How Does Modular Arithmetic Simplify Integer Divisibility Proofs?

[shamieh]

Let $a$, $b$, and $c$ be integers, where a $\ne$ 0. Then
$$
$$
(i) if $a$ | $b$ and $a$ | $c$, then $a$ | ($b+c$)
$$
$$
(ii) if $a$ | $b$ and $a$|$bc$ for all integers $c$;
$$
$$
(iii) if $a$ |$b$ and $b$|$c$, then $a$|$c$.

**Prove that if $a$|$b$ and $b$|$c$ then $a$|$c$ using a column proof that has steps in the first column
and the reason for the step in the second column.**

Here is what I was thinking.. Would this be sufficient enough?$(iii)\ \ \ \dfrac{b}a,\,\dfrac{c}b\in\Bbb Z\ \Rightarrow\ \dfrac{b}a\dfrac{c}b = \dfrac{c}a\in\Bbb Z$

 

[MarkFL]

I think I would state (where $k_i\in\mathbb{Z}$):

(iii) \(\displaystyle a|b\implies b=k_1a\,\land\,b|c\implies c=k_2b=k_1k_2a=k_3a\,\therefore\,a|c\)