Let A = Q × Q and let ∗ be a binary operation on A defined by (a,b)∗(c,d) = (ac, b+ad) for (a,b), (c,d) ∈ A. Determine whether ∗ is commutative and associative. Then, with respect to ∗ on A: (i) Find the identity element in A. (ii) Find the invertible element of A.

A=Q×Q... [Given]For any(a,b),(c,d)∈A,∗is defined by(a,b)∗(c,d)=(ac,b+ad). [Given]To check∗is commutative i.e. to check(a,b)∗(c,d)=(c,d)∗(a,b)for any(a,b),(c,d)∈ANow,(a,b)∗(c,d)=(ac,b+ad)(c,d)∗(a,b)=(ca,d+cb)=(ac,d+bc)≠(ac,b+ad)∴(a,b)∗(c,d)≠(c,d)∗(a,b)Thus,∗is not commutative .. (1)To check associativityLet(a,b),(c,d),(e,f)∈ANow,(a,b)∗((c,d)∗(e,f))=(a,b)∗(ce,d+cf)=(ace,b+a(d+cf))=(ace,b+ad+acf). (2)∴(a,b)∗((c,d)∗(e,f))=(ace,b+ad+acf)((a,b)∗(c,d))∗(e,f)=(ac,b+ad)∗(e,f)=(ace,b+ad+acf)=(a,b)∗((c,d)∗(e,f)). From(2)∴(a,b)∗((c,d)∗(e,f))=((a,b)∗(c,d))∗(e,f)Thus,∗is associative .. (3)(i) To find identity elementLete=(a′,b′)be identity element ofA⟹(a,b)∗(a′,b′)=(a,b)=(a′,b′)∗(a,b)As(a,b)∗(a′,b′)=(a,b)⟹(aa′,b+ab′)=(a,b) Using definition of∗⟹aa′=aandb+ab′=b⟹a′=1andb′=0We can verify it as follows(a′,b′)∗(a,b)=(a′a,b′+a′b)=(1⋅a,0+1⋅b)=(a,b)Similarly,(a,b)∗(a′,b′)=(a,b)Hence,e=(1,0)is the identity element inA (ii) To find inverse elementLetf=(c′,d′)be inverse element of(a,b)∈A⟹(a,b)∗(c′,d′)=(1,0)=(c′,d′)∗(a,b).. Using definition of inverse elementNow,(a,b)∗(c′,d′)=(1,0)⟹(ac′,b+ad′)=(1,0) [Using definition of∗]⟹ac′=1andb+ad′=0⟹c′=1aandd′=−baWe can verify it as follows(c′,d′)∗(a,b)=(c′a,d′+c′b)=(1a×a,−ba+1a×b)=(1,0)Hence,f=(1a,−ba)is the inverse element ofA