Here, S is a set of positive integers.

a b = 3a + 5b

The multiplication and addition of integers is always a positive integer.

Thus, for all a, b S, a b = 3a + 5b S

So, it is a binary operation.

- For closure, let m,n Z,

m n = n Z.

So, it is closed.

- For associative, let m,n and p Z.

(m * n ) * p = n * p = p Z

m * ( n * p) = m * p = p Z

(m * n ) * p = m * ( n * p)

So, the operation is associative.

-For communitative, let m,n Z.

m * n = n Z

n * m = m Z

m * n n * m

So, the operation is not communitaitve.

Here, Z= { 0, 1, 2, 3 }

Identity element of 2 is 0 because 2+40 = 2 and 0+42 =2

Identity element of 3 is 0 because 3+40 = 3 and 0+43 =3

Inverse element of 2 is 2 because 2+42 = 0

Inverse element of 3 is 1 because 3+41 = 0

Let e be the identity element of 3 under binary operation defined by,

m n = m + 1 + 1.

Then, 3 e = 3

or, 3 + e + 1 = 3 [m * n = m + n + 1]

So, e = -1.

Again, let e’ be the identity element of – 2.

Then, (-2) e’ = - 2.

or, (-2) + e’ + 1 = -2

or, e’ + 1 = 0

So, e’ = - 1.

So, - 1 is the required element of both 3 and – 2.

Again, let i be the inverse element of 3 under the given binary operation .

Then, 3 i = e

or, 3 + i + 1 = - 1 [m * n = m + n + 1, e = -1]

or, i = - 1 – 4 = - 5.

So, - 5 is the inverse...

From the table,

We see that the operation defined on any two elements of S gives an element of S itself.

i.e. For each a,b S, a*b = S

So, S is closed under operation *.

For each a,b S,

a * (b * c) = (a * b) * c

or, a * a = b * c

or, a = a (true)

So, * satisfies associative property.

a * a = a,

a * b = b * a= b

and, a * c = c * a = c

a is an identity element.

a * a = a,

a is the inverse of a.

Similarly, b * c = c * b = a,

b and c are the inverse elements of c...

Here,

which is closed.

Let a-1 and b-1 be the inverse elements of a and b in the communitative group (G, ) with the identity element e, then,

a a-1 = e = a-1 a

b b-1 = e = b-1 b

(a b) (a-1 b-1) = (a b) (b-1 a-1)

= a (b b-1) a-1

= a e a-1

= e (a a-1)

= e e

= e

Similarly, (a-1 b-1) (a b ) = e

Hence, (a b)-1 = a-1 b-1