set of integers.

The method of contrapositive relies on the logical equivalence between a statement "If P, then Q" and its contrapositive "If not Q, then not P". If we can prove that the contrapositive is true, then the original statement is also true.

First, we find the negation of the conclusion Q and the hypothesis P.

Negation of Q (not Q): $n$ is not odd. Since $n$ is an integer, if it's not odd, it must be even.

So, not Q: $n$ is even.

Negation of P (not P): It is not the case that ($n$ is an integer and $n^2$ is odd). Using De Morgan's law, this is ($n$ is not an integer) or ($n^2$ is not odd).

Since the premise of the original statement is about an integer $n$, the condition "$n$ is an integer" is part of the context. So, we focus on the part related to $n^2$. The negation of "$n^2$ is odd" is "$n^2$ is not odd", which means "$n^2$ is even".

So, not P: $n^2$ is even. (assuming $n$ is an integer)

The contrapositive statement is "If not Q, then not P", which is:

"If $n$ is even, then $n^2$ is even".

Now, we need to prove the contrapositive statement is true.

Assume the hypothesis of the contrapositive is true: $n$ is an integer and $n$ is even.

If $n$ is an even integer, then by definition of an even number, $n$ can be written in the form $n = 2k$ for some integer $k$.

Now, let's find the square of $n$, which is $n^2$:

$n^2 = (2k)^2$

$n^2 = 4k^2$

We can write $4k^2$ as $2 \cdot (2k^2)$.

$n^2 = 2 \cdot (2k^2)$

Since $k$ is an integer, $k^2$ is also an integer. Let $m = 2k^2$. Since $k$ is an integer, $m$ is also an integer.

So, we have $n^2 = 2m$, where $m$ is an integer.

By the definition of an even number, this means that $n^2$ is an even number.

Thus, we have shown that if $n$ is an integer and $n$ is even, then $n^2$ is even.

This proves that the contrapositive statement "If $n$ is an integer and $n$ is even, then $n^2$ is even" is true.

Since the contrapositive statement is true, the original statement, which is logically equivalent to its contrapositive, must also be true.

Conclusion:

The statement "If $n$ is an integer and $n^2$ is odd, then $n$ is odd" is true.

The given statement is P: "All prime numbers are odd".

Checking the truth value of the statement:

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Examples of prime numbers are 2, 3, 5, 7, 11, 13, ...

An odd number is an integer that is not divisible by 2.

The statement claims that every single prime number is odd.

Let's look at the first few prime numbers:

2 is a prime number.

Is 2 odd? No, 2 is divisible by 2, so it is an even number.

We have found a prime number (2) that is not odd.

Since the statement "All prime numbers are odd" claims a property holds for *every* prime number, and we found one example where it does not hold, the statement is false.

Therefore, the statement P is False.

Writing the negation of the statement:

The statement P is a universal statement of the form "For all X, Property Y holds", where X is "prime number" and Property Y is "is odd".

The negation of a universal statement is an existential statement: "There exists an X such that Property Y does not hold".

Applying this to our statement:

Negation of P: There exists a prime number that is not odd.

A number that is not odd is even.

The negation of the statement is: There exists a prime number which is even.

Justification for the negation:

The negation claims that there is at least one prime number that is even. We know that the number 2 is prime, and the number 2 is even. So, the existence claimed by the negation is true.

The negation is True.

This confirms that our original statement P was indeed False, as its negation is True.

The given statement is a conditional statement of the form "If P, then Q".

Let P be the statement: "$x$ is a prime number".

Let Q be the statement: "$x$ is not divisible by any number other than 1 and itself". (Note that this is part of the definition of a prime number).

The original statement is: If P, then Q.

Contrapositive:

The contrapositive of "If P, then Q" is "If not Q, then not P".

First, we find the negations of P and Q:

not P: "$x$ is not a prime number".

not Q: "$x$ is divisible by some number other than 1 and itself".

Now, we form the conditional statement "If not Q, then not P".

The contrapositive is: If $x$ is divisible by some number other than 1 and itself, then $x$ is not a prime number.

Converse:

The converse of "If P, then Q" is "If Q, then P".

We interchange the hypothesis (P) and the conclusion (Q) of the original statement.

The converse is: If $x$ is not divisible by any number other than 1 and itself, then $x$ is a prime number.

We want to check the validity of the statement "The product of two even integers is an even integer" using the method of direct proof.

Statement: If two integers are even, then their product is an even integer.

Method of Direct Proof:

In a direct proof, we assume the hypothesis is true and logically deduce that the conclusion must also be true.

Hypothesis: Let $m$ and $n$ be two even integers.

Conclusion: We want to show that the product $m \cdot n$ is an even integer.

By the definition of an even integer, an integer is even if it can be written in the form $2k$ for some integer $k$.

Since $m$ is an even integer, there exists some integer $k_1$ such that $m = 2k_1$.

Since $n$ is an even integer, there exists some integer $k_2$ such that $n = 2k_2$.

Now, consider the product of $m$ and $n$:

$m \cdot n = (2k_1) \cdot (2k_2)$

$m \cdot n = 4 k_1 k_2$

We can rewrite $4 k_1 k_2$ as $2 \cdot (2 k_1 k_2)$.

$m \cdot n = 2 \cdot (2 k_1 k_2)$

Let $K = 2 k_1 k_2$. Since $k_1$ and $k_2$ are integers, their product $k_1 k_2$ is an integer. The product of an integer ($k_1 k_2$) and 2 is also an integer. Therefore, $K = 2 k_1 k_2$ is an integer.

So, the product $m \cdot n$ can be written in the form $2K$, where $K$ is an integer.

By the definition of an even integer, this means that $m \cdot n$ is an even integer.

We have successfully started with the assumption that $m$ and $n$ are even integers and logically deduced that their product $m \cdot n$ is an even integer.

Conclusion:

The statement "The product of two even integers is an even integer" is valid (True).

We want to verify the truth of the statement "If a quadrilateral is a rhombus, then it is a parallelogram" by showing that its contrapositive is true.

Let the original statement be $S$. $S$ is of the form "If P, then Q", where:

P: A quadrilateral is a rhombus.

Q: It is a parallelogram.

The contrapositive of "If P, then Q" is "If not Q, then not P".

First, we find the negations of Q and P.

Negation of Q (not Q): It is not a parallelogram.

Negation of P (not P): A quadrilateral is not a rhombus.

The contrapositive statement is: If a quadrilateral is not a parallelogram, then it is not a rhombus.

Now, we need to prove the contrapositive statement is true.

Assume the hypothesis of the contrapositive is true: A quadrilateral is not a parallelogram.

Understanding Definitions:

A parallelogram is a quadrilateral with both pairs of opposite sides parallel.

A rhombus is a quadrilateral with all four sides equal in length. A rhombus is also defined as a parallelogram with all four sides equal in length.

From the definition, a rhombus is a type of parallelogram. This means that if a quadrilateral is a rhombus, it must also satisfy the definition of a parallelogram (i.e., having both pairs of opposite sides parallel). In other words, the property of being a rhombus implies the property of being a parallelogram.

Therefore, the set of rhombuses is a subset of the set of parallelograms.

Now, let's return to the hypothesis of the contrapositive: "A quadrilateral is not a parallelogram".

If a quadrilateral is not a parallelogram, it means it does not satisfy the definition of a parallelogram (i.e., it does not have both pairs of opposite sides parallel).

Since every rhombus is a parallelogram, if a quadrilateral is NOT a parallelogram, it cannot be a rhombus either (because if it were a rhombus, it would have to be a parallelogram).

Thus, if a quadrilateral is not a parallelogram, then it is not a rhombus.

We have shown that the contrapositive statement "If a quadrilateral is not a parallelogram, then it is not a rhombus" is true.

Since the contrapositive statement is true, the original statement, which is logically equivalent to its contrapositive, must also be true.

Conclusion:

The statement "If a quadrilateral is a rhombus, then it is a parallelogram" is true.

The given statement is a compound statement formed by the conjunction (using "and") of two conditional statements.

Let the first conditional statement be $S_1$: "If a number is divisible by 3, then the sum of its digits is divisible by 3".

Let the second conditional statement be $S_2$: "if the sum of its digits is divisible by 3, then the number is divisible by 3".

The original statement is $S_1$ and $S_2$.

Let P be the statement: "A number is divisible by 3".

Let Q be the statement: "The sum of its digits is divisible by 3".

Then $S_1$ is "If P, then Q".

And $S_2$ is "If Q, then P".

The structure of the combined statement "$S_1$ and $S_2$" is "(If P, then Q) and (If Q, then P)".

This compound statement is logically equivalent to the biconditional statement "P if and only if Q".

Using the "if and only if" connective, we combine the two component statements P and Q.

The rewritten statement is: A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

To write the negation of a statement involving quantifiers, we interchange universal quantifiers ($\forall$, "For every" / "All") with existential quantifiers ($\exists$, "There exists") and vice versa, and negate the remaining part of the statement.

(i) There exists a triangle that is equilateral and right-angled.

This statement is of the form "There exists X such that P(X)".

The quantifier is "There exists".

The property P(X) is "X is equilateral and X is right-angled". This property is a conjunction (P' and Q').

The negation of "There exists X such that P(X)" is "For all X, not P(X)".

The negation of P(X) ("P' and Q'") is "not P' or not Q'" (by De Morgan's Law).

not P': "X is not equilateral".

not Q': "X is not right-angled".

So, the negation of the property is "X is not equilateral or X is not right-angled".

The negation of the original statement is: For every triangle, it is not equilateral or it is not right-angled.

(ii) For every real number $x$, $x^2 \geq 0$.

This statement is of the form "For every X, P(X)".

The quantifier is "For every".

The property P(X) is "$x^2 \geq 0$".

The negation of "For every X, P(X)" is "There exists X such that not P(X)".

The negation of P(X) ("$x^2 \geq 0$") is "$x^2 < 0$".

The negation of the original statement is: There exists a real number $x$ such that $x^2 < 0$.

(iii) All students have passed the exam or submitted the assignment.

This statement is of the form "For all X, P(X)".

The quantifier is "All" (which means "For every").

The property P(X) for a student X is "X has passed the exam or X has submitted the assignment". This property is a disjunction (P'' or Q'').

The negation of "For all X, P(X)" is "There exists X such that not P(X)".

So, the negation starts with "There exists a student...".

The negation of P(X) ("P'' or Q''") is "not P'' and not Q''" (by De Morgan's Law).

not P'': "The student has not passed the exam".

not Q'': "The student has not submitted the assignment".

So, the negation of the property is "The student has not passed the exam and has not submitted the assignment".

The negation of the original statement is: There exists a student who has not passed the exam and has not submitted the assignment.

We need to check the validity of the given argument and identify its underlying logical structure.

Let's break down the argument into its constituent statements:

Premise 1: If it is cloudy, it will rain.

Premise 2: It is cloudy.

Conclusion: Therefore, it will rain.

Now, let's represent these statements using logical variables:

Let P be the statement: "It is cloudy."

Let Q be the statement: "It will rain."

In symbolic form, the argument can be written as:

Premise 1: $P \to Q$ (If P, then Q)

Premise 2: $P$

Conclusion: $Q$

This logical structure is a fundamental rule of inference known as Modus Ponens (Latin for "mode that affirms").

Modus Ponens states that if a conditional statement ($P \to Q$) is accepted as true, and the antecedent ($P$) of the conditional statement is also accepted as true, then the consequent ($Q$) must be true.

To check the validity of the argument, we can see if the conclusion logically follows from the premises. The argument is valid if and only if the statement $((P \to Q) \land P) \to Q$ is a tautology (always true).

We can construct a truth table:

Since the final column of the truth table is always True, the statement $((P \to Q) \land P) \to Q$ is a tautology. This means that whenever the premises are true, the conclusion is also true.

Conclusion:

The underlying logical structure is Modus Ponens.

The argument is valid because it follows the structure of Modus Ponens, which is a valid rule of inference.

A contradiction (or tautological contradiction) is a statement that is always false, regardless of the truth values of its individual components.

Example of a contradiction:

"The sky is blue and the sky is not blue."

Justification:

Let P be the statement "The sky is blue".

The given compound statement is of the form "P and not P".

We can analyze the truth value of this compound statement based on the possible truth values of the component statement P.

Case 1: Assume P is True. This means "The sky is blue" is True.

Then "not P" ("The sky is not blue") is False.

The compound statement is "True and False", which is always False.

Case 2: Assume P is False. This means "The sky is blue" is False.

Then "not P" ("The sky is not blue") is True.

The compound statement is "False and True", which is always False.

In both possible scenarios for the truth value of P, the compound statement "P and not P" is always false.

Therefore, the statement "The sky is blue and the sky is not blue" is a contradiction.