Absolute Value of Integer is not less than Divisors

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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 $c \ne 0$.

From Negative of Absolute Value:

$a \le \size a$

Then:

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

$\blacksquare$


Also see


Sources