\displaystyle \textbf{Question 1: }\text{Construct the composition table for }\times_4\text{ on the set }S=\{0,1,2,3\}.
\displaystyle \text{Answer:}
\displaystyle 1\times_4 1=\text{Remainder obtained by dividing }1\times1=1\text{ by }4=1.
\displaystyle 0\times_4 1=\text{Remainder obtained by dividing }0\times1=0\text{ by }4=0.
\displaystyle 2\times_4 3=\text{Remainder obtained by dividing }2\times3=6\text{ by }4=2.
\displaystyle 3\times_4 3=\text{Remainder obtained by dividing }3\times3=9\text{ by }4=1.
\displaystyle \text{Hence, the composition table is as follows:}
\displaystyle \begin{array}{|c|c|c|c|c|} \hline \times_4 & 0 & 1 & 2 & 3\\ \hline 0 & 0 & 0 & 0 & 0\\ \hline 1 & 0 & 1 & 2 & 3\\ \hline 2 & 0 & 2 & 0 & 2\\ \hline 3 & 0 & 3 & 2 & 1\\ \hline \end{array}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Construct the composition table for }+_5\text{ on the set }S=\{0,1,2,3,4\}.
\displaystyle \text{Answer:}
\displaystyle 1+_5 1=\text{Remainder obtained by dividing }1+1=2\text{ by }5=2.
\displaystyle 3+_5 4=\text{Remainder obtained by dividing }3+4=7\text{ by }5=2.
\displaystyle 4+_5 4=\text{Remainder obtained by dividing }4+4=8\text{ by }5=3.
\displaystyle \text{Hence, the composition table is as follows:}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline +_5 & 0 & 1 & 2 & 3 & 4\\ \hline 0 & 0 & 1 & 2 & 3 & 4\\ \hline 1 & 1 & 2 & 3 & 4 & 0\\ \hline 2 & 2 & 3 & 4 & 0 & 1\\ \hline 3 & 3 & 4 & 0 & 1 & 2\\ \hline 4 & 4 & 0 & 1 & 2 & 3\\ \hline \end{array}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Construct the composition table for }\times_6\text{ on the set }S=\{0,1,2,3,4,5\}.
\displaystyle \text{Answer:}
\displaystyle 1\times_6 1=\text{Remainder obtained by dividing }1\times1=1\text{ by }6=1.
\displaystyle 3\times_6 4=\text{Remainder obtained by dividing }3\times4=12\text{ by }6=0.
\displaystyle 4\times_6 5=\text{Remainder obtained by dividing }4\times5=20\text{ by }6=2.
\displaystyle \text{Hence, the composition table is as follows:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \times_6 & 0 & 1 & 2 & 3 & 4 & 5\\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0\\ \hline 1 & 0 & 1 & 2 & 3 & 4 & 5\\ \hline 2 & 0 & 2 & 4 & 0 & 2 & 4\\ \hline 3 & 0 & 3 & 0 & 3 & 0 & 3\\ \hline 4 & 0 & 4 & 2 & 0 & 4 & 2\\ \hline 5 & 0 & 5 & 4 & 3 & 2 & 1\\ \hline \end{array}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Construct the composition table for }\times_5\text{ on }Z_5=\{0,1,2,3,4\}.
\displaystyle \text{Answer:}
\displaystyle 1\times_5 1=\text{Remainder obtained by dividing }1\times1=1\text{ by }5=1.
\displaystyle 3\times_5 4=\text{Remainder obtained by dividing }3\times4=12\text{ by }5=2.
\displaystyle 4\times_5 4=\text{Remainder obtained by dividing }4\times4=16\text{ by }5=1.
\displaystyle \text{Hence, the composition table is as follows:}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \times_5 & 0 & 1 & 2 & 3 & 4\\ \hline 0 & 0 & 0 & 0 & 0 & 0\\ \hline 1 & 0 & 1 & 2 & 3 & 4\\ \hline 2 & 0 & 2 & 4 & 1 & 3\\ \hline 3 & 0 & 3 & 1 & 4 & 2\\ \hline 4 & 0 & 4 & 3 & 2 & 1\\ \hline \end{array}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{For the binary operation }\times_{10}\text{ on the set }S=\{1,3,7,9\},
\displaystyle \text{find the inverse of }3.
\displaystyle \text{Answer:}
\displaystyle 1\times_{10}1=\text{Remainder obtained by dividing }1\times1=1\text{ by }10=1.
\displaystyle 3\times_{10}7=\text{Remainder obtained by dividing }3\times7=21\text{ by }10=1.
\displaystyle 7\times_{10}9=\text{Remainder obtained by dividing }7\times9=63\text{ by }10=3.
\displaystyle \text{Hence, the composition table is as follows:}
\displaystyle \begin{array}{|c|c|c|c|c|} \hline \times_{10} & 1 & 3 & 7 & 9\\ \hline 1 & 1 & 3 & 7 & 9\\ \hline 3 & 3 & 9 & 1 & 7\\ \hline 7 & 7 & 1 & 9 & 3\\ \hline 9 & 9 & 7 & 3 & 1\\ \hline \end{array}
\displaystyle \text{We observe that the elements of the first row are the same as those of the top row.}
\displaystyle \therefore 1\in S\text{ is the identity element with respect to }\times_{10}.
\displaystyle \text{Finding the inverse of }3:
\displaystyle \text{From the above table, }3\times_{10}7=1\text{ and }7\times_{10}3=1.
\displaystyle \therefore \text{the inverse of }3\text{ is }7.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{For the binary operation }\times_7\text{ on the set }S=\{1,2,3,4,5,6\},
\displaystyle \text{compute }3^{-1}\times_7 4.
\displaystyle \text{Answer:}
\displaystyle 1\times_7 1=\text{Remainder obtained by dividing }1\times1=1\text{ by }7=1.
\displaystyle 3\times_7 4=\text{Remainder obtained by dividing }3\times4=12\text{ by }7=5.
\displaystyle 4\times_7 5=\text{Remainder obtained by dividing }4\times5=20\text{ by }7=6.
\displaystyle \text{Hence, the composition table is as follows:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \times_7 & 1 & 2 & 3 & 4 & 5 & 6\\ \hline 1 & 1 & 2 & 3 & 4 & 5 & 6\\ \hline 2 & 2 & 4 & 6 & 1 & 3 & 5\\ \hline 3 & 3 & 6 & 2 & 5 & 1 & 4\\ \hline 4 & 4 & 1 & 5 & 2 & 6 & 3\\ \hline 5 & 5 & 3 & 1 & 6 & 4 & 2\\ \hline 6 & 6 & 5 & 4 & 3 & 2 & 1\\ \hline \end{array}
\displaystyle \text{We observe that the elements of the first row are the same as those of the top row.}
\displaystyle \therefore 1\text{ is the identity element with respect to }\times_7.
\displaystyle 3\times_7 5=1=5\times_7 3.
\displaystyle \therefore 3^{-1}=5.
\displaystyle \text{Now, }3^{-1}\times_7 4=5\times_7 4=6.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the inverse of }5\text{ under multiplication modulo }11\text{ on }Z_{11}.
\displaystyle \text{Answer:}
\displaystyle 1\times_{11}1=\text{Remainder obtained by dividing }1\times1=1\text{ by }11=1.
\displaystyle 3\times_{11}4=\text{Remainder obtained by dividing }3\times4=12\text{ by }11=1.
\displaystyle 4\times_{11}5=\text{Remainder obtained by dividing }4\times5=20\text{ by }11=9.
\displaystyle \text{Hence, the composition table is as follows:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \times_{11} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\ \hline 1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\ \hline 2 & 2 & 4 & 6 & 8 & 10 & 1 & 3 & 5 & 7 & 9\\ \hline 3 & 3 & 6 & 9 & 1 & 4 & 7 & 10 & 2 & 5 & 8\\ \hline 4 & 4 & 8 & 1 & 5 & 9 & 2 & 6 & 10 & 3 & 7\\ \hline 5 & 5 & 10 & 4 & 9 & 3 & 8 & 2 & 7 & 1 & 6\\ \hline 6 & 6 & 1 & 7 & 2 & 8 & 3 & 9 & 4 & 10 & 5\\ \hline 7 & 7 & 3 & 10 & 6 & 2 & 9 & 5 & 1 & 8 & 4\\ \hline 8 & 8 & 5 & 2 & 10 & 7 & 4 & 1 & 9 & 6 & 3\\ \hline 9 & 9 & 7 & 5 & 3 & 1 & 10 & 8 & 6 & 4 & 2\\ \hline 10 & 10 & 9 & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1\\ \hline \end{array}
\displaystyle \text{We observe that the elements of the first row are the same as those of the top row.}
\displaystyle \therefore 1\text{ is the identity element with respect to }\times_{11}.
\displaystyle \text{Also, }5\times_{11}9=1\text{ and }9\times_{11}5=1.
\displaystyle \therefore 5^{-1}=9.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Write the multiplication table for the set of integers modulo }5.
\displaystyle \text{Answer:}
\displaystyle 1\times_5 1=\text{Remainder obtained by dividing }1\times1=1\text{ by }5=1.
\displaystyle 3\times_5 4=\text{Remainder obtained by dividing }3\times4=12\text{ by }5=2.
\displaystyle 4\times_5 4=\text{Remainder obtained by dividing }4\times4=16\text{ by }5=1.
\displaystyle \text{Hence, the multiplication table is as follows:}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \times_5 & 0 & 1 & 2 & 3 & 4\\ \hline 0 & 0 & 0 & 0 & 0 & 0\\ \hline 1 & 0 & 1 & 2 & 3 & 4\\ \hline 2 & 0 & 2 & 4 & 1 & 3\\ \hline 3 & 0 & 3 & 1 & 4 & 2\\ \hline 4 & 0 & 4 & 3 & 2 & 1\\ \hline \end{array}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Consider the binary operations }*\text{ and }\circ\text{ defined by the}
\displaystyle \text{following tables on the set }S=\{a,b,c,d\}.
\displaystyle \text{(i) }\begin{array}{|c|c|c|c|c|} \hline * & a & b & c & d\\ \hline a & a & b & c & d\\ \hline b & b & a & d & c\\ \hline c & c & d & a & b\\ \hline d & d & c & b & a\\ \hline \end{array}
\displaystyle \text{(ii) }\begin{array}{|c|c|c|c|c|} \hline \circ & a & b & c & d\\ \hline a & a & a & a & a\\ \hline b & a & b & c & d\\ \hline c & a & c & d & b\\ \hline d & a & d & b & c\\ \hline \end{array}
\displaystyle \text{Show that both binary operations are commutative and associative.}
\displaystyle \text{Write down their identity elements and list the inverses of the elements.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Commutativity of }*:
\displaystyle \text{The composition table is symmetric about its principal diagonal.}
\displaystyle \therefore x*y=y*x\text{ for all }x,y\in S.
\displaystyle \therefore *\text{ is commutative.}
\displaystyle \text{Associativity of }*:
\displaystyle \text{Represent }a,b,c,d\text{ respectively by }(0,0),(1,0),(0,1),(1,1).
\displaystyle \text{The operation }*\text{ corresponds to componentwise addition modulo }2.
\displaystyle \text{Thus, for }x,y,z\in S,
\displaystyle (x*y)*z=x*(y*z),
\displaystyle \text{since addition modulo }2\text{ is associative.}
\displaystyle \therefore *\text{ is associative.}
\displaystyle \text{Identity element:}
\displaystyle \text{The row and column corresponding to }a\text{ reproduce the elements }a,b,c,d.
\displaystyle \therefore a*x=x=x*a\text{ for every }x\in S.
\displaystyle \therefore a\text{ is the identity element with respect to }*.
\displaystyle \text{Inverses of the elements:}
\displaystyle a*a=a,\qquad b*b=a,\qquad c*c=a,\qquad d*d=a.
\displaystyle \therefore a^{-1}=a,\qquad b^{-1}=b,\qquad c^{-1}=c,\qquad d^{-1}=d.
\displaystyle \text{(ii) Commutativity of }\circ:
\displaystyle \text{The composition table is symmetric about its principal diagonal.}
\displaystyle \therefore x\circ y=y\circ x\text{ for all }x,y\in S.
\displaystyle \therefore \circ\text{ is commutative.}
\displaystyle \text{Associativity of }\circ:
\displaystyle \text{If any one of }x,y,z\text{ is }a,\text{ then}
\displaystyle (x\circ y)\circ z=a=x\circ(y\circ z),
\displaystyle \text{because }a\circ t=a=t\circ a\text{ for every }t\in S.
\displaystyle \text{For the elements }b,c,d,\text{ represent }b,c,d\text{ by }0,1,2\text{ respectively.}
\displaystyle \text{On }\{b,c,d\},\text{ the operation }\circ\text{ corresponds to addition modulo }3.
\displaystyle \text{Hence, for }x,y,z\in\{b,c,d\},
\displaystyle (x\circ y)\circ z=x\circ(y\circ z),
\displaystyle \text{since addition modulo }3\text{ is associative.}
\displaystyle \therefore \circ\text{ is associative on }S.
\displaystyle \text{Identity element:}
\displaystyle \text{The row and column corresponding to }b\text{ reproduce the elements }a,b,c,d.
\displaystyle \therefore b\circ x=x=x\circ b\text{ for every }x\in S.
\displaystyle \therefore b\text{ is the identity element with respect to }\circ.
\displaystyle \text{Inverses of the elements:}
\displaystyle b\circ b=b,\qquad c\circ d=b=d\circ c.
\displaystyle \therefore b^{-1}=b,\qquad c^{-1}=d,\qquad d^{-1}=c.
\displaystyle \text{Also, }a\circ x=a\neq b\text{ for every }x\in S.
\displaystyle \therefore a\text{ has no inverse with respect to }\circ.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Define a binary operation }*\text{ on the set }\{0,1,2,3,4,5\}\text{ as}
\displaystyle a*b=\left\{\begin{array}{ll} a+b, & \text{if }a+b<6,\\ a+b-6, & \text{if }a+b\ge6. \end{array}\right.
\displaystyle \text{Show that }0\text{ is the identity for this operation and each element }a\neq0\text{ is invertible}
\displaystyle \text{with }6-a\text{ being the inverse of }a.\hspace{0.3cm}\text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X=\{0,1,2,3,4,5\}.
\displaystyle \text{The operation }*\text{ on }X\text{ is defined by}
\displaystyle a*b=\left\{\begin{array}{ll} a+b, & \text{if }a+b<6,\\ a+b-6, & \text{if }a+b\ge6. \end{array}\right.
\displaystyle \text{An element }e\in X\text{ is the identity element if }a*e=a=e*a\text{ for every }a\in X.
\displaystyle \text{For any }a\in X,
\displaystyle a*0=a+0=a,\qquad(\text{since }a<6).
\displaystyle 0*a=0+a=a,\qquad(\text{since }a<6).
\displaystyle \therefore a*0=a=0*a\text{ for every }a\in X.
\displaystyle \therefore 0\text{ is the identity element for the operation }*.
\displaystyle \text{Now, an element }a\in X\text{ is invertible if there exists }b\in X\text{ such that}
\displaystyle a*b=0=b*a.
\displaystyle \text{If }a+b<6,\text{ then }a+b=0,
\displaystyle \text{which is possible only when }a=0\text{ and }b=0.
\displaystyle \text{For }a\neq0,\text{ we must have }a+b\ge6.
\displaystyle \therefore a+b-6=0.
\displaystyle \therefore b=6-a.
\displaystyle \text{Since }a\in\{1,2,3,4,5\},\;6-a\in X.
\displaystyle \therefore \text{the inverse of every non-zero element }a\text{ is }6-a.
\displaystyle \therefore a^{-1}=6-a,\qquad a\neq0.
\displaystyle \\

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