📺 Watch Video Explanation:
Given:
\( S = \mathbb{Q} \setminus \{1\}, \quad a*b = a + b – ab \)
i. Closure (Binary Operation):
\( a*b = a + b – ab = 1 – (1-a)(1-b) \)
If \( a, b \neq 1 \), then \( (1-a) \neq 0 \), \( (1-b) \neq 0 \)
Thus:
\( (1-a)(1-b) \neq 0 \Rightarrow a*b \neq 1 \)
✔ Closed ⇒ Binary operation on \( S \)
ii. Commutativity:
\( a*b = a + b – ab = b + a – ba = b*a \)
✔ Commutative
iii. Associativity:
LHS:
\( (a*b)*c = a + b + c – ab – bc – ca + abc \)
RHS:
\( a*(b*c) = a + b + c – ab – bc – ca + abc \)
Both sides are equal
✔ Associative
Conclusion:
✔ The operation is a binary operation and is both commutative and associative on \( S \).