Question:
Define a binary operation \( * \) on the set \( S = \{0,1,2,3,4,5\} \) 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 element and each element \( a \neq 0 \) is invertible with inverse \( 6 – a \).
Solution:
1. Identity Element:
Check:
\[ a * 0 = \begin{cases} a + 0 = a \end{cases} \quad \text{and} \quad 0 * a = \begin{cases} 0 + a = a \end{cases} \]
Thus, \( a * 0 = 0 * a = a \) for all \( a \in S \).
So, 0 is the identity element.
—2. Inverse of an Element:
We need \( b \) such that:
\[ a * b = 0 \]
Using definition:
\[ a + b \equiv 0 \pmod{6} \]
So,
\[ b = 6 – a \]
Verification:
- \( 1 * 5 = 6 \equiv 0 \)
- \( 2 * 4 = 6 \equiv 0 \)
- \( 3 * 3 = 6 \equiv 0 \)
- \( 4 * 2 = 6 \equiv 0 \)
- \( 5 * 1 = 6 \equiv 0 \)
Thus, each element has an inverse.
—Final Answer:
0 is the identity element and the inverse of each \( a \neq 0 \) is \( 6 – a \).