Q: Which of the following is not true?

Truth table \$p\$\$q\$ \$ p\leftrightarrow q\$\$ T\$ \$T\$\$T\$ \$T\$ \$F\$ \$F\$ \$F\$ \$T\$ \$F\$ \$F\$ \$F\$ \$T\$ Since the last column of the truth table has False and Truth.It cannot be tautology.

Q: The negation of the statement \$(p\rightarrow q)\wedge r\$ is

\$p\rightarrow q\$\$=\sim p\vee q---(1)\$\$\sim [(p\rightarrow q)\wedge r]\$\$\sim [\sim p\vee q\wedge r]\$...[form(i)]\$\sim (\sim p\vee q)\vee \sim r\$...[Demorgans low]\$\sim (\sim p)\wedge \sim q\vee q\sim r \$ \$ av(b\wedge c)\$\$p\wedge \sim q\vee \sim r \$ \$ =(a\vee b)\vee (a\vee c)]\$

Q: State whether the following statements are true or false. Give reasons for your answers.For any real number \$x, x^2 \geq 0\$.

The square of any real number is always positive.Let \$x\$ be a positive real number. We can see \$x^2\ge 0\$ as \$x \times x\ge 0\$If \$x\$ is negative real number, then, \$(-x)(-x)=xx=x^2 \ge 0\$So, the statement is aways TRUE.

Q: If \$p\rightarrow (q \vee r)\$ is false, then the truth values of \$p,q,r\$ are respectively

\$p \rightarrow q\$ is false only when \$p\$ is true and \$q\$ is false.\$\therefore p\rightarrow (q \vee r)\$ is false when \$p\$ is true and \$(q \vee r)\$ is false, and\$q \vee r\$ false when both \$q, r\$ are false.Hence T,F,F

Q: If p, q, r have truth values T, F, T respectively, then which of the following is \$\text{True}\$?

Since, p is true and q is false.\$\therefore p\rightarrow q\$ has truth value F.Statement r has truth value T.\$\therefore (p\rightarrow q)\wedge r\$ has truth value F.Also, \$(p\rightarrow q)\wedge \sim r\$ has truth value F.Now, \$p\wedge q\$ has truth value F and \$p\vee r\$ has truth value T.\$\therefore (p\wedge q)\wedge (p\vee q)\$ has truth value F.As, \$p\wedge r\$ has truth value T.Therefore, \$q\rightarrow (p\wedge r)\$ has truth value T.

Q: \$P\rightarrow (q\rightarrow r)\$ is logically equivalent to

\$ P \rightarrow (\sim q\rightarrow r)\$\$ = p \rightarrow (\sim q\vee r)\$\$ \sim p\vee (\sim q\vee r) \$\$ (\sim p \vee \sim q) \vee r\$\$ \sim (p\wedge q) \vee r\$\$ (p\wedge q) \rightarrow r\$

Q: Prove that \$3+\sqrt{7}\$ is irrational number.

We have to prove that \$3+\sqrt{7}\$ is irrational.Let us assume the opposite, that \$3+\sqrt{7}\$ is rational.Hence \$3+\sqrt{7}\$ can be written in the form \$\dfrac{a}{b}\$ where \$a\$ and \$b\$ are co-prime and \$b\neq 0\$Hence \$3+\sqrt{7}=\dfrac{a}{b}\$\$\Rightarrow \sqrt{7}=\dfrac{a}{b}-3\$\$\Rightarrow \sqrt{7}=\dfrac{a-3b}{b}\$where \$\sqrt{7}\$ is irrational and \$\dfrac{a-3b}{b}\$ is rational.Since,rational\$\neq\$ irrational.This is a contradiction.\$\therefore\$ Our assumption is incorrect.Hence \$3+\sqrt{7}\$ is irrational.Hence proved.

Q: State whether the following statement is True or False.All even numbers are composite numbers.

False \$2\$ is a prime number.A prime number is a whole number that only has two factors which are itself and one. A composite number has factors in addition to one and itself. The numbers 0 and 1 are neither prime nor composite. All even numbers are divisible by two and so all even numbers greater than two are composite numbers.

Question 1

Which of the following is not true?

Accepted Answer

Truth table $p$$q$ $ p\leftrightarrow q$$ T$ $T$$T$ $T$ $F$ $F$ $F$ $T$ $F$ $F$ $F$ $T$ Since the last column of the truth table has False and Truth.It cannot be tautology.

Question 2

The negation of the statement $(p\rightarrow q)\wedge r$ is

Accepted Answer

$p\rightarrow q$$=\sim p\vee q---(1)$$\sim [(p\rightarrow q)\wedge r]$$\sim [\sim p\vee q\wedge r]$...[form(i)]$\sim (\sim p\vee q)\vee \sim r$...[Demorgans low]$\sim (\sim p)\wedge \sim q\vee q\sim r $   $  av(b\wedge c)$$p\wedge \sim q\vee \sim r  $  $   =(a\vee b)\vee (a\vee c)]$

Question 3

State whether the following statements are true or false. Give reasons for your answers.For any real number $x, x^2 \geq 0$.

Accepted Answer

The square of any real number is always positive.Let $x$ be a positive real number. We can see $x^2\ge 0$     as     $x \times x\ge 0$If $x$ is negative real number, then,  $(-x)(-x)=xx=x^2 \ge 0$So, the statement is aways TRUE.

Question 4

If $p\rightarrow  (q \vee r)$ is false, then the truth values of $p,q,r$ are respectively

Accepted Answer

$p \rightarrow  q$ is false only when $p$ is true and $q$ is false.$\therefore  p\rightarrow (q \vee r)$ is false when $p$ is true and $(q \vee r)$ is false, and$q \vee r$ false when both $q, r$ are false.Hence T,F,F

Question 5

Which of the following is negation of the statement "All birds can fly".

Accepted Answer

To find the negation of the statement, we find the opposite of the conclusion. So, negation of the given statement is &#34; Some birds cannot fly&#34;

Question 6

If p, q, r have truth values T, F, T respectively, then which of the following is $\text{True}$?

Accepted Answer

Since, p is true and q is false.$\therefore p\rightarrow q$ has truth value F.Statement r has truth value T.$\therefore (p\rightarrow q)\wedge r$ has truth value F.Also, $(p\rightarrow q)\wedge \sim r$ has truth value F.Now, $p\wedge q$ has truth value F and $p\vee r$ has truth value T.$\therefore (p\wedge q)\wedge (p\vee q)$ has truth value F.As, $p\wedge r$ has truth value T.Therefore, $q\rightarrow (p\wedge r)$ has truth value T.

Question 7

$P\rightarrow (q\rightarrow r)$ is logically equivalent to

Accepted Answer

$ P \rightarrow (\sim q\rightarrow r)$$ = p \rightarrow (\sim q\vee r)$$ \sim p\vee (\sim q\vee r) $$ (\sim p \vee \sim q) \vee r$$ \sim (p\wedge q) \vee r$$ (p\wedge q) \rightarrow r$

Question 8

Which of the following is not proposition?

Accepted Answer

A proposition is a declaration that can be either true or false, but not both.A) $3$ is prime, its trueB) $\sqrt2$ is irrational, it is true.C) Mathematics is interesting, we cannot say anything about it.D) $5$ is an even integer, It is false.Hence, option C.

Question 9

Prove that $3+\sqrt{7}$ is irrational number.

Accepted Answer

We have to prove that $3+\sqrt{7}$ is irrational.Let us assume the opposite, that $3+\sqrt{7}$ is rational.Hence $3+\sqrt{7}$ can be written in the form $\dfrac{a}{b}$ where $a$ and $b$ are co-prime and $b\neq 0$Hence $3+\sqrt{7}=\dfrac{a}{b}$$\Rightarrow \sqrt{7}=\dfrac{a}{b}-3$$\Rightarrow \sqrt{7}=\dfrac{a-3b}{b}$where $\sqrt{7}$ is irrational and $\dfrac{a-3b}{b}$ is rational.Since,rational$\neq$ irrational.This is a contradiction.$\therefore$ Our assumption is incorrect.Hence $3+\sqrt{7}$ is irrational.Hence proved.

Question 10

State whether the following statement is True or False.All even numbers are composite numbers.

Accepted Answer

False $2$ is a prime number.A prime number is a whole number that only has two factors which are itself and one. A composite number has factors in addition to one and itself. The numbers 0 and 1 are neither prime nor composite. All even numbers are divisible by two and so all even numbers greater than two are composite numbers.