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Q.Test whether the relation R={(m,n):3 divides m−n}R=\{(m,n):3\text{ divides }m-n\} on {1,2,3,…,10}\{1,2,3,\ldots,10\} is reflexive, symmetric or transitive. What is the conclusion?

Odisha ChseOdisha CHSE +2 Science Board Exam 2026Subjective· 5mImportance★★★★★
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R={(m,n):3∣(m−n)}R=\{(m,n):3\mid(m-n)\} satisfies all three properties, so it is an equivalence relation.

Reflexivity: for any mm, m−m=0m-m=0, and 3∣03\mid0. So (m,m)∈R(m,m)\in R for all mm — RR is reflexive.

Symmetry: if (m,n)∈R(m,n)\in R, then 3∣(m−n)3\mid(m-n), i.e. m−n=3km-n=3k for some integer kk. Then n−m=−3k=3(−k)n-m=-3k=3(-k), so 3∣(n−m)3\mid(n-m), i.e. (n,m)∈R(n,m)\in R — RR is symmetric.

Transitivity: if (m,n)∈R(m,n)\in R and (n,p)∈R(n,p)\in R, then m−n=3k1m-n=3k_1 and n−p=3k2n-p=3k_2 for integers k1,k2k_1,k_2. Adding:

(m−n)+(n−p)=m−p=3(k1+k2)(m-n)+(n-p)=m-p=3(k_1+k_2)

so 3∣(m−p)3\mid(m-p), i.e. (m,p)∈R(m,p)\in R — RR is transitive.

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