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Given that ‘*’ is an operation that is valid in the Natural Numbers ‘N’ and it is defined as given: 

⇒ a*b = a^b, where a,b∈N 

Since a∈N and b∈N, 

According to the problem it is given that on applying the operation ‘*’ for two given natural numbers it gives a natural number as a result of the operation,

⇒ a*b∈N ...... (1) 

We also know that p^q>0 if p>0 and q>0. 

So, we can state that,

⇒ a^b>0 

⇒ a^b∈N ...... (2) 

From (1) and (2) we can see that both L.H.S and R.H.S gave only Natural numbers as a result. 

Thus we can clearly state that ‘*’ is a Binary Operation on ‘N’.

Given that ‘Ο’ is an operation that is valid in the Integers ‘Z’ and it is defined as given: 

⇒ aΟb = a^b, where a,b∈Z 

Since a∈Z and b∈Z, 

According to the problem it is given that on applying the operation ‘*’ for two given integers it gives Integers as a result of the operation, 

⇒ aΟb∈Z ...... (1) 

Let us values of a = 2 and b = – 2 on substituting in the R.H.S side we get, 

⇒ a^b = 2 ^{– 2} 

⇒ a^b = 1/4

⇒ a^b∉Z ...... (2) 

From (2), we can see that a^b doesn’t give only Integers as a result. So, this cannot be stated as a binary function. 

∴ The operation ‘Ο’ does not define a binary function on Z.

Given that ‘x₆’ is an operation that is valid for the numbers in the Set S = {1,2,3,4,5} and it is defined as given: 

⇒ ax₆b = Remainder when ab is divided by 6, where a,b∈S 

Since a∈S and b∈S, 

According to the problem it is given that on applying the operation ‘*’ for two given numbers in the set ‘S’ it gives one of the numbers in the set ‘S’ as a result of the operation, 

⇒ ax₆b∈S ...... (1) 

Let us take the values of a = 3, b = 4, 

⇒ ab = 3×4 

⇒ ab = 12 

We know that 12 is a multiple of 6. So, on dividing 12 with 6 we get 0 as remainder which is not in the given set ‘S’. 

The operation ‘x₆’ does not define a binary operation on set S.

i. We are given with the set R – {– 1}. 

A general binary operation is nothing but an association of any pair of elements a, b from an arbitrary set X to another element of X. This gives rise to a general definition as follows: 

A binary operation * on a set is a function * : A X A → A. We denote * (a, b) as a * b. 

Here the function *: R – {1}X R – {1} → R – {1} is given by a * b = a + b – ab 

For the ‘ * ’ to be commutative, a * b = b * a must be true for all a, b belong to R – {1}. Let’s check. 

1. a * b = a + b – ab2. b * a = b + a – ba = a + b – ab⇒ a * b = b * a (as shown by 1 and 2) 

Hence ‘ * ’ is commutative on R – {1} 

For the ‘ * ’ to be associative, a * (b * c) = (a * b) * c must hold for every a, b, c∈ R – {1}. 

3. a * (b * c) = a * (b + c – bc) 

= a + (b + c – bc) – a(b + c + bc) 

= a + b + c – ab – bc – ac + abc 

4. (a * b) * c = (a + b – ab) * c 

= a + b – ab + c – (a + b – ab)c 

= a + b + c – ab – bc – ac + abc⇒ 3. = 4. 

Hence ‘ * ’ is associative on R – {1} 

ii. Identity Element: Given a binary operation*: A X A → A, an element e ∈A, if it exists, is called an identity of the operation*, if a*e = a = e*a ∀ a ∈A.

Let e be the identity element of R – {1} and a be an element of R – {1}. 

Therefore, a * e = a⇒ a + e – ae = a⇒ e + ea = 0⇒ e(1 – a) = 0⇒ e = 0. 

(1 – a≠0 as the a cannot be equal to 1 as the operation is valid in R – {1}) 

iii. Given a binary operation * A X A → A with the identity element e in A, an element a ∈ A is said to be invertible with respect to the operation, if there exists an element b in A such that a * b = e = b * a and b is called the inverse of a and is denoted by a ^–1. 

Let us proceed with the solution. 

Let b R – {1} be the invertible element/s in R – {1} of a, here a ∈ R – {1}. 

∴a * b = e (We know the identity element from previous)

⇒ a + b – ab = 0⇒ b – ab = – a⇒ b(1 – a) = – a

⇒ b = -a/(1 -a) (Here a ≠ 1, b ≠ 1)

Given that ‘*’ is an operation that is valid in the Positive integers ‘Z ⁺ ’ and it is defined as given: 

⇒ a*b = |a – b|, where a,b∈Z⁺ , 

Since a∈Z ⁺ and b∈Z ⁺ , 

According to the problem it is given that on applying the operation ‘*’ for two given positive integers it gives a Positive integer as a result of the operation, 

⇒ a*b∈Z ...... (1) 

Let us take a = 2 and b = 2,

⇒ |a – b| = |2 – 2| 

⇒ |a – b| = |0|

⇒ |a – b| = 0∉Z ⁺ 

∴ The operation * does not define a binary function on Z ⁺ .

Given that ‘*’ is an operation that is valid in the Positive integers ‘Z ⁺ ’ and it is defined as given: ⇒ a*b = ab, where a,b∈Z⁺ , 

Since a∈Z⁺ and b∈Z⁺ , 

According to the problem it is given that on applying the operation ‘*’ for two given positive integers it gives a Positive integer as a result of the operation, 

⇒ a*b∈Z ^{+ –} ...... (1) 

We know that p q>0 if p>0 and q>0. 

⇒ ab>0∈N 

⇒ ab∈Z ⁺ 

∴ The operation * defines a binary operation on Z ⁺.

Given that ‘*’ is an operation that is valid in the Real Numbers ‘R’ and it is defined as given: 

⇒ a*b = ab², where a,b∈R 

Since a∈R and b∈R, 

According to the problem it is given that on applying the operation ‘*’ for two given real numbers it gives a real number as a result of the operation, 

⇒ a*b∈R ...... (1) 

We know that ab∈R if a∈R and b∈R 

∴ The operation * defines a binary operation on R.

It is given that a * b = (2a – b)²

We know that

3 * 5 = (2 × 3 – 5)² = 1

5 * 3 = (2 × 5 – 3)² = 49

Therefore, 3 * 5 ≠ 5 * 3

Given that ‘*’ is an operation that is valid on the set S which consists of all real numbers except – 1 i.e., R – { – 1} defined as a*b = a + b + ab 

Let us assume a + b + ab = – 1

⇒ a + ab + b + 1 = 0

⇒ a(1 + b) + (1 + b) = 0 

⇒ (a + 1)(b + 1) = 0 

⇒ a = – 1 or b = – 1 

But according to the problem, it is given that a≠ – 1 and b≠ – 1 so, 

a + b + ab≠ – 1, so we can say that the operation ‘*’ defines a binary operation on set ‘S’. 

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 = a + b + ab 

⇒ b*a = a*b

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

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 + bc + 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 ‘N’. We need to also solve for x in the given expression: 

⇒ (2*x)*3 = 7 

⇒ (2 + x + 2x)*3 = 7 

⇒ (2 + 3x)*3 = 7 

⇒ 2 + 3x + 3 + ((2 + 3x)×3) = 7 

⇒ 5 + 3x + 6 + 9x = 7 

⇒ 11 + 12x = 7 

⇒ 12x = – 4

⇒ x = \(\frac{4}{12}.\)

⇒ x = \(\frac{-1}{3}.\)

∴ The value of x is \(\frac{-1}{3}.\)

Given that * is an operation that is valid for the following Domain and Range R×R→R and is defined by a*b = 2ab. 

We need to find the value of (2*3)*4 

According to the problem the binary operation involving * is true for all real values of a and b. 

⇒ (2*3)*4 = ((2×2) + 3)*4 

⇒ (2*3)*4 = (4 + 3)*4 

⇒ (2*3)*4 = 7*4 

⇒ (2*3)*4 = (2×7) + 4 

⇒ (2*3)*4 = 14 + 4 

⇒ (2*3)*4 = 18 

∴ The value of (2*3)*4 is 18.

Given that * is a binary operation on N defined by a*b = 1 for all a,b∈N. 

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 = 1 

⇒ b*a = 1 

⇒ b*a = a*b 

∴ The commutative property holds for given binary operation ‘*’ on ‘N’. 

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

Let’s check the associativity of given binary operation: 

⇒ (a*b)*c = (1)*c 

⇒ (a*b)*c = 1*c 

⇒ (a*b)*c = 1 ...... (1) 

⇒ a*(b*c) = a*(1) 

⇒ a*(b*c) = a*1 

⇒ a*(b*c) = 1 ...... (2) 

From (1) and (2) we can clearly say that, 

Associative property holds for given binary operation ‘*’ on ‘N’

Let A be a non-empty set having n elements. Then, the total number of binary operations on this set, i.e., from A×A is \(n^{n^2}\)

Here n = 2 Thus, total number of possible binary operations are 2⁴ = 16.

The associative property of binary operations hold if, for a non-empty set A, we can write (a * b) *c = a*(b * c). 

Suppose N be the set of natural numbers and multiplication be the binary operation. Let a = 4, b = 5 c = 6. We can write (a × b) × c = 120 = a × (b × c).

A binary operation * on a set A is commutative if a * b = b * a, for all (a, b) ∈ A (non-empty set). 

Let addition be the operating binary operation for a = 8 and b = 9, a + b = 17 = b + a.

Binary operations on a set are calculations that combine two elements of the set (called operands) to produce another element of the same set. 

The binary operations * on a non-empty set A are functions from A × A to A. The binary operation, *: A × A → A. It is an operation of two elements of the set whose domains and co-domain are in the same set.

We observe the following: 

a * b = b * a = b 

c * a = a * c = c 

a * d = d * a = d 

b * c = c * b = d 

b * d = d * b = c 

c * d = d * c = b 

There ‘ * ’ is commutative. 

Also, 

a * (b * c) = a * (d) = d (From above) 

(a * b) * c = (b) * c = d (Also from above) 

Hence, ‘ * ’ is associative too. 

Therefore, to find the identity element, e for e belong to S, we need: 

a * e = e * a = a, a belong to S. 

Therefore, a * e = a 

e = a (since, a * a = a, from the given table) 

To find out the inverse, a * x = e = b * x, x belongs to S 

a * x = e 

x = a (From the given table) 

Therefore, the inverse of a is a, b is b, c is c and d is d.

It is given that

a * b = 2a + b

So we get

(2 * 3) * 4 

= (2 × 2 + 3) * 4

= 7 * 4 = 2 × 7 + 4 

= 18

We know that a * b = a + 4b²

It is given that

(-5) * (2 * 0) = (-5) * (2 + 4 × 0²) = – 5 * 2

In the same way

= – 5 + 4 × (-2)² = 11

The given binary operation is a*b = a+b+2 

In order to find the inverse of the relation, we have to find the identity element first.

Let that identity element be e then 

a*e = a 

From que. 

a*e = a+e+2 

So, from the above two relations we have 

a+e+2 = a 

or, e+2 = 0 

∴ e = -2 

Hence the identity element is -2 for this binary operation. 

Now let a’ be the inverse of this relation 

Then as per the definition of the inverse element 

a*a’ = e 

∴ a+a’+2 = -2 

∴ a’ = -4 -a 

And for 4 ,i.e. a = 4 

a’ = -4 – 4 

∴ a’ = -8 

Thus the inverse element of 4 is -8 for the given binary operation.

Given that ‘*’ is an operation that is valid on the set S which consists of all real numbers except 1 i.e., R – {1} defined as a*b = a + b – ab 

Let us assume a + b – ab = 1

⇒ a – ab + b – 1 = 0

⇒ a(1 – b) – 1(1 – b) = 0

⇒ (1 – a)(1 – b) = 0 

⇒ a = 1 or b = 1 

But according to the problem, it is given that a≠1 and b≠1 so, 

a + b + ab≠1, so we can say that the operation ‘*’ defines a binary operation on set ‘S’. 

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 = a + b – ab 

⇒ b*a = a*b 

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

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 – bc + 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 ‘S’.

(i) Consider e as an identity element.

We know that a * e = a Ɐ a ∈ Q {1}

So a + e – ae = a

It can be written as

e (1 – a) = 0

So e = 0 ∈ Q {1}

a * 0 = a + 0 = a

So 0 * a = 0 + a = a

Hence, 0 is the identity element in Q – {1}

(ii) Consider a ∈ Q – {1} where a ^-1 = b

We know that a * b = 0

It can be written as

a + b – ab = 0

So a = ab – b

a = (a – 1) b

We get b = a/ a – 1 ∈ Q – {1}

So a ^-1 = a/ a – 1 ∈ Q – {1}

Therefore, each a ∈ Q – {1} has its inverse.

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

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

We know that he Identity property is defined as follows: 

⇒ a*e = e*a = a 

⇒ a + e = a 

⇒ e = a – a 

⇒ e = 0 

∴ The required Identity element w.r.t * is 0.

Given that * is a binary operation on the set I of integers.

The operation is defined by a*b = 2a + b – 3. 

We need to find the value of 3*4. 

Since 3 and 4 belongs to the set of integers we can use the binary operation. 

⇒ 3*4 = (2×3) + 4–3 

⇒ 3*4 = 6 + 1 

⇒ 3*4 = 7 

∴ The value of 3*4 is 7.

Given that * is an operation that is valid for the natural numbers ‘N’ and is defined by a*b = LCM(a,b). 

We need to find the value of 5*7. 

According to the Problem, Binary operation is assumed to be true for the values of a and b to be natural. 

⇒ 5*7 = LCM(5,7) 

We know that LCM of two prime numbers is the product of that given two prime numbers. 

⇒ 5*7 = 5×7 

⇒ 5*7 = 35 

∴ The value of 5*7 is 35.

To find: (i) 

LCM of 12 and 16 

Prime factorizing 12 and 16 we get. 

20 = 2² × 3 

16 = 2⁴ 

⇒ LCM of 20 and 16 = 2⁴ × 3 = 48 

(ii) 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. 

(iii)let x ∈ N and x*1 = lcm(x,1) = x = lcm(1,x) 

1 is the identity element. 

(iv)let there exist y in n such that x*y = e = y*x 

Here e = 1, 

Lcm(x,y) = 1 

This happens only when x = y = 1. 

1 is the invertible element of n with respect to *.

Given that binary operation ‘*’ is valid for set of all rational numbers Q defined by a*b = a + b + ab for all a,b∈R. 

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

We know that he Identity property is defined as follows: 

⇒ a*e = e*a = a 

⇒ a + e + ea = a 

⇒ e + ae = a – a 

⇒ e(1 + a) = 0 

⇒ e = 0 

∴ The required Identity element w.r.t * is 0.

Given that * is an operation that is valid on all natural numbers ‘N’ and is defined by a*b = L.C.M(a,b) 

According to the problem, binary operation given is assumed to be true. 

Let us find the values of 2*4,3*5,1*6 

⇒ 2*4 = L.C.M(2,4) = 4 

⇒ 3*5 = L.C.m(3,5) 

⇒ 3*5 = 3×5 = 15 

⇒ 1*6 = L.C.M(1,6) 

⇒ 1*6 = 1×6 = 6 

The values of 2*4 is 4 , 3*5 is 15 and 1*6 is 6

(i) * is an operation as a*b = a+ b+ ab where a, b ∈ Q- {-1}. Let a = 1 and \(b=\frac{-3}{2}\)two rational numbers. 

\(a*b=1+\frac{-3}{2}+1.\frac{-3}{2}\)\(\Rightarrow \frac{2-3}{2}-\frac{3}{2}=\frac{-1-3}{2}\)\(\Rightarrow\frac{-4}{2}=-2\in Q-\{-1\}\)

So, * is a binary operation from Q - { -1 } x Q - { -1} → Q - {-1}. 

(ii) For commutative binary operation, a*b = b*a. 

\(b*a=\frac{-3}{2}+1+\frac{-3}{2}.1\)\(\Rightarrow \frac{-3+2}{2}-\frac{3}{2}\)\(=\frac{-1-3}{2}\Rightarrow \frac{-4}{2}\)\(-2\in Q-\{-1\}\)

Since a*b = b*a, hence * is a commutative binary operation. 

(iii) For associative binary operation, a*(b*c) = (a*b) *c 

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

= a+ b+ c+ bc+ ab+ ac+ abc 

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

= a+ b+ c+ ab+ ac+ bc+ abc 

Now as a*(b*c) = (a*b) *c, hence an associative binary operation. 

(iv) For a binary operation *, e identity element exists if a*e = e*a = a. As a*b = a+ b- ab 

a*e = a+ e+ ae  (1) 

e*a = e+ a +e a  (2) 

using a*e = a 

a+ e+ ae = a 

e+ ae = 0 

e(1+a) = 0 

either e = 0 or a = -1 as operation is on Q excluding -1 so a≠-1, hence e = 0. 

So identity element e = 0. 

(v) for a binary operation * if e is identity element then it is invertible with respect to * if for an element b, a*b = e = b*a where b is called inverse of * and denoted by a^-1. 

a*b = 0 

a+ b+ ab = 0 

b(1+a) = -a

\(b=\frac{-a}{(1+a)}\)

\(a^{-1}=\frac{-a}{(a+1)}\)

(i) * is an operation as \(a*b=\frac{1}{2}(a+b)\)where a, b ∈ Q+. Let a =1 and b = 2 two integers. 

\(a*b=\frac{1}{2}(1+2)\)\(\Rightarrow \frac{3}{2}\in Q ^+\)

So, * is a binary operation from Q⁺ x Q⁺ → Q⁺. 

(ii) For commutative binary operation, a*b = b*a. 

\(b*a=\frac{1}{2}(2+1)\)\(\Rightarrow \frac{3}{2}\in Q^+\)

Since a*b = b*a, hence * is a commutative binary operation. 

(iii) For associative binary operation, a*(b*c) = (a*b) *c. 

\(a*(b*c)=a*\frac{1}{2}(b+c)\)\(\Rightarrow \cfrac{1}{2}\left(a+\frac{b+c}{2}\right)\)\(=\frac{1}{4}(2a+b+c)\)

\((a*b)*c=\frac{1}{2}(a+b)*c\)\(\Rightarrow \frac{1}{2}\left(\frac{a+b}{2}+c\right)\)\(=\frac{1}{4}(a+b+2c)\)

As a*(b*c) ≠(a*b) *c, hence * is not associative binary operation.

(i) * is an operation as \(a*b=\frac{ab}{2}\) where a, b ∈ Q+. Let \(a=\frac{1}{2}\)and b = 2 two integers. 

\(a*b=\frac{1}{2}*2\Rightarrow 1\in Q^+\)

So, * is a binary operation from Q⁺ x Q⁺ → Q. 

(ii) For commutative binary operation, a*b = b*a. 

\(b*a=2.\frac{1}{2}\Rightarrow 1 \in Q^+\)

Since a*b = b*a, hence * is a commutative binary operation. 

(iii) For associative binary operation, a*(b*c) = (a*b) *c. 

\(a*(b*c)=a*\frac{bc}{2}\)\(\Rightarrow \frac{a\frac{bc}{2}}{2}=\frac{abc}{4}\)

\((a*b)*c=\frac{ab}{2}*c\)\(\Rightarrow \frac{\frac{ab}{2}c}{2}=\frac{abc}{4}\)

As a*(b*c) = (a*b) *c, hence * is an associative binary operation. For a binary operation *, e identity element exists if a*e = e*a = a. 

\(a*e=\frac{ae}{2}\) (1) 

\(a*e=\frac{ea}{2}\) (2) 

using a*e = a 

\(\frac{ae}{2}=a\Rightarrow \frac{ae}{2}-a=0\)\(\Rightarrow \frac{a}{2}(e-2)=0\)

Either a = 0 or e = 2 as given a≠0, so e = 2. 

For a binary operation * if e is identity element then it is invertible with respect to * if for an element b, a*b = e = b*a where b is called inverse of * and denoted by a^-1. 

a*b = 2

\(\frac{ab}{2}=2\Rightarrow b=\frac{4}{a}\)

\(a^{-1}=\frac{4}{a}\)

(i)Let a = 1, b = 2 ∈ Q ⁺ 

a*b = \(\frac{1}{2}(1+2)\)= 1.5 ∈ Q⁺ 

* is closed and is thus a binary operation on Q⁺ 

(ii) a*b = \(\frac{1}{2}(1+2)\) = 1.5  

And b*a = \(\frac{1}{2}(2+1)\) = 1.5 

Hence * is commutative. 

(iii)let c = 3. 

(a*b)*c = 1.5*c = \(\frac{1}{2}(1.5+3)=2.75\)

a*(b*c) = a*\(\frac{1}{2}(2+3)\) = 1*2.5 = \(\frac{1}{2}(1+2.5)\) = 1.75 

hence * is not associative.

We know that Z⁺ defined by a * b = | a – b | for a, a ∈ Z⁺

So a * a = |a – a| = 0 ∉ Z⁺

Therefore, * is not a binary operation.

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 = L.C.M(a,b) 

⇒ b*a = L.C.M(b,a) = L.C.M(a,b) 

⇒ b*a = a*b 

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

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

Let’s check the associativity of given binary operation: 

⇒ (a*b)*c = (L.C.M(a,b))*c 

⇒ (a*b)*c = L.C.M(a,b)*c 

⇒ (a*b)*c = L.C.M(L.C.M(a,b),c) 

⇒ (a*b)*c = L.C.M(a,b,c) ...... (1) 

⇒ a*(b*c) = a*(L.C.M(b,c)) 

⇒ a*(b*c) = a*L.C.M(b,c) 

⇒ a*(b*c) = L.C.M(a,L.C.M(b,c)) 

⇒ a*(b*c) = L.C.M(a,b,c) ...... (2) 

From(1) and (2) we can say that associative property holds for binary function ‘*’ on ‘N’

For a set to be closed for addition, 

For any 2 elements of the set,say a and b, a + b must also be a member of the given set, where a and b may be same or distinct 

In the given problem let a = 1 and b = 1 

a + b = 2 which is not in the given in set 

So the set is not closed for addition. 

Hence proved.

a*b = a - b if a>b 

= - (a - b) if b>a 

b*a = a - b if a>b 

= - (a - b) if b>a 

So a*b = b*a 

So * is commutative 

To show that * is associative we need to show 

(a*b)*c = a*(b*c) 

Or ||a - b| - c| = |a - |b - c|| 

Let us consider c>a>b 

Eg a = 1,b = - 1,c = 5 

LHS: 

|a - b| = |1 + 1| = 2 

||a - b| - c| = |2 - 5| = 3 

RHS 

|b - c| = | - 1 - 5| = 6 

|a - |b - c|| = |1 - 6| = | - 5| = 5 

As LHS is not equal to RHS * is not associative

let a = 1,b = 0 ∈ R - { - 1} 

a*b = \(\frac{1}{0+1}\) = 1 

And b*a = \(\frac{1}{0+1}\) = 0 

Hence * is not commutative. 

Let c = 3. 

(a*b)*c = 1*c = \(\frac{1}{3+1}=\frac{1}{4}\)

a*(b*c) = a*\(\frac{0}{3+1}\) = 1*0 = \(\frac{1}{0+1}\) = 1 

Hence * is not associative.

let a = 1,b = 2∈N 

a*b = 1³ + 2³ = 9 

And b*a = 2³ + 1³ = 9 

Hence * is commutative. 

Let c = 3 

(a*b)*c = 9*c = 9 ³ + 3³ 

a*(b*c) = a*(2³ + 3³) = 1*35 = 1³ + 35³ 

(a*b)*c ≠ a*(b*c) 

Hence * is not associative.

Commutative:

Consider two elements 1 and 2 of Z

Here, 1 * 2 = 1 – 2 + 1 × 2 = 1 and 2 * 1 = 2 – 1 + 2 × 1 = 3

Therefore, the operation is not commutative.

Associative:

Take 2, 3, 4 ∈ Z

We get

(2 * 3) * 4 = (2 – 3 + 2 × 3) * 4 = 5 * 4 = 5 – 4 + 5 × 4 = 21

2 * (3 * 4) = 2 * (3 – 4 + 12) = 2 * 11 = 2 – 11 + 2 × 11 = 13

Here, (2 * 3) * 4 ≠ 2 * (3 * 4)

Therefore, the operation is not associative.

Given that * is a binary operation on R defined by a*b = a + b – 7 for all a,b∈R. 

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 – 7 

⇒ b*a = b + a – 7 = a + b – 7 

⇒ b*a = a*b 

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

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 – 7)*c 

⇒ (a*b)*c = a + b – 7 + c – 7 

⇒ (a*b)*c = a + b + c – 14 ...... (1) 

⇒ a*(b*c) = a*(b + c – 7) 

⇒ a*(b*c) = a + b + c – 7 – 7 

⇒ a*(b*c) = a + b + c – 14 ...... (2) 

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

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

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 + ab 

⇒ b*a = b + ba = b + ab 

⇒ b*a≠a*b 

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

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 + ab)*c

⇒ (a*b)*c = a + ab + ((a + ab)×c) 

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

⇒ a*(b*c) = a*(b + bc) 

⇒ a*(b*c) = a + (a×(b + bc)) 

⇒ a*(b*c) = a + ab + abc ...... (2) 

From (1) and (2) we can clearly say that associativity doesn’t hold for the binary operation ‘*’ on ‘Q’.

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

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 

⇒ b*a = b = a 

⇒ b*a≠a*b 

∴ The commutative property doesn’t hold for given binary operation ‘*’ on ‘Q’. 

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)*c 

⇒ (a*b)*c = a – b – c ...... (1) 

⇒ a*(b*c) = a*(b – c) 

⇒ a*(b*c) = a – (b – c) 

⇒ a*(b*c) = a – b + c ...... (2) 

From (1) and (2) we can clearly say that associativity doesn’t hold for the binary operation ‘*’ on ‘Q’

Consider a, b ∈ Z where a * b = a + b – ab and b * a = b + a – ba

So we get a * b = b * a

Associative:

Consider a, b, c ∈ Z

Here,

(a * b) * c = (a + b – ab) * c = a + b – ab + c – (a + b – ab) c

We get

(a * b) * c = a + b + c – ab – bc – ca + abc

a * (b * c) = a * (b + c – bc) = a + b + c – bc – a (b + c – bc)

We get

a * (b * c) = a + b + c – an – bc – ca + abc

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

Therefore, operation on Z is associative.

Consider m and n ∈ N where m * n = LCM (m, n) = LCM (n, m) = n * m

* is commutative binary operation.

We know that

(m * n * p = [LCM of (m, n)] * p = [LCM of (m, n) and p] = LCM of (m, n, p)

Similarly

m * (n * p) = m * [LCM of (n, p)] = LCM of [m and LCM of (n, p] = LCM of (m, n, p)

So we get (m * n) * p = m * (n * p)

Therefore, the operation is associative.

(i) For a binary operation *, e identity element exists if a*e = e*a = a. As a*b = a+ b- ab 

a*e = a+ e- ae  (1) 

e*a = e+ a- e a  (2) 

using a*e = a 

a+ e- ae = a 

e-ae = 0 

e(1-a) = 0 

either e = 0 or a = 1 as operation is on Q excluding 1 so a≠1, hence e = 0. 

So identity element e = 0. 

(ii) for a binary operation * if e is identity element then it is invertible with respect to * if for an element b, a*b = e = b*a where b is called inverse of * and denoted by a^-1. 

a*b = 0 

a+ b- ab = 0 

b(1-a) = -a

\(b=\frac{-a}{(1-a)}\Rightarrow\frac{a}{(a-1)}\)

\(a^{-1}=\frac{a}{(a-1)}\)

To find: identity and inverse element 

For a binary operation if a*e = a, then e s called the right identity 

If e*a = a then e is called the left identity 

For the given binary operation, 

e*b = b 

⇒ e + b = b 

⇒ e = 0 which is less than 6. 

b*e = b 

⇒ b + e = b 

⇒ e = 0 which is less than 6 

For the 2^nd condition, 

e*b = b 

⇒ e + b - 6 = b 

⇒ e = 6 

But e = 6 does not belong to the given set (0,1,2,3,4,5) 

So the identity element is 0 

An element c is said to be the inverse of a, if a*c = e where e is the identity element (in our case it is 0) 

a*c = e 

⇒ a + c = e 

⇒ a + c = 0 

⇒ c = - a 

a belongs to (0,1,2,3,4,5) 

- a belongs to (0, - 1, - 2, - 3, - 4, - 5) 

So c belongs to (0, - 1, - 2, - 3, - 4, - 5) 

So c = - a is not the inverse for all elements a 

Putting in the 2^nd condition a*c = e 

⇒ a + c - 6 = 0 

⇒ c = 6 - a 

0≤a<6 ⇒ - 6≤ - a<0⇒ 0≤6 - a<60≤c<5 

So c belongs to the given set 

Hence the inverse of the element a is (6 - a) 

Hence proved

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 = 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 – bc + 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 – ac – bc + abc ...... (2)

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

Given that * is a binary operation on N defined by a*b = 2^ab for all a,b∈N. 

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 = 2^ab 

⇒ b*a = 2^ba = 2^ab 

⇒ b*a = a*b 

∴ The commutative property holds for given binary operation ‘*’ on ‘N’. 

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

Let’s check the associativity of given binary operation: 

⇒ (a*b)*c = (2^ab)*c

⇒ \((a * b ) * c = 2^{ab}.c\) ..........(1)

⇒ a*(b*c) = a*(2^bc)

⇒ \((a * b ) * c = 2^{a.2^{bc}}\) ...........(2)

From (1) and (2) we can clearly say that associativity doesn’t hold for the binary operation ‘*’ on ‘N’.

We know that a * b = LCM of a and b

So 20 * 16 = LCM of 20 and 16 = 80

Commutative:

Consider a and b ∈ N

So LCM of a and b = LCM of b and a

Here, a * b = b * a Ɐ a and b ∈ N

Therefore, * is commutative.

Associative:

(a * b) * c = LCM of a and b * c

Here,

= LCM of {(LCM of a and b) and C} = LCM of {LCM of a, b and c}

We get

= LCM of (a and [LCM of b and c]) = a * (LCM of b and c)

So

= a * (b * c)

We can write it as

a * (b * c) = (a * b) * c Ɐ a, b ∈ c ∈ N.

Therefore, * is associative.

* is an operation as m*n = LCM (m, n) where m, n ∈ N. Let m = 2 and b = 3 two natural numbers. 

m*n = 2*3 

= LCM (2, 3) 

= 6∈ N 

So, * is a binary operation from N x N → N. 

For commutative, 

n*m = 3*2 

= LCM (3, 2) 

= 6∈ N 

Since m*n = n*m, hence * is commutative operation. 

Again, for associative, let p = 4 

m*(n*p) = 2*LCM (3, 4) 

= 2*12 

= LCM (2, 12) 

= 12∈ N 

(m*n) *p = LCM (2, 3) *4 

= 6*4 

= LCM (6, 4) 

= 12∈ N 

As m*(n*p) = (m*n) *p, hence * an associative operation.

It is given that a * b = (a + 3b²)

We know that

2 * 4 = 2 + 3 × 4² = 50