Answer to Question #267855 in Discrete Mathematics for Jaishree

Question

Determine whether the relation R on the set of all integers

is reflexive, symmetric, antisymmetric, and/or transitive,

where (x, y) ∈ R if and only if

a) x ≠ y. b) xy ≥ 1.

c) x = y + 1 or x = y − 1.

d) x ≡ y (mod 7). e) x is a multiple of y.

f ) x and y are both negative or both nonnegative.

g) x = y2. h) x ≥ y2

Answer 1

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, $0\cdot0<1$

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 $|x|< y$ ,so y is not a multiple of x

transitive: if x is a multiple of y and y is a multiple of z then x is a multiple of z

f)

reflexive: x is negative or nonnegative

symmetric: if x and y are both negative or both nonnegative, then y and x are both negative or both nonnegative

not antisymmetric: if x and y are both negative or both nonnegative, then y and x are both negative or both nonnegative

not transitive: if x and y are both negative and y and z are both nonnegative then x and z are not both nonnegative or both negative