# Contents

## Definition

An even number is an integer $n \in \mathbb{Z}$ that is a multiple of 2, hence such that $n = 2 k$ for $k \in \mathbb{Z}$.

An odd number is an integer that is not an even number.

## Properties

###### Example

(adjoint modality of even and odd integers)

Regard the integers as a preordered set $(\mathbb{Z}, \leq)$ in the canonical way, and thus as a thin category.

Consider the full subcategory inclusions

$\array{ (\mathbb{Z}, \leq ) & \overset{even}{\hookrightarrow}& (\mathbb{Z},\leq) \\ n &\mapsto & 2 n } \phantom{AAAAA} \array{ (\mathbb{Z}, \leq ) & \overset{odd}{\hookrightarrow}& (\mathbb{Z},\leq) \\ n &\mapsto & 2 n + 1 }$

of the even and the odd integers, as well as the functor

$\array{ (\mathbb{Z}, \leq ) & \overset{\lfloor-/2\rfloor}{\longrightarrow}& (\mathbb{Z},\leq) \\ n &\mapsto& \lfloor n/2 \rfloor }$

which sends any $n$ to the floor $\lfloor n/2 \rfloor$ of $n/2$, hence to the largest integer which is smaller or equal to the rational number $n/2$.

These functors form an adjoint triple

$even \;\dashv\; \lfloor -/2 \rfloor \;\dashv\; odd$

and hence induce an adjoint modality

$Even \;\dashv\; Odd$

on $(\mathbb{Z}, \leq)$ with

1. $Even \coloneqq 2 \lfloor -/2 \rfloor$ sending any integer to its “even floor value”

2. $Odd \coloneqq 2 \lfloor -/2 \rfloor + 1$ sending any integer to its “odd ceiling value”.

###### Proof

Observe that for all $n \in \mathbb{Z}$ we have

$2 \lfloor n/2 \rfloor \overset{ \epsilon_n }{\leq} n \overset{ \eta_n }{\leq} 2 \lfloor n/2 \rfloor + 1 \,,$

where the first inequality is an equality precisely if $n$ is even, while the second is an equality precisely if $n$ is odd. Hence this provides candidate unit $\eta$ and counit.

Hence by this characterization of adjoint functors

1. the adjunction $\lfloor -/2 \rfloor \dashv odd$ is equivalent to the condition that

for every $n \leq 2 k + 1$ we have $2 \lfloor n/2 \rfloor + 1 \leq 2 k + 1$;

2. the adjunction $even \dashv \lfloor -/2 \rfloor$ is equivalent to the condition that

for every $2k \leq n$ we have $2k \leq 2 \lfloor n/2 \rfloor$,

which is readily seen to be the case

Last revised on July 30, 2018 at 08:39:54. See the history of this page for a list of all contributions to it.