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itz_yash·
Can someone Help me know why the Third Operation doesnot have an identity element ?. Cuz for a^z then for z=1 we will always get a , Ultimately resulting z=1 as the identity element then how come the answer is correct for Option A ?.
KG AI MentorKG AI Mentor
Quick AnswerGlad it helped — the reason is that an identity must work on both sides. For f(x,y)=x^y, 1 is only a right-identity (a^1 = a) but not a left-identity (1^a = 1 ≠ a for a>1), so no two-sided identity. Key StepsIdentity z must satisfy f(a,z)=a and f(z,a)=a for every a ∈ N. For f(x,y)=x^y: f(a,1)=a (right OK) but f(1,a)=1 (left fails; e.g. f(1,2)=1 ≠ 2). Hence only (I) f(x,y)=x+y−3 has z=3, and (II) f(x,y)=max(x,y) has z=1 (if N={1,2,…}), so option (A) is correct. Well done for checking the left/right cases! — KG AI Mentor 🤖
KG AI MentorKG AI Mentor
Quick AnswerThe third operation has no identity. z = 1 is only a right-identity (a^1 = a) but fails as a left-identity because 1^a = 1 ≠ a for a > 1. Key StepsIdentity z must satisfy f(a,z) = a = f(z,a) for all a ∈ N (both left and right). For f(x,y) = x^y: f(a,1) = a OK, but f(1,a) = 1 which is not a when a > 1 (e.g. f(2,1)=2 but f(1,2)=1), so no two-sided identity. For completeness: f(x,y) = x + y − 3 has identity z = 3 (a+3-3=a and 3+a-3=a), and f(x,y)=max(x,y) has identity equal to the minimum of N (usually 1), since max(a,1)=a=max(1,a). — KG AI Mentor 🤖
@kg_ai_mentor Ok i understood , thanks for solving my doubt