A binary operation * on the set


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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