Question
Define a binary operation * on the set {0, 1, 2, 3, 4, 5} as:
$\text{a}\times\text{b}=\begin{cases}\text{a + b},&\text{if }\text{a + b}<6\\\text{a + b}-6,&\text{if }\text{a + b}\geq6\end{cases}$
Show that 0 is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 − a being the inverse of a.

Answer

Let X = {0, 1, 2, 3, 4, 5}.
The operation * on X is defined as:
$\text{a}\times\text{b}=\begin{cases}\text{a + b},&\text{if }\text{a + b}<6\\\text{a + b}-6,&\text{if }\text{a + b}\geq6\end{cases}$
An element $\text{e}\in\text{X}$ is the identity element for the operation *, if 
a * e = a = e * a $\forall\text{ a}\in\text{X}$.
For $\text{a}\in\text{X}$, we observed that:
a * 0 = a + 0 = a $[\text{a}\in\text{X}\Rightarrow\text{a}+0<6]$
0 * a = a = 0 * a $[\text{a}\in\text{X}\Rightarrow0+\text{a}<6]$
$\therefore$ a * 0 = a = 0 * a $\forall\text{ a}\in\text{X}$
Thus, 0 is the identity element for the given operation *.
An element $\text{a}\in\text{X}$ is invertible if there exists be X such that a * b = 0 = b * a.
i.e.,$\begin{cases}\text{a + b}=0=\text{b + a},&\text{if }\text{a + b}<6\\\text{a + b}-6=0=\text{b + a}-6,&\text{if }\text{a + b}\geq6\end{cases}$
i.e.,
a = -b or b = 6 - a
But, X = {0, 1, 2, 3, 4, 5} and $\text{a, b}\in\text{X}$. Then, $\text{a}\neq-\text{b}$.
Therefore, b = 6 - a is the inverse of $\text{a}\in\text{X}$.
Hence, the inverse of an element $\text{a}\in\text{X},\text{a}\neq0$ is 6 - a i.e., a-1 = 6 - a.

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