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Given:
\( a*b = a + b + ab, \quad a,b \in \mathbb{Q} \setminus \{-1\} \)
i. Commutativity:
\( a*b = a + b + ab = b + a + ba = b*a \)
✔ Commutative
Associativity:
\( (a*b)*c = a + b + c + ab + bc + ca + abc \)
\( a*(b*c) = a + b + c + ab + bc + ca + abc \)
✔ Associative
ii. Identity Element:
\( a*e = a \Rightarrow a + e + ae = a \Rightarrow e(1+a)=0 \)
Since \( a \neq -1 \), \( (1+a)\neq 0 \), so:
\( e = 0 \)
✔ Identity = 0
iii. Inverse Element:
Find \( b \) such that:
\( a*b = 0 \)
\( a + b + ab = 0 \Rightarrow b(1+a) = -a \)
\( b = \frac{-a}{1+a} \)
✔ Inverse of \( a \) is \( \frac{-a}{1+a} \)
✔ Every element has an inverse
Conclusion:
✔ Commutative & Associative
✔ Identity = 0
✔ Inverse of \( a \) = \( \frac{-a}{1+a} \)