Show that the binary operation on A = R - -1 defined as a a b = a… - Vidyadip

Question

Show that the binary operation $\ast \text{ on A = R - {-1}}$ defined as a $\text{a} \ast \text{b} = \text{a + b + ab}$ for all $\text{a, b}\in \text{A}$ is communicative and associative on A. Also find the identity element of $\ast$ in A and prove that every element of a is invertible.

✓

Answer

Commutative: For any elements $\text{a, b}\in \text{A}$
$\text{a}\ast\text{b} = \text{a + b + ab = b + a + ba = b}\ast \text{a}.$ Hence $\ast$ is commutative
Associative: For any three elements $\text{a, b, c,}\in \text{A}$
$\text{a}\ast \text{(b}\ast\text{c}) = \text{a}\ast\text{(b + c + bc) = a + b + c + bc + ab + ac + abc}$
$\text{(a} \ast\text{b}) \ast\text{c} = \text{(a + b + ab)}\ast \text{c} = \text{a + b + ab + c + ac + bc + abc}$
$\therefore \text{a}{\ast} \text{(b}{\ast}\text{c}) = \text{(a}{\ast} \text{b}) \ast\text{c}, \text{Hence } \ast \text{ is Associative.}$
Identity element: let e $\in$ A be the identity element them $\text{a}\ast \text{e = e}\ast \text{a = a}$
$\Rightarrow\text{a + e + ae = e + a + ea = a}\Rightarrow\text{e(1 + a) = 0, as a} \neq -1 $
$\text{e = 0}$ is the identity element
Invertible: let $\text{a, b}\in \text{A}$ so that ‘b’ is inverse of a
$\therefore \text{a}\ast \text{b = b}\ast\text{a = e}$
$\Rightarrow\text{a + b + ab = b + a + ba = 0}$
$\text{As a}\neq -1, \text{b} = \frac{\text{-a}}{1 + \text{a}} \in \text{A}.$ Hence every element of A is invertible.

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