Absolute Value of Integer is not less than Divisors

Theorem

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

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

Corollary

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

Let $a \divides b$.

Then:

$a \le b$

Proof

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

$a \le \size a$

Then:

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

$\blacksquare$