Clearly 1 is the identity element of G
w² . a^-1 = e ⇒ w² . a^-1 = a^-1 = w
Since w² . w = w³ = 1
Inverse of w² is w.

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.

Let a be any element and e be the identity element.
The a * e = e * a = a
a * e = a ⇒ a + e -1 = a ⇒ e-1 = 0 ⇒ e = 1
∴ The identity element is 1

∴p v (q Λ r) is a contingency

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

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

(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 = {a+\(\sqrt5\) b:a,b ∈ z}
Let C = a+\(\sqrt5\) b
B = c+\(\sqrt5\)d∈A
where a, b, c, d ∈ Z
[∵ ac + 5bd∈Z and ad+bc ∈Z]
∴ B = (a+\(\sqrt5\)b).(c+\(\sqrt5\)d)
= ac+\(\sqrt5\)ad+cb\(\sqrt5\) + 5bd
= (ac+5bd)+\(\sqrt5\)(ad+bc)∈A
∴ C.B ∈A∀ a, b, c, d∈Z
∴ Usual multiplicaition is a binary operation on.

√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

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

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 V q) ∧ ~q

​​​​​​Truth Table for ~(p ∧ ~q)

Truth Table for ​​​​​​~p ∧ ~q