For each relation, indicate whether it is reflexive or anti-reflexive, symmetric or anti-symmetric, transitive or not transitive. a. The domain of the relation L is the set of all real numbers. For x, y ∈ R, xLy if x < y.

b. The domain for the relation D is the set of all integers. For any two integers, x and y, xDy if x evenly divides y. An integer x evenly divides y if there is another integer n such that y = xn. (Note that the domain is the set of all integers, not just positive integers.)

This is False

Example

|-3+7| = |-3| + |7|

|4| = 3+7

4 = 10

False

True because of the absolute value bars

(a)

L is not reflexive, L is anti-reflexive

L is not symmetric.

L is not anti-symmetric

L is transitive.

(b)

D is reflexive

D is not symmetric.

D is anti-symmetric

D is transitive.

Step-by-step explanation:

a)

Given that;

domain of the given relation L is the set of all real numbers

For x , y ∈ R , xLy if x less than y.

relation L, where xLy if x less than y, For x, y ∈ R

so For every x ∈ R, it is then false that x less than x.

That is (x, x) does not belongs to L.

∴ L is not reflexive, L is anti-reflexive.

For every x,y ∈ R, if (x,y) ∈ L (i.e. x < y), then (y, x) does not belongs to L, since it is false that y < x.

∴ L is not symmetric.

For every x ∈ R, we can say its false that x less than x. That is (x, x) does not belongs to L.

∴ L is not anti-symmetric.

For every x,y,z ∈ R, if (x, y) ∈ L(i.e. x < y) and (y, z) ∈ L(i.e. y < z), then (x, z) ∈ L, since it is true that x<z when x<y and y<z.

∴ L is transitive.

b)

Also lets consider a relation D, where xDy if there is an integer n such that y = xn, For x, y ∈ Z.

Now

For every x ∈ Z, it is true that x = x × 1. That is (x, x) belongs to D.

∴ D is reflexive,

For every x,y ∈ Z, if (x,y) ∈ P (i.e. y=x × n), then (y, x).

if (x,y) ∈ D, then there exist an integer n such that y=x × n.

Then x = y × (1/n).

Thus, (y,x) does not belongs to D, since 1/n is not an integer, but it is a real number.

∴ D is not symmetric.

For every x,y ∈ Z, if (x,y) ∈ D (i.e. y=x × n), then (y, x) also belongs to D only when x=y. where n=1.

∴ D is anti-symmetric.

For every x,y,z ∈ Z, if (x,y) ∈ D (i.e. x × n1= y), and (y,z) ∈ D (i.e. y × n2 = z), then (x,z)∈ D.

if (x,y) ∈ D, then there exist an integer n1 such that y=x × n1.

if (y,z) ∈ D, then there exist an integer n2 such that z=y × n2.

Thenz = (x × n1) × n2 ⇒ z=x × (n1 × n2).

where (n1 × n2) is an integer. Thus (x,z) ∈ Z

∴ D is transitive.

False

Step-by-step explanation:

|x+y| = |x| + |y|

To show this is false, all we have to do is find one example where it is false

Let x = -1 and y = 4

|-1+4| = |-1| + |4|

|3| = |1| + |4|

3 = 5

This is false so we have a set of integers where the statement is not true