$a$ 米 $b= \begin{cases}a+b, & \text { if } a+b<6 \\ a+b-6 & \text { if } a+b \geq 6\end{cases}$ Show that zero is the identity for this operation and each element $a \neq 0$ of the set is invertible with 6 - a being the inverse of $a$.

Solution:

Let $X=\{0,1,2,3,4,5\}$.

The operation * on $\mathrm{X}$ is defined as:

$a * b= \begin{cases}a+b & \text { if } a+b\lt6 \\\\ a+b-6 & \text { if } a+b \geq 6\end{cases}$

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:

$ \begin{aligned} &a * 0=a+0=a \quad[a \in X \Rightarrow a+0\lt6] \\\\ &0 * a=0+a=a \quad[a \in X \Rightarrow 0+a\lt6] \\\\ &\therefore a * 0=a=0 * a \forall a \in X \\ \end{aligned} $

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., $\begin{cases}a+b=0=b+a, & \text { if } a+b\lt6 \\\\ a+b-6=0=b+a-6, & \text { if } a+b \geq 6\end{cases}$

i.e.,

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

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

$\therefore b=6-a$ is the inverse of $a \square a \in X$.

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