# All Factors Divide Integer then Whole Divides Integer

## Theorem

Let $S = \set {a_1, a_2, \ldots, a_r} \subseteq \Z$ be a finite subset of the integers.

Let $S$ be pairwise coprime.

Let:

$\forall j \in \set {1, 2, \ldots, r}: a_r \divides b$

where $\divides$ denotes divisibility.

Then:

$\displaystyle \prod_{j \mathop = 1}^r a_j \divides b$

## Proof

Proof by induction:

In the following, it is assumed at all times that $S = \set {a_1, a_2, \ldots, a_r} \subseteq \Z$ is pairwise coprime.

For all $r \in \N_{> 1}$, let $\map P r$ be the proposition:

$\displaystyle \prod_{j \mathop = 1}^r a_j \divides b$

### Basis for the Induction

$\map P 2$ is the case:

$a_1 a_2 \divides b$

By definition of divisibility:

$b = a_1 q_1 = a_2 q_2$

where $q_1, q_2 \in \Z$.

$\exists x, y \in \Z: a_1 x + a_2 y = 1$

So:

 $\displaystyle b$ $=$ $\displaystyle a_2 q_2 \cdot a_1 x + a_1 q_1 \cdot a_2 y$ $\displaystyle$ $=$ $\displaystyle a_1 a_2 \paren {q_2 x + q_1 y}$

and so $a_1 a_2 \divides b$.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$\displaystyle \prod_{j \mathop = 1}^k a_j \divides b$

Then we need to show:

$\displaystyle \prod_{j \mathop = 1}^{k + 1} a_j \divides b$

### Induction Step

This is our induction step:

$a_{k + 1} \divides b$
$\displaystyle a_{k + 1} \perp \prod_{j \mathop = 1}^k a_j$

By the induction hypothesis:

$\displaystyle \prod_{j \mathop = 1}^k a_j \divides b$

So by the basis for the induction:

$\displaystyle \prod_{j \mathop = 1}^{k + 1} a_j \divides b$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore: $\displaystyle r \in \N_{> 1}: \prod_{j \mathop = 1}^r a_j \divides b$

$\blacksquare$