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Question

# Show that each of the relation R in the set A={x∈Z:0≤x≤12}, given by(i) R={(a,b):|a−b|is a multiple of 4}(ii) R={(a,b):a=b}is an equivalence relation. Find the set of all elements related to 1 in each case.

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Solution

## (i)A={x∈Z:0≤x≤12}={0,1,2,3,4,5,6,7,8,9,10,11,12}R={(a,b):|a−b|is a multiple of 4}For any element a∈A, we have (a,a)∈R as |a−a|=0 is a multiple of 4.∴R is reflexive.Now, let (a,b)∈R⇒|a−b| is a multiple of 4.⇒|−(a−b)|=|b−a| is a multiple of 4.⇒(b,a)∈R∴ is symmetric.Now, let (a,b),(b,c)∈R⇒|a−b| is a multiple of 4 and |b−c| is a multiple of 4.⇒(a−b) is a multiple of 4 and (b−c) is a multiple of 4.⇒(a−c)=(a−b)+(b−c) is a multiple of 4.⇒|a−c| is a multiple of 4.⇒(a,c)∈R.∴R is transitive.Hence, R is an equivalence relation.The set of elements related to 1 is {1,5,9} since |1−1|=0 is a multiple of 4,|5−1|=4 is a multiple of 4, and |9−1|=8 is a multiple of 4Hence, R is an equivalence relation.(ii)A={x∈Z:0≤x≤12}={0,1,2,3,4,......,11,12}R={(a,b):a=b}={(0,),(1,1),(2,2),........,(11,11),(12,12)}For any element a∈A, we have (a,a)∈R, since a=a.∴R is reflexive.Now, let (a,b)∈R.⇒a=b⇒b=a⇒(b,a)∈R∴R is symmetric.Now, let (a,b)∈R and (b,c)∈R.⇒a=b and b=c⇒a=c⇒(a,c)∈R∴R is transitive.Hence, R is an equivalence relation.The elements in R that are related to 1 will be those elements from set A which are equal to 1.Hence, the set of elements related to 1 is {1}.

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