By definition `a|b` if and only if `b=an,` `n in ZZ`.
A relation `circ` is transitive if `(a circ b ^^b circ c)=>(a circ c)`. So we need to prove that if `a|b` and `b|c` then `a|c`.
If `a|b` then `b=an`
If `b|c` then `c= bm=anm`
Since `m` and `n` are integers then `mn` is also integer which means that `a|b.`
Let `a,b>0` and `a|b`.
Since `a|b` it follows that `b=an` for some integer `n` and since `b` is positive `an` must be positive and since `a` is positive `n` must be positive as well. So `n >=1`. If `n=1` then `a=b`.
If `n>1` then `a=b/n` and since both `b` and `n` are positive `b/n<b`.
Let a divides b i.e. `b/a ,b>=a ,b= ma ,`
and m is an integer.
Let b divides c i.e `c/b ,c>=b ,c=nb ,`
and n is an integer.
c=n(ma)=(nm)a ,mn is an integer , this mean a is multiple of c
i.e. a divides c i.e a/c
this proves transitivity.