$a | b$, 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?

  • 2
    $\begingroup$ Suppose $a$ does divide $c-b$, then $a$ divides $(c-b)+b$, which gives a contradiction. $\endgroup$ – Student Sep 23 '17 at 16:06
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    $\begingroup$ Viewed $\bmod a\!:\,\ b\equiv 0\,\Rightarrow\, c-b\equiv c\not\equiv 0.\ $ What could be simpler? $\endgroup$ – Bill Dubuque Sep 23 '17 at 16:28
  • $\begingroup$ @BillDubuque very powerful the congruence way $\endgroup$ – 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.


Here is another way (uglier than 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 $b-c$.


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