Find the identity element for the binary operation on set $Q$ of rational numbers defined as follows:
(i) $a * b = a^{2} + b^{2}$ (ii) $a * b = (a - b)^{2}$
(ii) $a * b = a b^{2}$

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Solution

$i)$ Let $e$ be an identity element & $a \times e = e \times a ⟹ a^{2} + e^{2} = e^{2} + a^{2} .$
$∴$ There is no possible value of $e$ where $a^{2} + e^{2} = e^{2} + a^{2} .$

$i i)$ Let $e$ be an identity element & $a \times e = e \times a ⟹ {(a - e)}^{2} = {(e - a)}^{2} ⟹ a^{2} + e^{2} - 2 a e = e^{2} + a^{2} - 2 a e .$
There is no possible value of $e$ where $a^{2} + e^{2} - 2 a e = e^{2} + a^{2} - 2 a e .$

$i i i)$ Let $e$ be an identity element & $a \times e = e \times a = a ⟹ {a e}^{2} = {e a}^{2} = a ⟹ a e (e - a) = 0$
$∴ e = 0$ or $e = a .$

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Similar questions

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Q

a * b = a - b

a * b = a^{2} + b^{2}

a * b = a + a b

a * b = (a - b)^{2}

a * b = \frac{a b}{4}

a * b = a b^{2}

a * b = \sqrt{a^{2} + b^{2}} for all a, b \in R .

Let $*$ be a binary operation on the set Q of rational number as follows:
$(i) a * b = a^{2} + b^{2}$
Show that none of the opeartion has identity.

Let $*$ be a binary operation on the set Q of rational number as follows:
$(i) a * b = a - b$

$(i i) a * b = a^{2} + b^{2}$

$(i i i) a * b = a + a b$

$(i v) a * b = (a - b)^{2}$

$(v) a * b = \frac{a b}{4}$

$(v i) a * b = a b^{2}$
Show that none of the operation has identity.

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