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

Answer:

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

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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' 

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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]} 

Answer:

\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'. 

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\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'.

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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 ?

Answer:

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' .

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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

Answer:

\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 '.

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\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'.

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\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 '  

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\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 '.

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\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'.

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\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'

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\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'.

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\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'.

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\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 '.

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\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'.

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\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'.

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\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'.

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\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 ' .

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\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'.

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\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'.

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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].

Answer:

\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\}

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

Answer:

\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.

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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.

Answer:

\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. 

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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.

Answer:

\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}

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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. }

Answer:

\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 ' .

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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 .

Answer:

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.

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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.

Answer:

\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'.

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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

Answer:

\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'.

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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.

Answer:

\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'.

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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]

Answer:

(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.