Solution to a) not reflexive: x = x symmetric: if x ≠ y then y ≠ x … - Sikademy
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Archangel Macsika

a) not reflexive: x = x symmetric: if x ≠ y then y ≠ x not antisymmetric: if x ≠ y then y ≠ x not transitive: for example, if 5 ≠ 6 and 6 ≠ 5 then 5=5 b) not reflexive: for example, 0\cdot0<10⋅0<1 Is symmetric because we have xy = yx. Not antisymmetric because we have xy = yx. Is transitive because if we have (a, b) ∈ R and that (b, c) ∈ R, it follows that (a, c) ∈ R. c) Not reflexive because we can’t have (1, 1) Is symmetric because we have x = y + 1 and y = x − 1. They are equivalent equations. Not antisymmetric because of the same reason above. Not transitive because if we have (1, 2) and (2, 1) in the relation, (1, 1) is not in relation. g) Not reflexive because (2, 2) does not satisfy. Not symmetric because although we can have (9, 3), we can’t have (3, 9). Is antisymmetric because each integer will map to another integer but not in reverse (besides 0 and 1). Not transitive because if we have (16, 4) and (4, 2), it’s not the case that 16 = 22 . h) Not reflexive because we can’t have (2, 2). Not symmetric because if we have (9, 3), we can’t have (3, 9). Is antisymmetric, because each integer will map to another integer but not in reverse (besides 0 and 1). Is transitive because if x ≥ y2 and y ≥ z2 , then x ≥ z2 d) reflexive: x - x = 0 is divided by 7 symmetric: if x-y is divided by 7 then y-x is divided by 7 not antisymmetric: if x-y is divided by 7 then y-x is divided by 7 transitive: if x-y is divided by 7 and y-z is divided by 7 then x-z is divided by 7 e) reflexive: x is a multiple of x not symmetric: for example, 2 is a multiple of 4 but 4 is a multiple of 2 antisymmetric: if x ≠ y then |x|< y∣x∣

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