Question
Define a binary operation * on the set {0, 1, 2, 3, 4, 5} as $\text{a}*\text{b}=\begin{cases}\text{a + b},&\text{if a + b}<6\\\text{a + b}-6&\text{if a + b}\geq6\end{cases}$
Show that zero is the identity for this operation and each element a 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:
$\text{a}*\text{b}=\begin{cases}\text{a + b},&\text{if a + b}<6\\\text{a + b}-6&\text{if a + b}\geq6\end{cases}$
An element $\text{e}\in\text{X}$ is the identity element for the operation *, if $\text{a}*\text{e}=\text{a}=\text{e}*\text{a}\ \forall\text{a}\in\text{X}.$
For $\text{a}\in\text{X},$ we observed that:
$\text{a}*0=\text{a}+0=\text{a}\ \ \ [\text{a}\in\text{X}\Rightarrow\text{a}+0<6]$
$0*\text{a}=0+\text{a}=\text{a}\ \ \ [\text{a}\in\text{X}\Rightarrow0+\text{a}<6]$
$\therefore\text{A}*0=\text{a}=0*\text{a}\ \forall\text{a}\in\text{X}$
Thus, 0 is the identity element for the given operation*.
An element $\text{a}\in\text{X}$ is invertible if there exists $\text{b}\in\text{X}$ such that a * b = 0 = b * a.
$\text{i.e.},\begin{cases}\text{a + b}=0=\text{b + a},&\text{if a + b}<6\\\text{a + b}-6=0=\text{b + a}-6,&\text{if a + b}\geq6\end{cases}$
i.e.,
a = -b or b = 6 - a
But, X = {0, 1, 2, 3, 4, 5} and $\text{a},\text{b}\in\text{X}.\ \text{Then},\text{a}\neq-\text{b}.$
$\therefore$ b = 6 - a is the inverse of $\text{a }\Box\text{ a}\in\text{X}.$
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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