Integer Absolute Value not less than Divisors

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Theorem

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

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


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 $c \ne 0$.

From Negative of Absolute Value:

$a \le \size a$

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 $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \size c\) \(=\) \(\displaystyle \size a \size q\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \size a \size q \ge \size a \times 1\) \(=\) \(\displaystyle \size a\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle a \le \size a\) \(\le\) \(\displaystyle \size c\) $\quad$ $\quad$

$\blacksquare$


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