\displaystyle \textbf{Question 1: }\text{Determine whether the following operation defines a binary operation}
\displaystyle \text{on }N:\;a\ast b=a^b,\text{ for all }a,b\in N.
\displaystyle \text{Answer:}
\displaystyle \text{(i) For }a,b\in N,\text{ the operation is defined by }a\ast b=a^b.
\displaystyle \text{Since }a\in N\text{ and }b\in N,\text{ we have }a^b\in N.
\displaystyle \therefore a\ast b\in N\text{ for all }a,b\in N.
\displaystyle \text{Hence, }\ast\text{ is a binary operation on }N.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Determine whether the following operation defines a binary operation}
\displaystyle \text{on }Z:\;a\bigcirc b=a^b,\text{ for all }a,b\in Z.
\displaystyle \text{Answer:}
\displaystyle \text{(ii) The operation is defined by }a\bigcirc b=a^b,\text{ where }a,b\in Z.
\displaystyle \text{Consider }a=2\text{ and }b=-2.
\displaystyle \therefore a\bigcirc b=2^{-2}=\frac{1}{4}.
\displaystyle \text{Since }\frac{1}{4}\notin Z,\text{ we have }a\bigcirc b\notin Z.
\displaystyle \text{Hence, the operation is not closed on }Z.
\displaystyle \therefore \bigcirc\text{ does not define a binary operation on }Z.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Determine whether the following operation defines a binary operation}
\displaystyle \text{on }N:\;a\ast b=a+b-2,\text{ for all }a,b\in N.
\displaystyle \text{Answer:}
\displaystyle \text{(iii) The operation is defined by }a\ast b=a+b-2,\text{ where }a,b\in N.
\displaystyle \text{Take }a=1\text{ and }b=1.
\displaystyle \therefore a\ast b=1+1-2=0.
\displaystyle \text{Since }0\notin N,\text{ we have }a\ast b\notin N.
\displaystyle \text{Hence, the operation is not closed on }N.
\displaystyle \therefore \ast\text{ does not define a binary operation on }N.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Determine whether the following operation defines a binary operation}
\displaystyle \text{(iv) on }S=\{1,2,3,4,5\}\text{ defined by }a\times_6 b=\text{remainder when }ab
\displaystyle \text{is divided by }6.
\displaystyle \text{Answer:}
\displaystyle \text{The operation is defined by }a\times_6 b=\text{remainder when }ab\text{ is divided by }6,
\displaystyle \text{where }a,b\in S.
\displaystyle \text{Take }a=3\text{ and }b=4.
\displaystyle \therefore ab=3\times4=12.
\displaystyle \text{The remainder when }12\text{ is divided by }6\text{ is }0.
\displaystyle \text{Since }0\notin S,\text{ we have }a\times_6 b\notin S.
\displaystyle \text{Hence, the operation is not closed on }S.
\displaystyle \therefore \times_6\text{ does not define a binary operation on }S.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Determine whether the following operation defines a binary operation}
\displaystyle \text{on }S=\{0,1,2,3,4,5\}\text{ defined by }a+_6b=\begin{cases}a+b,&\text{if }a+b<6,\\a+b-6,&\text{if }a+b\ge6.\end{cases}
\displaystyle \text{Answer:}
\displaystyle \text{(v) Let }a,b\in S.
\displaystyle \text{If }a+b<6,\text{ then }a+_6b=a+b\in S.
\displaystyle \text{If }a+b\ge6,\text{ then }a+_6b=a+b-6.
\displaystyle \text{Since }0\le a+b-6\le4,\text{ we have }a+b-6\in S.
\displaystyle \therefore a+_6b\in S\text{ for all }a,b\in S.
\displaystyle \therefore +_6\text{ defines a binary operation on }S.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Determine whether the following operation defines a binary operation}
\displaystyle \text{(vi) on }N\text{ defined by }a\odot b=a^b+b^a,\text{ for all }a,b\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }a,b\in N.
\displaystyle \text{Since }a,b\in N,\text{ both }a^b\in N\text{ and }b^a\in N.
\displaystyle \text{Also, the sum of two natural numbers is a natural number.}
\displaystyle \therefore a^b+b^a\in N.
\displaystyle \therefore a\odot b\in N\text{ for all }a,b\in N.
\displaystyle \text{Hence, }\odot\text{ defines a binary operation on }N.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Determine whether the following operation defines a binary operation}
\displaystyle \text{(vii) on }Q\text{ defined by }a\ast b=\frac{a-1}{b+1},\text{ for all }a,b\in Q.
\displaystyle \text{Answer:}
\displaystyle \text{The operation is defined by }a\ast b=\frac{a-1}{b+1},\text{ where }a,b\in Q.
\displaystyle \text{Take }a=2\text{ and }b=-1.
\displaystyle \therefore a\ast b=\frac{2-1}{-1+1}=\frac{1}{0},\text{ which is undefined.}
\displaystyle \text{Thus, }a\ast b\notin Q\text{ for this ordered pair.}
\displaystyle \text{Hence, the operation is not defined for every pair in }Q\times Q.
\displaystyle \therefore \ast\text{ does not define a binary operation on }Q.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Determine whether the following definition gives a binary operation.}
\displaystyle \text{If not, justify your answer.}
\displaystyle \text{(i) On }Z^+,\text{ defined by }a\ast b=a-b,\text{ where }Z^+\text{ denotes the set of}
\displaystyle \text{all non-negative integers.}
\displaystyle \text{Answer:}
\displaystyle \text{The operation is defined by }a\ast b=a-b,\text{ where }a,b\in Z^+.
\displaystyle \text{Take }a=1\text{ and }b=2.
\displaystyle \therefore a\ast b=1-2=-1.
\displaystyle \text{Since }-1\notin Z^+,\text{ we have }a\ast b\notin Z^+.
\displaystyle \text{Hence, the operation is not closed on }Z^+.
\displaystyle \therefore \ast\text{ does not define a binary operation on }Z^+.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Determine whether the following definition gives a binary operation.}
\displaystyle \text{If not, justify your answer.}
\displaystyle \text{(ii) On }Z^+,\text{ defined by }a\ast b=ab,\text{ where }Z^+\text{ denotes the set of}
\displaystyle \text{all non-negative integers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }a,b\in Z^+.
\displaystyle \text{The product of two non-negative integers is a non-negative integer.}
\displaystyle \therefore ab\in Z^+.
\displaystyle \therefore a\ast b\in Z^+\text{ for all }a,b\in Z^+.
\displaystyle \text{Hence, }\ast\text{ defines a binary operation on }Z^+.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Determine whether the following definition gives a binary operation.}
\displaystyle \text{If not, justify your answer.}
\displaystyle \text{(iii) On }R,\text{ defined by }a\ast b=ab^2.
\displaystyle \text{Answer:}
\displaystyle \text{Let }a,b\in R.
\displaystyle \text{Since }b^2\in R\text{ and the product of two real numbers is a real number,}
\displaystyle ab^2\in R.
\displaystyle \therefore a\ast b\in R\text{ for all }a,b\in R.
\displaystyle \text{Hence, }\ast\text{ defines a binary operation on }R.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Determine whether the following definition gives a binary operation.}
\displaystyle \text{If not, justify your answer.}
\displaystyle \text{(iv) On }Z^+\text{ defined by }a\ast b=|a-b|,\text{ where }Z^+\text{ denotes the set of}
\displaystyle \text{all non-negative integers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }a,b\in Z^+.
\displaystyle \text{Since }|a-b|\text{ is always a non-negative integer, }|a-b|\in Z^+.
\displaystyle \therefore a\ast b\in Z^+\text{ for all }a,b\in Z^+.
\displaystyle \text{Hence, }\ast\text{ defines a binary operation on }Z^+.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Determine whether the following definition gives a binary operation.}
\displaystyle \text{If not, justify your answer.}
\displaystyle \text{(v) On }Z^+\text{ defined by }a\ast b=a,\text{ where }Z^+\text{ denotes the set of}
\displaystyle \text{all non-negative integers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }a,b\in Z^+.
\displaystyle \text{Since }a\in Z^+,\text{ we have }a\ast b=a\in Z^+.
\displaystyle \therefore a\ast b\in Z^+\text{ for all }a,b\in Z^+.
\displaystyle \text{Hence, }\ast\text{ defines a binary operation on }Z^+.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Determine whether the following definition gives a binary operation.}
\displaystyle \text{If not, justify your answer.}
\displaystyle \text{(vi) On }R,\text{ defined by }a\ast b=a+4b^2.
\displaystyle \text{Answer:}
\displaystyle \text{Let }a,b\in R.
\displaystyle \text{Since }b^2\in R,\text{ we have }4b^2\in R.
\displaystyle \text{Also, the sum of two real numbers is a real number.}
\displaystyle \therefore a+4b^2\in R.
\displaystyle \therefore a\ast b\in R\text{ for all }a,b\in R.
\displaystyle \text{Hence, }\ast\text{ defines a binary operation on }R.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Let }\ast\text{ be a binary operation on the set }I\text{ of integers,}
\displaystyle \text{defined by }a\ast b=2a+b-3.\text{ Find the value of }3\ast4.\hfill\text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given }a\ast b=2a+b-3.
\displaystyle \therefore 3\ast4=(2\times3)+4-3.
\displaystyle =6+4-3=7.
\displaystyle \therefore 3\ast4=7.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Is }\ast\text{ defined on the set }\{1,2,3,4,5\}\text{ by }a\ast b=
\displaystyle \text{LCM of }a\text{ and }b\text{ a binary operation? Justify your answer.}
\displaystyle \text{Answer:}
\displaystyle \text{Take }a=2\text{ and }b=3.
\displaystyle \text{Then }a\ast b=\mathrm{LCM}(2,3)=6.
\displaystyle \text{Since }6\notin\{1,2,3,4,5\},\text{ we have }a\ast b\notin\{1,2,3,4,5\}.
\displaystyle \text{Hence, the operation is not closed on }\{1,2,3,4,5\}.
\displaystyle \therefore \ast\text{ does not define a binary operation on }\{1,2,3,4,5\}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Let }S=\{a,b,c\}.\text{ Find the total number of binary operations}
\displaystyle \text{on }S.
\displaystyle \text{Answer:}
\displaystyle \text{The number of binary operations on a set with }n\text{ elements is }n^{n^2}.
\displaystyle \text{Here }n=3.
\displaystyle \therefore n^{n^2}=3^{3^2}=3^9=19683.
\displaystyle \therefore \text{The total number of binary operations on }S\text{ is }19683.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the total number of binary operations on }\{a,b\}.
\displaystyle \text{Answer:}
\displaystyle \text{The number of binary operations on a set with }n\text{ elements is }n^{n^2}.
\displaystyle \text{Here }n=2.
\displaystyle \therefore n^{n^2}=2^{2^2}=2^4=16.
\displaystyle \therefore \text{The total number of binary operations on }\{a,b\}\text{ is }16.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Prove that the operation }\ast\text{ on the set}
\displaystyle M=\left\{\begin{bmatrix}a&0\\0&b\end{bmatrix}:a,b\in R-\{0\}\right\}
\displaystyle \text{defined by }A\ast B=AB\text{ is a binary operation.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A,B\in M.
\displaystyle \text{Then }A=\begin{bmatrix}a&0\\0&b\end{bmatrix}\text{ and }B=\begin{bmatrix}c&0\\0&d\end{bmatrix},
\displaystyle \text{where }a,b,c,d\in R-\{0\}.
\displaystyle A\ast B=AB=\begin{bmatrix}a&0\\0&b\end{bmatrix}\begin{bmatrix}c&0\\0&d\end{bmatrix}
\displaystyle =\begin{bmatrix}ac&0\\0&bd\end{bmatrix}.
\displaystyle \text{Since }a,c\in R-\{0\},\text{ we have }ac\in R-\{0\}.
\displaystyle \text{Similarly, }bd\in R-\{0\}.
\displaystyle \therefore AB=\begin{bmatrix}ac&0\\0&bd\end{bmatrix}\in M.
\displaystyle \therefore A\ast B\in M\text{ for all }A,B\in M.
\displaystyle \text{Hence, }\ast\text{ defines a binary operation on }M.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Let }S\text{ be the set of all rational numbers of the form }\frac{m}{n},
\displaystyle \text{where }m\in Z\text{ and }n=1,2,3.\text{ Prove that }\ast\text{ on }S\text{ defined by}
\displaystyle a\ast b=ab\text{ is not a binary operation.}
\displaystyle \text{Answer:}
\displaystyle \text{Take }a=\frac{5}{3}\text{ and }b=\frac{8}{3},\text{ which belong to }S.
\displaystyle \therefore a\ast b=\frac{5}{3}\times\frac{8}{3}=\frac{40}{9}.
\displaystyle \text{Since }\frac{40}{9}\text{ is not of the form }\frac{m}{n},\text{ where }n=1,2\text{ or }3,
\displaystyle \frac{40}{9}\notin S.
\displaystyle \text{Hence, the operation is not closed on }S.
\displaystyle \therefore \ast\text{ does not define a binary operation on }S.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The binary operation }\ast:R\times R\rightarrow R\text{ is defined by}
\displaystyle a\ast b=2a+b.\text{ Find }(2\ast3)\ast4.\hfill\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given }a\ast b=2a+b.
\displaystyle 2\ast3=2(2)+3=7.
\displaystyle \therefore (2\ast3)\ast4=7\ast4=2(7)+4=18.
\displaystyle \therefore (2\ast3)\ast4=18.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Let }\ast\text{ be a binary operation on }N\text{ given by }a\ast b=
\displaystyle \mathrm{LCM}(a,b),\text{ for all }a,b\in N.\text{ Find }5\ast7.\hfill\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given }a\ast b=\mathrm{LCM}(a,b).
\displaystyle 5\ast7=\mathrm{LCM}(5,7).
\displaystyle \text{Since }5\text{ and }7\text{ are prime numbers, }\mathrm{LCM}(5,7)=5\times7=35.
\displaystyle \therefore 5\ast7=35.
\displaystyle \\

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