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Given that * is a binary operation on Z defined by a*b = a + b + ab for all a,b∈Z. 

We know that commutative property is p*q = q*p, where * is a binary operation. 

Let’s check the commutativity of given binary operation: 

⇒ a*b = a + b + ab 

⇒ b*a = b + a + ba 

⇒ b*a = a + b + ab 

⇒ b*a = a*b 

∴ Commutative property holds for given binary operation ‘*’ on ‘Z’. 

We know that associative property is (p*q)*r = p*(q*r) 

Let’s check the associativity of given binary operation: 

⇒ (a*b)*c = (a + b + ab)*c 

⇒ (a*b)*c = (a + b + ab + c + ((a + b + ab)×c)) 

⇒ (a*b)*c = a + b + c + ab + ac + ac + abc ...... (1) 

⇒ a*(b*c) = a*(b + c + bc) 

⇒ a*(b*c) = (a + b + c + bc + (a×(b + c + bc))) 

⇒ a*(b*c) = a + b + c + ab + bc + ac + abc ...... (2) 

From (1) and (2) we can clearly say that associativity holds for the binary operation ‘*’ on ‘Z’.

To prove: * is not a binary operation 

Given: a and b are defined on positive integer set 

And a*b = |a - b| 

⇒ a*b = (a - b), when a>b 

= b - a when b>a 

= 0 when a = b 

But 0 is neither positive nor negative. 

So 0 does not belong to Z⁺ . 

So a*b = |a - b| does not belong to Z⁺ for all values of a and b 

So * is not a binary operation.

Hence proved

Given that * is a binary operation on Z defined by a*b = 3a + 7b for all a,b∈Z

We know that commutative property is p*q = q*p, where * is a binary operation. 

Let’s check the commutativity of given binary operation: 

⇒ a*b = 3a + 7b 

⇒ b*a = 3b + 7a 

⇒ b*a≠a*b 

∴ Commutative property doesn’t holds for given binary operation ‘*’ on ‘Z’.

To find: 2*4 

Given: a*b = a + 3b² 

⇒ 2*4 = (2 + 3 × 4² ) = 2 + 48 = 50

To find: LCM of 20 and 16 

Prime factorizing 20 and 16 we get. 

20 = 2² × 5 16 = 24 

⇒ LCM of 20 and 16 = 2⁴ × 5 = 80 

(i) To find LCM highest power of each prime factor has been taken from both the numbers and multiplied. 

So it is irrelevant in which order the number are taken as their prime factors will remain the same. 

So LCM(a,b) = LCM(b,a) 

So * is commutative 

(ii) Let us assume that * is associative 

⇒ LCM[LCM(a,b),c] = LCM[a,LCM(b,c)] 

Let the prime factors of a be p₁,p₂ 

Let the prime factors of b be p₂,p₃ 

Let the prime factors of c be p₃,p₄ 

Let the higher factor of p_i be q_i for i = 1,2,3,4 

LCM (a,b) = p₁ ^q1 x p₂^q2 x p₃ ^q3 

LCM[LCM(a,b),c] = p₁^q1 x p₂^q2 x p₃^q3 x p₄^q4 

LCM (b,c) = p₂^q2 x p₃^q3 x p₄^q4 

LCM[a,LCM(b,c)] = p₁^q1 x p₂^q2 x p₃^q3 x p₄^q4  

⇒ * is associative

Given that * is a binary operation on set A defined by a*b = b for all a,b∈A.

We know that commutative property is p*q = q*p, where * is a binary operation. 

Let’s check the commutativity of given binary operation: 

⇒ a*b = b 

⇒ b*a = a 

⇒ b*a≠a*b 

∴ The commutative property does not hold for given binary operation ‘*’ on ‘A’. 

We know that associative property is (p*q)*r = p*(q*r) 

Let’s check the associativity of given binary operation: 

⇒ (a*b)*c = (b)*c 

⇒ (a*b)*c = b*c 

⇒ (a*b)*c = c ...... (1) 

⇒ a*(b*c) = a*(c) 

⇒ a*(b*c) = a*c 

⇒ a*(b*c) = c ...... (2) 

From (1) and (2) we can clearly say that associativity holds for the binary operation ‘*’ on ‘A’.

To find: 3*5 and 5*3 

Given: a*b = (2a - b)² 

⇒ 3*5 = (6 - 5)² = 1 

Now 5*3 = (10 - 3)² = 49 

⇒ 3*5 is not equal to 5*3

To find: 4*5 

a*b = 3a + 4b - 2 

Here a = 4 and b = 5 

⇒ 4*5 = 3 × 4 + 4 × 5 - 2 = 12 + 20 - 2 = 30 

⇒ 4*5 = 30

Given that binary operation ‘*’ is valid for set ‘Z’ of integers defined by a*b = a + b for all a,b∈Z. 

Let us assume a∈Z and the identity element that we need to compute be e∈Z. 

We know that he Identity property is defined as follows: 

⇒ a*e = e*a = a 

⇒ a + e + 2 = a 

⇒ e + 2 = a – a 

⇒ e = – 2 

∴ The required Identity element w.r.t * is – 2.

Given that binary operation ‘*’ is valid for set ‘Z’ defined by a*b = a + b – 5 for all a,b∈Z. 

Let us assume a∈Z and the identity element that we need to compute be e∈Z. 

We know that he Identity property is defined as follows: 

⇒ a*e = e*a = a 

⇒ a + e – 5 = a 

⇒ e – 5 = a – a 

⇒ e = 5 

∴ The required Identity element w.r.t * is 5.

To find: (2*3)*4 

Given: a*b = 2a + b 

⇒ 2*3 = 2 × 2 + 3 = 7 

Now 7*4 = 2 × 7 + 4 = 14 + 4 = 18 

⇒ (2*3)*4 = 18

It is given that a * b = ab/4

(i) For a, b, c ∈ Q₀

We know that

a * b = ab/4 = ba/4 = b * a

(a * b) * c = ab/ 4 * c = [ab/4 * c]/ 4 = (ab) c/ 16

a * (b * c) = a * bc/4 = [a(bc/4)]/ 4 = a (bc)/16

Here, (ab) c = a (bc)

Therefore, (a * b) * c = a * (b * c)

(ii) Consider e as the identity element and a ∈ Q₀

Here, a * e = a

So we get

ae/4 = a where e = 4

Hence, 4 is the identity element in Q.

(iii) Consider a ∈ Q₀ which is inverse b

a * b = e

So we get

ab/4 = 4

Here, b = 16/a ∈ Q₀

Hence, a ∈ Q₀ has 16/a as inverse.

To find: ( - 5)*(2*0) 

Given: a*b = a + 4b² 

⇒ (2*0) = 2 + 4 × 0² = 2 

Now ( - 5)*2 = - 5 + 4 × 2² = - 5 + 16 = 11

The given binary operation is a*b = ab/2

And from the definition of identity element e, 

a*e = a ........(i) 

We have, 

a*e = ae/2……….(ii) 

Thus from (i) & (ii) ,

ae/2 = a 

or, e/2 = 1 

∴ e = 2 

Hence the identity element for this binary operation is 2.

To find: value of x 

Given: a*b = \(\frac{ab}{5}\)

⇒ x*5 = \(\frac{5x}{5}\) = x 

Now (2*x) = \(\frac{2x}{5}\)

⇒ \(\frac{2x}{5}\) = 10⇒ x = 25

It is given that a * b = ab/5

So we know that 2 * (x * 5) = 10

We get

2 * ([x × 5]/ 5) = 10

On further calculation

2 * x = 10

So (2 × x)/ 5 = 10

We get x = 25