C) A = {1, 2, 3, 4, 5, 6}
R = {(i, j) | |i - j| = 2}.
Find whether R is an equivalence relation or not?
Answer (2)
The relation R is not an equivalence relation because it fails to meet the reflexivity and transitivity properties. While it is symmetric, the lack of these two key properties means it cannot be classified as an equivalence relation. Thus, R does not qualify as an equivalence relation.
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To determine if the relation R = {( i , j ) ∣ ∣ i − j ∣ = 2 } is an equivalence relation on the set A = { 1 , 2 , 3 , 4 , 5 , 6 } , we need to check if it satisfies the three properties of equivalence relations: reflexivity, symmetry, and transitivity.
Reflexivity: A relation R is reflexive if for every element a in A , the pair ( a , a ) is in R . In this case, for R to be reflexive, ∣ a − a ∣ = 0 , which is not equal to 2. Therefore, R is not reflexive.
Symmetry: A relation R is symmetric if whenever ( i , j ) ∈ R , ( j , i ) is also in R . Let us see if ∣ i − j ∣ = 2 implies ∣ j − i ∣ = 2 . Indeed, since ∣ i − j ∣ = ∣ j − i ∣ , R satisfies the symmetry property.
Transitivity: A relation R is transitive if whenever ( i , j ) ∈ R and ( j , k ) ∈ R , then ( i , k ) is also in R . Suppose ∣ i − j ∣ = 2 and ∣ j − k ∣ = 2 . For transitivity, we would need ∣ i − k ∣ = 2 . However, this is not generally true. For example, if i = 1 , j = 3 , and k = 5 , then ∣ i − k ∣ = 4 , which is not 2. Therefore, R is not transitive.
Since the relation R does not satisfy reflexivity and transitivity, R is not an equivalence relation.
