\displaystyle \textbf{Question 1: }\text{Let }\ast\text{ be a binary operation on }N\text{ defined by }a\ast b=
\displaystyle \mathrm{LCM}(a,b),\text{ for all }a,b\in N.
\displaystyle \text{(i) Find }2\ast4,\;3\ast5,\;1\ast6.
\displaystyle \text{(ii) Check the commutativity and associativity of }\ast\text{ on }N.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }2\ast4=\mathrm{LCM}(2,4)=4.
\displaystyle 3\ast5=\mathrm{LCM}(3,5)=15.
\displaystyle 1\ast6=\mathrm{LCM}(1,6)=6.
\displaystyle \therefore 2\ast4=4,\;3\ast5=15,\text{ and }1\ast6=6.
\displaystyle \text{(ii) Commutativity:}
\displaystyle a\ast b=\mathrm{LCM}(a,b)=\mathrm{LCM}(b,a)=b\ast a,\text{ for all }a,b\in N.
\displaystyle \therefore \ast\text{ is commutative on }N.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=\mathrm{LCM}(\mathrm{LCM}(a,b),c)=\mathrm{LCM}(a,b,c).
\displaystyle a\ast(b\ast c)=\mathrm{LCM}(a,\mathrm{LCM}(b,c))=\mathrm{LCM}(a,b,c).
\displaystyle \therefore (a\ast b)\ast c=a\ast(b\ast c)\text{ for all }a,b,c\in N.
\displaystyle \therefore \ast\text{ is associative on }N.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Determine which of the following binary operations are}
\displaystyle \text{associative and which are commutative.}
\displaystyle \text{(i) }\ast\text{ on }N\text{ defined by }a\ast b=1,\text{ for all }a,b\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=1=b\ast a,\text{ for all }a,b\in N.
\displaystyle \therefore \ast\text{ is commutative on }N.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=1\ast c=1.
\displaystyle a\ast(b\ast c)=a\ast1=1.
\displaystyle \therefore (a\ast b)\ast c=a\ast(b\ast c)\text{ for all }a,b,c\in N.
\displaystyle \therefore \ast\text{ is associative on }N.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{(ii) }\ast\text{ on }Q\text{ defined by }a\ast b=\frac{a+b}{2},
\displaystyle \text{for all }a,b\in Q.\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=\frac{a+b}{2}=\frac{b+a}{2}=b\ast a,\text{ for all }a,b\in Q.
\displaystyle \therefore \ast\text{ is commutative on }Q.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=\left(\frac{a+b}{2}\right)\ast c=\frac{\frac{a+b}{2}+c}{2}=\frac{a+b+2c}{4}. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a\ast\left(\frac{b+c}{2}\right)=\frac{a+\frac{b+c}{2}}{2}=\frac{2a+b+c}{4}. \hspace{1cm}\ldots(ii)
\displaystyle \text{Since }\frac{a+b+2c}{4}\ne\frac{2a+b+c}{4}\text{ in general,}
\displaystyle (a\ast b)\ast c\ne a\ast(b\ast c).
\displaystyle \therefore \ast\text{ is not associative on }Q.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Let }A\text{ be any set containing more than one element. Let }\ast\text{ be a}
\displaystyle \text{binary operation on }A\text{ defined by }a\ast b=b,\text{ for all }a,b\in A.
\displaystyle \text{Is }\ast\text{ commutative or associative on }A?
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=b\text{ and }b\ast a=a.
\displaystyle \text{Since }A\text{ contains more than one element, }a\ne b\text{ for some }a,b\in A.
\displaystyle \therefore a\ast b\ne b\ast a.
\displaystyle \therefore \ast\text{ is not commutative on }A.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=b\ast c=c.
\displaystyle a\ast(b\ast c)=a\ast c=c.
\displaystyle \therefore (a\ast b)\ast c=a\ast(b\ast c)\text{ for all }a,b,c\in A.
\displaystyle \therefore \ast\text{ is associative on }A.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Check the commutativity and associativity of the following}
\displaystyle \text{binary operation:}
\displaystyle \text{(i) }\ast\text{ on }Z\text{ defined by }a\ast b=a+b+ab,\text{ for all }a,b\in Z.
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=a+b+ab.
\displaystyle b\ast a=b+a+ba=a+b+ab=a\ast b.
\displaystyle \therefore \ast\text{ is commutative on }Z.
\displaystyle \text{Associativity:}
\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(i)
\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(ii)
\displaystyle \text{From (i) and (ii), }(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 \\

\displaystyle \textbf{Question 4: }\text{Check the commutativity and associativity of the following}
\displaystyle \text{binary operation:}
\displaystyle \text{(ii) }\ast\text{ on }N\text{ defined by }a\ast b=2^{ab},\text{ for all }a,b\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=2^{ab}=2^{ba}=b\ast a.
\displaystyle \therefore \ast\text{ is commutative on }N.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=(2^{ab})\ast c=2^{(2^{ab})c}=2^{c\,2^{ab}}. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a\ast(2^{bc})=2^{a(2^{bc})}=2^{a\,2^{bc}}. \hspace{1cm}\ldots(ii)
\displaystyle \text{Since }2^{c\,2^{ab}}\ne2^{a\,2^{bc}}\text{ in general,}
\displaystyle (a\ast b)\ast c\ne a\ast(b\ast c).
\displaystyle \therefore \ast\text{ is not associative on }N.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Check the commutativity and associativity of the following}
\displaystyle \text{binary operation:}
\displaystyle \text{(iii) }\ast\text{ on }Q\text{ defined by }a\ast b=a-b,\text{ for all }a,b\in Q.
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=a-b,\qquad b\ast a=b-a.
\displaystyle \text{Since }a-b\ne b-a\text{ in general,}
\displaystyle \therefore \ast\text{ is not commutative on }Q.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=(a-b)-c=a-b-c. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a-(b-c)=a-b+c. \hspace{1cm}\ldots(ii)
\displaystyle \text{Since }a-b-c\ne a-b+c\text{ in general,}
\displaystyle (a\ast b)\ast c\ne a\ast(b\ast c).
\displaystyle \therefore \ast\text{ is not associative on }Q.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Check the commutativity and associativity of the following}
\displaystyle \text{binary operation:}
\displaystyle \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{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\odot b=a^2+b^2=b^2+a^2=b\odot a.
\displaystyle \therefore \odot\text{ is commutative on }Q.
\displaystyle \text{Associativity:}
\displaystyle (a\odot b)\odot c=(a^2+b^2)\odot c.
\displaystyle =(a^2+b^2)^2+c^2.
\displaystyle =a^4+b^4+2a^2b^2+c^2. \hspace{1cm}\ldots(i)
\displaystyle a\odot(b\odot c)=a\odot(b^2+c^2).
\displaystyle =a^2+(b^2+c^2)^2.
\displaystyle =a^2+b^4+c^4+2b^2c^2. \hspace{1cm}\ldots(ii)
\displaystyle \text{Since }a^4+b^4+2a^2b^2+c^2\ne a^2+b^4+c^4+2b^2c^2\text{ in general,}
\displaystyle (a\odot b)\odot c\ne a\odot(b\odot c).
\displaystyle \therefore \odot\text{ is not associative on }Q.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Check the commutativity and associativity of the following}
\displaystyle \text{binary operation:}
\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{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\bigcirc b=\frac{ab}{2}=\frac{ba}{2}=b\bigcirc a.
\displaystyle \therefore \bigcirc\text{ is commutative on }Q.
\displaystyle \text{Associativity:}
\displaystyle (a\bigcirc b)\bigcirc c=\left(\frac{ab}{2}\right)\bigcirc c.
\displaystyle =\frac{\left(\frac{ab}{2}\right)c}{2}=\frac{abc}{4}. \hspace{1cm}\ldots(i)
\displaystyle a\bigcirc(b\bigcirc c)=a\bigcirc\left(\frac{bc}{2}\right).
\displaystyle =\frac{a\left(\frac{bc}{2}\right)}{2}=\frac{abc}{4}. \hspace{1cm}\ldots(ii)
\displaystyle \text{From (i) and (ii), }(a\bigcirc b)\bigcirc c=a\bigcirc(b\bigcirc c).
\displaystyle \therefore \bigcirc\text{ is associative on }Q.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Check the commutativity and associativity of the following}
\displaystyle \text{binary operation:}
\displaystyle \text{(vi) }\ast\text{ on }Q\text{ defined by }a\ast b=ab^2,\text{ for all }a,b\in Q.
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=ab^2,\qquad b\ast a=ba^2.
\displaystyle \text{Since }ab^2\ne ba^2\text{ in general,}
\displaystyle \therefore \ast\text{ is not commutative on }Q.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=(ab^2)\ast c=ab^2c^2. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a\ast(bc^2)=a(bc^2)^2=ab^2c^4. \hspace{1cm}\ldots(ii)
\displaystyle \text{Since }ab^2c^2\ne ab^2c^4\text{ in general,}
\displaystyle (a\ast b)\ast c\ne a\ast(b\ast c).
\displaystyle \therefore \ast\text{ is not associative on }Q.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Check the commutativity and associativity of the following}
\displaystyle \text{binary operation:}
\displaystyle \text{(vii) }\ast\text{ on }Q\text{ defined by }a\ast b=a+ab,\text{ for all }a,b\in Q.
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=a+ab,\qquad b\ast a=b+ab.
\displaystyle \text{Since }a+ab\ne b+ab\text{ in general,}
\displaystyle \therefore \ast\text{ is not commutative on }Q.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=(a+ab)\ast c=(a+ab)+(a+ab)c.
\displaystyle =a+ab+ac+abc. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a\ast(b+bc)=a+a(b+bc).
\displaystyle =a+ab+abc. \hspace{1cm}\ldots(ii)
\displaystyle \text{Since }a+ab+ac+abc\ne a+ab+abc\text{ in general,}
\displaystyle (a\ast b)\ast c\ne a\ast(b\ast c).
\displaystyle \therefore \ast\text{ is not associative on }Q.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Check the commutativity and associativity of the following}
\displaystyle \text{binary operation:}
\displaystyle \text{(viii) }\ast\text{ on }R\text{ defined by }a\ast b=a+b-7,\text{ for all }a,b\in R.
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=a+b-7=b+a-7=b\ast a.
\displaystyle \therefore \ast\text{ is commutative on }R.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=(a+b-7)\ast c=(a+b-7)+c-7.
\displaystyle =a+b+c-14. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a\ast(b+c-7)=a+(b+c-7)-7.
\displaystyle =a+b+c-14. \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.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Check the commutativity and associativity of the following}
\displaystyle \text{binary operation:}
\displaystyle \text{(ix) }\ast\text{ on }Q\text{ defined by }a\ast b=(a-b)^2,\text{ for all }a,b\in Q.
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=(a-b)^2=(b-a)^2=b\ast a.
\displaystyle \therefore \ast\text{ is commutative on }Q.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=((a-b)^2)\ast c=\left((a-b)^2-c\right)^2. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a\ast((b-c)^2)=\left(a-(b-c)^2\right)^2. \hspace{1cm}\ldots(ii)
\displaystyle \text{Since }\left((a-b)^2-c\right)^2\ne\left(a-(b-c)^2\right)^2\text{ in general,}
\displaystyle (a\ast b)\ast c\ne a\ast(b\ast c).
\displaystyle \therefore \ast\text{ is not associative on }Q.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Check the commutativity and associativity of the following}
\displaystyle \text{binary operation:}
\displaystyle \text{(x) }\ast\text{ on }Q\text{ defined by }a\ast b=ab+1,\text{ for all }a,b\in Q.
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=ab+1=ba+1=b\ast a.
\displaystyle \therefore \ast\text{ is commutative on }Q.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=(ab+1)\ast c=(ab+1)c+1.
\displaystyle =abc+c+1. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a\ast(bc+1)=a(bc+1)+1.
\displaystyle =abc+a+1. \hspace{1cm}\ldots(ii)
\displaystyle \text{Since }abc+c+1\ne abc+a+1\text{ in general,}
\displaystyle (a\ast b)\ast c\ne a\ast(b\ast c).
\displaystyle \therefore \ast\text{ is not associative on }Q.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Check the commutativity and associativity of the following}
\displaystyle \text{binary operation:}
\displaystyle \text{(xi) }\ast\text{ on }N\text{ defined by }a\ast b=a^b,\text{ for all }a,b\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=a^b,\qquad b\ast a=b^a.
\displaystyle \text{Since }a^b\ne b^a\text{ in general,}
\displaystyle \therefore \ast\text{ is not commutative on }N.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=(a^b)\ast c=(a^b)^c=a^{bc}. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a\ast(b^c)=a^{\,b^c}. \hspace{1cm}\ldots(ii)
\displaystyle \text{Since }a^{bc}\ne a^{\,b^c}\text{ in general,}
\displaystyle (a\ast b)\ast c\ne a\ast(b\ast c).
\displaystyle \therefore \ast\text{ is not associative on }N.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Check the commutativity and associativity of the following}
\displaystyle \text{binary operation:}
\displaystyle \text{(xii) }\ast\text{ on }Z\text{ defined by }a\ast b=a-b,\text{ for all }a,b\in Z.
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=a-b,\qquad b\ast a=b-a.
\displaystyle \text{Since }a-b\ne b-a\text{ in general,}
\displaystyle \therefore \ast\text{ is not commutative on }Z.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=(a-b)\ast c=(a-b)-c=a-b-c. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a\ast(b-c)=a-(b-c)=a-b+c. \hspace{1cm}\ldots(ii)
\displaystyle \text{Since }a-b-c\ne a-b+c\text{ in general,}
\displaystyle (a\ast b)\ast c\ne a\ast(b\ast c).
\displaystyle \therefore \ast\text{ is not associative on }Z.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Check the commutativity and associativity of the following}
\displaystyle \text{binary operation:}
\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{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=\frac{ab}{4}=\frac{ba}{4}=b\ast a.
\displaystyle \therefore \ast\text{ is commutative on }Q.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=\left(\frac{ab}{4}\right)\ast c=\frac{\left(\frac{ab}{4}\right)c}{4}=\frac{abc}{16}. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a\ast\left(\frac{bc}{4}\right)=\frac{a\left(\frac{bc}{4}\right)}{4}=\frac{abc}{16}. \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.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Check the commutativity and associativity of the following}
\displaystyle \text{binary operation:}
\displaystyle \text{(xiv) }\ast\text{ on }Z\text{ defined by }a\ast b=a+b-ab,\text{ for all }a,b\in Z.
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=a+b-ab=b+a-ba=b\ast a.
\displaystyle \therefore \ast\text{ is commutative on }Z.
\displaystyle \text{Associativity:}
\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(i)
\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(ii)
\displaystyle \text{From (i) and (ii), }(a\ast b)\ast c=a\ast(b\ast c).
\displaystyle \therefore \ast\text{ is associative on }Z.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Check the commutativity and associativity of the following}
\displaystyle \text{binary operation:}
\displaystyle \text{(xv) }\ast\text{ on }N\text{ defined by }a\ast b=\gcd(a,b),\text{ for all }a,b\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=\gcd(a,b)=\gcd(b,a)=b\ast a.
\displaystyle \therefore \ast\text{ is commutative on }N.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=\gcd(\gcd(a,b),c)=\gcd(a,b,c). \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=\gcd(a,\gcd(b,c))=\gcd(a,b,c). \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 }N.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If the binary operation }\circ\text{ is defined by }a\circ b=a+b-ab
\displaystyle \text{on the set }Q-\{-1\}\text{ of all rational numbers other than }-1,
\displaystyle \text{show that }\circ\text{ is commutative on }Q-\{-1\}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }a,b\in Q-\{-1\}.
\displaystyle a\circ b=a+b-ab.
\displaystyle b\circ a=b+a-ba=a+b-ab.
\displaystyle \therefore a\circ b=b\circ a,\text{ for all }a,b\in Q-\{-1\}.
\displaystyle \therefore \circ\text{ is commutative on }Q-\{-1\}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Show that the binary operation }\ast\text{ on }Z\text{ defined by}
\displaystyle a\ast b=3a+7b\text{ is not commutative.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }a=1\text{ and }b=2.
\displaystyle 1\ast2=3(1)+7(2)=3+14=17.
\displaystyle 2\ast1=3(2)+7(1)=6+7=13.
\displaystyle \text{Since }1\ast2\ne2\ast1,
\displaystyle \therefore \ast\text{ is not commutative on }Z.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{On the set }Z\text{ of integers, a binary operation }\ast\text{ is defined by}
\displaystyle a\ast b=ab+1,\text{ for all }a,b\in Z.\text{ Prove that }\ast\text{ is not associative on }Z.
\displaystyle \text{Answer:}
\displaystyle \text{Let }a,b,c\in Z.
\displaystyle a\ast(b\ast c)=a\ast(bc+1)=a(bc+1)+1=abc+a+1.
\displaystyle (a\ast b)\ast c=(ab+1)\ast c=(ab+1)c+1=abc+c+1.
\displaystyle \text{Take }a=1,\;b=2,\;c=3.
\displaystyle 1\ast(2\ast3)=1\ast7=8.
\displaystyle (1\ast2)\ast3=3\ast3=10.
\displaystyle \text{Since }1\ast(2\ast3)\ne(1\ast2)\ast3,
\displaystyle \therefore \ast\text{ is not associative on }Z.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Let }S\text{ be the set of all real numbers except }-1\text{ and let }\ast
\displaystyle \text{be defined by }a\ast b=a+b+ab,\text{ for all }a,b\in S.
\displaystyle \text{Determine whether }\ast\text{ is a binary operation on }S.
\displaystyle \text{If yes, check its commutativity and associativity. Also, solve}
\displaystyle (2\ast x)\ast3=7.
\displaystyle \text{Answer:}
\displaystyle \text{Let }a,b\in S.
\displaystyle \text{Suppose }a\ast b=-1.
\displaystyle a+b+ab=-1.
\displaystyle a+b+ab+1=0.
\displaystyle (a+1)(b+1)=0.
\displaystyle \therefore a=-1\text{ or }b=-1,
\displaystyle \text{which is impossible since }a,b\in S.
\displaystyle \therefore a\ast b\ne-1,\text{ and hence }a\ast b\in S.
\displaystyle \therefore \ast\text{ is a binary operation on }S.
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=a+b+ab=b+a+ba=b\ast a.
\displaystyle \therefore \ast\text{ is commutative on }S.
\displaystyle \text{Associativity:}
\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(i)
\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(ii)
\displaystyle \text{From (i) and (ii), }(a\ast b)\ast c=a\ast(b\ast c).
\displaystyle \therefore \ast\text{ is associative on }S.
\displaystyle \text{Now, }(2\ast x)\ast3=7.
\displaystyle 2\ast x=2+x+2x=2+3x.
\displaystyle (2+3x)\ast3=7.
\displaystyle (2+3x)+3+3(2+3x)=7.
\displaystyle 11+12x=7.
\displaystyle 12x=-4.
\displaystyle x=-\frac13.
\displaystyle \text{Since }-\frac13\in S,\text{ the required solution is }x=-\frac13.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{On }Q,\text{ the set of all rational numbers, }\ast\text{ is defined by}
\displaystyle a\ast b=\frac{a-b}{2}.\text{ Show that }\ast\text{ is not associative.}
\displaystyle \text{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.
\displaystyle \text{To check associativity, we compare }(a\ast b)\ast c\text{ and }a\ast(b\ast c).
\displaystyle (a\ast b)\ast c=\left(\frac{a-b}{2}\right)\ast c.
\displaystyle =\frac{\frac{a-b}{2}-c}{2}=\frac{a-b-2c}{4}. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a\ast\left(\frac{b-c}{2}\right).
\displaystyle =\frac{a-\frac{b-c}{2}}{2}=\frac{2a-b+c}{4}. \hspace{1cm}\ldots(ii)
\displaystyle \text{Since }\frac{a-b-2c}{4}\ne\frac{2a-b+c}{4}\text{ in general,}
\displaystyle (a\ast b)\ast c\ne a\ast(b\ast c).
\displaystyle \therefore \ast\text{ is not associative on }Q.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{On }Z,\text{ the set of all integers, a binary operation }\ast\text{ is defined by}
\displaystyle a\ast b=a+3b-4.\text{ Prove that }\ast\text{ is neither commutative nor associative on }Z.
\displaystyle \text{Answer:}
\displaystyle \text{Commutativity:}
\displaystyle a\ast b=a+3b-4,\qquad b\ast a=b+3a-4.
\displaystyle \text{Take }a=1,\;b=2.
\displaystyle 1\ast2=1+3(2)-4=3.
\displaystyle 2\ast1=2+3(1)-4=1.
\displaystyle \text{Since }1\ast2\ne2\ast1,
\displaystyle \therefore \ast\text{ is not commutative on }Z.
\displaystyle \text{Associativity:}
\displaystyle (a\ast b)\ast c=(a+3b-4)\ast c.
\displaystyle =(a+3b-4)+3c-4=a+3b+3c-8. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a\ast(b+3c-4).
\displaystyle =a+3(b+3c-4)-4=a+3b+9c-16. \hspace{1cm}\ldots(ii)
\displaystyle \text{Since }a+3b+3c-8\ne a+3b+9c-16\text{ in general,}
\displaystyle (a\ast b)\ast c\ne a\ast(b\ast c).
\displaystyle \therefore \ast\text{ is not associative on }Z.
\displaystyle \therefore \ast\text{ is neither commutative nor associative on }Z.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{On the set }Q\text{ of all rational numbers, a binary operation }\ast
\displaystyle \text{is defined by }a\ast b=\frac{ab}{5}.\text{ Prove that }\ast\text{ is associative on }Q.
\displaystyle \text{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.
\displaystyle \text{To check associativity, we compare }(a\ast b)\ast c\text{ and }a\ast(b\ast c).
\displaystyle (a\ast b)\ast c=\left(\frac{ab}{5}\right)\ast c.
\displaystyle =\frac{\left(\frac{ab}{5}\right)c}{5}=\frac{abc}{25}. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a\ast\left(\frac{bc}{5}\right).
\displaystyle =\frac{a\left(\frac{bc}{5}\right)}{5}=\frac{abc}{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.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The binary operation }\ast\text{ is defined by }a\ast b=\frac{ab}{7}
\displaystyle \text{on the set }Q\text{ of all rational numbers. Show that }\ast\text{ is associative.}
\displaystyle \text{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.
\displaystyle \text{To check associativity, we compare }(a\ast b)\ast c\text{ and }a\ast(b\ast c).
\displaystyle (a\ast b)\ast c=\left(\frac{ab}{7}\right)\ast c.
\displaystyle =\frac{\left(\frac{ab}{7}\right)c}{7}=\frac{abc}{49}. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a\ast\left(\frac{bc}{7}\right).
\displaystyle =\frac{a\left(\frac{bc}{7}\right)}{7}=\frac{abc}{49}. \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.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{On }Q,\text{ the set of all rational numbers, a binary operation }\ast
\displaystyle \text{is defined by }a\ast b=\frac{a+b}{2}.\text{ Show that }\ast\text{ is not associative on }Q.
\displaystyle \text{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.
\displaystyle \text{To check associativity, we compare }(a\ast b)\ast c\text{ and }a\ast(b\ast c).
\displaystyle (a\ast b)\ast c=\left(\frac{a+b}{2}\right)\ast c.
\displaystyle =\frac{\frac{a+b}{2}+c}{2}=\frac{a+b+2c}{4}. \hspace{1cm}\ldots(i)
\displaystyle a\ast(b\ast c)=a\ast\left(\frac{b+c}{2}\right).
\displaystyle =\frac{a+\frac{b+c}{2}}{2}=\frac{2a+b+c}{4}. \hspace{1cm}\ldots(ii)
\displaystyle \text{Since }\frac{a+b+2c}{4}\ne\frac{2a+b+c}{4}\text{ in general,}
\displaystyle (a\ast b)\ast c\ne a\ast(b\ast c).
\displaystyle \therefore \ast\text{ is not associative on }Q.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Let }S\text{ be the set of all rational numbers except }1\text{ and let }\ast
\displaystyle \text{be defined on }S\text{ by }a\ast b=a+b-ab,\text{ for all }a,b\in S.
\displaystyle \text{Prove that:}
\displaystyle \text{(i) }\ast\text{ is a binary operation on }S.
\displaystyle \text{(ii) }\ast\text{ is commutative as well as associative.}\qquad\text{[CBSE 2014]}
\displaystyle \text{Answer:}

\displaystyle \text{(i) Let }a,b\in S.
\displaystyle \text{Since }a,b\in Q,\text{ we have }a+b-ab\in Q.
\displaystyle \text{Suppose }a\ast b=1.
\displaystyle a+b-ab=1.
\displaystyle ab-a-b+1=0.
\displaystyle (a-1)(b-1)=0.
\displaystyle \therefore a=1\text{ or }b=1,
\displaystyle \text{which is impossible since }a,b\in S.
\displaystyle \therefore a\ast b\ne1,\text{ and hence }a\ast b\in S.
\displaystyle \text{Also, }a\ast b=a+b-ab\text{ is uniquely determined for every }a,b\in S.
\displaystyle \therefore \ast\text{ is a binary operation on }S.

\displaystyle \text{(ii) Commutativity:}
\displaystyle a\ast b=a+b-ab=b+a-ba=b\ast a.
\displaystyle \therefore \ast\text{ is commutative on }S.
\displaystyle \text{Associativity:}
\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 }S.
\displaystyle \therefore \ast\text{ is commutative as well as associative on }S.
\displaystyle \\

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