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

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.

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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 $\text{a},\text{b}\in\text{X}$ 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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