Answer:
Assuming a being a divisor of c and [tex]a,b,c \in \mathbb{Z}[/tex]
Step-by-step explanation:
We are told that [tex]a \mid c[/tex]
and that [tex]a+b =c[/tex]
which means that
[tex]\exists k \in \mathbb{N}[/tex] so that [tex]c = k.a[/tex]
so we can rewrite [tex]a+b[/tex] as
[tex]a+b = c = k.a[/tex]
[tex]b = k.a-a = (k-1).a[/tex]
and as [tex]k \in \mathbb{N}[/tex]
We have that either
Showing that [tex]a \mid c[/tex]
