Define a binary operation $^*$ on the set $\{0, 1, 2, 3, 4, 5\}$ as $a \,^* \, b = \begin{cases} a+b, & \text{If } a+b < 6 \\ a+b-6, & \text{If } a+b \geq 6 \end{cases}$. Show that $0$ is the identity for this operation and each element $a \neq 0$ of the set is invertible with $6-a$ being the inverse of $a$.

(A) 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 $a \,^* \, e = a = e \,^* \, a$ for all $a \in X$.
For $a \in X$,we have:
$a \,^* \, 0 = a + 0 = a$ (since $a \in X \Rightarrow a+0 < 6$)
$0 \,^* \, a = 0 + a = a$ (since $a \in X \Rightarrow 0+a < 6$)
Therefore,$a \,^* \, 0 = a = 0 \,^* \, a$ for all $a \in X$.
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 $a \,^* \, b = 0 = b \,^* \, a$.
If $a \neq 0$,let $b = 6-a$. Since $a \in \{1, 2, 3, 4, 5\}$,$b \in \{5, 4, 3, 2, 1\} \subset X$.
Then $a \,^* \, b = a + (6-a) - 6 = 0$ (since $a+b = 6 \geq 6$).
Similarly,$b \,^* \, a = (6-a) + a - 6 = 0$.
Thus,$6-a$ is the inverse of $a$ for all $a \in X, a \neq 0$.