Question 1: Let \ast be a binary operation on Z defined by a \ast b = a + b - 4 for all a, b \in Z.  

(i) Show that ' \ast ' is both commutative and associative.

(ii) Find the identity element in Z.

(iii) Find the invertible elements in Z.

Answer:

(i)     Let us prove the commutativity of \ast

\text{Let } a, b \in Z. \text{ Then, }

a \ast b = a + b - 4 = b + a - 4 = b \ast a

\text{So, } a \ast b = b \ast a, \ \forall \ a, b \in Z

\text{Thus,  } \ast \text{ is commutative on } Z.

Now, let us prove the associativity of Z.

\text{Let } a, b, c \in Z. \text{ Then, } 

a \ast (b \ast c) = a \ast (b + c - 4) = a + b + c - 4 - 4 = a + b + c - 8

(a \ast b) \ast c = (a + b - 4) \ast c = a + b - 4 + c - 4 = a + b + c - 8

\text{So, } a \ast (b \ast c) = (a \ast b) \ast c, \text{ for all } a, b, c \in Z

\text{Thus,  } \ast \text{ is associative on } Z. 

(ii)    Let e be the identity element in Z with respect to \ast  

\text{Such that, } a \ast e = a = e \ast a \ \forall \  a \in Z

a \ast e = a \text{ and } e \ast a = a, \ \forall \  a \in Z

a + e - 4 = a \text{ and } e + a - 4 = a, \ \forall \ a \in Z

e = 4, \forall a \in Z

Thus, 4 is the identity element in Z with respect to \ast .

(iii)  \text{Let } a \text{ and } b \in Z \text{ be the inverse of } a.

\text{Then, } a \ast b = e = b \ast a

a \ast b = e \text{ and } b \ast a = e

a + b - 4 = 4 \text{ and } b + a - 4 = 4

b = 8 - a \in Z

\text{So, } 8 - a \text{ is the inverse of } a \in Z

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Question 2: Let \ast  be a binary operation on Q_0 (set of non-zero rational numbers) defined by a \ast b = \frac{3ab}{5} \text{ or all } a,b \in Q_0.   Show that \ast  is commutative as well as associative. Also, find its identity element, if it exists

Answer:

Let us prove the commutativity of \ast 

\text{Let } a, b \in Q_0

\displaystyle a \ast b = \frac{3ab}{5} = \frac{3ba}{5} = b \ast a

\text{So, } a \ast b = b \ast a, \text{ for all } a, b \in Q_0

Let us prove the associativity of \ast

\text{ Let } a, b, c \in Q_0

\displaystyle a \ast (b \ast c) = a \ast \frac{3bc}{5} = \frac{a \cdot \frac{3bc}{5}}{5} = \frac{3abc}{25}

\displaystyle (a \ast b) \ast c = \frac{3ab}{5} \ast c = \frac{\frac{3ab}{5}}{5} = \frac{3abc}{25}

\displaystyle \text{So, }  a \ast (b \ast c) = (a \ast b) \ast c, \text{ for all } a, b, c \in Q_0

Thus \ast is associative on Q_0

Let us find the identity element 

Let e be the identity element in Z with respect to \ast

\text{Such that, } a \ast e = a = e \ast a \ \forall \  a \in Q_0

a \ast e = a \text{ and } e \ast a = a, \ \forall \  a \in Q_0

\displaystyle \frac{3ae}{5} = a \text{ and } \frac{3ea}{5} = a, \ \forall \  a \in Q_0

\displaystyle e = \frac{5}{3} \ \forall \  a \in Q_0 \ \ \ \ [ \text{ because } a ! = 0]

\displaystyle \text{Hence, } \frac{5}{3} \text{ is the identity element in } Q_0 \text{ with respect to } \ast .

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Question 3: Let \ast  be a binary operation on Q - \{- 1 \} defined by a \ast b = a + b + ab for all a, b \in Q - \{- 1 \}. Then,

(i) Show that ' \ast ' is both commutative and associative on Q -  \{-1 \}.

(ii) Find the identity element in Q -  \{-1 \}

(iii) Show that every element of Q -  \{-1 \} is invertible. Also, find the inverse of an arbitrary element.

Answer:

(i) Let us check the commutativity of \ast 

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

\text{Then } a \ast b = a + b + ab = b + a + ba = b \ast a

\text{So, } a \ast b = b \ast a, \ \forall \ a, b \in Q - \{-1\}

Let us prove that associativity of \ast $

\text{Let } a, b, c \in Q - \{-1\}, then,

a \ast (b \ast c) = a \ast (b + c + b c) = a + (b + c + b c) + a (b + c + b c) = a + b + c + b c + a b + a c + a b c

(a \ast b) \ast c = (a + b + a b) \ast c = a + b + a b + c + (a + b + a b) c = a + b + a b + c + a c + b c + a b c

\text{So, } a \ast (b \ast c) = (a \ast b) \ast c, \ \forall \ a, b, c \in Q - \{-1\}

\text{So, } \ast \text{ is associative on } Q - \{-1\}. 

(ii) Let e be the identity element in I^+ with respect to \ast

\text{Such that, } a \ast e = a = e \ast a, \ \forall \ a \in Q - \{-1\}

a \ast e = a \text{ and } e \ast a = a, \ \forall \ a \in Q - \{-1\}  

a + e + ae = a \text{ and } e + a + ea = a, \ \forall \ a \in Q - \{-1\}

e + ae = 0 \text{ and } e + ea = 0, \ \forall \ a \in Q - \{-1\}

e (1 + a) = 0 \text{ and } e (1 + a) = 0, \ \forall \ a \in Q - \{-1\}

e = 0, \ \forall \ a \in Q - \{-1\} \ \ \ \ [ \text{ because } a ! = -1]

\text{Thus, } 0 \text{is the identity element in } Q - \{-1\} \text{  with respect to } \ast . 

(iii) \text{Let } a \text{ and } b \in Q - \{-1\} \text{ be the inverse of } a. 

\text{Then, } a \ast b = e = b \ast a  

a \ast b = e \text{ and } b \ast a = e  

a + b + ab = 0 \text{ and } b + a + ba = 0  

b (1 + a) = - a Q - \{-1\}  

\displaystyle b = \frac{-a}{1} + a Q - \{-1\} \ \ \ [ \text{ because } a ! = -1]  

\displaystyle \text{So, } \frac{-a}{1} + a \text{ is the inverse of } a \in Q - \{-1\} 

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Question 4: Let A = R_0 \times R, \text{ where } R_0 denote the set of all non-zero real numbers. A binary operation '\bigcirc' is defined on A as follows: (a, b) \bigcirc (c, d) =(ac,bc + d) \text{ for all } (a,b),(c, d) \in R_0 \times R.

(i) Show that '\bigcirc' is commutative and associative on A

(ii) Find the identity element in A

(iii) Find the invertible elements in A.  

Answer:

(i) \text{Let } X = (a, b) \text{ and } Y = (c, d)  \in  A, \ \forall \  a, c  \in  R_0 \text{ and } b, d  \in  R

\text{Then, } X  \bigcirc  Y = (ac, bc + d)

\text{And } Y  \bigcirc  X = (ca, da + b)

\text{So, } X  \bigcirc  Y = Y  \bigcirc  X, \ \forall \  X, Y  \in  A

\text{Thus, }  \bigcirc  \text{commutative on } A.

\text{Let us check the associativity of }  \bigcirc 

\text{Let } X = (a, b), Y = (c, d) \text{and } Z = (e, f), \ \forall \  a, c, e  \in  R_0 \text{and } b, d, f  \in  R

X  \bigcirc  (Y  \bigcirc  Z) = (a, b)  \bigcirc  (ce, de + f) = (ace, bce + de + f)

(X  \bigcirc  Y)  \bigcirc  Z = (ac, bc + d)  \bigcirc  (e, f) = (ace, (bc + d) e + f) = (ace, bce + de + f)

\text{So, } X  \bigcirc  (Y  \bigcirc  Z) = (X  \bigcirc  Y)  \bigcirc  Z, \ \forall \  X, Y, Z  \in  A 

\text{(ii) Let } E = (x, y) \text{ be the identity element in } A \text{ with respect to }  \bigcirc , \ \forall \  x  \in  R_0 \text{ and } y  \in  R  

Such that, 

X \bigcirc E = X = E \bigcirc X, \ \forall \  X  \in  A  

X \bigcirc E = X \text{ and } E \bigcirc X = X  

(ax, bx +y) = (a, b) \text{ and } (xa, ya + b) = (a, b)  

\text{Conside }r (ax, bx + y) = (a, b)  

ax = a x = 1  

\text{And } bx + y = b  

y = 0 \ \ \ \ [\text{ since } x = 1]  

\text{Consider } (xa, ya + b) = (a, b)  

xa = a  

x = 1  

\text{And } ya + b = b  

y = 0 \ \ \ \ [\text{ since } x = 1] \text{So, } (1, 0) \text{is the identity element in } A \text{with respect to }  \bigcirc . 

\text{(iii) Let } F = (m, n) \text{ be the inverse in } A \ \forall \  m  \in  R_0 \text{ and } n  \in  R

X  \bigcirc  F = E \text{ and } F  \bigcirc  X = E

(am, bm + n) = (1, 0) \text{ and } (ma, na + b) = (1, 0)

\text{Consider } (am, bm + n) = (1, 0)

\displaystyle am = 1 \Rightarrow m = \frac{1}{a}

\text{And } bm + n = 0

\displaystyle n = \frac{-b}{a} \ \ \ \ [\text{ since } m = \frac{1}{a}]

\text{Consider } (ma, na + b) = (1, 0)

\displaystyle ma = 1 \Rightarrow m = \frac{1}{a}

\displaystyle \text{And } na + b = 0 \Rightarrow n = \frac{-b}{a}

\displaystyle \text{So, the inverse of } (a, b)  \in  A \text{ with respect to }  \bigcirc  \text{ is } \Big( \frac{1}{a}, \frac{-1}{a} \Big)

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Question 5: Let 'o' be a binary operation on the set Q_0 of all non-zero \text{ rational numbers defined by } a \circ b = \frac{ab}{2}  \text{ for all } a, b \in Q_0.

(i) Show that '\circ' is both commutative and associate.

(ii) Find the identity element in Q_0

(iii) Find the invertible elements of Q_0

Answer:

(i) Commutativity:

\text{Let } a, b \in Q_0. \text{ Then } 

\displaystyle a \circ b = \frac{ab}{2} = \frac{ba}{2} = b \circ a

\text{ Therefore, } a \circ b = b \circ \ \forall \ a ,  b \in Q_0

\text{Thus, }  \ \forall \ a \in Q_0  \text{ is commutative on } Q_0

Associativity:

\text{Let } a, b \in Q_0. \text{ Then } 

\displaystyle a \circ ( b \circ  c) = a \circ  \Big( \frac{bc}{2} \Big) = \frac{a \cdot \frac{bc}{2}}{2} = \frac{abc}{4}

\displaystyle (a \circ  b) \circ  c = \Big( \frac{ab}{2} \Big) \circ c = \frac{ \frac{ab}{2} \cdot c}{2} = \frac{abc}{4}

\text{Therefore } a \circ ( b \circ  c) = (a \circ  b) \circ  c, \ \forall \ a, b, c, \in Q_o

\text{Thus, } \circ \text{ is associative on } Q_0

\text{(ii) Let } e \text{ be the identity element in } Q_o \text{ with respect to } \ast \text{ such that: }

a \circ e = a = e \circ a \ \forall \ a \in Q_0

a \circ e = a \text{ and } e \circ a = a, \ \forall \ a \in Q_0

\displaystyle \Rightarrow \frac{ae}{2} = a \text{ and } \frac{ea}{2} = a, \ \forall \ a \in Q_0

e = 2 \in Q_0,  \ \forall \ a \in Q_0

\text{Thus, } 2 \text{ is the identify element in } Q_0 \text{ with respect to } \circ.

\text{(iii) Let } a \in Q_0 \text{ and } b \in Q_0 \text{ be the inverse of } a. \text{ Then, }

a \circ b = e = b \circ a

\Rightarrow a \circ  b = e \text{ and } b \circ  a = e

\displaystyle \Rightarrow \frac{ab}{2} = 2 \text{ and } \frac{ba}{2} = 2

\displaystyle \Rightarrow b = \frac{4}{a} \in Q_0

\displaystyle \text{Thus } \frac{4}{a} \text{ is the inverse of } a \in Q_0

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Question 6: On R- \{ 1 \}, a binary operation \ast is defined by a \ast b=a+b-ab. Prove that \ast  is commutative and associative. Find the identity element for \ast \text{ on } R - \{1\}. Also, prove that every element of R - \{1\} is invertible.

Answer:

\text{(i) We are given 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 \ast on a set is a function \ast : A X A \rightarrow A. We denote \ast (a, b) \text{ as } a \ast b.

\text{Here the function } \ast : R - \{1\}X R - \{1\} \rightarrow R - \{1\} \text{ is given by } a \ast b = a + b - ab 

\text{For the } ' \ast ' \text{ to be commutative, } a \ast b = b \ast a \text{ must be true for all } \\ \\ a, b \text{ belong to } R - \{ 1 \}. \text{Lets check. }  

\text{1. } a \ast b = a + b - ab \\ \\ \text{2. } b \ast a = b + a - ba = a + b - ab \\ \\ \Rightarrow a \ast b = b \ast a \text{ (as shown by 1 and 2) }  

\text{Hence } ' \ast ' \text{ is commutative on } R - \{1\} 

\text{For the } '\ast ' \text{ to be associative, } a \ast (b \ast c) = (a \ast b) \ast c \text{ must hold for every } a, b, c \in R - \{1\}.

\text{3.} a \ast (b \ast c) = a \ast (b + c - bc) \\ = a + (b + c - bc) - a(b + c + bc) = a + b + c - ab - bc - ac + abc 

\text{4.} (a \ast b) \ast c = (a + b - ab) \ast c \\ = a + b - ab + c - (a + b - ab)c = a + b + c - ab - bc - ac + abc \\ \\ \Rightarrow 3. = 4. 

\text{Hence } ' \ast ' \text{ is associative on } R - \{1\} 

\text{(ii) Identity Element: Given a binary operation } \ast : A X A \rightarrow A, \text{ an element } \\ \\ e \in A, \text{ if it exists, is called an identity of the operation } \ast , \text{ if } \\ \\ a \ast e = a = e \ast a \ \forall \ a \in A. 

\text{Let } e \text{ be the identity element of } R - \{1\} \text{ and a be an element of } R - \{1\}. 

\text{Therefore, } a \ast e = a \Rightarrow a + e - ae = a \Rightarrow e + ea = 0 \Rightarrow e(1 - a) = 0 \Rightarrow e = 0.

(1 - a \neq 0  \text{ as } a \text{ cannot be equal to } 1 \text{ as the operation is valid in } R - \{1\})  

\text{(iii.) Given a binary operation } \ast A X A \rightarrow A \text{ with the identity element } e \text{ in } A, \\ \\  \text{ an element }a \in A \text{ is said to be invertible with respect to the operation, if there  } \\ \\ \text{ exists an element } b \text{ in } A \text{ such that } a \ast b = e = b \ast a \text{ and } b \text{ is called the inverse of }a \\ \\ \text{ and is denoted by } a^{-1}. 

Let us proceed with the solution. 

\text{Let } b \in R - \{1\} \text{ be the invertible element in } R - \{1\} \text{ of a, here } a \in R - \{1\}. 

\text{Therefore } a \ast b = e \text{ (We know the identity element from previous) } 

\Rightarrow a + b - ab = 0 \Rightarrow b - ab = - a \Rightarrow b(1 - a) = - a 

\displaystyle \Rightarrow b = \frac{-a}{1-a} = \frac{a}{a-1} (\text{ Here } a \neq 1, b \neq 1)

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Question 7: Let R_0 denote the set of all non-zero real numbers and let A =R_0 \times R_0. If  ' \ast ' is a binary operation on A \text{ defined by } (a,b) \ast (c, d) =(ac,bd) \text{ for all } (a,b), (c, d) \in A.

(i) Show that ' \ast ' is both commutative and associative on A

(ii) Find the identity element in A

(iii) Find the invertible element in A 

Answer:

(i) Show that ' \ast ' is both commutative and associative on A

Commutativity:

\text{ Let } (a, b) \text{ and } (c, d) \in A \ \forall  \ a, b, c, d, \in R_o. \text{ Then, }

(a, b) \ast ( c, d) = ( ac, bd) = ( ca, db) = ( c, d) \ast ( a, b)

\text{ Therefore } (a, b) \ast ( c, d)  = ( c, d) \ast ( a, b)

\text{ Therefore } \ast \text{ is commutative on } A

Associativity:

\text{ Let } (a, b), (c, d) \text{ and } ( e, f) \in A \ \forall \ a, b, c, d, e, f, \in R_o. \text{ Then, }

(a, b) \ast ((c, d) \ast ( e, f) ) = ( a, b) \ast ( ce, df) = ( ace, bdf)

((a, b) \ast ( c, d) ) \ast ( e, f) = ( ac, bd) \ast ( e, f) = ( ace, bdf)

\text{ Therefore } (a, b) \ast ((c, d) \ast ( e, f) ) = ((a, b) \ast ( c, d) ) \ast ( e, f)

\text{ Therefore } \ast \text{ is associative on } A

(ii) Find the identity element in A

\text{ Let } (cx, y) \text{ be the identify element in } A \ \forall \ (x, y) \in A. \text{ Then, } 

(a, b) \ast ( x, y) = ( a, b) = ( x. y) \ast ( a, b)

\Rightarrow (a, b) \ast ( x, y) = ( a, b)  \text{ and } ( x, y) \ast (a, b) = ( a, b)

\Rightarrow  (ax, by) = ( a. b) \text{ and } ( xa, yb) = ( a, b)

\Rightarrow  x= 1 \text{ and } y = 1

\text{ Thus } (1, 1) \text{ is the identity element of } A

(iii) Find the invertible element in A 

\text{Let } (m, n) \text{ be the inverse of } (a, b) \ \forall \ (a, b) \in A. \text{ Then, } 

(a, b) \ast ( m, n) = ( 1, 1)

\Rightarrow ( am, bn) = ( 1, 1)

\Rightarrow am = 1 \text{ and } bn = 1 

\displaystyle \Rightarrow m = \frac{1}{a} \text{ and } n = \frac{1}{b}

\displaystyle \text{Thus, } \Big(  \frac{1}{a}, \frac{1}{b} \Big) \text{ is the inverse of } (a, b) \ \forall \ (a, b) \in A.

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Question 8: Let \ast  be the binary operation on N defined by a \ast b = HCF of a and b. Does there exist identity for this binary operation on N ?

Answer:

\text{The binary operation } \ast \text{ on } N \text{ defined as: }  

a \ast b = \text{ H.C.F. of } a \text{ and } b, \ a, b  \in  N.  

Therefore, 

b \ast a = \text{ H.C.F. of } b \text{ and } a = \text{ H.C.F. of } a \text{ and } b  

\text{Hence, } \ast \text{ is commutative on } N.  

\text{Let } a, b, c  \in  N.  

a \ast (b \ast c) = a \ast (\text{ H.C.F. of } b \text{ and } c) = \text{ H.C.F. of } a, b \text{ and } c  

(a \ast b) \ast c = (\text{ H.C.F. of } a \text{ and } b) \ast c = \text{ H.C.F. of } a, b \text{ and } c  

\text{For example, take the numbers } 2, 4, 7  \in  N. 

 Here, if the operation is applied to the above numbers as follows: 

(2 \ast 4) \ast 7 = 1 \ast 7 = 1 (1 \text{is the } \text{ H.C.F. of } 2, 4 \text{ and } \text{ H.C.F. of } 1, 7 \text{ is } 1).  

\text{Also, } 2 \ast (4 \ast 7) = 2 \ast 1 = 1 (\text{Similar as the above reason } )  

\text{Therefore, } \ast \text{ is associative. }  

Now, Identity Element: 

\text{Given a binary operation } \ast : N \times N \rightarrow N, \text{ an element } e  \in A, \text{ if it exists, is  } \\ \\  \text{called an identity of the operation } \ast , \text{ if } a \ast e = a = e \ast a \ \forall \  a  \in A.  

\text{But there does not exist any value of } e  \in N \text{ such that } a \ast e = a = e \ast a 

Therefore, this operation does not have any identity.

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Question 9: Let A = R \times R \text{ and } \ast be a binary operation on A defined by (a, b) \ast (c, d) = (a + c, b + d). Show that \ast is commutative and associative. Find the binary element for \ast \text{ on } A, if any.     [CBSE 2017]

Answer:

A = R  \times  R

\text{For the } '  \ast  ' \text{ to be commutative, } p  \ast  q = p  \ast  q \text{ must be true for all } p, q \text{ belong to } A. 

Let’s check. Note: Here p, q represent the ordered pairs (a, b) \text{ and } (c, d) respectively.

p  \ast  q = (a, b)  \ast  (c, d) = (a + c, b + d) \\ \\ q  \ast  p = (c, d)  \ast  (a, b) = (c + a, d + b) = (a + c, b + d) \Rightarrow  p  \ast  q = q  \ast  p \\ \\ \Rightarrow \text{ (binary operation } \ast  \text{is commutative) }  

\text{For the } '  \ast ' \text{ to be associative, } a  \ast  (b  \ast  c) = (a  \ast  b)  \ast  c \text{ must hold for every } a, b, c \in A. \text{ Here } r = (e, f) 

p  \ast  (q  \ast  r) = (a, b)  \ast  ((c, d)  \ast  (e, f)) = (a, b)  \ast  (c + e, d + f) = (a + c + e, b + d + f) \\ \\ (p  \ast  q)  \ast  r = ((a, b)  \ast  (c, d))  \ast  (e, f) = (a + c, b + d)  \ast  (e, f) = (a + c + e, b + d + f) \\ \\ \Rightarrow  p  \ast  (q  \ast  r) = (p  \ast  q)  \ast  r \text{ (Associative) }   

Binary elements:  

\text{Identity Element: Given a binary operation } \ast : A  \times  A  \rightarrow  A, \text{ an element } e \in A, \text{ if it exists, is called an identity of the operation } \ast , \text{ if } p \ast e = p = e \ast p \forall p \in A. 

\text{Here } p = (a, b) \text{ and } e = (x, y).

\text{For the element } e \text{ to exist, }  (a, b)  \ast  (x, y) = (a, b)\Rightarrow  (a + x, b + y) = (a, b)\Rightarrow  a + x = a, b + y = b

\text{Since, ordered pairs are only equal when both the first and second terms are equal } \Rightarrow  x = 0, y = 0

\text{Hence the identity element } e = (x, y) = (0, 0) 

\text{Given a binary operation }  \ast  : A  \times  A  \rightarrow  A \text{ with the identity element } e \text{ in } A, \text{ an element } a \text{ in  } \\ \\  A\text{ is said to be invertible with respect to the operation, if there exists an element } b \text{ in } \\ \\   A \text{ such that } a  \ast  b = e = b  \ast  a \text{ and } b \text{ is called the inverse of } a \text{ and is denoted by } a^{ -1}. 

\text{Let } i = (r, s) \text{ be the inverse of } p = (a, b) \text{ in } A. 

\text{Therefore, } (r, s)  \ast  (a, b) = e = (0, 0)\Rightarrow  (r + a, s + b) = (0, 0) \\ \\ \therefore r + a = 0, r + b = 0  \Rightarrow r = - a, s = - b 

\text{Therefore, } (r, s) = ( - a, - b) \text{ is the inverse pair. }