The Problem: Let $a,b,c \in \mathbb{Z}$. Show that if $b \mid a$ and $b \nmid (a + c)$, then $b \nmid c$.
My Answer So Far:
By definition, we have $ a = bq_1$ and $\nexists q_2 \in \mathbb{Z}$ such that $a+c= bq_2$ for $ q_1, q_2 \in \mathbb{Z}$. Substituting $a = bq_1$ gives $\nexists q_2 \in \mathbb{Z}$ such that $bq_1 + c = bq_2$ or $\nexists q_2 \in \mathbb{Z}$ such that $c = bq_2-bq_1$. By factoring out $b$ we have $\nexists q_2 \in \mathbb{Z}$ such that $c = b(q_2-q_1)$. Substituting $q_3=q_2-q_1$. we have $\nexists q_3 \in \mathbb{Z}$ such that $c = bq_3$.
$\therefore \; b \nmid c$ if $b \mid a$ and $ b \nmid (a+c)$ for $a,b,c \in \mathbb{Z}$.
My Question:
How does my proof look? I now know that it is easier to prove the contrapositive, but is this proof just as valid?