# Integer Absolute Value not less than Divisors

## Theorem

A (non-zero) integer is greater than or equal to its divisors in magnitude:

$\forall c \in \Z_{\ne 0}: a \divides c \implies a \le \size a \le \size c$

It follows that a non-zero integer can have only a finite number of divisors, since they must all be less than or equal to it.

### Corollary

Let $a, b \in \Z_{>0}$ be (strictly) positive integers.

Let $a \mathrel \backslash b$.

Then:

$a \le b$

## Proof

Suppose $a \divides c$ for some $c \ne 0$.

$a \le \size a$

Then:

 $\displaystyle a$ $\divides$ $\displaystyle c$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle \exists q \in \Z: \ \$ $\displaystyle c$ $=$ $\displaystyle a q$ $\quad$ Definition of Divisor of Integer $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle \size c$ $=$ $\displaystyle \size a \size q$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle \size a \size q \ge \size a \times 1$ $=$ $\displaystyle \size a$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle a \le \size a$ $\le$ $\displaystyle \size c$ $\quad$ $\quad$

$\blacksquare$