a chia hết cho b => a=k.b, k thuộc Z
b chia hết cho c => b=m.c, m thuộc Z
Suy ra: a=k.b=k.m.c chia hết cho c
\(a⋮b\Rightarrow a=bk\)\(\left(k\inℕ\right)\)\(\left(1\right)\)
\(b⋮c\Rightarrow b=cq\)\(\left(q\inℕ\right)\)\(\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow a=cqk\)
\(\Rightarrow c\inƯ\left(a\right)\)
\(\Rightarrow a⋮c\left(đpcm\right)\)