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# Define a binary operation $\ast$ on the set $\{0, 1, 2, 3, 4, 5\}$ as $a \ast b = \left\{ \begin{array} {1 1} a+b, & \quad \text{ if a+b < 6} \\ a+b-6, & \quad \text{ if a+b \geq 6} \\ \end{array} \right.$ Show that zero is the identity for this operation and each element $a\neq0$ of the set is invertible with $6-a$ being the inverse of $a$.

Toolbox:
• An element $e \in N$ is an identify element for operation * if $a*e=e*a$ for all $a \in N$
• The element $a \in X$ is invertible if there exist $b \in X$ such that $a*b=e=b*a$
Given the set $X=\{0,1,2,3,4,5\}$ where the binary operation $\ast$ is defined by $a * b= \left\{ \begin{array}{1 1} a+b & \quad if\;a+b < 6\\ a+b-6 & \quad if a+b \geq 6 \end{array} \right.$
$\textbf {Step 1: Checking if zero is the identity}$:
An element $e \in N$ is an identify element for operation * if $a*e=e*a$ for all $a \in N$
To check if zero is the identity, we see that $a*0=a+0=a \qquad for\;a \in x$ and also $0*a=0+a=a \qquad for \;a \in x$
Given $a \in X, \qquad a+0 < 6\;$ and also $\;0+a < 6$
$\Rightarrow 0$ is the identify element for the given given operation
$\textbf {Step 2: Proving that the inverse of a is 6-a}$:
The element $a \in X$ is invertible if there exist $b \in X$ such that $a*b=e=b*a$
In this case, $e=0 \rightarrow a*b=0=b*a$.
$\Rightarrow a*b = \left\{ \begin{array}{1 1} a+b=0=b+a & \quad if\;a+b < 6\\ a+b-6=0=b+a-6 & \quad a+b \geq 6 \end{array} \right.$
ie $a=-b \;or\; b=6-a$
but since $a,b \in X=\{0,1,2,3,4,5\}$, $\;a \neq -b$
Hence $b=6-a\;$ is the inverse of $a$, i.e., $a^{-1}=6-a, \;\forall a \in \{1,2,3,4,5\}$
edited Mar 20, 2013