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Q.Show that the relation R in the set Z of integers given by R={(a,b):2 divides (a−b)}R = \{(a,b) : 2 \text{ divides } (a-b)\} is an equivalence relation.

(OR)
In a set of natural numbers N, a binary operation ∗*, defined by a∗b=a * b = L.C.M. of a and b. Is ∗* associative? Find the identity element of ∗* in N.
Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2022Subjective· 4mImportance★★★★★
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Check reflexivity, symmetry, transitivity directly from the definition 2∣(a−b)2\mid(a-b). (OR: use associativity of LCM; identity is the element ee with lcm(a,e)=a\text{lcm}(a,e)=a.)

Main part. R={(a,b):2 divides (a−b)}R=\{(a,b): 2 \text{ divides } (a-b)\} on Z\mathbf Z.

Reflexive: For any a∈Za\in\mathbf Z, a−a=0a-a=0 and 2∣02\mid 0, so (a,a)∈R(a,a)\in R for all aa. Reflexive.

Symmetric: If (a,b)∈R(a,b)\in R, then 2∣(a−b)2\mid(a-b), i.e. a−b=2ka-b=2k for some integer kk. Then b−a=−2k=2(−k)b-a=-2k=2(-k), so 2∣(b−a)2\mid(b-a), giving (b,a)∈R(b,a)\in R. Symmetric.

Transitive: If (a,b)∈R(a,b)\in R and (b,c)∈R(b,c)\in R, then a−b=2ka-b=2k and b−c=2mb-c=2m for integers k,mk,m. Adding, a−c=(a−b)+(b−c)=2(k+m)a-c=(a-b)+(b-c)=2(k+m), so 2∣(a−c)2\mid(a-c), giving (a,c)∈R(a,c)\in R. Transitive.

Since R is reflexive, symmetric and transitive, R is an equivalence relation.

OR. a∗b=LCM(a,b)a*b=\text{LCM}(a,b) on N.

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