Question

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

$\left(ii\right)a\ast b=a^2+b^2$

$\left(iii\right)a\ast b=a+ab$

$\left(iv\right)a\ast b=(a-b{)}^2$

$\left(v\right)a\ast b=\frac{ab}{4}$

$\left(vi\right)a\ast b=a{b}^2$
Show that none of the operation has identity.

Answer 1

An element $e\in Q$ will be the identity element for the operation $\ast$ if
$a\ast e=a=e\ast a,\forall a\in Q$
$\left(i\right)a\ast b=a-b$
If $a\ast e=a,a\ne 0\Rightarrow a-e=a,a\ne 0\Rightarrow e=0$
Also, $e\ast a=a\Rightarrow e-a=a\Rightarrow e=2a$
$\therefore e=0=2a,a\ne 0$
But the identity is unique. Hence this operation has no identity.

An element $e\in Q$ will be the identity element for the operation $\ast$ if
$a\ast e=a=e\ast a,\forall a\in Qa\ast b=a^2+b^2$
If $a\ast e=a$, then $a^2+e^2=a$
For $a=-2,(-2{)}^4+e^2=4+e^2\ne -2$
Hence, there is no identity element.

An element $e\in Q$ will be the identity element for the operation $\ast$ if
$a\ast b=a+ab$
If a $\ast e=a\Rightarrow a+ae=a\Rightarrow ae=0\Rightarrow e=0,a\ne 0$
Also if
$\Rightarrow e\ast a=a\Rightarrow e+ea=a\Rightarrow e=\frac{a}{1-a},a\ne 1$
$\therefore e=0=\frac{a}{1-a},a\ne 0$
But the identity is unique. Hence this operation has no identity.

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

An element $e\in Q$ will be the identity element for the operation $\ast$ if
$a\ast b=\frac{ab}{4}$
If $a\ast e=a,then\frac{ae}{4}=a$. Hence, e =4 is the identity element.
$\therefore a\ast 4=4\ast a=\frac{4a}{4}=a$

An element $e\in Q$ will be the identity element for the operation $\ast$ if
$a\ast b=a{b}^2$
If $a\ast e=a\Rightarrow a{e}^2=a$
$\Rightarrow e^2=1\Rightarrow e=\pm 1$
But identity is unique. Hence this operation has no identity.
Therefore only part (v) has an identity element.