Let $ a \in \mathbb{N}$, $ b \in \mathbb{N}_0 $, $ r \in \mathbb{N}_0 $, $ q\in \mathbb{N} $ and $ b \leq a$ so that $$a = bq + r $$with $$ 0 \leq r \lt b$$
Proof:
Let $ D = \{d \in \mathbb{Z} \vert ~ d \mid a \text{ and } d \mid b \}$ and Let $ D' = \{d \in \mathbb{Z} \vert ~ d \mid b \text{ and } d \mid r \}$
Let $ x \in D $, so that $x \mid a$ and $x \mid b$:
$$ a = bq +r $$ $$ a - bq = r $$
So $x \mid a-bq=r$ so $x \mid r$
Let $$ x \in D', $$ so that $x \mid b$ and $x \mid r$:
$$ a = bq +r $$
So $x \mid b$ and $x \mid r$ but also $x \mid a$ and since $$ a \in D$$
$$ \gcd(a,b) = \gcd(b,r)$$
Any critique? Did i proved it somehow in a correct way? If not, could you show me what i did wrong or how to do it properly? Thanks in advance!