Question

Define a binary operation *on the set {0,1,2,3,4,5} as
\(a*b =   \begin{cases}     a+b       & \quad \text{if } a+b<6 \\     a+b-6  & \quad \text{if } a+b\geq6    \end{cases}\)
Show that zero is the identity for this operation and each element a≠0 of the set is invertible with 6−a being the inverse of a.

Answer 1

Let X = {0, 1, 2, 3, 4, 5}. 
The operation * on X is defined as:
\(a*b =   \begin{cases}     a+b       & \quad \text{if } a+b<6 \\     a+b-6  & \quad \text{if } a+b\geq6    \end{cases}\)
An element e ∈ X is the identity element for the operation *, 
if \(a*e=a=e*a \,\forall \,a \in X.\)
For a ∈ X, we observed that:
\(a*0=a+0=   \:a\,[a\in X\geq a+0<6]\)
\(0*a=0+a=\:a[a\in X \geq0+a<6]\)
therefore \(\therefore a*0=a=0*a\,\forall\,a\in X.\)
Thus, 0 is the identity element for the given operation*.
An element a ∈ X is invertible if there exists b∈ X such that a * b = 0 = b * a
i.e a+b=0=b+a,
if a+b<6.
a+b-6=b+a-6
i.e., a = −b or b = 6 − a
But, X = {0, 1, 2, 3, 4, 5} and a, b ∈ X. 
Then, a ≠ −b.
∴ b = 6 − a is the inverse of a \(\forall\) a ∈ X.

Hence, the inverse of an element a ∈ X, a ≠ 0 is 6 − a i.e., \(a^{-1}=6-a.\)