Proof that if $a$ divides $b$ and $a$ does not divide $c$ then $a$ does not divide $c - b$

$a | b$, therefore $b = ka$ where $k$ is some integer.

$a ∤ c$.

The conclusion $a ∤ (c-b)$ is obvious since $b = ka$

and therefore $a ∤ (c - (ka))$

The statement makes sense to me, but I have difficulty with formally proving it. Could someone please help?

• Suppose $a$ does divide $c-b$, then $a$ divides $(c-b)+b$, which gives a contradiction. – Student Sep 23 '17 at 16:06
• Viewed $\bmod a\!:\,\ b\equiv 0\,\Rightarrow\, c-b\equiv c\not\equiv 0.\$ What could be simpler? – Bill Dubuque Sep 23 '17 at 16:28
• @BillDubuque very powerful the congruence way – Maman Sep 23 '17 at 21:46

Assume by contradiction that $a|c-ka$, then $c-ka=ha$, so $c=ha+ka=(h+k)a$, i.e. $a|c$, that is a contradiction.
$a$ divides $b$ so there exists $k\in \mathbb{Z}$ such dat : $b=ka$
$a$ does not divide $c$ so using euclidean division (here it's possible) we can write $c=aq+r$ with $0< r <\mid a\mid$.
Then $c-b=aq+r-ka=a(q-k)+r$ and moreover by conditions we have dat $a$ cannot divide $r$. So $a$ does not divide $b-c$.