i broke up in a question.
The relation "divide" is Anti-Symmetric relation.
∀a,b ∈ Z : if a|b and a≠b, then b∤a.
Proof by contradiction, assuming that:
(I) a|b
(II) a≠b
(III) b|a
How a|b, by def. of "divide":
b = a.q , for q ∈ Z
developing a|b:
b = a.q
[For here, i dont know what i do]
Thank you very much in advance.
As A. Deshmukh notes, you're trying to prove something that's a bit to strong, because $1 \neq -1$ and $1 | -1$ and $-1 | 1$. However the only exceptions are of the form $a = -b$. To see this, suppose that $a \neq b$ and $a | b$ and $b | a$. Let $a = bc$ and $b = ad$. Then $a = adc$, so $dc = 1$. The only integer solutions of $dc = 1$ are $d = c = 1$ and $d = c = -1$. Since $a \neq b$, $d = c = 1$ is ruled out. Therefore $c = -1$, so $a = -b$. Q.E.D.