Define a binary operation *on the set {0, 1, 2, 3, 4, 5} as

Let X = {0, 1, 2, 3, 4, 5}.

The operation * on X is defined as:

$a * b= \begin{cases}a+b & \text { if } a+b<6 \\ a+b-6 & \text { if } a+b \geq 6\end{cases}$

An element $e \in X$ is the identity element for the operation *, if $a * e=a=e * a \forall a \in X$.

For $a \in X$, we observed that:

$a * 0=a+0=a \quad[a \in X \Rightarrow a+0<6]$

$0 * a=0+a=a \quad[a \in X \Rightarrow 0+a<6]$

$\therefore a * 0=a=0 * a \forall a \in X$

Thus, 0 is the identity element for the given operation *.

An element $a \in X$ is invertible if there exists $b \in X$ such that $a^{*} b=0=b^{*} a$.

i.e. $\begin{cases}a+b=0=b+a, & \text { if } a+b<6 \\ a+b-6=0=b+a-6, & \text { if } a+b \geq 6\end{cases}$

i.e.,

$a=-b$ or $b=6-a$

But, $X=\{0,1,2,3,4,5\}$ and $a, b \in X .$ Then, $a \neq-b .$

∴b = 6 − a is the inverse of a &mnForE; a ∈ X.

Hence, the inverse of an element $a \in X, a \neq 0$ is $6-a$ i.e., $a^{-1}=6-a$.