Question:
For the binary operation \( \times_7 \) on the set \( S = \{1,2,3,4,5,6\} \), compute \( 3^{-1} \times_7 4 \).
Concept:
The inverse of \( a \) under modulo 7 satisfies:
\[ a \times a^{-1} \equiv 1 \pmod{7} \]
Then compute the required operation using modulo 7.
Solution:
Step 1: Find inverse of 3 modulo 7
- \( 3 \times 1 = 3 \not\equiv 1 \)
- \( 3 \times 2 = 6 \not\equiv 1 \)
- \( 3 \times 3 = 9 \equiv 2 \)
- \( 3 \times 4 = 12 \equiv 5 \)
- \( 3 \times 5 = 15 \equiv 1 \) ✅
So, \( 3^{-1} = 5 \) (mod 7)
Step 2: Compute the expression
\[ 3^{-1} \times_7 4 = 5 \times_7 4 \]
\[ 5 \times 4 = 20 \equiv 6 \pmod{7} \]
Final Answer:
\[ \boxed{6} \]