Define a binary operation ^* on the set 0, 1, 2, 3, 4, 5 as: a = … - Vidyadip

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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