📺 Watch Video Explanation:
Given:
\( (a,b) \circ (c,d) = (ac, bc + d) \)
i. Commutativity:
\( (a,b)\circ(c,d) = (ac, bc + d) \)
\( (c,d)\circ(a,b) = (ca, da + b) \)
Since:
\( bc + d \neq da + b \ \text{(in general)} \)
❌ NOT commutative
Associativity:
LHS:
\( [(a,b)\circ(c,d)]\circ(e,f) = (ac, bc + d)\circ(e,f) \)
\( = (ace, (bc+d)e + f) = (ace, bce + de + f) \)
RHS:
\( (a,b)\circ[(c,d)\circ(e,f)] = (a,b)\circ(ce, de + f) \)
\( = (ace, b(ce) + (de+f)) = (ace, bce + de + f) \)
✔ Associative
ii. Identity Element:
Let identity be \( (x,y) \)
\( (a,b)\circ(x,y) = (a,b) \)
\( (ax, bx + y) = (a,b) \)
Thus:
\( ax = a \Rightarrow x = 1 \)
\( bx + y = b \Rightarrow b(1)+y=b \Rightarrow y=0 \)
✔ Identity = \( (1,0) \)
iii. Inverse Element:
Let inverse of \( (a,b) \) be \( (c,d) \)
\( (a,b)\circ(c,d) = (1,0) \)
\( (ac, bc + d) = (1,0) \)
Thus:
\( ac = 1 \Rightarrow c = \frac{1}{a} \)
\( bc + d = 0 \Rightarrow d = -\frac{b}{a} \)
✔ Inverse of \( (a,b) = \left(\frac{1}{a}, -\frac{b}{a}\right) \)
Conclusion:
❌ Not commutative
✔ Associative
✔ Identity = (1,0)
✔ Every element invertible