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 ’ $pvq$ ‘. Since, the statement ’ $pvq$ ’ 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 $2.$ [false]

$q:57$ is divisible by $3.$ [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 $6.$ [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) $2+7\ne 6$.

(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 $3x^2-x-10=0$.

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

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

(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.

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

(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 $x^2-5x+6=0$.

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

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

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

(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 $p\to q$ 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 $AP$, then $2b=a+c$.

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.

Solution

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

(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 $xy$ 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 $0<x<1$.

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 $2x=3y$, 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

Solution

Here, two cases arise

Case I When $n$ is even,

Let

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

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

Case II When $n$ is odd,

Let

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

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.

$\Rightarrow p\vee r$ 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.

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

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

Then, our supposition is wrong.

Hence, $p$ is true.

Solution

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

On squaring both sides,

Hence, we have the result.

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

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.

Solution

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

Solution

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

Solution

(d) Connective word is ‘and’.

Solution

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

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

Solution

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

$\sim p:7$ is not 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 .

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

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

Solution

(b) Let $p$ : Plants take in $C{O}_2$ and give out $O_2$.

Let $q$ : Plants take in $C{O}_2$.

$r$ : Plants give out $O_2$.

$\sim q$ : Plants do not take in $C{O}_2$.

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

$\sim (q\wedge r)$ : Plants do not take in $C{O}_2$ or do not give out $O_2$.

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.

Solution

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

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

Solution

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

and $q:8$ is greater than 6 .

$\therefore p\to q$

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

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

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

Solution

(b) Let

$\begin{array}{c}p:x>y\\ q:x+a>y+a\\ p\to q\end{array}$

Converse of the above statement is

$q\to p$

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

Solution

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

and $q$ : Sky is filled with clouds.

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

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

Solution

(c) $p\to q$

If $p$, then $q$

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

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

Solution

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

and $q:x$ is not even.

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

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

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 $p\to q$ is

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

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

Solution

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

Solution

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

Solution

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

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

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.