Divides is Partial Ordering on Positive Integers

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Theorem

Divides is a partial ordering of $\Z_{>0}$.


Proof

Checking in turn each of the critera for an ordering:


Divides is Reflexive

$\forall n \in \Z: n \backslash n$ from Every Integer Divides Itself‎.

$\blacksquare$


Divides is Transitive

$\forall x, y, z \in \Z: x \backslash y \land y \backslash z \implies x \backslash z$


\(\displaystyle \) \(\displaystyle x \backslash y\) \(\implies\) \(\displaystyle \exists q_1 \in \Z: q_1 x = y\) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle y \backslash z\) \(\implies\) \(\displaystyle \exists q_2 \in \Z: q_2 y = z\) \(\displaystyle \)                    
\(\displaystyle \implies\) \(\displaystyle q_2 \left({q_1 x}\right)\) \(=\) \(\displaystyle z\) \(\displaystyle \)                    
\(\displaystyle \implies\) \(\displaystyle \left({q_2 q_1 x}\right)\) \(=\) \(\displaystyle z\) \(\displaystyle \)                    
\(\displaystyle \implies\) \(\displaystyle q \in \Z: q x\) \(=\) \(\displaystyle z\) \(\displaystyle \)                    
\(\displaystyle \implies\) \(\displaystyle x \backslash z\) \(\) \(\displaystyle \) \(\displaystyle \)                    


$\blacksquare$


Divides is Antisymmetric

We have $\forall a, b \in \Z: a \backslash b \land b \backslash a \implies \left\vert{a}\right\vert = \left\vert{b}\right\vert$ which follows from Integer Absolute Value Greater than Divisors:

\(\displaystyle \) \(\displaystyle a \backslash b\) \(\implies\) \(\displaystyle \left\vert{a}\right\vert \le \left\vert{b}\right\vert\) \(\displaystyle \)          Integer Absolute Value Greater than Divisors          
\(\displaystyle \) \(\displaystyle b \backslash a\) \(\implies\) \(\displaystyle \left\vert{b}\right\vert \le \left\vert{a}\right\vert\) \(\displaystyle \)          Integer Absolute Value Greater than Divisors          
\(\displaystyle \) \(\displaystyle \) \(\implies\) \(\displaystyle \left\vert{a}\right\vert = \left\vert{b}\right\vert\) \(\displaystyle \)                    


If we restrict ourselves to the domain of positive integers, we can see:

$\forall a, b \in \Z_{>0}: a \backslash b \land b \backslash a \implies a = b$

Hence the result.

$\blacksquare$



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