Answer:
(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.
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