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.
$\blacksquare$
Sources
- Seth Warner: Modern Algebra (1965): $\S 14$: Example $14.4$
- A.N. Kolmogorov and S.V. Fomin: Introductory Real Analysis (1968): $\S 3.1$: Example $4$
- T.S. Blyth: Set Theory and Abstract Algebra (1975):$\S 7$: Example $7.2$