Is my proof correct? 'let a,b​​∈ Z. We write A | B if A divides B. Is the relation |, symmetric, transitive and/or reflexive?' The relationship is not symmetrical. When a relationship is symmetrical: if xRy implies yRx for all x, y ∈ A (where A is a non-empty set, and R is a relation in A)
If a, b ​​∈ Z, and as a | b means that there is an integer k such that b = a k ∙. When we take a = 1 and b = 2 applies: b = a ∙ k → k = 2. So for a = 1 and b = 2 holds a | b, for k ∈ Z.
If a, b ​​∈ Z, and if b | a is that there is an integer k such that a = b ∙ k. When we take again a = 1 and b = 2, a = b ∙ k → k = 1/2. So for a = 1 and b = 2 is not that b | a ┤ because k ∉ Z.
This implies that f is not symmetrical.
The relationship is transitive. When a, b, c ∈ Z, and as a | b and b | a, is that there is an integer k such that b = a ∙ k and l is an integer that there exists such that c = b ∙ l. This means that
c = a b ∙ ∙ l. Because a, b, c ∈ Z means that the b ∙ l ∈ Z (an integer multiplied by an integer indicates an integer). This means that there exists an integer m (where m = a ∙ b) such that c = m ∙ A. It follows that a | c, so the relationship is the transition.
The relationship is reflexive when (∀a ∈ Z)a|a. because every a ∈ Z can be divided by itself, we say that it is reflexive.

After Looking it through again I now have my proof like this:
(a) Since 0 does not divide 0,  "|" is not reﬂexive.
(b) 2 divides 4 so 2 | 4. But 4 does not divide 2, so 4 does not divide 2. Thus, "|" is not symmetric.
(c) To see that  is transitive, let a, b, c be integers. Suppose that a| b and b |c. Thus,
a divides b and b divides c so there exist integers k and l such that b = ak and c = bl. This
gives c = bl = (ak)l = a(kl). Therefore, a divides c so a | c.
Is this prove correct?
 A: The relation "|" is reflexive, since $n|n$ holds for all $n$, even $n = 0$. 

The relation "|" is not symmetric. Notice that $2$ divides $6$ (as $6 = 2 \cdot 3$), but $6$ does not divide $2$ (there is no integer $k$ such that $2 = 6 \cdot k$). 

Transitive ? Suppose $a|b$ and $b|c$. By definition, we have $k_1, k_2$ such that $b = a \cdot k_1$ and $c = b \cdot k_2$. 
We want to show that $a|c$, or that $c = a \cdot k$ for some $k$. Can you find a $k$ that works?
A: Answer: Not reflexive, not symmetric, not anti-symmetric, transitive
Reason:

*

*Reflexive: for all x ∈ Z, R(x,x) is reflexive, but here R(0,0) is a violation (0/0 is undefined), as 0 belongs to the set of integers but does not satisfy this relation.


*Symmetric: for all x ∈ Z, R(x,y)and R(y,x) is symmetric and clearly this relation cannot be symmetric. Considering the example {1,2}, for this to be symmetric (1,2)(2,1) both ordered pair should be present but it is also a violation (2 can be divided by 1 but 1 cannot be divided by 2).


*Anti-symmetric: for all x ∈ Z, R(x,y) and R(y,x) then x=y is anti-symmetric. But here, we can easily make a violation as R(-2,2) and R(2,-2) are not anti-symmetric (-2/2 or 2/-2 is -2).


*Transitivity: for all x ∈ Z , R(x,y), R(y,z) then R(x,z). This is always true for the above divides relation.
So the given relation "|" is not reflexive, not symmetric, not anti-symmetric but transitive only.
A: The relationship is transitive. A relation is transitive if xRy and yRz implies that xRz for all a,b,c∈A ( with A as a non empty set and R as a relation to A).
When a,b,c∈ ℤ, and if a|b and b|b, is that there exists an integer k so b= a ∙ k and that there exists an integer l so c= b ∙ l. This means that
c = a ∙ b ∙ l. Because a,b,c∈ ℤ this means that b ∙ l ∈ ℤ (an integer multiplied by an integer produces an integer). This means that there exists an integer m (where m = a ∙ b) so that c = m ∙ a. It follows that a|c, so the relationship is transitive.
