# Divisibility Theorems

Divisibility
Let $$a, b \in \mathbb{Z}$$ $$b\vert a$$ iff $$\exists k \in\mathbb{Z} \ni a=bk$$.
I. $$ac \vert b \Rightarrow a \vert b$$ and $$ac \vert b \Rightarrow c \vert b$$
Proof: By definition $$ac \vert b$$ iff $$\exists n \in \mathbb{Z} \ni b=acn$$. If $$cn=k \in \mathbb{Z}$$ then $$b=acn=ak$$, so $$a \vert b$$
Proof: By definition $$ac \vert b$$ iff $$\exists n \in \mathbb{Z} \ni b=acn$$. If $$an=k \in \mathbb{Z}$$ then $$b=acn=ck$$, so $$c \vert b$$
II. If $$a \vert b$$ and $$a \vert c$$ then $$a \vert \left( b + c \right)$$
Proof: By definition $$a \vert b$$ iff $$\exists k \in \mathbb{Z} \ni b=ak$$ and $$a \vert c$$ iff $$\exists n \in \mathbb{Z} \ni c=an$$. Since $$ak + an = b + c = a \left( k + n \right)$$ and $$\left( k+ n \right)\in\mathbb{Z}$$ then by definition $$a \vert \left(b+c \right)$$.
III. If $$a \vert b$$ and $$a \vert c$$ then $$a \vert \left( b – c \right)$$
Proof: By definition $$a \vert b$$ iff $$\exists k \in \mathbb{Z} \ni b=ak$$ and $$a \vert c$$ iff $$\exists n \in \mathbb{Z} \ni c=an$$. Since $$ak – an = b – c = a \left( k – n \right)$$ and $$\left( k – n \right)\in\mathbb{Z}$$ then by definition $$a \vert \left(b-c \right)$$.
IV. If $$a \vert b$$ and $$a \vert c$$ then $$a \vert bc$$
Proof: By definition $$a \vert b$$ iff $$\exists k \in \mathbb{Z} \ni b=ak$$ and $$a \vert c$$ iff $$\exists n \in \mathbb{Z} \ni c=an$$. Since $$bc =ak \cdot an =at$$ and $$kan=t\in\mathbb{Z}$$ then by definition $$a \vert bc$$.
V. $$a \vert b \Rightarrow a \vert bc$$
Proof: By definition $$a \vert b$$ iff $$\exists k \in \mathbb{Z} \ni b=ak$$. Since $$bc=akc=am$$ and $$kc=m\in\mathbb{Z}$$ then by definition $$a \vert bc$$.
VI. If $$a \vert b$$ and $$a \vert c$$ then $$a^2 \vert bc$$
Proof: By definition $$a \vert b$$ iff $$\exists k \in \mathbb{Z} \ni b=ak$$ and $$a \vert c$$ iff $$\exists n \in \mathbb{Z} \ni c=an$$. Since $$bc=ak \cdot an=a^2t$$ where $$kn=t\in\mathbb{Z}$$ then by definition $$a^2 \vert bc$$.