\displaystyle \textbf{Question 1: }\text{Let }\ast\text{ be a binary operation on }Z\text{ defined by }a\ast b=a+b-4,
\displaystyle \text{for all }a,b\in Z.
\displaystyle \text{(i) Show that }\ast\text{ is both commutative and associative.}
\displaystyle \text{(ii) Find the identity element in }Z.
\displaystyle \text{(iii) Find the invertible elements in }Z.
\displaystyle \text{Answer:}

\displaystyle \text{(i) Commutativity:}
\displaystyle \text{Let }a,b\in Z.
\displaystyle a\ast b=a+b-4=b+a-4=b\ast a.
\displaystyle \therefore \ast\text{ is commutative on }Z.
\displaystyle \text{Associativity:}
\displaystyle \text{Let }a,b,c\in Z.
\displaystyle a\ast(b\ast c)=a\ast(b+c-4)=a+b+c-4-4=a+b+c-8.
\displaystyle (a\ast b)\ast c=(a+b-4)\ast c=a+b-4+c-4=a+b+c-8.
\displaystyle \therefore a\ast(b\ast c)=(a\ast b)\ast c,\text{ for all }a,b,c\in Z.
\displaystyle \therefore \ast\text{ is associative on }Z.

\displaystyle \text{(ii) Let }e\in Z\text{ be the identity element with respect to }\ast.
\displaystyle \text{Then }a\ast e=a=e\ast a,\text{ for all }a\in Z.
\displaystyle a\ast e=a.
\displaystyle a+e-4=a.
\displaystyle \therefore e=4.
\displaystyle e\ast a=a.
\displaystyle e+a-4=a.
\displaystyle \therefore e=4.
\displaystyle \therefore 4\text{ is the identity element in }Z\text{ with respect to }\ast.

\displaystyle \text{(iii) Let }b\in Z\text{ be the inverse of }a\in Z.
\displaystyle \text{Then }a\ast b=e=b\ast a.
\displaystyle a\ast b=4.
\displaystyle a+b-4=4.
\displaystyle a+b=8.
\displaystyle \therefore b=8-a.
\displaystyle \text{Since }a\in Z,\;8-a\in Z.
\displaystyle \therefore \text{ every element }a\in Z\text{ is invertible and its inverse is }8-a.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Let }\ast\text{ be a binary operation on }Q_0\text{ (the set of non-zero}
\displaystyle \text{rational numbers) defined by }a\ast b=\frac{3ab}{5},\text{ for all }a,b\in Q_0.
\displaystyle \text{Show that }\ast\text{ is commutative as well as associative. Also, find its}
\displaystyle \text{identity element, if it exists.}
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle \text{Let }a,b\in Q_0.
\displaystyle a\ast b=\frac{3ab}{5}=\frac{3ba}{5}=b\ast a.
\displaystyle \therefore \ast\text{ is commutative on }Q_0.
\displaystyle \text{Associativity:}
\displaystyle \text{Let }a,b,c\in Q_0.
\displaystyle a\ast(b\ast c)=a\ast\left(\frac{3bc}{5}\right)=\frac{3\left(a\cdot\frac{3bc}{5}\right)}{5}=\frac{9abc}{25}. \hspace{1cm}\ldots(i)
\displaystyle (a\ast b)\ast c=\left(\frac{3ab}{5}\right)\ast c=\frac{3\left(\frac{3ab}{5}\right)c}{5}=\frac{9abc}{25}. \hspace{1cm}\ldots(ii)
\displaystyle \text{From (i) and (ii), }a\ast(b\ast c)=(a\ast b)\ast c.
\displaystyle \therefore \ast\text{ is associative on }Q_0.
\displaystyle \text{Identity element:}
\displaystyle \text{Let }e\in Q_0\text{ be the identity element with respect to }\ast.
\displaystyle \text{Then }a\ast e=a=e\ast a,\text{ for all }a\in Q_0.
\displaystyle a\ast e=a.
\displaystyle \frac{3ae}{5}=a.
\displaystyle \therefore e=\frac{5}{3},\qquad(\because a\ne0).
\displaystyle e\ast a=a.
\displaystyle \frac{3ea}{5}=a.
\displaystyle \therefore e=\frac{5}{3}.
\displaystyle \therefore \frac{5}{3}\text{ is the identity element in }Q_0\text{ with respect to }\ast.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Let }\ast\text{ be a binary operation on }Q-\{-1\}\text{ defined by}
\displaystyle a\ast b=a+b+ab,\text{ for all }a,b\in Q-\{-1\}.
\displaystyle \text{(i) Show that }\ast\text{ is both commutative and associative on }Q-\{-1\}.
\displaystyle \text{(ii) Find the identity element in }Q-\{-1\}.
\displaystyle \text{(iii) Show that every element of }Q-\{-1\}\text{ is invertible. Also, find the}
\displaystyle \text{inverse of an arbitrary element.}
\displaystyle \text{Answer:}

\displaystyle \text{(i) Commutativity:}
\displaystyle \text{Let }a,b\in Q-\{-1\}.
\displaystyle a\ast b=a+b+ab=b+a+ba=b\ast a.
\displaystyle \therefore \ast\text{ is commutative on }Q-\{-1\}.
\displaystyle \text{Associativity:}
\displaystyle \text{Let }a,b,c\in Q-\{-1\}.
\displaystyle a\ast(b\ast c)=a\ast(b+c+bc).
\displaystyle =a+b+c+bc+a(b+c+bc).
\displaystyle =a+b+c+ab+ac+bc+abc. \hspace{1cm}\ldots(i)
\displaystyle (a\ast b)\ast c=(a+b+ab)\ast c.
\displaystyle =a+b+ab+c+(a+b+ab)c.
\displaystyle =a+b+c+ab+ac+bc+abc. \hspace{1cm}\ldots(ii)
\displaystyle \text{From (i) and (ii), }a\ast(b\ast c)=(a\ast b)\ast c.
\displaystyle \therefore \ast\text{ is associative on }Q-\{-1\}.

\displaystyle \text{(ii) Let }e\in Q-\{-1\}\text{ be the identity element with respect to }\ast.
\displaystyle \text{Then }a\ast e=a=e\ast a,\text{ for all }a\in Q-\{-1\}.
\displaystyle a\ast e=a.
\displaystyle a+e+ae=a.
\displaystyle e(1+a)=0.
\displaystyle \text{Since }a\ne-1,\;1+a\ne0.
\displaystyle \therefore e=0.
\displaystyle e\ast a=0+a+0=a.
\displaystyle \text{Also, }0\in Q-\{-1\}.
\displaystyle \therefore 0\text{ is the identity element in }Q-\{-1\}\text{ with respect to }\ast.

\displaystyle \text{(iii) Let }a\in Q-\{-1\}\text{ and let }b\in Q-\{-1\}\text{ be its inverse.}
\displaystyle \text{Then }a\ast b=0=b\ast a.
\displaystyle a+b+ab=0.
\displaystyle b(1+a)=-a.
\displaystyle \therefore b=\frac{-a}{a+1},\qquad(\because a\ne-1).
\displaystyle \text{Now, }\frac{-a}{a+1}\in Q.
\displaystyle \text{Also, suppose }\frac{-a}{a+1}=-1.
\displaystyle -a=-(a+1).
\displaystyle -a=-a-1,
\displaystyle \text{which is impossible.}
\displaystyle \therefore \frac{-a}{a+1}\ne-1,\text{ and hence }\frac{-a}{a+1}\in Q-\{-1\}.
\displaystyle \therefore \text{ every element }a\in Q-\{-1\}\text{ is invertible and its inverse is }\frac{-a}{a+1}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }A=R_0\times R,\text{ where }R_0\text{ denotes the set of all non-zero}
\displaystyle \text{real numbers. A binary operation }\bigcirc\text{ is defined on }A\text{ by}
\displaystyle (a,b)\bigcirc(c,d)=(ac,bc+d),\text{ for all }(a,b),(c,d)\in R_0\times R.
\displaystyle \text{(i) Check whether }\bigcirc\text{ is commutative and associative on }A.
\displaystyle \text{(ii) Find the identity element in }A.
\displaystyle \text{(iii) Find the invertible elements in }A.
\displaystyle \text{Answer:}

\displaystyle \text{(i) Commutativity:}
\displaystyle \text{Let }X=(a,b),\;Y=(c,d)\in A.
\displaystyle X\bigcirc Y=(a,b)\bigcirc(c,d)=(ac,bc+d).
\displaystyle Y\bigcirc X=(c,d)\bigcirc(a,b)=(ca,da+b).
\displaystyle \text{In general, }bc+d\ne da+b.
\displaystyle \text{For example, let }X=(1,2)\text{ and }Y=(3,4).
\displaystyle X\bigcirc Y=(1,2)\bigcirc(3,4)=(3,10).
\displaystyle Y\bigcirc X=(3,4)\bigcirc(1,2)=(3,6).
\displaystyle \text{Since }X\bigcirc Y\ne Y\bigcirc X,
\displaystyle \therefore \bigcirc\text{ is not commutative on }A.
\displaystyle \text{Associativity:}
\displaystyle \text{Let }X=(a,b),\;Y=(c,d),\;Z=(e,f)\in A.
\displaystyle X\bigcirc(Y\bigcirc Z)=(a,b)\bigcirc(ce,de+f).
\displaystyle =(ace,bce+de+f). \hspace{1cm}\ldots(i)
\displaystyle (X\bigcirc Y)\bigcirc Z=(ac,bc+d)\bigcirc(e,f).
\displaystyle =(ace,(bc+d)e+f).
\displaystyle =(ace,bce+de+f). \hspace{1cm}\ldots(ii)
\displaystyle \text{From (i) and (ii), }X\bigcirc(Y\bigcirc Z)=(X\bigcirc Y)\bigcirc Z.
\displaystyle \therefore \bigcirc\text{ is associative on }A.

\displaystyle \text{(ii) Let }E=(x,y)\in A\text{ be the identity element with respect to }\bigcirc.
\displaystyle \text{Then }X\bigcirc E=X=E\bigcirc X,\text{ for all }X=(a,b)\in A.
\displaystyle X\bigcirc E=(a,b)\bigcirc(x,y)=(ax,bx+y)=(a,b).
\displaystyle \therefore ax=a\text{ and }bx+y=b.
\displaystyle \text{Since }a\ne0,\;x=1.
\displaystyle \text{Substituting }x=1,\text{ we get }b+y=b,
\displaystyle \therefore y=0.
\displaystyle E\bigcirc X=(1,0)\bigcirc(a,b)=(a,b).
\displaystyle \therefore (1,0)\text{ is the identity element in }A.

\displaystyle \text{(iii) Let }X=(a,b)\in A\text{ and let }F=(m,n)\in A\text{ be its inverse.}
\displaystyle \text{Then }X\bigcirc F=E=F\bigcirc X,\text{ where }E=(1,0).
\displaystyle X\bigcirc F=(a,b)\bigcirc(m,n)=(am,bm+n)=(1,0).
\displaystyle \therefore am=1\text{ and }bm+n=0.
\displaystyle \therefore m=\frac1a.
\displaystyle \therefore n=-bm=-\frac ba.
\displaystyle F\bigcirc X=\left(\frac1a,-\frac ba\right)\bigcirc(a,b).
\displaystyle =\left(1,-\frac ba\cdot a+b\right)=(1,0).
\displaystyle \text{Since }a\ne0,\;\frac1a\in R_0\text{ and }-\frac ba\in R.
\displaystyle \therefore \left(\frac1a,-\frac ba\right)\in A.
\displaystyle \therefore \text{ every element }(a,b)\in A\text{ is invertible, and its inverse is}
\displaystyle \left(\frac1a,-\frac ba\right).
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Let }\circ\text{ be a binary operation on the set }Q_0\text{ of all non-zero}
\displaystyle \text{rational numbers defined by }a\circ b=\frac{ab}{2},\text{ for all }a,b\in Q_0.
\displaystyle \text{(i) Show that }\circ\text{ is both commutative and associative.}
\displaystyle \text{(ii) Find the identity element in }Q_0.
\displaystyle \text{(iii) Find the invertible elements of }Q_0.
\displaystyle \text{Answer:}

\displaystyle \text{(i) Commutativity:}
\displaystyle \text{Let }a,b\in Q_0.
\displaystyle a\circ b=\frac{ab}{2}=\frac{ba}{2}=b\circ a.
\displaystyle \therefore \circ\text{ is commutative on }Q_0.
\displaystyle \text{Associativity:}
\displaystyle \text{Let }a,b,c\in Q_0.
\displaystyle a\circ(b\circ c)=a\circ\left(\frac{bc}{2}\right)=\frac{a\left(\frac{bc}{2}\right)}{2}=\frac{abc}{4}. \hspace{1cm}\ldots(i)
\displaystyle (a\circ b)\circ c=\left(\frac{ab}{2}\right)\circ c=\frac{\left(\frac{ab}{2}\right)c}{2}=\frac{abc}{4}. \hspace{1cm}\ldots(ii)
\displaystyle \text{From (i) and (ii), }a\circ(b\circ c)=(a\circ b)\circ c.
\displaystyle \therefore \circ\text{ is associative on }Q_0.

\displaystyle \text{(ii) Let }e\in Q_0\text{ be the identity element with respect to }\circ.
\displaystyle \text{Then }a\circ e=a=e\circ a,\text{ for all }a\in Q_0.
\displaystyle a\circ e=a.
\displaystyle \frac{ae}{2}=a.
\displaystyle \therefore e=2.
\displaystyle e\circ a=a.
\displaystyle \frac{ea}{2}=a.
\displaystyle \therefore e=2.
\displaystyle \therefore 2\text{ is the identity element in }Q_0\text{ with respect to }\circ.

\displaystyle \text{(iii) Let }a\in Q_0\text{ and let }b\in Q_0\text{ be the inverse of }a.
\displaystyle \text{Then }a\circ b=e=b\circ a,\text{ where }e=2.
\displaystyle a\circ b=2.
\displaystyle \frac{ab}{2}=2.
\displaystyle ab=4.
\displaystyle \therefore b=\frac4a.
\displaystyle \text{Since }a\ne0,\;\frac4a\in Q_0.
\displaystyle \therefore \text{ every element }a\in Q_0\text{ is invertible and its inverse is }\frac4a.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{On }R-\{1\},\text{ a binary operation }\ast\text{ is defined by}
\displaystyle a\ast b=a+b-ab.
\displaystyle \text{Prove that }\ast\text{ is commutative and associative. Find the identity element}
\displaystyle \text{for }\ast\text{ on }R-\{1\}.\text{ Also, prove that every element of }R-\{1\}
\displaystyle \text{is invertible.}
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle \text{Let }a,b\in R-\{1\}.
\displaystyle a\ast b=a+b-ab=b+a-ba=b\ast a.
\displaystyle \therefore \ast\text{ is commutative on }R-\{1\}.
\displaystyle \text{Associativity:}
\displaystyle \text{Let }a,b,c\in R-\{1\}.
\displaystyle a\ast(b\ast c)=a\ast(b+c-bc).
\displaystyle =a+b+c-bc-a(b+c-bc).
\displaystyle =a+b+c-ab-ac-bc+abc. \hspace{1cm}\ldots(i)
\displaystyle (a\ast b)\ast c=(a+b-ab)\ast c.
\displaystyle =a+b-ab+c-(a+b-ab)c.
\displaystyle =a+b+c-ab-ac-bc+abc. \hspace{1cm}\ldots(ii)
\displaystyle \text{From (i) and (ii), }a\ast(b\ast c)=(a\ast b)\ast c.
\displaystyle \therefore \ast\text{ is associative on }R-\{1\}.
\displaystyle \text{Identity element:}
\displaystyle \text{Let }e\in R-\{1\}\text{ be the identity element with respect to }\ast.
\displaystyle \text{Then }a\ast e=a=e\ast a,\text{ for all }a\in R-\{1\}.
\displaystyle a\ast e=a.
\displaystyle a+e-ae=a.
\displaystyle e-ae=0.
\displaystyle e(1-a)=0.
\displaystyle \text{Since }a\ne1,\;1-a\ne0.
\displaystyle \therefore e=0.
\displaystyle e\ast a=0+a-0=a.
\displaystyle \text{Also, }0\in R-\{1\}.
\displaystyle \therefore 0\text{ is the identity element in }R-\{1\}\text{ with respect to }\ast.
\displaystyle \text{Invertible elements:}
\displaystyle \text{Let }a\in R-\{1\}\text{ and let }b\in R-\{1\}\text{ be its inverse.}
\displaystyle \text{Then }a\ast b=0=b\ast a.
\displaystyle a\ast b=0.
\displaystyle a+b-ab=0.
\displaystyle b(1-a)=-a.
\displaystyle \therefore b=\frac{-a}{1-a}=\frac{a}{a-1},\qquad(\because a\ne1).
\displaystyle \text{Now, }\frac{a}{a-1}\in R.
\displaystyle \text{Also, suppose }\frac{a}{a-1}=1.
\displaystyle a=a-1,
\displaystyle \text{which is impossible.}
\displaystyle \therefore \frac{a}{a-1}\ne1,\text{ and hence }\frac{a}{a-1}\in R-\{1\}.
\displaystyle a\ast\frac{a}{a-1}=a+\frac{a}{a-1}-\frac{a^2}{a-1}=0.
\displaystyle \frac{a}{a-1}\ast a=\frac{a}{a-1}+a-\frac{a^2}{a-1}=0.
\displaystyle \therefore \text{ every element }a\in R-\{1\}\text{ is invertible and its inverse is }\frac{a}{a-1}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Let }R_0\text{ denote the set of all non-zero real numbers and let}
\displaystyle A=R_0\times R_0.\text{ A binary operation }\ast\text{ on }A\text{ is defined by}
\displaystyle (a,b)\ast(c,d)=(ac,bd),\text{ for all }(a,b),(c,d)\in A.
\displaystyle \text{(i) Show that }\ast\text{ is both commutative and associative on }A.
\displaystyle \text{(ii) Find the identity element in }A.
\displaystyle \text{(iii) Find the invertible elements in }A.
\displaystyle \text{Answer:}

\displaystyle \text{(i) Commutativity:}
\displaystyle \text{Let }(a,b),(c,d)\in A.
\displaystyle (a,b)\ast(c,d)=(ac,bd)=(ca,db)=(c,d)\ast(a,b).
\displaystyle \therefore \ast\text{ is commutative on }A.
\displaystyle \text{Associativity:}
\displaystyle \text{Let }(a,b),(c,d),(e,f)\in A.
\displaystyle (a,b)\ast\big((c,d)\ast(e,f)\big)=(a,b)\ast(ce,df).
\displaystyle =(ace,bdf). \hspace{1cm}\ldots(i)
\displaystyle \big((a,b)\ast(c,d)\big)\ast(e,f)=(ac,bd)\ast(e,f).
\displaystyle =(ace,bdf). \hspace{1cm}\ldots(ii)
\displaystyle \text{From (i) and (ii),}
\displaystyle (a,b)\ast\big((c,d)\ast(e,f)\big)=\big((a,b)\ast(c,d)\big)\ast(e,f).
\displaystyle \therefore \ast\text{ is associative on }A.

\displaystyle \text{(ii) Let }E=(x,y)\in A\text{ be the identity element with respect to }\ast.
\displaystyle \text{Then }(a,b)\ast(x,y)=(a,b)=(x,y)\ast(a,b),
\displaystyle \text{for all }(a,b)\in A.
\displaystyle (a,b)\ast(x,y)=(ax,by)=(a,b).
\displaystyle \therefore ax=a\text{ and }by=b.
\displaystyle \text{Since }a\ne0\text{ and }b\ne0,\;x=1\text{ and }y=1.
\displaystyle (1,1)\ast(a,b)=(a,b).
\displaystyle \therefore (1,1)\text{ is the identity element in }A.

\displaystyle \text{(iii) Let }(a,b)\in A\text{ and let }(m,n)\in A\text{ be its inverse.}
\displaystyle \text{Then }(a,b)\ast(m,n)=(1,1)=(m,n)\ast(a,b).
\displaystyle (a,b)\ast(m,n)=(am,bn)=(1,1).
\displaystyle \therefore am=1\text{ and }bn=1.
\displaystyle \therefore m=\frac1a\text{ and }n=\frac1b.
\displaystyle (m,n)\ast(a,b)=\left(\frac1a,\frac1b\right)\ast(a,b).
\displaystyle =\left(\frac1a\cdot a,\frac1b\cdot b\right)=(1,1).
\displaystyle \text{Since }a,b\ne0,\;\frac1a,\frac1b\in R_0.
\displaystyle \therefore \left(\frac1a,\frac1b\right)\in A.
\displaystyle \therefore \text{ every element }(a,b)\in A\text{ is invertible and its inverse is}
\displaystyle \left(\frac1a,\frac1b\right).
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Let }\ast\text{ be the binary operation on }N\text{ defined by}
\displaystyle a\ast b=\text{H.C.F. of }a\text{ and }b.\text{ Does there exist an identity element}
\displaystyle \text{for this binary operation on }N?
\displaystyle \text{Answer:}
\displaystyle \text{Let }e\in N\text{ be the identity element with respect to }\ast.
\displaystyle \text{Then }a\ast e=a=e\ast a,\text{ for all }a\in N.
\displaystyle \text{Since }a\ast e=\text{H.C.F. of }a\text{ and }e,
\displaystyle \text{we must have }\mathrm{HCF}(a,e)=a,\text{ for every }a\in N.
\displaystyle \text{Thus, }a\text{ divides }e,\text{ for every }a\in N.
\displaystyle \text{Hence }e\text{ must be a common multiple of all positive integers.}
\displaystyle \text{But no positive integer is a multiple of every positive integer.}
\displaystyle \therefore \text{ no identity element exists for this binary operation on }N.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Let }A=R\times R\text{ and }\ast\text{ be a binary operation on }A
\displaystyle \text{defined by }(a,b)\ast(c,d)=(a+c,b+d).
\displaystyle \text{Show that }\ast\text{ is commutative and associative. Find the identity element}
\displaystyle \text{for }\ast\text{ on }A,\text{ if any.}\qquad[\text{CBSE 2017}]
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle \text{Let }p=(a,b)\text{ and }q=(c,d)\in A.
\displaystyle p\ast q=(a,b)\ast(c,d)=(a+c,b+d).
\displaystyle q\ast p=(c,d)\ast(a,b)=(c+a,d+b).
\displaystyle =(a+c,b+d).
\displaystyle \therefore p\ast q=q\ast p,\text{ for all }p,q\in A.
\displaystyle \therefore \ast\text{ is commutative on }A.
\displaystyle \text{Associativity:}
\displaystyle \text{Let }p=(a,b),\;q=(c,d)\text{ and }r=(e,f)\in A.
\displaystyle p\ast(q\ast r)=(a,b)\ast\big((c,d)\ast(e,f)\big).
\displaystyle =(a,b)\ast(c+e,d+f).
\displaystyle =(a+c+e,b+d+f). \hspace{1cm}\ldots(i)
\displaystyle (p\ast q)\ast r=\big((a,b)\ast(c,d)\big)\ast(e,f).
\displaystyle =(a+c,b+d)\ast(e,f).
\displaystyle =(a+c+e,b+d+f). \hspace{1cm}\ldots(ii)
\displaystyle \text{From (i) and (ii), }p\ast(q\ast r)=(p\ast q)\ast r.
\displaystyle \therefore \ast\text{ is associative on }A.
\displaystyle \text{Identity element:}
\displaystyle \text{Let }E=(x,y)\in A\text{ be the identity element with respect to }\ast.
\displaystyle \text{Then }p\ast E=p=E\ast p,\text{ for all }p=(a,b)\in A.
\displaystyle p\ast E=(a,b)\ast(x,y)=(a+x,b+y)=(a,b).
\displaystyle \therefore a+x=a\text{ and }b+y=b.
\displaystyle \therefore x=0\text{ and }y=0.
\displaystyle E=(0,0).
\displaystyle E\ast p=(0,0)\ast(a,b)=(a,b).
\displaystyle \therefore (0,0)\text{ is the identity element in }A\text{ with respect to }\ast.
\displaystyle \\

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