The quaternions $\mathbb{H}$: real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1,i,j,k$

Let $\mathbb{R}$ be the field of real numbers (The real numbers, The reals form a field, Field) and let $4=\{0,1,2,3\}$ be the von Neumann natural number $4$ (The natural numbers $\mathbb{N}$ (von Neumann), On $\mathbb{N}$ the order is membership: $m<n⟺m\in n$). The set of quaternions is the set of all functions $4\to \mathbb{R}$,

$\mathbb{H}:={\mathbb{R}}^4=\{x:4\to \mathbb{R}\},$

whose members are written as quadruples $x=(x_0,x_1,x_2,x_3)$ with $x_p\in \mathbb{R}$ for $p\in 4$. Note that the coordinates are indexed from $0$, because $4$ is the set $\{0,1,2,3\}$.

Addition is componentwise:

$x+y:=(x_0+y_0,x_1+y_1,x_2+y_2,x_3+y_3)\mathrm{.}$

Multiplication is given by the explicit formula

$xy:=(x_0{y}_0-x_1{y}_1-x_2{y}_2-x_3{y}_3,x_0{y}_1+x_1{y}_0+x_2{y}_3-x_3{y}_2,x_0{y}_2+x_2{y}_0+x_3{y}_1-x_1{y}_3,x_0{y}_3+x_3{y}_0+x_1{y}_2-x_2{y}_1)\mathrm{.}$

The distinguished elements are

$0_{\mathbb{H}}:=(0,0,0,0),1_{\mathbb{H}}:=(1,0,0,0)\mathrm{.}$

Both rules are functions $\mathbb{H}\times \mathbb{H}\to \mathbb{H}$, since every coordinate on the right is a real number formed from real numbers by the field operations of $\mathbb{R}$; so each is a binary operation on $\mathbb{H}$ (Binary operation on a set; associativity, commutativity, and a subset closed under the operation), and no separate well-definedness question arises. That these data satisfy the ring axioms (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides), with $1_{\mathbb{H}}$ a two-sided identity (Left identity, right identity, and two-sided identity for a binary operation), is proved in $\mathbb{H}$ is a division ring that is not commutative, hence not a field: $q^{-1}=\bar{q}\mathrm{/}N(q)$ for $q\mathrm{\ne }0$, while $ij=k$ and $ji=-k$ and is not assumed here.

The basis quaternions and the table. Write

$e_0=1=(1,0,0,0),e_1=i=(0,1,0,0),e_2=j=(0,0,1,0),e_3=k=(0,0,0,1),$

and for a real $\lambda$ write $\widehat{\lambda }:=(\lambda ,0,0,0)$, so that $\widehat{1}=1$ and $\widehat{0}=0_{\mathbb{H}}$. Evaluating the multiplication formula on these sixteen pairs gives the table

$i^2=j^2=k^2=-1,ij=k,jk=i,ki=j,ji=-k,kj=-i,ik=-j,$

together with $1x=x1=x$ for $x\in \{1,i,j,k\}$, where $-1$ abbreviates $\widehat{-1}=(-1,0,0,0)$ and similarly for $-i$, $-j$, $-k$. In the usual notation $x=x_0+x_{1}i+x_{2}j+x_{3}k$, and the displayed product formula is precisely what the table forces once products are expanded and real coefficients are collected; but the formula, not the table, is the definition, so nothing is "extended by linearity" and no unproved notion is used.

Conjugate and norm. For $x\in \mathbb{H}$ put

$\bar{x}:=(x_0,-x_1,-x_2,-x_3),N(x):=x_0^2+x_1^2+x_2^2+x_3^2\in \mathbb{R}\mathrm{.}$

Both are defined by explicit real formulas; $\bar{x}\in \mathbb{H}$ and $N(x)\in \mathbb{R}$, and $N$ is a function $\mathbb{H}\to \mathbb{R}$, not a quaternion.

• Why $\mathbb{H}$ is defined on this page rather than among the examples. It is a division ring that is not a field ($\mathbb{H}$ is a division ring that is not commutative, hence not a field: $q^{-1}=\bar{q}\mathrm{/}N(q)$ for $q\mathrm{\ne }0$, while $ij=k$ and $ji=-k$), so it is the witness that Division ring: a ring with $1\mathrm{\ne }0$ in which every nonzero element is a unit is strictly weaker than Field, and it is a ring whose multiplication is not commutative. Companion pages are leaves in the reading order, so no later page may cite an item homed on one; $\mathbb{H}$ is placed here so that later pages can use it.

• Indexing from $0$ is not a stylistic choice. A quaternion is a function on the von Neumann natural $4$, which is the set $\{0,1,2,3\}$ (On $\mathbb{N}$ the order is membership: $m<n⟺m\in n$), so its coordinates are $x_0,x_1,x_2,x_3$ and the real part is $x_0$. Every sum over basis indices below runs over $p\in 4$, that is over $p=0,1,2,3$.

• The map $\lambda \mapsto \widehat{\lambda }$ is how real scalars enter. There is no separate scalar multiplication in the definition: multiplying by a real number means multiplying by the quaternion $\widehat{\lambda }$, and the formula shows at once that $\widehat{\lambda }x=x\widehat{\lambda }=(\lambda x_0,\lambda x_1,\lambda x_2,\lambda x_3)$. This is used throughout the proof of $\mathbb{H}$ is a division ring that is not commutative, hence not a field: $q^{-1}=\bar{q}\mathrm{/}N(q)$ for $q\mathrm{\ne }0$, while $ij=k$ and $ji=-k$ and is what keeps that proof free of any notion of linearity.