Subset Product/Examples/Congruence Modulo Initial Segment of Natural Numbers

From ProofWiki
Jump to navigation Jump to search

Example of Subset Product

Let $m \in \Z_{>0}$ be a (strictly) positive integer.

Let $\N_{<m}$ denote the initial segment of the natural numbers $\N$:

$\N_{<m} = \set {0, 1, \ldots, m - 1}$

Let $\RR_m$ denote the equivalence relation:

$\forall x, y \in \Z: x \mathrel {\RR_m} y \iff \exists k \in \Z: x = y + k m$

For each $a \in \N_{<m}$, let $\eqclass a m$ be the equivalence class of $a \in \N_{<m}$ under $\RR_m$:

$\eqclass a m := \set {a + z m: z \in \Z}$

Let $\Z_m$ be the set defined as:

$\Z_m := \set {\eqclass a m: a \in \N_{<m} }$

Let $+_\PP$ denote the operation induced on $\powerset \Z$ by integer addition.

Then the algebraic structure $\struct {\Z_m, +_\PP}$ is closed in the sense:

$\forall \eqclass a m, \eqclass b m \in \Z_m: \eqclass a m +_\PP \eqclass b m \in \Z_m$


Proof

From Equivalence Class Example: Congruence Modulo $\N_{<m}$:

$\RR_m$ is an equivalence relation
$\eqclass a m$ is the equivalence class of $a \in \N_{<m}$ under $\RR_m$.

Let $\eqclass a m, \eqclass b m \in \RR_m$ be arbitrary.

Then we have:

\(\ds \eqclass a m +_\PP \eqclass b m\) \(=\) \(\ds \set {a + z_1 m + b + z_2 m: z_1, z_2 \in \Z}\) Definition of $\eqclass a m$
\(\ds \) \(=\) \(\ds \set {a + b + \paren {z_1 + z_2} m: z_1, z_2 \in \Z}\) rearranging
\(\ds \) \(=\) \(\ds \set {a + b + z m: z \in \Z}\) where $z = z_1 + z_2 \in \Z$
\(\ds \) \(=\) \(\ds \eqclass {a + b} m\) Definition of $\eqclass {a + b} m$


It also follows that:

$\forall a, b \in \N_{<m}: \eqclass a m = \eqclass b m \iff a = b$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Subset_Product/Examples/Congruence_Modulo_Initial_Segment_of_Natural_Numbers&oldid=562581"