Solution

As we know, a statement is a sentence which is either true or false but not both simultaneously.

(i) It is true statement.

(ii) It is true statement.

(iii) It is false statement.

(iv) It is false statement.

(v) It is false statement.

(vi) $y+9=7$

It is not considered as a statement, since the value of $y$ is not given.

(vii) It is a question, so it is not a statement.

(viii) It is a true statement.

(ix) It is true statement.

(x) It is false statement.

Solution

(i) $p$ : Number 7 is prime.

$q$ : Number 7 is odd.

(ii) $P$ : Chennai is in India.

$q$ : Chennai is capital of Tamil Nadu.

(iii) $p: 100$ is divisible by 3 .

$q: 100$ is divisible by 11 .

$r: 100$ is divisible by 5 .

(iv) $p$ : Chandigarh is capital of Haryana.

$q$ : Chandigarh is capital of UP.

(v) $p: \sqrt{7}$ is a rational number.

$q: \sqrt{7}$ is an irrational number.

(vi) $p: 0$ is less than every positive integer.

$q: 0$ is less than every negative integer.

(vii) $p$ : Plants use sunlight for photosysthesis.

$q$ : Plants use water for photosynthesis.

$r$ : Plants use carbon dioxide for photosysthesis.

(viii) $p$ : Two lines in a plane intersect at one point.

$q$ : Two lines in a plane are parallel.

(ix) $p:$ A rectangli, is a quadrilateral.

$q$ : A rectangle is a 5 -sided polygon.

Solution

(i) Given compound statement is of the form ’ $p v q$ ‘. Since, the statement ’ $p v q$ ’ has the truth value $T$ whenever either $p$ or q or both have the truth value $T$.

So, it is true statement.

Its component statements are

$p: 57$ is divisible by $2 . \quad$ [false]

$q: 57$ is divisible by $3 . \quad$ [true]

(ii) Given compound statement is of the form ’ $p \wedge q$ ‘. Since, the statement ’ $p \wedge q$ ’ have the truth value $T$ whenever both $p$ and $q$ have the truth value $T$.

So, it is a true statement.

Its component statements are

$p: 24$ is multiple of 4 [true]

$q: 24$ is multiple of $6 . \quad$ [true]

(iii) It is a false statement. Since ’ $p \wedge q$ ’ has truth value $F$ whenever either $p$ or $q$ or both have the truth value $F$.

Its component statements are

$p$ : All living things have two eyes. [false]

$q$ : All living things have two legs. [false]

(iv) It is a true statement.

Its component statements are

$p: 2$ is an even number. [true]

$q: 2$ is a prime number. [true]

Solution

(i) The number 17 is not prime.

(ii) $2+7 \neq 6$.

(iii) Violets are not blue.

(iv) $\sqrt{5}$ is not a rational number.

(v) 2 is a prime number.

(vi) Every real number is not an irrational number.

(vii) Cow has not four legs.

(viii) A leap year has not 366 days.

(ix) There exist similar triangles which are not congruent.

(x) Area of a circle is not same as the perimeter of the circle.

Solution

(i) $p$ : Rahul passed in Hindi.

$q$ : Rahul passed in English.

$p \wedge q$ : Rahul passed in Hindi and English.

(ii) $p: x$ is even integers.

$q: y$ is even integers.

$p \cap q: x$ and $y$ are even integers.

(iii) $p: 2$ is factor of 12 .

$q: 3$ is factor of 12.

$r: 6$ is factor of 12.

$p \wedge q \wedge r: 2,3$ and 6 are factor of 12 .

(iv) $p: x$ is an odd integer.

$q:(x+1)$ is an odd integer.

$p \vee q$ : Either $x$ or $(x+1)$ is an odd integer.

(v) $p:$ A number is divisible by 2 .

$q$ : A number is divisible by 3.

$p \vee q$ : A number is either divisible by 2 or 3 .

(vi) $p: x=2$ is a root of $3 x^{2}-x-10=0$.

$q: x=3$ is a root of $3 x^{2}-x-10=0$.

$p \vee q$ : Either $x=2$ or $x=3$ is a root of $3 x^{2}-x-10=0$.

(vii) $p$ : Students can take Hindi as an optional paper.

$q$ : Students can take English as an optional subject.

$p \vee q$ : Students can take Hindi or English as an optional paper.

Thinking Process

Use (i) $\sim(p \wedge q)=\sim p \mathbf{V} \sim q$

(ii) $\sim(p \vee q)=\sim p \wedge \sim q$

Solution

(i) Let $p:$ All rational numbers are real.

$q$ : All rational numbers are complex.

$\sim p$ : All rational number are not real.

$\sim q$ : All rational numbers are not complex.

$\sim(p \wedge q)$ : All rational numbers are not real or not complex. $\quad[\because \sim(p \wedge q)=\sim p \vee \sim q]$

(ii) Let $p:$ All real numbers are rationals.

$q$ : All real numbers are irrational.

Then, the negation of the above statement is given by

$\sim(p \vee q)$ : All real numbers are not rational and all real numbers are not irrational.

$[\because \sim(p \vee q)=\sim p \wedge \sim q]$

(iii) Let $p: x=2$ is root of quadratic equation $x^{2}-5 x+6=0$.

$q: x=3$ is root of quadratic equation $x^{2}-5 x+6=0$.

Then, the negation of conjunction of above statement is given by

$\sim(p \wedge q): x=2$ is not a root of quadratic equation $x^{2}-5 x+6=0$ or $x=3$ is not a root of the quadratic equation $x^{2}-5 x+6=0$.

(iv) Let $p:$ A triangle has 3 -sides.

$q$ : A triangle has 4 -sides.

Then, negation of disjunction of the above statement is given by

$\sim(p \vee q)$ : A triangle has neither 3 -sides nor 4 -sides.

(v) Let $p: 35$ is a prime number.

$q: 35$ is a composite number.

Then, negation of disjunction of the above statement is given by

$\sim(p \vee q): 35$ is not a prime number and it is not a composite number.

(vi) Let $p:$ All prime integers are even.

$q:$ All prime integers are odd.

Then negation of disjunction of the above statement is given by

$\sim(p \vee q)$ : All prime integers are not even and all prime integers are not odd.

(vii) Let $p:|x|$ is equal to $x$.

$q:|x|$ is equal to $-x$.

Then negation of disjunction of the above statement is given by

$\sim(p \vee q):|x|$ is not equal to $x$ and it is not equal to $-x$.

(viii) Let $p: 6$ is divisible by 2 .

$q: 6$ is divisible by 3 .

Then, negation of conjunction of above statement is given by

$\sim(p \wedge q)$ : 6 is not divisible by 2 or it is not divisible by 3

Solution

We know that, some of the common expressions of conditional statement $p \rightarrow q$ are

(i) if $p$, then $q$

(ii) $q$ if $p$

(iii) $p$ only if $q$

(iv) $p$ is sufficient for $q$

(v) $q$ is necesary for $p$

(vi) $\sim q$ implies $\sim p$

So, use above information to get the answer

(i) If the number is odd number, then its square is odd number.

(ii) If you take the dinner, then you will get sweet dish.

(iii) If you will not study, then you will fail.

(iv) If an integer is divisible by 5 , then its unit digits are 0 or 5 .

(v) If the number is prime, then its square is not prime.

(vi) If $a, b$ and $c$ are in $A P$, then $2 b=a+c$.

Solution

(i) $p \leftrightarrow q$ : The unit digit of on integer is zero, if and only if it is divisible by 5 .

(ii) $p \leftrightarrow q$ : A natural number no odd if and only if it is not divisible by 2 .

(iii) $p \leftrightarrow q$ : A triangle is an equilateral triangle if and only if all three sides of triangle are equal.

Thinking Process

We know that, the statement $(\sim q) \rightarrow(\sim p)$ is called contrapositive of the statement $p \rightarrow q$.

Solution

(i) If $x \neq 3$, then $x \neq y$ or $y \neq 3$.

(ii) If $n$ is not an integer, then it is not a natural number.

(iii) If the triangle is not equilateral, then all three sides of the triangle are not equal.

(iv) If $x y$ is not positive integer, then either $x$ or $y$ is not negative integer.

(v) If natural number $n$ is not divisible by 2 or 3 , then $n$ is not divisible by 6 .

(vi) The weather will not be cold, if it does not snow.

(vii) If $x^{2} \nless 1$, then $x$ is not a real number such that $0<x<1$.

Thinking Process

We know that, the converse of the statement " $p arrow q$ " is " $(q) arrow(p)$ “.

Solution

(i) If thes rectangle ’ $R$ ’ is rhombus, then it is square.

(ii) If tomorrow is Tuesday, then today is Monday.

(iii) If you must visit Taj Mahal, you go to Agra.

(iv) If the triangle is right angle, then sum of squares of two sides of a triangle is equal to the square of third side.

(v) If the triangle is equilateral, then all three angles of triangle are equal. (vi) If $2 x=3 y$, then $x: y=3: 2$

(vii) If the opposite angles of a quadrilateral are supplementary, then $S$ is cyclic.

(viii) If $x$ is neither positive nor negative, then $x$ is 0 .

(ix) If the ratio of corresponding sides of two triangles are equal, then triangles are similar.

Solution

Quantifier are the phrases like ‘There exist’ and ‘For every’, ‘For all’ etc.

(i) There exists

(ii) For all

(iii) There exists

(iv) For every

(v) For all

(vi) There exists

(vii) For all

(viii) There exists

(ix) There exists

(x) There exists

Thinking Process

We know that, in direct method to show a statement, if $p$ then $q$ is true, we assume $p$ is true and show $q$ is true i.e., $p arrow q$.

Solution

Here, two cases arise

Case I When $n$ is even,

Let

$ \begin{aligned} n & =2 K, K \in N \\ \Rightarrow\quad n^3-n & =(2 K)^{3}-(2 K)=2 K(4 K^{2}-1) \\ & =2 \lambda \text {, where } \lambda=K(4 K^{2}-1) \end{aligned} $

Thus, $(n^{3}-n)$ is even when $n$ is even.

Case II When $n$ is odd,

Let

$ \begin{aligned} n & =2 K+1, K \in N \\ \Rightarrow \quad n^{3}-n & =(2 K+1)^{3}-(2 K+1) \\ & =(2 K+1)[(2 K+1)^{2}-1] \\ & =(2 K+1)[4 K^{2}+1+4 K-1] \\ & =(2 K+1)(4 K^{2}+4 K) \\ & =4 K(2 K+1)(K+1) \\ & =2 \mu, \text { when } \mu=2 K(K+1)(2 K+1) \end{aligned} $

Then, $n^{3}-n$ is even when $n$ is odd

So, $n^{3}-n$ is always even.

Solution

(i) $p: 125$ is divisible by 5 and 7 .

Let $q$ : 125 is divisible by 5 .

$r: 125$ is divisible by 7 .

$q$ is true, $r$ is false.

$\Rightarrow q \wedge r$ is false.

[since, $p \wedge q$ has the truth value $F$ (false) whenever either $p$ or $q$ or both have the truth value]

Hence, $p$ is not valid.

(ii) $p: 131$ is a multiple of 3 or 11 .

Let $q: 131$ is multiple of 3.

$r: 131$ is a multiple of 11.

$p$ is true, $r$ is false.

$\Rightarrow \quad p \vee r$ is true.

[since, $p \vee q$ has the truth value $T$ (true) whenever either $p$ or $q$ or both have the truth value T]

Hence, $q$ is valid.

Solution

Let $p$ is false i.e., sum of an irrational and a rational number is rational.

Let $\sqrt{m}$ is irrational and $n$ is rational number.

$\begin{matrix} \Rightarrow & \sqrt{m}+n=r & \text { [rational] } \\ \Rightarrow & \sqrt{m}=r-n & \end{matrix} $

$\sqrt{m}$ is irrational, where as $(r-n)$ is rational. This is contradiction.

Then, our supposition is wrong.

Hence, $p$ is true.

Thinking Process

In direct method assume $p$ is true and show q is true i.e., $p \Rightarrow q$.

Solution

Let $p: x=y, \quad x, y \in R$

On squaring both sides,

Hence, we have the result.

$ \begin{gathered} x^{2}=y^{2}: q \\ p \Rightarrow q \end{gathered} $

Thinking Process

In contrapositive method assume $\sim q$ is true and show $\sim$ p is true i.e., $\sim q \Rightarrow \sim p$.

Solution

Let $p: n^{2}$ is an even integer.

$q: n$ is also an even integer.

Let $\sim p$ is true i.e., $n$ is not an even integer.

$\Rightarrow n^{2}$ is not an even integer.

[since, square of an odd integer is odd]

$\Rightarrow \sim p$ is true.

Therefore, $\sim q$ is true $\Rightarrow \sim p$ is true.

Hence proved.

Solution

(c) As we know a statement is a sentence which is either true or false.

So, 6 is a natural number, which is true.

Hence, it is a statement.

• (a) “$x$ is a real number” is not a statement because it is an open sentence; it depends on the value of $x$ and is not definitively true or false without additional information.
• (b) “Switch off the fan” is not a statement because it is an imperative sentence, which is a command and cannot be classified as true or false.
• (d) “Let me go” is not a statement because it is a request or plea, and such sentences cannot be classified as true or false.

Solution

(d) ‘Come here’ is not a statement. Since, no sentence can be called a statement, if it is an order.

• (a) “Smoking is injurious to health” is a statement because it is a declarative sentence that can be evaluated as true or false.
• (b) “$2+2=4$” is a statement because it is a mathematical equation that can be evaluated as true.
• (c) “2 is the only even prime number” is a statement because it is a declarative sentence that can be evaluated as true.

Solution

(b) In ’ $2+7>9$ or $2+7<9$ ‘, or is the connective.

• (a) “and” is incorrect because the statement uses “or” to connect the two conditions, not “and”.
• (c) “>” is incorrect because it is a relational operator comparing two values, not a connective.
• (d) “<” is incorrect because it is also a relational operator comparing two values, not a connective.

Solution

(d) Connective word is ‘and’.

• (a) “or” is incorrect because the statement uses “and” to connect two clauses, not “or”.
• (b) “Earth” is incorrect because “Earth” is a noun in the statement, not a connective word.
• (c) “Sun” is incorrect because “Sun” is also a noun in the statement, not a connective word.

Solution

(c) Let $p$ : A circle is an ellipse.

$\sim p$ : A circle is not an ellipse.

• (a) An ellipse is a circle: This statement does not negate the original statement “A circle is an ellipse.” Instead, it incorrectly implies that all ellipses are circles, which is not logically equivalent to the negation of the original statement.

• (b) An ellipse is not a circle: This statement is not the direct negation of “A circle is an ellipse.” It is a different statement that does not address the original statement directly. The correct negation should specifically state that a circle is not an ellipse.

• (d) A circle is an ellipse: This is the original statement itself and not its negation. Therefore, it does not serve as the negation of the statement “A circle is an ellipse.”

Solution

(b) Let $p: 7$ is greater than 8 .

$\sim p: 7$ is not greater than 8 .

• Option (a) “7 is equal to 8” is incorrect because the negation of “7 is greater than 8” does not imply equality; it only implies that 7 is not greater than 8, which includes the possibility of 7 being less than 8.
• Option (c) “8 is less than 7” is incorrect because it is logically equivalent to the original statement “7 is greater than 8,” not its negation.
• Option (d) “None of these” is incorrect because option (b) correctly represents the negation of the statement “7 is greater than 8.”

Solution

(b) Let $p: 72$ is divisible by 2 and 3 .

Let $q: 72$ is divisible by 2 .

$r: 72$ is divisible by 3 .

$\sim q: 72$ is not divisible by 2 .

$\sim r: 72$ is not divisible by 3 .

$\sim(q \wedge r): \sim q v \sim r$

$\Rightarrow 72$ is not divisible by 2 or 72 is not divisible by 3 .

• Option (b) is incorrect because it states that 72 is not divisible by 2 and 72 is not divisible by 3, which is the conjunction of the negations of both conditions. The correct negation of “72 is divisible by 2 and 3” should be the disjunction of the negations, not the conjunction.

• Option (c) is incorrect because it states that 72 is divisible by 2 and 72 is not divisible by 3. This does not correctly represent the negation of the original statement, as it only negates one part of the conjunction and affirms the other.

• Option (d) is incorrect because it states that 72 is not divisible by 2 and 72 is divisible by 3. This also does not correctly represent the negation of the original statement, as it only negates one part of the conjunction and affirms the other.

Solution

(b) Let $p$ : Plants take in $CO_2$ and give out $O_2$.

Let $q$ : Plants take in $CO_2$.

$r$ : Plants give out $O_2$.

$\sim q$ : Plants do not take in $CO_2$.

$\sim r$ : Plants do not give out $O_2$.

$\sim(q \wedge r)$ : Plants do not take in $CO_2$ or do not give out $O_2$.

• Option (a): “Plants do not take in $CO_2$ and do not give out $O_2$” is incorrect because it represents the negation of both parts of the statement simultaneously, which is a conjunction ($\sim q \wedge \sim r$). The correct negation should be a disjunction ($\sim q \vee \sim r$).

• Option (c): “Plants take in $CO_2$ and do not give out $O_2$” is incorrect because it only negates the second part of the statement while keeping the first part unchanged. The correct negation should negate the entire conjunction, not just one part.

• Option (d): “Plants take in $CO_2$ or do not give out $O_2$” is incorrect because it does not properly negate the original conjunction. The correct negation should be a disjunction of the negations of both parts, not a disjunction where one part remains unchanged.

Solution

(c) Let $p$ : Rajesh or Rajni lived in Bengaluru.

and $q$ : Rajesh lived in Bengaluru.

$r$ : Rajni lived in Bengaluru.

$\sim q$ : Rajesh did not live in Bengaluru.

$\sim r$ : Rajni did not live in Bengaluru.

$\sim(q \vee r)$ : Rajesh did not live in Bengaluru and Rajni did not live in Bengaluru.

• Option (a): “Rajesh did not live in Bengaluru or Rajni lives in Bengaluru” is incorrect because it does not represent the negation of the original statement. The original statement is “Rajesh or Rajni lived in Bengaluru,” which means at least one of them lived in Bengaluru. The negation should state that neither of them lived in Bengaluru, not that one did not live and the other did.

• Option (b): “Rajesh lives in Bengaluru and Rajni did not live in Bengaluru” is incorrect because it partially negates the original statement but not completely. The original statement implies that at least one of them lived in Bengaluru. The negation should indicate that neither of them lived in Bengaluru, not that one did and the other did not.

• Option (d): “Rajesh did not live in Bengaluru or Rajni did not live in Bengaluru” is incorrect because it does not fully negate the original statement. The original statement “Rajesh or Rajni lived in Bengaluru” means at least one of them lived in Bengaluru. The negation should state that neither of them lived in Bengaluru, not that at least one of them did not live in Bengaluru.

Solution

(a) Let $p$ : 101 is not a multiple of 3.

$\sim p: 101$ is a multiple of 3.

• Option (b) “101 is a multiple of 2” is incorrect because the negation of the statement “101 is not a multiple of 3” should directly address the multiple of 3, not 2. The statement about being a multiple of 2 is irrelevant to the original statement about multiples of 3.

• Option (c) “101 is an odd number” is incorrect because it does not address the multiple of 3. The original statement is about whether 101 is a multiple of 3, and whether 101 is odd or even is unrelated to this.

• Option (d) “101 is an even number” is incorrect because it does not address the multiple of 3. The original statement is about whether 101 is a multiple of 3, and whether 101 is even or odd is irrelevant to this.

Solution

(c) Let $p: 7$ is greater than 5 .

and $q: 8$ is greater than 6 .

$\therefore p \rightarrow q$

$\sim p: 7$ is not greater than 5 .

$\sim q: 8$ is not greater than 6 .

$(\sim q) \rightarrow(\sim p)$ i.e., If 8 is not greater than 6 , then 7 is not greater than 5 .

• Option (a): This option states “If 8 is greater than 6, then 7 is greater than 5.” This is simply the original statement reversed, not the contrapositive. The contrapositive requires negating both the hypothesis and the conclusion and then reversing them.

• Option (b): This option states “If 8 is not greater than 6, then 7 is greater than 5.” This is incorrect because the contrapositive should negate both the hypothesis and the conclusion of the original statement. Here, only the hypothesis is negated, not the conclusion.

• Option (d): This option states “If 8 is greater than 6, then 7 is not greater than 5.” This is incorrect because it negates the conclusion of the original statement without negating the hypothesis. The contrapositive requires both the hypothesis and the conclusion to be negated and then reversed.

Solution

(b) Let

$ \begin{gathered} p: x>y \\ q: x+a>y+a \\ p \rightarrow q \end{gathered} $

Converse of the above statement is

$ q \rightarrow p $

i.e., If $x+a>y+a$, then $x>y$.

• Option (a) is incorrect because it states “If ( x < y ), then ( x + a < y + a )”. This is not the converse of the original statement. Instead, it is a restatement of the original statement with ( x ) and ( y ) swapped and the inequality reversed, which does not logically follow from the original statement.

• Option (c) is incorrect because it is identical to option (a) and thus suffers from the same issue. It is not the converse of the original statement but rather a restatement with swapped variables and reversed inequality.

• Option (d) is incorrect because it states “If ( x > y ), then ( x + a < y + a )”. This directly contradicts the original statement, which asserts that ( x + a > y + a ) when ( x > y ). Therefore, it cannot be the converse of the original statement.

Solution

(a) Let $p$ : Sun is not shining.

and $q$ : Sky is filled with clouds.

Converse of the above statement $p \rightarrow q$ is $q \rightarrow p$.

If sky is filled with clouds, then the Sun is not shining.

• Option (b): This option states “If Sun is shining, then sky is filled with clouds.” This is incorrect because it is the inverse of the original statement, not the converse. The inverse of “If p, then q” is “If not p, then not q.”

• Option (c): This option states “If sky is clear, then Sun is shining.” This is incorrect because it is the contrapositive of the original statement, not the converse. The contrapositive of “If p, then q” is “If not q, then not p.”

• Option (d): This option states “If Sun is not shining, then sky is not filled with clouds.” This is incorrect because it is the negation of the original statement, not the converse. The negation of “If p, then q” is “If p, then not q.”

Solution

(c) $p \rightarrow q$

If $p$, then $q$

Contrapositive of the statement $p \rightarrow q$ is $(\sim q) \rightarrow(\sim p)$.

If $\sim q$, then $\sim p$.

• Option (a) “if ( q ), then ( p )” is incorrect because it represents the converse of the original statement ( p \rightarrow q ), not the contrapositive. The converse of a statement ( p \rightarrow q ) is ( q \rightarrow p ).

• Option (b) “if ( p ), then ( \sim q )” is incorrect because it represents the negation of the consequent in the original statement ( p \rightarrow q ), which is not logically equivalent to the contrapositive. The negation of the consequent would be ( p \rightarrow \sim q ).

• Option (d) “if ( \sim p ), then ( \sim q )” is incorrect because it represents the inverse of the original statement ( p \rightarrow q ), not the contrapositive. The inverse of a statement ( p \rightarrow q ) is ( \sim p \rightarrow \sim q ).

Solution

(b) Let $p: x^{2}$ is not even.

and $q: x$ is not even.

Converse of the statement $p \rightarrow q$ is $q \rightarrow p$.

i.e., If $x$ is not even, then $x^{2}$ is not even.

• Option (a) is incorrect because the statement “If ( x^2 ) is odd, then ( x ) is even” is not logically related to the given statement. The correct converse would involve the same conditions but reversed, and this option does not fit that criteria.

• Option (c) is incorrect because the statement “If ( x ) is even, then ( x^2 ) is even” is actually the contrapositive of the given statement. The contrapositive of “If ( x^2 ) is not even, then ( x ) is not even” is “If ( x ) is even, then ( x^2 ) is even,” which is logically equivalent to the original statement, not its converse.

• Option (d) is incorrect because the statement “If ( x ) is odd, then ( x^2 ) is even” is false. If ( x ) is odd, then ( x^2 ) is also odd, not even. This option does not relate to the given statement in any logical manner.

Solution

(a) Let $p$ : Chandigarh is capital of Punjab.

and $q$ : Chandigarh is in India.

$\sim p$ : Chandigarh is not capital of Punjab.

$\sim q$ : Chandigarh is not in India.

Contrapositive of the statement $p \rightarrow q$ is

if $(\sim q)$, then $(\sim p)$.

If Chandigarh is not in India, then Chandigarh is not the capital of Punjab.

• Option (b): This option states “if Chandigarh is in India, then Chandigarh is Capital of Punjab.” This is simply the converse of the original statement, not the contrapositive. The converse of a statement ( p \rightarrow q ) is ( q \rightarrow p ), which is not logically equivalent to the contrapositive.

• Option (c): This option states “if Chandigarh is not capital of Punjab, then Chandigarh is not capital of India.” This statement introduces a new concept, “capital of India,” which was not part of the original statement. Therefore, it is irrelevant and incorrect in the context of finding the contrapositive.

• Option (d): This option states “if Chandigarh is capital of Punjab, then Chandigarh is not in India.” This is the negation of the original statement, not the contrapositive. The negation of ( p \rightarrow q ) is ( p \rightarrow \sim q ), which is not logically equivalent to the contrapositive.

Solution

(c) ’ $p \rightarrow q$ ’ is same as ’ $p$ only if $q$ ‘.

• (a) “$q$ is sufficient for $p$” is incorrect because it implies that if $q$ is true, then $p$ must be true, which is the converse of the conditional $p \rightarrow q$.
• (b) “$p$ is necessary for $q$” is incorrect because it implies that $q$ cannot be true unless $p$ is true, which is the converse of the conditional $p \rightarrow q$.
• (d) “if $q$ then $p$” is incorrect because it directly states the converse of the conditional $p \rightarrow q$.

Solution

(a) The negation of the above statement is ‘It is false that the product of 3 and 4 is 9 ‘.

• (b) The product of 3 and 4 is 12: This statement is a factual correction of the original statement, not a negation. Negation involves stating that the original statement is false, not providing the correct information.

• (c) The product of 3 and 4 is not 12: This statement is incorrect because it introduces a new false statement. The original statement is about the product being 9, not 12, so this does not serve as a proper negation.

• (d) It is false that the product of 3 and 4 is not 9: This statement is a double negation, which actually affirms the original statement. It means that the product of 3 and 4 is indeed 9, which is not the intended negation.

Solution

(c) The false negation of the given statement is “It is false that a natural number is not greater than zero”.

• Option (a) “A natural number is not greater than zero” is a correct negation because it directly contradicts the original statement “A natural number is greater than zero.”

• Option (b) “It is false that a natural number is greater than zero” is also a correct negation because it explicitly states that the original statement is false.

• Option (d) “None of the above” is incorrect because both options (a) and (b) are valid negations of the original statement.

Solution

(d) If two simple statements $p$ and $q$ are connected by the word ‘and’, then the resulting compound statement $p$ and $q$ is called a conjuction of $p$ and $q$ Here, none of the given statement is conjunction

• Option (a): “Ram and Shyam are friends” is a simple statement, not a conjunction. It does not combine two separate statements using the word ‘and’.

• Option (b): “Both Ram and Shyam are tall” is a single statement describing the height of both individuals, not a conjunction of two separate statements.

• Option (c): “Both Ram and Shyam are enemies” is a single statement describing the relationship between the two individuals, not a conjunction of two separate statements.

Solution

(i) It is a statement.

(ii) It is a statement.

(iii) It is not a statement, since it is an exclamations.

(iv) It is a statement.

(v) It is not a statement, since it is a questions.

• (iii) It is not a statement, since it is an exclamation.
• (v) It is not a statement, since it is a question.