nLab even number

An even number is that which is divisible into two equal parts.

An odd number is that which is not divisible into two equal parts, or that which differs by a monad [=unit] from an even number.

[Euclid, Def. 6 & 7 of Elements Book VII (~ 400-300 BC), see here]

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.