Let a \(\in \) S and e be the identity element.
Then a * e = a ⇒ \(\frac { ae }{ 2 } \) = a
⇒ ae = 2a ⇒ e = 2
Let a^-1 be the inverse of \(\frac { 1 }{ 2 } \)
Then \(\frac { 1 }{ 2 } \) * a^-1 = e = \(\frac { 1 }{ 2 } \) * a^-1 = e
⇒ \(\frac { { a }^{ -1 } }{ 2 } \frac { { a }^{ -1 } }{ 2 } \) = 2 ⇒ a^-1 = 8.

The last column contains only T
The given statement is a tautology

Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \end{matrix} \right) \vee \left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\((A\wedge B)\vee C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)

Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & 1 \\ 1 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 1 \\ 1 & 1 \\ 0 & 1 \end{matrix} \right) \wedge \left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) =\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) \)

(i) If 6 + 2 = 5, then the milk is white.
Let p: 6 + 2 = 5 (F)
q: Milk is white (T)  
p ➝ q is having the truth value T
(ii) China is in Europe or \(\sqrt3\) is an integer.
p: China is in Europe (F)
q: \(\sqrt3\) is an integer (F)
p v q is having the truth value (F).
(iii) It is not true time 5 + 5 = 9 or Earth is a planet.
Let P: 5 + 5 = 9 is not true (T)
q: Earth is a planet (T
~p ∨ q is having the truth value T
(iv) 11 is a prime number and all the sides of a rectangle are equal.
p:11 is a prime number (T)
q: Allthe sides of arectangle areequal (F)
p _^ q is having the truth value F

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

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”.

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

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.

Given s is a non-empty set and x 0 y = x, x,y \(\in \) s y₀x = y
x₀u ≠ y₀x ⇒ is not commutative
Now, x₀(y₀z) = x₀y = x
and (x₀y) 0 z = x₀z = x
x₀(y₀z) = (y₀z) 0 z
0 is associative.

From the table, it is clear that the column of p➝(¬q V ~r) and ~pV(~q V r) are identical
∴  p➝(¬q V ~r) ≡ ~pV(~q V r) 
Hence proved.

given A = {Q\{1}}
A is defined on A by x*y = x+y-xy
Let x,y ≠ 1
∴ x*y = x + y - xy
Now to prove that x + y - xy ≠ 1
Let us assume that x + y - xy = 1
x+y-xy-1 = 0
(x-1)-y(x-1) = 0
(x-1)(1-y) = 0
x =1 or y = 1 which is a false [∵x, y ≠ 1]
∴ is a binary operation on A.
Commutative property:
Let x,y ∈A ⇒ x, y≠1
∴x*y = x+y-xy
and y*x = y+x-yx
⇒x+y = y*x∀x, y∈A
A has commutative property under *
Associative property:
Let x,y,z ∊A ⇒x,y,z≠1
Consider (x*y)*z = (x+y-xy)*z
= x +y~xy +z- (x +y-xy)z
= x +y-xy+ z-xz- yz + xyz
= x +y+z-xy-yz-zx +xyz ...(1)
= x +y+z - yz - x (y + z - yz)
= x +y +z-yz-zy-xz +xyz ..(2)
From (1) & (2), (x*y)*z = x*(y*z)
A has associative property under *.

The entries in column (4) and (6) are identical  ~(p↔️q) ≡ p↔️~q

The entries in the columns corresponding to p ↔ q and ( p ⟶ q) ∧ (q ⟶ p) are identical and hence they are equivalent

Consider the set of all even integers Z_e = {2k | k ∈ Z} = {...,−6, −4, −2, 0, 2, 4, 6,...}.
Let us verify the properties satisfied by + on Z_e.
(i) The sum of any two even integers is also an even integer.
Because x, y∈Z_e, ⇒ x = 2m and y = 2n , m,n∈Z.
So (x + y) = 2m + 2n = 2(m+n)∈Z_e. Hence + is a binary operation on Z_e.
(ii) ∀ x, y∈Z_e, (x + y) = 2(m + n) = 2(m + n) = 2(n + m) = (2n + 2m) = (y + x).
So + has commutative property
(iii) Similarly it can be seen that ∀x, y, z∈Z_e, (x + y) + z = x + ( y + z).
Hence the associative property is true.
(iv) Now take x = 2k , then 2k + e = e + 2k = 2k ⇒ e = 0.
Thus ∀ x ∈ Z_e, ヨ0∈Z_e, ⋺x+0 = 0+x = x.
So, 0 is the identity element.
(v) Taking x = 2k and x′ as its inverse, we have 2k+x' = 0 = x'+2k ⇒ x' = −2k. i.e., x' = −x.
Thus ∀x ∈ Z_e, ヨ-x∈Z_e ⋺x + (−x) = (−x) + x = 0
Hence -x is the inverse of x ∈Z_e.

Let a = 2, b = -5
∴ a * b = a^b
⇒ 2 * -5 = 2^-5 = \(\frac{1}{2^5}\) ∉ z
* is not a binary operation on z.