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: $m < n \iff m \in n$). 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:

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

Multiplication is given by the explicit formula

$xy \;:=\; \bigl(\, x_0y_0 - x_1y_1 - x_2y_2 - x_3y_3,\;\; x_0y_1 + x_1y_0 + x_2y_3 - x_3y_2,\;\; x_0y_2 + x_2y_0 + x_3y_1 - x_1y_3,\;\; x_0y_3 + x_3y_0 + x_1y_2 - x_2y_1 \,\bigr).$

The distinguished elements are

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

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: $q^{-1} = \bar q / N(q)$ for $q \ne 0$, while $ij = k$ and $ji = -k$ and is not assumed here.

The basis quaternions and the table. Write

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

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

$i^{2} = j^{2} = k^{2} = -1, \qquad ij = k, \quad jk = i, \quad ki = j, \qquad ji = -k, \quad kj = -i, \quad ik = -j,$

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

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

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