Determine whether or not each of the definitions of $*$ given below gives a binary operation. In the event that $*$ is not a binary operation,give justification for this. On $Z^{+}$,define $*$ by $a * b = a$.
(A) On $Z^{+}$,the operation $*$ is defined by $a * b = a$.
For any two elements $a, b \in Z^{+}$,the result of the operation $a * b$ is $a$.
Since $a$ is an element of $Z^{+}$,it follows that for every pair $(a, b) \in Z^{+} \times Z^{+}$,there exists a unique element $a * b$ in $Z^{+}$.
By definition,a binary operation on a set $S$ is a function from $S \times S$ to $S$.
Since $a * b = a \in Z^{+}$ for all $a, b \in Z^{+}$,the operation $*$ satisfies the condition of being a binary operation.
Therefore,$*$ is a binary operation on $Z^{+}$.
For any two elements $a, b \in Z^{+}$,the result of the operation $a * b$ is $a$.
Since $a$ is an element of $Z^{+}$,it follows that for every pair $(a, b) \in Z^{+} \times Z^{+}$,there exists a unique element $a * b$ in $Z^{+}$.
By definition,a binary operation on a set $S$ is a function from $S \times S$ to $S$.
Since $a * b = a \in Z^{+}$ for all $a, b \in Z^{+}$,the operation $*$ satisfies the condition of being a binary operation.
Therefore,$*$ is a binary operation on $Z^{+}$.
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Similar Questions
Let $^*$ be a binary operation on the set $Q$ of rational numbers defined as $a * b = (a - b)^2$. Determine whether the operation is commutative and associative.
Medium
View SolutionConsider a binary operation $*$ on the set $\{1,2,3,4,5\}$ given by the following multiplication table. Is $^*$ commutative?
(Hint: use the following table)
$^*$ $1$ $2$ $3$ $4$ $5$ $1$ $1$ $1$ $1$ $1$ $1$ $2$ $1$ $2$ $2$ $2$ $2$ $3$ $1$ $2$ $3$ $3$ $3$ $4$ $1$ $2$ $3$ $4$ $4$ $5$ $1$ $2$ $3$ $4$ $5$
(Hint: use the following table)
| $^*$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $1$ | $1$ | $1$ | $1$ | $1$ | $1$ |
| $2$ | $1$ | $2$ | $2$ | $2$ | $2$ |
| $3$ | $1$ | $2$ | $3$ | $3$ | $3$ |
| $4$ | $1$ | $2$ | $3$ | $4$ | $4$ |
| $5$ | $1$ | $2$ | $3$ | $4$ | $5$ |
Medium
View SolutionConsider the binary operations $^*: R \times R \rightarrow R$ and $o: R \times R \rightarrow R$ defined as $a \,^*\, b = |a-b|$ and $a \,o\, b = a$,$\forall \, a, b \in R$. Show that $^*$ is commutative but not associative,and $o$ is associative but not commutative. Further,show that $\forall \, a, b, c \in R, a \,^*\, (b \,o\, c) = (a \,^*\, b) \,o\, (a \,^*\, c)$. [If it is so,we say that the operation $^*$ distributes over the operation $o$]. Does $o$ distribute over $^*$? Justify your answer.
Difficult
View SolutionLet $^*$ be the binary operation on $N$ given by $a \, ^* \, b = \text{L.C.M. of } a \text{ and } b$. Find the identity of $^*$ in $N$.
Medium
View SolutionIn the set of integers $(Z, *)$,if $a * b = a + b - n, \forall a, b \in Z$,where $n$ is a fixed integer,then the inverse of $(-n)$ is:
DifficultKCET 2013
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