Let $\text{A} = \Re\times\Re$ and$\ast$ be the binary operation on A defined by $\text(a, b) \ast \text{(c, d)} = \text{(a + c, b + d)}. $ Prove that $\ast$ is commutative and associative. Find the identity element for $\ast$ on A. Also write the inverse element of the element (3, – 5) in A.
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Answer
$\forall \text{a, b, c, d, e, f}\in \Re$
$\text{((a, b)}\ast \text{(c,d))}\ast(\text{e, f})= \text{a + c, b + d)}\ast\text{(e, f)}$
$= \text{(a + c + e, b + d + f)} \rightarrow\text{(3)}$
$\text{((a, b)}\ast \text{(c,d))}\ast(\text{e, f}) = \text{(a + b)}\ast\text{(c + e, d + f)}$
$ = \text{(a + c = e, b + d + f)} \rightarrow \text{(4)}$
$\therefore \ast \text { is Associative}$
Let (x, y) be on identity element in $\Re\times\Re$
$\Rightarrow \text{(a, b)}\ast\text{(x, y)} = \text{(a, b)} = \text{(x, y)}\ast \text{(a,b)}$
$\Rightarrow \text{a + x = a, b + y = b}$
$\text{x = 0, y = 0}$
$\therefore \text{(0, 0) is identity element} $
Let the inverse element of $\text{(3, -5) be } \text{(x}_{1},\text{y}_{1},) $
$\Rightarrow (3, - 5)\ast \text{(x}_{1},\text{y}_{1},) = (0, 0) = \text{(x}_{1},\text{y}_{1},) \ast(3, -5)$
$\text{3 + x}_{1} = 0,\text{-5 + y}_{1} = 0 $
$\text{x}_{1} = -3, \text{y}_{1} = 5 $
$\Rightarrow (-3, 5) \text{is an inverse of (3, -5)}$
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