(i) It is true that a*b = a^b ∈N; ∀a,b∈N. So * is a binary operation on N.
(ii) a*b = a^b and b*a = b^a. Put, a = 2 and b = 3. Then a*b = 23 = 8 but b*a = 3² = 9 So a*b need not be equal to b*a. Hence * does not have commutative property.
(iii) Next consider a*(b*c) = a*(b^c) = a^_(bc₎ Take a = 2, b = 3 and c = 4. 
Then a*(b*c) = 2*(*4) = 2^₃4 = 2⁸¹
But (a*b)*c = (a^b)*c = (a^b)^c = a(bc) = a^(bc) = 2¹²
Hence a*(b*c) ≠ (a*b)*c. So * does not have associative property on N.

Consider the set Z_o of all odd integers Z_o = {2k+1 : k∈Z} = {.., -5,-3,-1,1,3,5,...}. + is not a binary operation on Z_o because when x = 2m+1, y = 2n +1, x + y = 2(m+ n) + 2 is even for all m and n. For instance, consider the two odd numbers 3, 7 ∈Z_o. Their sum 3 + 7 = 10 is an even number. In general, if x, y∈Z₀, then (x+y)∉Z_o. Other properties need not be checked as it is not a binary operation.

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.

i) Though - is not binary on N; it is binary on Z. To check the validity of any more properties satisfied by – on Z, it is better to check them for some particular simple values.
ii) Take m = 4 , n = 5 and (m− n) = (4 − 5) = −1and (n −m) = (5 − 4) = 1.
Hence (m− n) ≠ (n −m). So the operation - is not commutative on Z.
iii) In order to check the associative property, let us put m = 4, n = 5 and p = 7 in both (m- n) - p and m- (n - p).
(m−n)− p = (4−5)−7 = (−1−7) = −8 …(1)
m−(n− p) = 4−(5−7) = (4+2) = 6  …(2)
From (1) and (2), it follows that (m - n) - p m - (n - p).
Hence – is not associative on Z.
iv) Identity does not exist (why?).
v) Inverse does not exist (why?).

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

Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) \vee \left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1\wedge 0 & 0\wedge 1 \\ 0\wedge 1 & 1\wedge 0 \\ 1\wedge 1 & 0\wedge 0 \end{matrix}\begin{matrix} 1\wedge 0 & 0\wedge 1 \\ 0\wedge 1 & 1\wedge 0 \\ 0\wedge 0 & 1\wedge 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \end{matrix} \right) \) [∵ a៱b=max(a,b)]

Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) \vee \left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1\vee 0 & 0\vee 1 \\ 0\vee 1 & 1\vee 0 \\ 1\vee 1 & 0\vee 0 \end{matrix}\begin{matrix} 1\vee 0 & 0\vee 1 \\ 0\vee 1 & 1\vee 0 \\ 0\vee 0 & 1\vee 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1 & 1 \\ 1 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 1 \\ 1 & 1 \\ 0 & 1 \end{matrix} \right) \) [∵ a∨b=max(a,b)]

Given A = {a, b, c, d} and * is defined as follows.

1) Commulative Property:
\(a * b=c\)  but  \(b \times a=d \Rightarrow a * b \neq b * a\)
\(a * c=b\) but \(c * a=c \Rightarrow a * c \neq c * a\) 
\(\therefore\) \(\text { * }\) is not commutative on A..
2) Associative Property: 
\((a+b) * c=c * c=a\) 
\(a *(b * c)=a * b=c\) 
\(\therefore(a * b) * c \neq a *(b * c)\) 
\(\text { * }\) is not associative on A.

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.

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 = \(\frac{-7}{15}\)]
\(=3-\frac { 7 }{ 15 } -\frac { 21 }{ 15 } -7\)
\(\therefore 3*\left( \frac { -7 }{ 15 } \right) =\frac { -88 }{ 15 } \)

Given m*n = mⁿ + n^m ∀m, n∈Z
Let us take - 2, 2∈Z
Consider m = -3, n = 2
(m * n) = (-2 * 2) = (-2)² + (2)^-2
\(=4+\frac { 1 }{ 4 } =\frac { 17 }{ 4 } \notin Z\)
∴ * is not a binary operation on Z

√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