Question

Prove that if d|m and d|n, then d|(m+n)

Answer 1

$d\mid m$ means to say there is some $k_1$ such that $m=dk_1$. Similarly, $d\mid n$ is equivalent to saying there exists $k_2[/te]x such that [tex]n=dk_2$.

Now, $m+n=dk_1+dk_2=d(k_1+k_2)=dk_3$, i.e. there is some $k_3$ such that $m+n$ is a multiple of $d$, i.e. $d\mid(m+n)$.