Check whether the $\ast$ operation defined on the set $ \text{A = R} \times \text{R as} $
$\text{(a, b)} \ast \text{(c, d)} = \text{(a + c, b + d)}$
is a binary operation or not, where R is the set of all real numbers. If it is a binary operation, is it commutative and associative too? Also find the identity element of $\ast$.
✓
Answer
$\text{(a, b)} \ast \text{(c, d)} = \text{(a + c, b + d)} \ \forall \ \text{a, b, c, d}\in \text{R}$
Since $\text{a + c}\in \text{R}$ and $\text{b + d}\in \text{R} \Rightarrow \text{(a + c, b+ d)} \in \text{R}\times\text{R}$
$\text{i.e} \ast \text{is binary operation}$
For commutative
$\text{consider (c, d)}\ast\text{(a, b}) = \text{(c + a, d + b)}$
$=\text{(a + c, b + d)}$
$\Rightarrow'\ast' \text{is commutative}$
For Associateive
$\text{(a, b), (c, d), (e, f)}\in \text{R}\times\text{R = A}$
$\text{[(a, b)}\ast \text{(c, d)}\ast\text{(e, f)} = \text{(a + c, b + d)}\ast \text{(e, f)} $
$= \text{(a + c + e, b + d + f)}\dots\dots\dots\dots\text{(i)}$
$\text{again (a, b)}\ast \text{[(c, d)]}\ast\text{(e, f)]} = \text{(a, b)}\ast \text{(c + e, d + f)}$
$= \text{(a + c + e, b + d + f)}\dots\dots\dots\text{(ii)}$
$\text{(i)} \&\text{(ii)} \Rightarrow '\ast' \text{is associtative}$
$\text{For identity element}$
$\text{Let (e}_1,\text{e}_{2}) \in \text{R}\times\text{R}$ be the identity element (if exists)
then $\text{a , b)}\ast \text{(e}_{1}, \text{e}_{2}) = \text{(a, b)} = \text{(e}_{1}, \text{e}_{2})\ast\text{(a, b)}$
$\Rightarrow\text({e}_{1}, \text{e}_{2}) = \text{0 , 0)} \in \text{R}\times\text{R} $
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.