A binary operation $*$ on the set $(0,1,2,3,4,5)$ is defined as
$a * b= \begin{cases}a+b ; & \text { if } a+b<6 \\ a+b-6 ; & \text { if } a+b \geq 6\end{cases}$
Show that 0 is the identity for this operation and each element a has an inverse (6 - a)
To find: identity and inverse element
For a binary operation if $a * e=a$, then e $s$ called the right identity
If $\mathrm{e}^{*} \mathrm{a}=\mathrm{a}$ then $\mathrm{e}$ is called the left identity
For the given binary operation,
$e^{*} b=b$
⇒ e + b = b
⇒ e = 0 which is less than 6.
b*e = b
⇒ b + e = b
⇒ e = 0 which is less than 6
For the $2^{\text {nd }}$ condition,
e*b = b
$\Rightarrow \mathrm{e}+\mathrm{b}-6=\mathrm{b}$
$\Rightarrow \mathrm{e}=6$
But $e=6$ does not belong to the given set $(0,1,2,3,4,5)$
So the identity element is 0
An element $c$ is said to be the inverse of $a$, if $a * c=e$ where $e$ is the identity element (in our case it is $0)$
$a^{*} c=e$
$\Rightarrow \mathrm{a}+\mathrm{c}=\mathrm{e}$
$\Rightarrow \mathrm{a}+\mathrm{c}=0$
$\Rightarrow \mathrm{c}=-\mathrm{a}$
a belongs to $(0,1,2,3,4,5)$
- a belongs to $(0,-1,-2,-3,-4,-5)$
So $c$ belongs to $(0,-1,-2,-3,-4,-5)$
So $c=-a$ is not the inverse for all elements a
Putting in the $2^{\text {nd }}$ condition
$a^{*} c=e$
$\Rightarrow a+c-6=0$
$\Rightarrow c=6-a$
$0 \leq a<6$
$\Rightarrow-6 \leq-a<0 \Rightarrow 0 \leq 6-a<60 \leq c<5$
So $c$ belongs to the given set
Hence the inverse of the element $\mathrm{a}$ is $(6-\mathrm{a})$
Hence proved
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