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Question 83 of 84

Q.Verify

(i) Closure property
(ii) Associative property and
(iii) Existence of identity for the following operation on the given set : m∗n=m+n−mn; m,n∈Zm*n=m+n-mn;\ m, n\in Z
Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2026Subjective· 3mImportance★★★★★
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Concept understanding — Properties of Binary Operations (Closure, Commutative, Associative, Identity, Inverse)

Once an operation ∗* is confirmed binary on a set SS (so closure automatically holds), four more properties decide how "arithmetic-like" it behaves.

Commutative property. ∗* is commutative if a∗b=b∗aa*b=b*a for every a,b∈Sa,b\in S -- order does not matter. Ordinary +,×+,\times on numbers are commutative; ordinary −- is not (4−5≠5−44-5\ne5-4).

Associative property. ∗* is associative if (a∗b)∗c=a∗(b∗c)(a*b)*c=a*(b*c) for every a,b,c∈Sa,b,c\in S -- grouping does not matter, so a chain a∗b∗ca*b*c is unambiguous. Ordinary −- fails this too: (4−5)−7=−8(4-5)-7=-8 but 4−(5−7)=64-(5-7)=6.

Existence of identity. An element e∈Se\in S is an identity element for ∗* if a∗e=a=e∗aa*e=a=e*a for every a∈Sa\in S. For ++ on Z\mathbb Z, e=0e=0; for ×\times on Q\mathbb Q, e=1e=1.

Existence of inverse. If an identity ee exists, then b∈Sb\in S is the inverse of aa (written b=a−1b=a^{-1}) if a∗b=e=b∗aa*b=e=b*a. For ++ on Z\mathbb Z, the inverse of mm is −m-m; for ×\times on Q\mathbb Q, the inverse of a nonzero xx is 1x\tfrac1x. (The notation a−1a^{-1} names an element, not the fraction 1a\tfrac1a.)

Uniqueness is guaranteed, not assumed.

Theorem 12.1 (Uniqueness of Identity). If an algebraic structure (S,∗)(S,*) has an identity element, it has only one. Proof idea: if e1,e2e_1,e_2 are both identities, treat e1e_1 as the identity acting on e2e_2 to get e1∗e2=e2e_1*e_2=e_2, then treat e2e_2 as the identity acting on e1e_1 to get e1∗e2=e1e_1*e_2=e_1; comparing gives e1=e2e_1=e_2.

Theorem 12.2 (Uniqueness of Inverse). If a∈Sa\in S has an inverse, it has only one. Proof idea: if a1,a2a_1,a_2 are both inverses of aa, then a1=a1∗e=a1∗(a∗a2)=(a1∗a)∗a2=e∗a2=a2a_1=a_1*e=a_1*(a*a_2)=(a_1*a)*a_2=e*a_2=a_2. …

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