See if | defines an order in ℤ

To show this we have to check a|b (a divides b ) satisfy  order axioms or not

Order axioms on < are

1. ∀a,b ∈ ℝ;exactly one of a = b; a < b; b < a holds: (Trichotomy)
2. ∀a,b,c ∈ ℝ a < b and b < c implies a < c: Transitivity
3. ∀a,b,c ∈ ℝ,a < b implies a + c < b + c: operations with addition
4. ∀a,b ∈ ℝ,a < b and 0 < c implies ac < bc: operations with multiplication

Now order axioms on | (if satisfy..) 

1. ∀a,b ∈ ℝ;exactly one of not(a | b); a | b; b | a holds: (Trichotomy)
2. ∀a,b,c ∈ ℝ a | b and b | c implies a | c: Transitivity
3. ∀a,b,c ∈ ℝ,a |b implies a + c | b + c: operations with addition
4. ∀a,b ∈ ℝ,a | b and 0 | c implies ac | bc: operations with multiplication