If a quadrilateral is a square then it is a rectangle.
Converse statement :
If a quadrilateral is a rectangle, then it a square.
Inverse statement :
If a quadrilateral is not a square then it is not a rectangle.
Contrapositive statement :
If a quadrilateral is not a rectangle, the it is not a square.

If x and y are numbers such that x = y, then x² = y²
Converse statement :
If x and y are numbers such that x² = y² then x = y
Inverse statement :
If x and y are numbers such that x ≠ y then x² ≠ y²
Contrapositive statement :
If x and y are numbers such that x²≠ y² then x ≠ y

Given * on A is commutative
Given b * a = c ⇒ a * b = c
Given c * a = a ⇒ a * c = a
Given b * c = a ⇒ c * b = a
Hence

√b is not defined for negative values, b which also ∈ R.
Hence, a√b is not defined for all a, b ∈R
* is not a binary operation on R

a *b = min (a, b) on A = {1,2,3,4, 5}  Let a,b ∈A
A = {1,2, 3, 4, 5}
a*b = min {(a, b)}
Now, 1,2 ∈ A \(\Rightarrow\)1 * 2 = 1 ∈ A
3, 5 ∈ A \(\Rightarrow\) 3 * 5 = 3 ∈ A
Hence * is a binary operation on A

Given a*b = a.|b| on R
a,b ∈ R \(\Rightarrow\) a.|b| ∈R as a ∈ R and |b|∈R.
Hence * is a binary operation on R

Then the four types of conditional statements corresponding to p and q are respectively listed below.
(i) p →q : (conditional statement) “If the number of primes is infinite then Ooty is in Kerala”.
(ii) q → p : (converse statement) “If Ooty is in Kerala then the number of primes is infinite”
(iii) ¬p → ¬q (inverse statement) “If the number of primes is not infinite then Ooty is not in Kerala”.
(iv) ¬q → ¬p (contrapositive statement) “If Ooty is not in Kerala then the number of primes is not infinite”.

Here the component statements p and q are given by,
p: Today is Monday; q: 4 + 4 = 8.
The truth value of p→q is T because the conclusion q is T.
An important point is that p→q should not be treated by actually considering the meanings of p and q in English. Also it is not necessary that p should be related to q at all.

(( p ∧ q) ∨ (¬r ∨¬s)) ∧ (¬ t ∧ v)) contains 6 variables p, q, r, s, t, and v. Hence the corresponding truth table will contain 2⁶ = 64 rows.

p ∨ ¬ t ( p ∨ ¬s) contains 3 variables p, s, and t. Hence the corresponding truth table will contain 2³ = 8 rows

Then A∨ B =\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\vee \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\vee 1 & 1\vee 1 \\ 1\vee 0 & 1\vee 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\)
\(A\wedge B=\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\wedge \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\wedge 1 & 1\wedge 1 \\ 1\wedge 0 & 1\wedge 1 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}\)

(1) ¬p: It is not cold.
(2) p ∧ q: It is cold and raining.
(3) p ∨ q: It is cold or raining.
(4) q ∨ ¬p: It is raining or it is not cold
Observe that the statement formula ¬ p has only 1 variable p and its truth table has 2 = ( 2¹ ) rows. Each of the statement formulae p ∧ q and p ∨ q has two variables p and q. The truth table corresponding to each of them has 4 = (2² ) rows. In general, it follows that if a statement formula involves n variables, then its truth table will contain 2ⁿ rows.

(1) Mount Everest is the highest mountain of the world.
(2) 3+ 4 = 8 .
(3) 7 + 5 >10 .
(4) Give me that book.
(5) (10 − x) = 7.
(6) How beautiful this flower is!
(7) Where are you going?
(8) Wish you all success.
(9) This is the beginning of the end.
The truth value of the sentences (1) and (3) are T, while that of (2) is F. Hence they are statements.
The sentence (5) is true for x = 3 and false for x ≠ 3 and hence it may be true or false but not both. So it is also a statement.
The sentences (4), (6), (7), (8) are not statements, because (4) is a command, (6) is an exclamatory, (7) is a question while (8) is a sentence expressing one’s wishes and (9) is a paradox.

In this problem a ∗ b is in the quotient form. Since the division by 0 is undefined, the denominator b -1 must be nonzero.
It is clear that b −1 = 0 if b = 1. As 1∈Q, ∗ is not a binary operation on the whole of Q. However it can be found that by omitting 1 from Q, the output a ∗b exists in Q\{1}. Hence ∗ is a binary operation on Q\{1}.

Since × is binary operation on Z, a,b ∈ Z⇒ a × b = ab∈Z and b × b = b²∈Z ...(1)
The fact that + is binary operation on Z and (1) ⇒ 3ab = (ab + ab + ab) ∈Z and 5b²= (b²+b²+b²+b²+b²)∈Z     ...(2)
Also a∈Z and 3ab ∈Z implies a+3ab∈Z   ...(3)
(2),(3), the closure property of -on Z yield a * b = (a+3ab-5b²)∈Z. Since a * b belongs to Z, * is a binary operation on Z.

Truth Table for (~p ⟶ r) ∧ ( p ↔️ q)

Truth Table for ( p V q) ∧ ~q

​​​​​​Truth Table for ~(p ∧ ~q)

Truth Table for ​​​​​​~p ∧ ~q