There are $a,b \in \mathbb Z, a \neq 0$ and $d$ which is the smallest number in $\mathbb N$ for that $d = l \cdot a + k \cdot b$ is true. ($k,l \in \mathbb Z$)
I think that $d|b$ and $d|a$ is true, but I can't find proof for it, can you?
There are $a,b \in \mathbb Z, a \neq 0$ and $d$ which is the smallest number in $\mathbb N$ for that $d = l \cdot a + k \cdot b$ is true. ($k,l \in \mathbb Z$)
I think that $d|b$ and $d|a$ is true, but I can't find proof for it, can you?
Divide $a$ by $d$,so $$ \begin{cases} a = (l a + k b) q + r\\ 0 \le r < d \end{cases} $$ so $r = a(1 - l q) + b (-k q)$ has the same form as $d$. If $r > 0$, this contradicts the definition of $d$. Hence $r = 0$ and $d$ divides $a$.