Divides is Transitive
From ProofWiki
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
- C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 3.10$: Theorem $16 \ \text{(i)}$
- C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 6.26$: Theorem $49 \ \text{(i)}$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 11.3$