Question 1: Let $'\ast'$ be a binary operation on $N$ defined by $a\ast b = L.C.M (a,b) \text{ for all } a,b \in N.$

$\text{(i) Find } 2 \ast 4, 3 \ast 5, 1 \ast 6.$

$\text{(ii) Check the commutativity and associativity of } '\ast' \text{ on } N$

Given that $\ast$ is an operation that is valid on all natural numbers $'N'$ and is defined by $a \ast b = \text{L.C.M}(a, b)$

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

(i)     Let us find the values of $2 \ast 4,3 \ast 5,1 \ast 6$

$\Rightarrow 2 \ast 4 = \text{L.C.M}(2,4) = 4$

$\Rightarrow 3 \ast 5 = \text{L.C.M}(3,5)$

$\Rightarrow 3 \ast 5 = 3 \times 5 = 15$

$\Rightarrow 1 \ast 6 = \text{L.C.M}(1,6)$

$\Rightarrow 1 \ast 6 = 1\times 6 = 6$

The values of $2 \ast 4 \text{ is } 4 , 3 \ast 5 \text{ is } 15 \text{ and } 1 \ast 6 is 6$

$\\$

We know that commutative property is $p \ast q = q \ast p, \text{ where } \ast$  is a binary operation.

(ii)    Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = \text{L.C.M}(a,b)$

$\Rightarrow b \ast a = \text{L.C.M}(b,a) = \text{L.C.M}(a,b)$

$\Rightarrow b \ast a = a \ast b$

Therefore Commutative property holds for given binary operation $'\ast ' \text{ on } 'N'$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b) \ast c = (\text{L.C.M}(a,b)) \ast c$

$\Rightarrow (a \ast b) \ast c = \text{L.C.M}(a,b) \ast c$

$\Rightarrow (a \ast b) \ast c = \text{L.C.M}(\text{L.C.M}(a,b),c)$

$\Rightarrow (a \ast b) \ast c = \text{L.C.M}(a,b,c) \ \ldots \ \ldots \ (1)$

$\Rightarrow a \ast (b \ast c) = a \ast (\text{L.C.M}(b,c))$

$\Rightarrow a \ast (b \ast c) = a \ast \text{L.C.M}(b,c)$

$\Rightarrow a \ast (b \ast c) = \text{L.C.M}(a, \text{L.C.M}(b,c))$

$\Rightarrow a \ast (b \ast c) = \text{L.C.M}(a,b,c) \ \ldots \ \ldots \ (2)$

From(1) and (2) we can say that associative property holds for binary function $'\ast ' \text{ on } 'N'$

$\\$

Question 2: Determine which of the following binary operations are associative and which are
commutative:
$\text{(i) } \ast \text{ on } N \text{ defined by } a \ast b =1 \text{ for all } a,b \in N$

$\displaystyle \text{(ii) } \ast \text{ on } Q \text{ defined by } a \ast b = \frac{a+b}{2} \text{ for all } a, b \in Q \hspace{2.0cm} \textbf{[CBSE 2014]}$

$\text{(i) } \ast \text{ on } N \text{ defined by } a \ast b =1 \text{ for all } a,b \in N$

Given that  $\ast$  is a binary operation on $N$ defined by $a \ast b = 1 \text{ for all } a,b \in N.$

We know that commutative property is $p \ast q = q \ast p,$ where  $\ast$ is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = 1$

$\Rightarrow b \ast a = 1$

$\Rightarrow b \ast a = a \ast b$

Therefore The commutative property holds for given binary operation $' \ast ' \text{ on } 'N'.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b) \ast c = (1) \ast c$

$\Rightarrow (a \ast b) \ast c = 1 \ast c$

$\Rightarrow (a \ast b) \ast c = 1 \ \ldots \ \ldots \ (1)$

$\Rightarrow a \ast (b \ast c) = a \ast (1)$

$\Rightarrow a \ast (b \ast c) = a \ast 1$

$\Rightarrow a \ast (b \ast c) = 1 \ \ldots \ \ldots \ (2)$

From (1) and (2) we can clearly say that,  Associative property holds for given binary operation $' \ast ' \text{ on } 'N'.$

$\\$

$\displaystyle \text{(ii) } \ast \text{ on } Q \text{ defined by } a \ast b = \frac{a+b}{2} \text{ for all } a, b \in Q$

$\displaystyle \text{Given that } \ast \text{ is a binary operation on } N \text{ defined by } \\ a \ast b = \frac{a+b}{2} \text{ for all } a, b \in N$

We know that commutative property is $p \ast q = q \ast p,$ where $\ast$ is a binary operation.

Let’s check the commutative property of the given binary operation:

$\displaystyle \Rightarrow a \ast b = \frac{a+b}{2}$

$\displaystyle \Rightarrow b \ast a = \frac{b+a}{2}= \frac{a+b}{2}$

$\Rightarrow b \ast a = a \ast b$

Therefore the commutative property holds for given binary operation $' \ast ' \text{ on } ' N ' .$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\displaystyle \Rightarrow ( a \ast b) \ast c = \Big( \frac{a+b}{2} \Big) \ast c$

$\displaystyle \Rightarrow ( a \ast b) \ast c = \frac{\frac{a+b}{2}+c}{2}$

$\displaystyle \Rightarrow ( a \ast b) \ast c = \frac{a+b+2c}{4} \hspace{1.0cm}\ldots (i)$

$\displaystyle \Rightarrow ( a \ast b) \ast c = a \ast \Big( \frac{b+c}{2} \Big)$

$\displaystyle \Rightarrow ( a \ast b) \ast c = \frac{a+ \frac{b+c}{2}}{2}$

$\displaystyle \Rightarrow ( a \ast b) \ast c = \frac{2a+b+c}{4} \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly see that associative property does not hold for the binary operation $'\ast' \text{ on } 'N'.$

$\\$

Question 3: Let $A$ be any set containing more than one element. Let $'\ast'$ be a binary operation on $A$ defined by $a \ast b =b$ for all $a, b \in A.$

Is $'\ast'$ commutative or associative on $A$ ?

Given that $\ast$ is a binary operation on set $A$ defined by $a \ast b = b \text{ for all } a, b \in A.$

We know that commutative property is $p \ast q = q \ast p,$ where $\ast$ is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = b$

$\Rightarrow b \ast a = a$

$\Rightarrow b \ast a \neq a \ast b$

Therefore the commutative property does not hold for given binary operation $' \ast ' \text{ on } 'A' .$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b) \ast c = (b) \ast c$

$\Rightarrow (a \ast b) \ast c = b \ast c$

$\Rightarrow (a \ast b) \ast c = c \hspace{1.0cm}\ldots (i)$

$\Rightarrow a \ast (b \ast c) = a \ast (c)$

$\Rightarrow a \ast (b \ast c) = a \ast c$

$\Rightarrow a \ast (b \ast c) = c \hspace{1.0cm} \ldots (ii)$

From (i) and (ii) we can clearly say that associativity holds for the binary operation $' \ast ' \text{ on } 'A' .$

$\\$

Question 4: Check the commutativity and associativity of each of the following binary operations:

$\text{(i) } ' \ast ' \text{ on } Z \text{ defined by } a \ast b = a +b + ab \text{ for all } a, b \in Z.$

$\text{(ii) } ' \ast ' \text{ on } N \text{ defined by } a \ast b=2^{ab} \text{ for all } a, b \in N$

$\text{(iii) } ' \ast ' \text{ on } Q \text{ defined by } a \ast b = a -b \text{ for all } a, b \in Q$

$\text{(iv) } '\odot' \text{ on } Q \text{ defined by } a \odot b = a^2 + b^2 \text{ for all } a, b \in Q.$

$\displaystyle \text{(v) } '\bigcirc' \text{ on } Q \text{ defined by } a \bigcirc b = \frac{ab}{2} \text{ for all } a, b \in Q$

$\text{(vi) } ' \ast ' \text{ on } Q \text{ defined by } a \ast b =ab^2 \text{ for all } a, b \in Q$

$\text{(vii) } ' \ast ' \text{ on } Q \text{ defined by } a \ast b = a + ab \text{ for all } a, b \in Q$

$\text{(viii) } ' \ast ' \text{ on } R \text{ defined by }a \ast b = a +b -7 \text{ for all } a, b \in Q$

$\text{(ix) } ' \ast ' \text{ on } Q \text{ defined by } a \ast b =(a -b)^2 \text{ for all } a, b \in Q$

$\text{(x) } ' \ast ' \text{ on } Q \text{ defined by } a \ast b = ab + 1 \text{ for all } a, b \in Q$

$\text{(xi) } ' \ast ' \text{ on } N, \text{ defined by } a \ast b = a^b \text{ for all } a, b \in N$

$\text{(xii) } ' \ast ' \text{ on } Z a \ast b= a - b \text{ for all } a, b \in Z$

$\displaystyle \text{(xiii) } ' \ast ' \text{ on } Q \text{ defined by } a \ast b =\frac{ab}{4}, \text{ for all } a , b \in Q$

$\text{(xiv) } ' \ast ' \text{ on } Z \text{ defined by } a \ast b = a+b -ab \text{ for all } a, b \in Z$

$\text{(xv) } ' \ast ' \text{ on } Q \text{ defined by } a \ast b =gcd(a, b) \text{ for all } a,b \in N$

$\text{(i) } ' \ast ' \text{ on } Z \text{ defined by } a \ast b = a +b + ab \text{ for all } a, b \in Z.$

Given that  $\ast$  is a binary operation on $Z$ defined by $a \ast b = a + b + ab \text{ for all } a, b \in Z.$

We know that commutative property is $p \ast q = q \ast p,$ where  $\ast$ is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = a + b + ab$

$\Rightarrow b \ast a = b + a + ba$

$\Rightarrow b \ast a = a + b + ab$

$\Rightarrow b \ast a = a \ast b$

Therefore Commutative property holds for a given binary operation $' \ast ' \text{ on } ' Z '.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b) \ast c = (a + b + ab) \ast c$

$\Rightarrow (a \ast b) \ast c = (a + b + ab + c + ((a + b + ab) \times c))$

$\Rightarrow (a \ast b) \ast c = a + b + c + ab + ac + ac + abc \hspace{1.0cm}\ldots (i)$

$\Rightarrow a \ast (b \ast c) = a \ast (b + c + bc)$

$\Rightarrow a \ast (b \ast c) = (a + b + c + bc + (a \times (b + c + bc)))$

$\Rightarrow a \ast (b \ast c) = a + b + c + ab + bc + ac + abc \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity holds for the binary operation $' \ast ' \text{ on } ' Z '.$

$\\$

$\text{(ii) } ' \ast ' \text{ on } N \text{ defined by } a \ast b=2^{ab} \text{ for all } a, b \in N$

Given that  $\ast$  is a binary operation on $N$ defined by $a \ast b = 2^{ab} \text{ for all } a,b \in N.$

We know that the commutative property is $p \ast q = q \ast p,$ where  $\ast$  is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = 2^{ab}$

$\Rightarrow b \ast a = 2^{ba} = 2^{ab}$

$\Rightarrow b \ast a = a \ast b$

Therefore the commutative property holds for given binary operation $' \ast ' \text{ on } 'N'.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b) \ast c = (2^{ab}) \ast c$

$\Rightarrow (a \ast b) \ast c = 2^{ab} \cdot c \hspace{1.0cm}\ldots (i)$

$\Rightarrow a \ast (b \ast c) = a \ast (2^{bc})$

$\Rightarrow (a \ast b) \ast c = 2^{a \cdot 2^{bc}} \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity doesn’t hold for the binary operation $'\ast ' \text{ on } 'N'.$

$\\$

$\text{(iii) } ' \ast ' \text{ on } Q \text{ defined by } a \ast b = a -b \text{ for all } a, b \in Q$

Given that $\ast$ is a binary operation on $Q$ defined by $a\ast b = a - b \text{ for all } a,b \in Q.$

We know that the commutative property is $p\ast q = q\ast p,$ where $\ast$ is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a\ast b = a - b$

$\Rightarrow b\ast a = b - a$

$\Rightarrow b\ast a \neq a\ast b$

Therefore the commutative property doesn’t hold for a given binary operation $'\ast ' \text{ on } ' Q '.$

We know that associative property is $(p\ast q)\ast r = p\ast (q\ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a\ast b)\ast c = (a - b)\ast c$

$\Rightarrow (a\ast b)\ast c = a - b - c \hspace{1.0cm}\ldots (i)$

$\Rightarrow a\ast (b\ast c) = a\ast (b - c)$

$\Rightarrow a\ast (b\ast c) = a - (b - c)$

$\Rightarrow a\ast (b\ast c) = a - b + c . \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity doesn’t hold for the binary operation  $' \ast ' \text{ on } ' Q '$

$\\$

$\text{(iv) } '\odot' \text{ on } Q \text{ defined by } a \odot b = a^2 + b^2 \text{ for all } a, b \in Q.$

Given that $\odot$ is a binary operation on $Q$ defined by $a \odot b = a^2 + b^2 \text{ for all } a,b \in Q.$

We know that commutative property is $p \odot q = q \odot p,$ where $\odot$ is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a \odot b = a^2 + b^2$

$\Rightarrow b \odot a = b^2 + a^2 = a^2 + b^2$

$\Rightarrow b \odot a = a \odot b$

Therefore Commutative property holds for given binary operation $' \odot ' \text{ on } 'Q'.$

We know that associative property is $(p \odot q) \odot r = p \odot (q \odot r)$

Let’s check the associativity of given binary operation:

$\Rightarrow ( a \odot b) \odot c = ( a^2 + b^2) \odot c$

$\Rightarrow ( a \odot b) \odot c = ( a^2 + b^2)^2 + c^2$

$\Rightarrow ( a \odot b) \odot c = a^4 + b^4 + 2 a^2 b^2 + c^2 \hspace{1.0cm}\ldots (i)$

$\Rightarrow ( a \odot b) \odot c = a \odot ( b^2 + c^2)$

$\Rightarrow ( a \odot b) \odot c = a^2 + ( b^2 + c^2) ^2$

$\Rightarrow ( a \odot b) \odot c = a^2 + b^4 + c^4 + 2 b^2 c^2 \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity doesn’t hold for the binary operation $' \ast ' \text{ on } ' Q '.$

$\\$

$\displaystyle \text{(v) } '\bigcirc' \text{ on } Q \text{ defined by } a \bigcirc b = \frac{ab}{2} \text{ for all } a, b \in Q$

$\displaystyle \text{Given that } \bigcirc \text{ is a binary operation on } Q \text{ defined by } \\ a \bigcirc b = \frac{ab}{2} \text{ for all } a,b \in Q.$

We know that commutative property is $p \bigcirc q = q \bigcirc p,$ where $\bigcirc$  is a binary operation.

Let’s check the comutativity of given binary operation:

$\displaystyle \Rightarrow a \bigcirc b = \frac{ab}{2}$

$\displaystyle \Rightarrow b \bigcirc a = \frac{ba}{2} =\frac{ab}{2}$

$\Rightarrow b \ast a = a \ast b$

Therefore the commutative property holds for given binary operation $' \bigcirc ' \text{ on } 'Q'.$

We know that associative property is $(p \bigcirc q) \bigcirc r = p \bigcirc (q \bigcirc r)$

Let’s check the associativity of given binary operation:

$\displaystyle \Rightarrow (a \bigcirc b ) \bigcirc c = \frac{ab}{2} \bigcirc c$

$\displaystyle \Rightarrow (a \bigcirc b ) \bigcirc c = \frac{\frac{ab}{2} c }{2}$

$\displaystyle \Rightarrow (a \bigcirc b ) \bigcirc c = \frac{abc}{4} \hspace{1.0cm}\ldots (i)$

$\displaystyle \Rightarrow (a \bigcirc b ) \bigcirc c = a \bigcirc \frac{bc}{2}$

$\displaystyle \Rightarrow (a \bigcirc b ) \bigcirc c = \frac{a \frac{bc}{2}}{2}$

$\displaystyle \Rightarrow (a \bigcirc b ) \bigcirc c = \frac{abc}{4} \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity hold for the binary operation $' \bigcirc ' \text{ on } 'Q'.$

$\\$

$\text{(vi) } ' \ast ' \text{ on } Q \text{ defined by } a \ast b =ab^2 \text{ for all } a, b \in Q$

Given that  $\ast$  is a binary operation on $Q$ defined by $a \ast b = ab^2 \text{ for all } a, b \in Q.$

We know that commutative property is $p \ast q = q \ast p,$ where  $\ast$  is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = ab^2$

$\Rightarrow b \ast a = ba^2$

$\Rightarrow b \ast a \neq a \ast b$

Therefore Commutative property doesn’t holds for a given binary operation $' \ast ' \text{ on } ' Q '.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b) \ast c = (ab^2) \ast c$

$\Rightarrow (a \ast b) \ast c = ab^2c^2 \hspace{1.0cm}\ldots (i)$

$\Rightarrow a \ast (b \ast c) = a \ast (bc^2)$

$\Rightarrow a \ast (b \ast c) = a(bc^2)^2$

$\Rightarrow a \ast (b \ast c) = ab^2c^4 \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity doesn’t hold for the binary operation $' \ast ' \text{ on } 'Q'$

$\\$

$\text{(vii) } ' \ast ' \text{ on } Q \text{ defined by } a \ast b = a + ab \text{ for all } a, b \in Q$

Given that  $\ast$  is a binary operation on $Q$ defined by $a \ast b = a + ab \text{ for all } a,b \in Q.$

We know that the commutative property is $p \ast q = q \ast p,$ where  $\ast$  is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = a + ab$

$\Rightarrow b \ast a = b + ba = b + ab$

$\Rightarrow b \ast a \neq a \ast b$

Therefore Commutative property doesn’t holds for a given binary operation $' \ast ' \text{ on } 'Q'.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b) \ast c = (a + ab) \ast c$

$\Rightarrow (a \ast b) \ast c = a + ab + ((a + ab) \times c)$

$\Rightarrow (a \ast b) \ast c = a + ab + ac + abc \hspace{1.0cm}\ldots (i)$

$\Rightarrow a \ast (b \ast c) = a \ast (b + bc)$

$\Rightarrow a \ast (b \ast c) = a + (a \times (b + bc))$

$\Rightarrow a \ast (b \ast c) = a + ab + abc \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity doesn’t hold for the binary operation $' \ast ' \text{ on } 'Q'.$

$\\$

$\text{(viii) } ' \ast ' \text{ on } R \text{ defined by }a \ast b = a +b -7 \text{ for all } a, b \in Q$

Given that  $\ast$ is a binary operation on $R$ defined by $a \ast b = a + b - 7 \text{ for all } a,b \in R.$

We know that the commutative property is $p \ast q = q \ast p,$ where  $\ast$ is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = a + b - 7$

$\Rightarrow b \ast a = b + a - 7 = a + b - 7$

$\Rightarrow b \ast a = a \ast b$

Therefore Commutative property holds for a given binary operation $' \ast ' \text{ on } 'R'.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b) \ast c = (a + b - 7) \ast c$

$\Rightarrow (a \ast b) \ast c = a + b - 7 + c - 7$

$\Rightarrow (a \ast b) \ast c = a + b + c - 14 \hspace{1.0cm}\ldots (i)$

$\Rightarrow a \ast (b \ast c) = a \ast (b + c - 7)$

$\Rightarrow a \ast (b \ast c) = a + b + c - 7 - 7$

$\Rightarrow a \ast (b \ast c) = a + b + c - 14 \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity holds for the binary operation $' \ast ' \text{ on } 'R'.$

$\\$

$\text{(ix) } ' \ast ' \text{ on } Q \text{ defined by } a \ast b =(a -b)^2 \text{ for all } a, b \in Q$

Given that $\ast$ is a binary operation on $Q$ defined by $a \ast b = (a - b)^2 \text{ for all } a,b \in Q.$

We know that commutative property is $p \ast q = q \ast p,$ where $\ast$  is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = (a - b)^2$

$\Rightarrow b \ast a = (b - a)^2 = (a- b)^2$

$\Rightarrow b \ast a = a \ast b$

Therefore Commutative property holds for given binary operation $' \ast ' \text{ on } 'Q'.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$  Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b ) \ast c = ((a - b)^2) \ast c$

$\Rightarrow (a \ast b ) \ast c = (( a - b)^2 - c)^2$

$\Rightarrow (a \ast b ) \ast c = ( a^2 + b^2 - 2ab - c)^2 \hspace{1.0cm}\ldots (i)$

$\Rightarrow (a \ast b ) \ast c = a \ast (( b - c)^2)$

$\Rightarrow (a \ast b ) \ast c = ( a^2 - ( b-c)^2 )^2$

$\Rightarrow (a \ast b ) \ast c = ( a^2 - b^2 - c^2 + 2bc)^2 \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity doesn’t hold for the binary operation $' \ast ' \text{ on } ' Q '.$

$\\$

$\text{(x) } ' \ast ' \text{ on } Q \text{ defined by } a \ast b = ab + 1 \text{ for all } a, b \in Q$

Given that  $\ast$  is a binary operation on $Q$ defined by $a \ast b = ab + 1 \text{ for all } a,b \in Q.$

We know that the commutative property is $p \ast q = q \ast p,$ where  $\ast$  is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = ab + 1$

$\Rightarrow b \ast a = ba + 1 = ab + 1$

$\Rightarrow b \ast a = a \ast b$

Therefore Commutative property holds for a given binary operation $' \ast ' \text{ on } 'Q'.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b) \ast c = (ab + 1) \ast c$

$\Rightarrow (a \ast b) \ast c = ((ab + 1) \times c) + 1$

$\Rightarrow (a \ast b) \ast c = abc + c + 1 \hspace{1.0cm}\ldots (i)$

$\Rightarrow a \ast (b \ast c) = a \ast (bc + 1)$

$\Rightarrow a \ast (b \ast c) = (a \times (bc + 1)) + 1$

$\Rightarrow a \ast (b \ast c) = abc + a + 1 \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity doesn’t hold for the binary operation $' \ast ' \text{ on } 'Q'.$

$\\$

$\text{(xi) } ' \ast ' \text{ on } N, \text{ defined by } a \ast b = a^b \text{ for all } a, b \in N$

Given that  $\ast$  is a binary operation on $N$ defined by $a \ast b = a b \text{ for all } a,b \in N.$

We know that the commutative property is $p \ast q = q \ast p,$ where  $\ast$  is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = a^b$

$\Rightarrow b \ast a = b^a$

$\Rightarrow b \ast a \neq a \ast b$

Therefore Commutative property doesn’t hold for a given binary operation $' \ast ' \text{ on } 'N'.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b) \ast c = (a^b) \ast c$

$\Rightarrow (a \ast b) \ast c = (a^b)c$

$\Rightarrow (a \ast b) \ast c = a^{bc} \hspace{1.0cm}\ldots (i)$

$\Rightarrow a \ast (b \ast c) = a \ast (b^c)$

$\Rightarrow a \ast (b \ast c) = a^{bc} \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity doesn’t hold for the binary operation $' \ast ' \text{ on } 'N'.$

$\\$

$\text{(xii) } ' \ast ' \text{ on } Z a \ast b= a - b \text{ for all } a, b \in Z$

Given that  $\ast$  is a binary operation on $Z$ defined by $a \ast b = a - b \text{ for all } a, b \in Z.$

We know that commutative property is $p \ast q = q \ast p,$ where  $\ast$  is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = a - b$

$\Rightarrow b \ast a = b - a$

$\Rightarrow b \ast a \neq a \ast b$

Therefore  Commutative property doesn’t holds for given binary operation $' \ast ' \text{ on } 'Z'.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b) \ast c = (a - b) \ast c$

$\Rightarrow (a \ast b) \ast c = (a - b) - c$

$\Rightarrow (a \ast b) \ast c = a - b - c \hspace{1.0cm}\ldots (i)$

$\Rightarrow a \ast (b \ast c) = a \ast (b - c)$

$\Rightarrow a \ast (b \ast c) = a - (b - c)$

$\Rightarrow a \ast (b \ast c) = a - b + c \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity doesn’t hold for the binary operation $' \ast ' \text{ on } 'Z'.$

$\\$

$\displaystyle \text{(xiii) } ' \ast ' \text{ on } Q \text{ defined by } a \ast b =\frac{ab}{4}, \text{ for all } a , b \in Q$

$\displaystyle \text{Given that } \ast \text{ is a binary operation on } Q \text{ defined by } \\ a \ast b = \frac{ab}{4} \text{ for all } a,b \in Q.$

We know that commutative property is $p \ast q = q \ast p,$ where $\ast$  is a binary operation.

Let’s check the commutativity of given binary operation:

$\displaystyle \Rightarrow a \ast b = \frac{ab}{4}$

$\displaystyle \Rightarrow b \ast a = \frac{ba}{4} = \frac{ab}{4}$

$\displaystyle \Rightarrow b \ast a = a \ast b$

Commutative property holds for given binary operation $' \ast ' \text{ on } ' Q ' .$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\displaystyle \Rightarrow (a \ast b) \ast c = \frac{ab}{4} \ast c$

$\displaystyle \Rightarrow (a \ast b) \ast c = \frac{\frac{ab}{4} \cdot c}{4}$

$\displaystyle \Rightarrow (a \ast b) \ast c = \frac{abc}{16} \hspace{1.0cm}\ldots (i)$

$\displaystyle \Rightarrow (a \ast b) \ast c =a \ast \frac{bc}{4}$

$\displaystyle \Rightarrow (a \ast b) \ast c = \frac{a \cdot \frac{bc}{4}}{4}$

$\displaystyle \Rightarrow (a \ast b) \ast c = \frac{abc}{16} \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity hold for the binary operation $' \ast ' \text{ on } ' Q ' .$

$\\$

$\text{(xiv) } ' \ast ' \text{ on } Z \text{ defined by } a \ast b = a+b -ab \text{ for all } a, b \in Z$

Given that  $\ast$  is a binary operation on $Z$ defined by $a \ast b = a + b - ab \text{ for all } a,b \in Z.$

We know that the commutative property is $p \ast q = q \ast p,$ where  $\ast$  is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = a + b - ab$

$\Rightarrow b \ast a = b + a - ba = a + b - ab$

$\Rightarrow b \ast a = a \ast b$

Therefore Commutative property holds for a given binary operation $' \ast ' \text{ on } 'Z'.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b) \ast c = (a + b - ab) \ast c$

$\Rightarrow (a \ast b) \ast c = a + b - ab + c - ((a + b - ab) \times c)$

$\Rightarrow (a \ast b) \ast c = a + b + c - ab - ac - bc + abc\hspace{1.0cm}\ldots (i)$

$\Rightarrow a \ast (b \ast c) = a \ast (b + c - bc)$

$\Rightarrow a \ast (b \ast c) = a + b + c - bc - (a \times (b + c - bc))$

$\Rightarrow a \ast (b \ast c) = a + b + c - ab - ac - bc + abc \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity holds for the binary operation $' \ast ' \text{ on } 'Z'.$

$\\$

$\text{(xv) } ' \ast ' \text{ on } Q \text{ defined by } a \ast b =gcd(a, b) \text{ for all } a,b \in N$

Given that  $\ast$  is a binary operation on $Q$ defined by $a \ast b = g.c.d(a,b) \text{ for all } a,b \in Q.$

We know that the commutative property is $p \ast q = q \ast p,$ where  $\ast$  is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = g.c.d(a,b)$

$\Rightarrow b \ast a = g.c.d(b,a) = g.c.d(a,b)$

$\Rightarrow b \ast a = a \ast b$

Therefore Commutative property holds for given binary operation $' \ast ' \text{ on } 'Q'.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b) \ast c = (g.c.d(a,b)) \ast c$

$\Rightarrow (a \ast b) \ast c = g.c.d(g.c.d(a,b),c)$

$\Rightarrow (a \ast b) \ast c = g.c.d(a,b,c) \hspace{1.0cm}\ldots (i)$

$\Rightarrow a \ast (b \ast c) = a \ast (g.c.d(b,c))$

$\Rightarrow a \ast (b \ast c) = g.c.d(a,g.c.d(b,c))$

$\Rightarrow a \ast (b \ast c) = g.c.d(a,b,c) \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity holds for the binary operation $' \ast ' \text{ on } 'Q'.$

$\\$

Question 5: If the binary operation $\circ$ is defined by $a \circ b =a+b -ab$ on the set $Q - \{ -1 \}$ of all rational numbers other than 1, show that $\circ$ is commutative on $Q - [1].$

$\text{Let } a, b \in Q - \{-1\}.$

$\text{Now, } a \circ b = a + b - ab = b + a - ba = b \circ a$

$\text{ So, } a \circ b = b \circ a \text{ for all } a, b \in Q - \{-1\}$

$\text{Hence, } \circ \text{ is commutative on } Q - \{-1\}$

$\\$

Question 6: Show that the binary operation $\ast$ on $Z$ defined by $a \ast b = 3a + 7b$ is not commutative.

$\text{Let } a, b \in Z$

$a \ast b = 3a + 7b$

$b \ast a = 3b + 7a$

$\text{Now, } a \ast b \neq b \ast a$

$\text{Let } a = 1 \text{ and } b = 2$

$1 \ast 2 = 3 \times 1 + 7 \times 2 = 3 + 14 = 17$

$2 \ast 1 = 3 \times 2 + 7 \times 1 = 6 + 7 = 13$

$\text{So, there exist } a = 1, b = 2 \in Z \text{ such that } a \ast b \neq b \ast a$

$\text{Hence, } \ast \text{is not commutative on } Z.$

$\\$

Question 7: On the set $Z$ of integers a binary operation $\ast$ is defined by $a \ast b =ab +1$ for all $a, b \in Z.$ Prove that $\ast$ is not associative on $Z.$

$\text{Let } a, b, c \in Z$

$a \ast (b \ast c) = a \ast (bc + 1) = a(bc + 1) + 1 = abc + a + 1$

$(a \ast b) \ast c = (ab + 1) \ast c = (ab + 1)c + 1 = abc + c + 1$

$\text{So, } a \ast (b \ast c) \neq (a \ast b) \ast c$

$\text{Hence, } \ast \text{ is not associative on } Z.$

$\\$

Question 8: Let $S$ be the set of all real numbers except $-1$ and let $' \ast '$ be an operation defined by $a \ast b =a+b + ab$ for all $a, b \in S.$ Determine whether $' \ast '$ is a binary operation on $S.$ If yes, check its commutativity and associativity. Also, solve the equation $(2 \ast x) \ast 3 = 7.$

$\text{Given that } ' \ast ' \text{ is an operation that is valid on the set } S \text{ which consists of } \\ \text{ all real numbers except }- 1 \text{ i.e., } R - \{ - 1\} \text{ defined as } a \ast b = a + b + ab$

$\text{Let us assume } a + b + ab = - 1$

$\Rightarrow a + ab + b + 1 = 0$

$\Rightarrow a(1 + b) + (1 + b) = 0$

$\Rightarrow (a + 1)(b + 1) = 0$

$\Rightarrow a = - 1 \text{ or } b = - 1$

$\text{But according to the problem, it is given that } a \neq - 1 \text{ and } b \neq - 1$

$\text{So, } a + b + ab \neq - 1, \text{ so we can say that the operation } ' \ast ' \\ \text{ defines a binary operation on set } 'S'.$

$\text{We know that the commutative property is } p \ast q = q \ast p, \text{ where } \ast \\ \text{ is a binary operation. }$

Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = a + b + ab$

$\Rightarrow b \ast a = b + a + ba = a + b + ab$

$\Rightarrow b \ast a = a \ast b$

$\text{Therefore Commutative property holds for given binary operation } ' \ast ' \text{ on } 'S'.$

$\text{ We know that associative property } is (p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b) \ast c = (a + b + ab) \ast c$

$\Rightarrow (a \ast b) \ast c = a + b + ab + c + ((a + b + ab) \times c)$

$\Rightarrow (a \ast b) \ast c = a + b + c + ab + ac + bc + abc \hspace{1.0cm}\ldots (i)$

$\Rightarrow a \ast (b \ast c) = a \ast (b + c + bc)$

$\Rightarrow a \ast (b \ast c) = a + b + c + bc + (a \times (b + c + bc))$

$\Rightarrow a \ast (b \ast c) = a + b + c + ab + bc + ac + abc \hspace{1.0cm}\ldots (ii)$

$\text{From (i) and (ii) we can clearly say that associativity holds } \\ \text{for the binary operation } ' \ast ' \text{ on } 'N'.$

$\text{We need to also solve for } x \text{ in the given } expression:$

$\Rightarrow (2 \ast x ) \ast 3 = 7$

$\Rightarrow (2 + x + 2x ) \ast 3 = 7$

$\Rightarrow (2 + 3 x ) \ast 3 = 7$

$\Rightarrow 2 + 3 x + 3 + ((2 + 3 x ) \times 3) = 7$

$\Rightarrow 5 + 3 x + 6 + 9 x = 7$

$\Rightarrow 11 + 12 x = 7$

$\Rightarrow 12 x = - 4$

$\displaystyle \Rightarrow x = \frac{-4}{12} = \frac{-1}{3}$

$\displaystyle \text{Therefore, the value of } x \text{ is} = \frac{-1}{3}$

$\\$

Question 9: On $Q,$ the set of all rational numbers, $\ast$  is defined by   $\displaystyle a \ast b = \frac{a-b}{2}, \text{ show that } \ast \text{ is not associative. }$

$\displaystyle \text{Given that } \ast \text{ is a binary operation on } Q \text{ defined by } a \ast b = \frac{a-b}{2} \text{ for all } a,b \in Q.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{a-b}{2} \ast c$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{\frac{a-b}{2} - c}{2}$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{(a-b)-2c}{4}$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{a-b-2c}{4} \hspace{1.0cm}\ldots (i)$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = a \ast \frac{b-c}{2}$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{a - \frac{b-c}{2}}{2}$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{2a-(b-c)}{4}$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{2a-b+c}{4} \hspace{1.0cm}\ldots (ii)$

From (1) and (ii) we can clearly say that associativity doesn’t hold for the binary operation $' \ast ' \text{ on } ' Q ' .$

$\\$

Question 10: On $Z,$ the set of all integers, a binary operation $\ast$ is defined by $a \ast b =a+ 3b - 4.$ Prove that $\ast$ is neither commutative nor associative on $Z .$

Given that $\ast$ is a binary operation on $Z$ defined by $a \ast b = a + 3b - 4 \text{ for all } a,b \in Z.$

We know that commutative property is $p \ast q = q \ast p,$ where $\ast$ is a binary operation.

Let’s check the commutativity of given binary operation:

$\Rightarrow a \ast b = a + 3b - 4$

$\Rightarrow b \ast a = b + 3a - 4$

$\Rightarrow b \ast a \neq a \ast b$

Therefore The commutative property doesn’t hold for a given binary operation on $'Z'.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\Rightarrow (a \ast b) \ast c = (a + 3b - 4) \ast c$

$\Rightarrow (a \ast b) \ast c = a + 3b - 4 + 3c - 4$

$\Rightarrow (a \ast b) \ast c = a + 3b + 3c - 8 \hspace{1.0cm}\ldots (i)$

$\Rightarrow a \ast (b \ast c) = a \ast (b + 3c - 4)$

$\Rightarrow a \ast (b \ast c) = a + (3 \times (b + 3c - 4)) - 4$

$\Rightarrow a \ast (b \ast c) = a + 3b + 9c - 12 - 4$

$\Rightarrow a \ast (b \ast c) = a + 3b + 9c - 16 \hspace{1.0cm}\ldots (ii)$

Hence from (i) and (ii) we can clearly say that associativity doesn’t hold for the binary operation on $Z.$

$\\$

Question 11: On the set $Q$ of all rational numbers if a binary operation $\ast$  $\displaystyle \text{ is defined by } a \ast b=\frac{ab}{5}, \text{ prove that } \ast \text{ is associative on } Q.$

$\displaystyle \text{Given that } \ast \text{ is a binary operation on } Q \text{ defined by } \\ a \ast b = \frac{ab}{5} \text{ for all } a,b \in Q.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{ab}{5} \ast c$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{\frac{ab}{5} \cdot c}{5}$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{abc}{25} \hspace{1.0cm}\ldots (i)$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = a \ast \frac{bc}{5}$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{a \cdot \frac{bc}{5}}{5}$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{abc}{25} \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity hold for the binary operation $' \ast ' \text{ on } 'Q'.$

$\\$

Question 12: The binary operation $\ast$ is defined by $\displaystyle a \ast b =\frac{ab}{7} \text{ on the set } Q$ of all rational numbers. Show that $\ast$ is associative

$\displaystyle \text{Given that } \ast \text{ is a binary operation on } Q \text{ defined by } \\ a \ast b = \frac{ab}{7} \text{ for all } a,b \in Q.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{ab}{7} \ast c$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{\frac{ab}{7} \cdot c}{7}$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{abc}{49} \hspace{1.0cm}\ldots (i)$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = a \ast \frac{bc}{7}$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{a \cdot \frac{bc}{7}}{7}$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{abc}{49} \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity hold for the binary operation $' \ast ' \text{ on } 'Q'.$

$\\$

Question 13: On $Q,$ the set of all rational numbers a binary operation $\ast$  $\displaystyle \text{ is defined by } a * b =\frac{a+b}{2}. \text{ Show that } * \text{ is not associative on } Q.$

$\displaystyle \text{Given that } \ast \text{ is a binary operation on } Q \text{ defined by } \\ a \ast b = \frac{a+b}{2} \text{ for all } a,b \in Q.$

We know that associative property is $(p \ast q) \ast r = p \ast (q \ast r)$

Let’s check the associativity of given binary operation:

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{a+b}{2} \ast c$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{\frac{a+b}{2} + c}{2}$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{a+b+2c}{4} \hspace{1.0cm}\ldots (i)$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = a \ast \frac{b+c}{2}$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{a + \frac{b+c}{2}}{2}$

$\displaystyle \Rightarrow ( a \ast b ) \ast c = \frac{2a+b+c}{4} \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we can clearly say that associativity does’nt hold for the binary operation $' \ast ' \text{ on } 'Q'.$

$\\$

Question 14: Let $S$ be the set of all rational numbers except $1$ and $\ast$ be defined on $S$ by  $a \ast b =a+b-ab,$ for all, $a,b \in S.$

Prove that : (i) $\ast$ is a binary operation on $S$

(ii) $\ast$ is commutative as well as associative     [CBSE 2014]

(i) Sum, difference and product of rational numbers is a unique rational number.

Therefore for each $(a, b) \in S \times S,$ there exists a unique image $( a + b - ab) \text{ in } S.$

$\Rightarrow *$ is a function

$\Rightarrow *$ is a binary operation on $S$

(ii) $a \ast b = a + b - ab = b + a - ba = b \ast a$

Therefore $\ast$ is commutative

$\Rightarrow a \ast ( b \ast c ) = a \ast ( b + c - bc)$

$\Rightarrow a \ast ( b \ast c ) = a + ( b + c - bc) - a ( b + c - bc)$

$\Rightarrow a \ast ( b \ast c ) = a + b + c - bc - ab - ac + abc \hspace{1.0cm}\ldots (i)$

$\Rightarrow( a \ast b ) * c = ( a + b - ab ) \ast c$

$\Rightarrow( a \ast b ) * c = ( a + b - ab) + c - ( a + b - ab) c$

$\Rightarrow( a \ast b ) * c = a + b - ab + c - ac -bc+abc \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we get $a \ast ( b \ast c ) = ( a \ast b ) * c$

Therefore $\ast$ is associative.