Q:

Prove divisibility is a partial order relation over natural numbers

I have proceeded like this so far:
(this natural number set includes $0$)
Reflexivity:
Proof: $a = k \cdot a,$ since $k$ also belongs to natural number, it is proved.
Anti-symmetry:
$a \mid b$ means $a = k_1 \cdot b$ and $b \mid a$ means $b = k_2 \cdot a$ now, $a = k1\cdot k2\cdot a.$ this means $a = b.$
Transitivity:
Not sure how to prove this.
I know it holds but not sure how it can be shown mathematically.

A:

Well if $a\mid b$ and $b\mid c,$ then $b=da$ and $c=eb$ for some natural numbers $d,e,$ and so $c=....$ can you take it from there?
Your reflexivity proof needs fixed, though. What is $k$?
Also, I would add a bit to your antisymmetry proof. Since $a=k_1k_2a,$ then $k_1k_2=1$ if $a\ne 0,$ so since $k_1,k_2$ are naturals, then $k_1=1,$ and so $a=b$ as desired. Now, in the case that $a=0,$ we must proceed a bit differently: since $a\mid b,$ then we have $b=k_2a=0$ directly.

