Question

Define a binary operation $\ast$ on the set $\{0,1,2,3,4,5\}$ as
$a\ast b=\{\begin{array}{c}a+b,ifa+b<6\\ a+b-6ifa+b\ge 6\end{array}$

Show that zero is the identity for this operation and each element $a\ne 0$ of the set is invertible with $6-a$ being the inverse of $a$.

Answer 1

Given the set $X=0,1,2,3,4,5$ where the binary operation $\ast$ is defined by

$\Rightarrow a\ast b=\{a+bifa+b<6a+b-6ifa+b\ge 6\}$

An element $e\in N$ is an identity element for operation

$\ast$ if $a\ast e=e\ast a$ for all $a\in N$

To check if zero is the identity, we see that

$a\ast 0=a+0=afora\in x$ and also

$0\ast a=0+aa\in x$

Given $a\in X$ $a+0<6$ and also $0+a<6$

$\Rightarrow 0$ is the identity element for the given operation

The element $a\in X$ is invertible if there exist $b\in X$ such that

$a\ast b=e=b\ast a$

In this case, $e=0\to a\ast b=0=b\ast a$

$\Rightarrow a\ast b=\{a+b=0=b+aifa+b<6a+b-6=0=b+a-6ifa+b\ge 6\}$

i.e $a=-borb=6-a$

But since $a,b\in X=0,1,2,3,4,5,a\ne -b$

Hence $b=6-a$is the inverse of a i.e

$a^{-1}=6-a,\forall a\in 1,2,3,4,5$