Prove divisibility is a partial order relation over natural numbers

Question

I have proceeded like this so far:

(The 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, $a = k_1\cdot k_2\cdot a.$ This means $a = b.$

Transitivity: Not sure how to prove this.

I know it holds but not sure how it can be shown mathematically.

Answer 1

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.

Answer 2

Since $a \mid b \implies a=sb, s \in \mathbb{N} \tag{1}$ And $b \mid c \implies b=tc t \in \mathbb{N} \tag{2}$ From $(1)$ and $(2)$ $$a=sb=s(tc) = (st)c = uc$$ , where $u = st \in \mathbb{N}$. Hence $a=uc \implies a \mid c$.