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 $Z.$

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

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

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

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

$b = 8 - a \in Z$

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

$\\$

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

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 $\ast$

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

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

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

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

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

$\\$

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.

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

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

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

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

$b (1 + a) = - a Q - \{-1\}$

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

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

$\\$

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 $A.$

(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,

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

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

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

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

$ax = a x = 1$

$\text{And } bx + y = b$

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

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

$xa = a$

$x = 1$

$\text{And } ya + b = b$

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

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

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

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

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

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

$\text{And } bm + n = 0$

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

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

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

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

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

$\\$

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 $Q_0$

(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:

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

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

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

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

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

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

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

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

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

$e = 2 \in Q_0, \ \forall \ a \in Q_0$

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

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

$a \circ b = e = b \circ a$

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

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

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

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

$\\$

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.

$\text{(i) We are given the set } R - \{- 1\}.$

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.

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

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

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

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

$\\$

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 $A$

(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 $A$

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

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

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

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

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

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

$\\$

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 $N$?

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

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

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.

$\\$

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]

$A = R \times R$

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

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:

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

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

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

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

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

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

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

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

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