Is $^*$ defined on the set $\{1, 2, 3, 4, 5\}$ by $a \,^*\, b = \text{L.C.M. of } a \text{ and } b$ a binary operation? Justify your answer.
(D) binary operation $^*$ on a set $A$ is a function $^*: A \times A \to A$. This means that for all $a, b \in A$,the result $a \,^*\, b$ must also be an element of $A$.
Let $A = \{1, 2, 3, 4, 5\}$. The operation is defined as $a \,^*\, b = \text{L.C.M. of } a \text{ and } b$.
For the operation to be a binary operation,$a \,^*\, b$ must belong to $A$ for all $a, b \in A$.
Let us test this with some elements from the set $A$:
Consider $a = 2$ and $b = 3$. Both $2, 3 \in A$.
$2 \,^*\, 3 = \text{L.C.M. of } 2 \text{ and } 3 = 6$.
Since $6 \notin \{1, 2, 3, 4, 5\}$,the result of the operation is not in the set $A$.
Similarly,$2 \,^*\, 5 = 10 \notin A$,$3 \,^*\, 4 = 12 \notin A$,etc.
Since there exist elements $a, b \in A$ such that $a \,^*\, b \notin A$,the operation $^*$ is not a binary operation on the set $\{1, 2, 3, 4, 5\}$.
Let $A = \{1, 2, 3, 4, 5\}$. The operation is defined as $a \,^*\, b = \text{L.C.M. of } a \text{ and } b$.
For the operation to be a binary operation,$a \,^*\, b$ must belong to $A$ for all $a, b \in A$.
Let us test this with some elements from the set $A$:
Consider $a = 2$ and $b = 3$. Both $2, 3 \in A$.
$2 \,^*\, 3 = \text{L.C.M. of } 2 \text{ and } 3 = 6$.
Since $6 \notin \{1, 2, 3, 4, 5\}$,the result of the operation is not in the set $A$.
Similarly,$2 \,^*\, 5 = 10 \notin A$,$3 \,^*\, 4 = 12 \notin A$,etc.
Since there exist elements $a, b \in A$ such that $a \,^*\, b \notin A$,the operation $^*$ is not a binary operation on the set $\{1, 2, 3, 4, 5\}$.
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Similar Questions
Which of the following statements is false?
Determine whether or not the definition of $*$ given below gives a binary operation. In the event that $*$ is not a binary operation,give justification for this. On $Z^{+}$,define $*$ by $a * b = a - b$.
Medium
View SolutionConsider a binary operation $*$ on the set $\{1, 2, 3, 4, 5\}$ given by the following multiplication table. Compute $(2 \,^* \,3) \,^* \,4$ and $2 \,^* \,(3 \,^* \,4)$.
$^*$ $1$ $2$ $3$ $4$ $5$ $1$ $1$ $1$ $1$ $1$ $1$ $2$ $1$ $2$ $2$ $2$ $2$ $3$ $1$ $2$ $3$ $3$ $3$ $4$ $1$ $2$ $3$ $4$ $4$ $5$ $1$ $2$ $3$ $4$ $5$
| $^*$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $1$ | $1$ | $1$ | $1$ | $1$ | $1$ |
| $2$ | $1$ | $2$ | $2$ | $2$ | $2$ |
| $3$ | $1$ | $2$ | $3$ | $3$ | $3$ |
| $4$ | $1$ | $2$ | $3$ | $4$ | $4$ |
| $5$ | $1$ | $2$ | $3$ | $4$ | $5$ |
Medium
View SolutionGiven a non-empty set $X$,let $^*: P(X) \times P(X) \rightarrow P(X)$ be defined as $A \,^*\, B = (A - B) \cup (B - A)$,$\forall A, B \in P(X)$. Show that the empty set $\Phi$ is the identity for the operation $^*$ and all the elements $A$ of $P(X)$ are invertible with $A^{-1} = A$.
Difficult
View SolutionConsider a binary operation $*$ on $N$ defined as $a * b = a^{3} + b^{3}$. Choose the correct answer.
Medium
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