Given a*b = a + b + ab -7, ∀ a,b ∈R
If a ∈R, b∈R then ab ∈ R
(a*b) = a +b+ ab - 7 ∈R
For example, let 1, 2 ∈ R
(1*2) = 1+2+(1)(2)-7
= 2 ∈ R
* a binary operation on R

[Here a = 3, b = \(\frac{-7}{15}\)]
\(=3-\frac { 7 }{ 15 } -\frac { 21 }{ 15 } -7\)
\(\therefore 3*\left( \frac { -7 }{ 15 } \right) =\frac { -88 }{ 15 } \)

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 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.

Prove that p⟶(q⟶r) = (p ∧ q)⟶r without using truth table. From example we know that p⟶ q = ~p V q
Consider LHS = p⟶(q⟶r)
= p ⟶ (~q V r) [Implication Law]
= ~p V (~q V r) [Implication Law]
= (~p V ~q) V r [associative property]
= ~(p ∧ q) V r [using Demorgan's law]
= (p ∧ q) ⟶ r [Implication Law]
Hence proved.

Given M = \(\left\{ \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) :x\in R-\{ 0\} \right\} \) and ∗ be the matrix multiplication.
Let A = \(\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)and 
B = \(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)∈M
Where x, y ∈R-{0}.
\(A*B=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)
\(=\left( \begin{matrix} xy+xy & xy+xy \\ xy+xy & xy+xy \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xy & xy \\ 2xy & 2xy \end{matrix} \right) \in M\\ \)
[∵ 2xy∈R-{0}]
∴ M is closed under M
Identity:
Since identity of 2\(\times\)2 matrices is I =\(\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) \)∉M 
∴ M has no identity under *.
Inverse:
Since it has no identity, it won't have inverse also.

Given \(a*b=\frac { a+b }{ 2 } \forall \in Q\)
i) Closure property:
Let a, b ∈ Q
∴ a*b = \(\frac{a+b}{2}\)∈Q
[∵ addition and division are closed on Q]
* is closed on Q.
(ii) Commutative property:
Let a, b ∈ Q
Then a+b \(=\frac { a+b }{ 2 } =\frac { b+a }{ 2 } =b*a\)
∴ a*b = b*a ∀a,b∈Q
∴ * is commutative on Q.
(iii) Associative property :
Let a, b, c ∈ Q
a*(b*c) = (a*b)*c
Let a = 2, b = 3, c-5
∴ a*(b*c) = 2*(3*-5)
\(=*\left( \frac { 3-5 }{ 2 } \right) \)
\(=2*(-1)=\frac { 2+(-1) }{ 2 } \)
\(=\frac { 1 }{ 2 } \quad \quad ...(1)\)
Now (a*b)*c = (2*3)*(-5)
\(=\left( \frac { 2+3 }{ 2 } \right) *(-5)\)
\(=\frac { 5 }{ 2 } *-5=\frac { \frac { 5 }{ 2 } +(-5) }{ 2 } \)
\(=\frac { 5-10 }{ 4 } =\frac { -5 }{ 4 } ...(2)\)
From (1) & (2), a*(b*c) ≠ (a*b)*c
∴ * is not associative on Q.

It can be obtained by using examples 12.15 and 12.16 that
p↔q ≡ (¬ p ∨ q)  ∧ (¬q ∨ p) ... (1)
≡ (¬p∨q)  ∧ ( p∨ ¬q) (by Commutative Law) ... (2)
≡ (¬p ∧ ( p ∨ ¬q)) ∨ (q ∧ ( p ∨ ¬q)) (by Distributive Law)
≡ (¬p ∧ p) ∨ (¬p ∧ ¬q) ∨ (q ∧ p) ∨ (q ∧ ¬q) (by Distributive Law)
≡ F ∨ (¬p ∧ ¬q) ∨ (q ∧ p) ∨ F; (by Complement Law)
≡ (¬p ∧ ¬q) ∨ (q ∧ p) ; (by Identity Law)
≡ ( p ∧ q) ∨ (¬p ∧ ¬q) ; (by Commutative Law)
Finally (1) becomes p ↔️ q ≡ ( p ∧ q) v (ㄱp ∧ ㄱq)

(a) Let a, b \(\in \) 2.
The a * b = a + b + z = b + a + z = b * a
(z, *) has commutative property.
(b) Let a, b, c \(\in \) G
(a * b) * c = (a+b+2) * c = a + b + 2 + c + 2
= a + b + c + 4
∴ z is associate under *.

The entries in the columns corresponding q ➝ p and ~p ➝ ~q are identical and hence they are equivalent.
q ➝ p ≡ ~p ➝ ~q
Hence proved

Truth Table for ( p V q) ∧ ~q