Show that $*: R \times R \rightarrow R$ given by $(a, b) \rightarrow a+4 b^{2}$ is a binary operation.
(N/A) binary operation $*$ on a set $R$ is a function $*: R \times R \rightarrow R$.
For any pair $(a, b) \in R \times R$,the expression $a + 4b^{2}$ is a well-defined real number because $a$ and $b$ are real numbers,and the set of real numbers $R$ is closed under addition and multiplication.
Since for every pair $(a, b) \in R \times R$,there exists a unique element $a + 4b^{2} \in R$,the operation $*$ is a binary operation on $R$.
For any pair $(a, b) \in R \times R$,the expression $a + 4b^{2}$ is a well-defined real number because $a$ and $b$ are real numbers,and the set of real numbers $R$ is closed under addition and multiplication.
Since for every pair $(a, b) \in R \times R$,there exists a unique element $a + 4b^{2} \in R$,the operation $*$ is a binary operation on $R$.
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Similar Questions
Consider 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 SolutionLet $^*$ be the binary operation on $N$ given by $a ^* b = \text{L.C.M. of } a \text{ and } b$. Find which elements of $N$ are invertible for the operation $^*$?
Difficult
View SolutionIn a group $(G, *)$,for some element $a$ of $G$,if $a^{2}=e$,where $e$ is the identity element,then
MediumKCET 2013
View SolutionShow that the number of binary operations on $\{1, 2\}$ having $1$ as identity and having $2$ as the inverse of $2$ is exactly one.
Difficult
View SolutionState whether the following statement is true or false and justify your answer: If $^*$ is a commutative binary operation on $N$,then $a ^* (b ^* c) = (c ^* b) ^* a$.
Medium
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