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

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Solution

An element $e \in Q$ will be the identity element for the operation $$ if
$a b = (a - b)^{2}$
If $a * e = a$ , then $(a - e)^{2} = a$ , A square is always positive. so for
$a = - 2, (- 2 - e)^{2} \neq - 2$
Hence, there is no identity element.

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Let ∗ be a binary operation on the set Q of rational number as follows: (iv)a∗b=(a−b)2 Show that none of the operations has an identity.

An element e∈Q will be the identity element for the operation ∗ if a∗b=(a−b)2 If a∗e=a, then (a−e)2=a, A square is always positive. so for a=−2,(−2−e)2≠−2 Hence, there is no identity element.

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

An element $e \in Q$ will be the identity element for the operation $$ if
$a b = (a - b)^{2}$
If $a * e = a$ , then $(a - e)^{2} = a$ , A square is always positive. so for
$a = - 2, (- 2 - e)^{2} \neq - 2$
Hence, there is no identity element.