Divides is Transitive

From ProofWiki
Jump to: navigation, search

Contents

Theorem

Let $\left({D, +, \circ}\right)$ be an integral domain.

Let $x, y, z \in D$.


Then:

$x \backslash y \land y \backslash z \implies x \backslash z$


Corollary

"Divides" is a transitive relation on $\Z$, the set of integers.


Proof

Let $x \backslash y \land y \backslash z$.

Then from the definition of divisor, we have:

  • $x \backslash y \iff \exists s \in D: y = s \circ x$
  • $y \backslash z \iff \exists t \in D: z = t \circ y$


Then:

$z = t \circ \left({s \circ x}\right) = \left({t \circ s}\right) \circ x$

Thus:

$\exists \left({t \circ s}\right) \in D: z = \left({t \circ s}\right) \circ x$

and the result follows.

$\blacksquare$


Proof of Corollary

Follows directly from the fact that Integers form Integral Domain.

$\blacksquare$


Alternatively:

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

This follows because:

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


$\blacksquare$


Sources

Personal tools
Namespaces
Variants
Actions
Navigation
ProofWiki.org
ToDo
Toolbox
Google AdSense