Let 𝑅 be a relation on 𝑍 × 𝑍 defined by (𝑎, 𝑏)𝑅(𝑐, 𝑑) if and only if 𝑎𝑑 − 𝑏𝑐 is divisible by 5 . Then 𝑅 is

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(d) : Reflexive : For (𝑎, 𝑏)𝑅(𝑎, 𝑏) nn⇒ 𝑎𝑏 − 𝑎𝑏 = 0 is divisible by 5 . So, (𝑎, 𝑏)𝑅(𝑎, 𝑏)∀𝑎, 𝑏 ∈ 𝑍 nn∴ 𝑅 is reflexive. nnSymmetric : For (𝑎, 𝑏)𝑅(𝑐, 𝑑), if 𝑎𝑑 − 𝑏𝑐 is divisible by 5 . Then, 𝑏𝑐 − 𝑎𝑑 nnis also divisible by 5 . nnSo, (𝑐, 𝑑)𝑅(𝑎, 𝑏)∀𝑎, 𝑏, 𝑐, 𝑑 ∈ 𝑍 nn∴ 𝑅 is symmetric. nnTransitive : For (𝑎, 𝑏)𝑅(𝑐, 𝑑) ⇒ 𝑎𝑑 − 𝑏𝑐 is divisible by 5 and nn(𝑐, 𝑑)𝑅(𝑒, 𝑓) ⇒ 𝑐𝑓 − 𝑑𝑒 is divisible by 5 nnLet 𝑎𝑑 − 𝑏𝑐 = 5𝑘1 and 𝑐𝑓 − 𝑑𝑒 = 5𝑘2, where 𝑘1 and 𝑘2 are integers. nn∴ 𝑎𝑑𝑓 − 𝑏𝑐𝑓 = 5𝑘1𝑓 (𝑖) nnand 𝑐𝑓𝑏 − 𝑑𝑒𝑏 = 5𝑘2𝑏 nnSolving (i) and (ii), we get nn𝑎𝑑𝑓 − 𝑏𝑐𝑓 + 𝑐𝑓𝑏 − 𝑑𝑒𝑏 = 5𝑘1𝑓 + 5𝑘2𝑏 (𝑖𝑖) nn⇒ 𝑎𝑑𝑓 − 𝑑𝑒𝑏 = 5(𝑘1𝑓 + 𝑘2𝑏) ⇒ 𝑑(𝑎𝑓 − 𝑏𝑒) = 5(𝑘1𝑓 + 𝑘2𝑏) nn𝑎𝑓 - be is not divisible by 5 for ∀𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓 ∈ 𝑍 nnSo, 𝑅 is not transitive. nn∴ 𝑅 is Reflexive and symmetric but not transitive. nn

Explanation

If 𝑅 is the smallest equivalence relation on the set {1,2,3,4} such that {(1,2), (1,3)} ⊂ 𝑅, then the number of elements in 𝑅 is .

Let 𝐴 = {1,3,4,6,9} and 𝐵 = {2,4,5,8,10}. Let 𝑅 be a relation defined on 𝐴 × 𝐵 such that 𝑅 = {((𝑎1, 𝑏1), (𝑎2, 𝑏2))∶ 𝑎1 ≤ 𝑏2 and 𝑏1 ≤ 𝑎2}. Then the number of elements in the set 𝑅 is

Let 𝐴 = {2,3,6,8,9,11} and 𝐵 = {1,4,5,10,15}. Let 𝑅 be a relation on 𝐴 × 𝐵 defined by (𝑎, 𝑏)𝑅(𝑐, 𝑑) if and only if 3𝑎𝑑 − 7𝑏𝑐 is an even integer. Then the relation 𝑅 is

Let 𝐴 = {1,2,3,4,5,6,7}. Then the relation 𝑅 = {(𝑥, 𝑦) ∈ 𝐴 × 𝐴∶ 𝑥 + 𝑦 = 7} is

Let a relation 𝑅 on 𝑁 × 𝑁 be defined as: (𝑥1, 𝑦1)𝑅(𝑥2, 𝑦2) if and only if 𝑥1 ≤ 𝑥2 or 𝑦1 ≤ 𝑦2. Consider the two statements : (I) 𝑅 is reflexive but not symmetric. (II) 𝑅 is transitive Then which one of the following is true?

For x ? 0, the least value of K, for which 4^{1 + x} + 4^{1 - x},\frac{K}{2},16^{x} + 16^{- x} are three consecutive terms of an A.P., is equal to :

The minimum number of elements that must be added to the relation R = {(a,b),(b,c),(b,d)} on the set { a,b,c,d} so that it is an equivalence relation is

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