Worked Examples in the Dual-Number Algebra

Introduction

This article collects explicit computations in the dual-number algebra $\mathbb{D}'_R = R[\varepsilon]/(\varepsilon^2)$, in the spirit of the worked computation that accompanies the algebra of a number system. It follows Dual-Numbers Algebra for the conventions, Dual-Numbers Norm and Invertibility for the criterion for invertibility, Dual-Numbers Ideals and the Maximal Ideal for the ideal structure, Dual-Numbers Zero Divisors for the zero-divisor classification. Every computation was recomputed in exact rational arithmetic; where a floating-point value appears it is flagged.

The treatment is mathematically honest: every claim is either proved or stated as a definition. No physics is invoked. Unless a statement is labelled otherwise, the base is $R = \mathbb{R}$ and the algebra is $\mathbb{D}' = \mathbb{D}'_{\mathbb{R}}$. A general dual number is

$$ A = a + \varepsilon a', \qquad a, a' \in \mathbb{R}, $$

with $a = \operatorname{Re} A$, $a' = \operatorname{Inf} A$, dual conjugation $\bar A = a - \varepsilon a'$, unit criterion $a \neq 0$, and maximal ideal $\mathrm{M} = (\varepsilon)$. The two distinguished submodules are $R_{\mathbb{D}'} = \mathbb{R}\cdot 1$ and $\varepsilon R_{\mathbb{D}'} = \varepsilon\mathbb{R}$.

Each computation below is a worked instance of a general result proved in a companion article; the instance is stated with its reference, so that the article remains a table of examples rather than a second proof of the theory.

The Algebra in Coordinates

The Basis and the Product

The algebra has the basis $\{1, \varepsilon\}$ with the multiplication table

$\cdot$	$1$	$\varepsilon$
$1$	$1$	$\varepsilon$
$\varepsilon$	$\varepsilon$	$0$

Example (product, worked). For $A = 2 + 3\varepsilon$ and $B = 4 + 5\varepsilon$,

$$ AB = (2\cdot 4) + (2\cdot 5 + 3\cdot 4)\varepsilon = 8 + 22\varepsilon. $$

The real parts multiply, and the infinitesimal part is $a\,b' + a'\,b$; the cross terms do not interact, because $\varepsilon^2 = 0$.

Example (squaring, worked). For $A = a + \varepsilon a'$, the square is

$$ A^2 = a^2 + 2aa'\,\varepsilon. $$

In particular $(2 + 3\varepsilon)^2 = 4 + 12\varepsilon$ and $(\varepsilon)^2 = 0$.

The Conjugate

Example (worked). $\overline{2 + 3\varepsilon} = 2 - 3\varepsilon$, and

$$ (2 + 3\varepsilon)(2 - 3\varepsilon) = 4 + (-6 + 6)\varepsilon = 4. $$

The Two Involutions

The Identity Involution

The identity map $\operatorname{id}(A) = A$ is an involutive algebra automorphism. Its fixed-point submodule is the whole algebra $\mathbb{D}'$, and the associated $\pm1$-eigenspace decomposition is the trivial one, with eigenvalue $+1$ everywhere.

The Dual Conjugation

Dual conjugation is

$$ \bar{\cdot} : a + \varepsilon a' \longmapsto a - \varepsilon a'. $$

Example (worked). $\overline{2 + 3\varepsilon} = 2 - 3\varepsilon$; applying it twice returns $2 + 3\varepsilon$, so $\bar{\bar A} = A$. It is an algebra automorphism: $\overline{AB} = \bar A\bar B$, checked on the worked product by $\overline{8 + 22\varepsilon} = 8 - 22\varepsilon$ and $(2 - 3\varepsilon)(4 - 5\varepsilon) = 8 - 22\varepsilon$.

The Fixed-Point Submodules

Proposition. The $+1$-eigenspace of dual conjugation is the real submodule

$$ R_{\mathbb{D}'} = \{A : \bar A = A\} = \{a : a \in \mathbb{R}\}, $$

and the $-1$-eigenspace is the infinitesimal submodule

$$ \varepsilon R_{\mathbb{D}'} = \{A : \bar A = -A\} = \{\varepsilon a' : a' \in \mathbb{R}\}. $$

Proof. $\bar{a + \varepsilon a'} = a - \varepsilon a'$ equals $a + \varepsilon a'$ exactly when $a' = 0$, and equals $-(a + \varepsilon a')$ exactly when $a = 0$; the two sets are as displayed.

Example (worked projections). For $A = 2 + 3\varepsilon$,

$$ \tfrac{1}{2}(A + \bar A) = 2, \qquad \tfrac{1}{2}(A - \bar A) = 3\varepsilon, $$

and indeed $2 + 3\varepsilon = A$.

The Intersection

Proposition. $R_{\mathbb{D}'} \cap \varepsilon R_{\mathbb{D}'} = \{0\}$, and $\mathbb{D}' = R_{\mathbb{D}'} \oplus \varepsilon R_{\mathbb{D}'}$.

Proof. An element of the intersection satisfies $a' = 0$ and $a = 0$, so it is zero. The sum decomposition is the worked projection identity above.

So the algebra has exactly one nontrivial involution, and the two submodules are the eigenspaces. The biquaternion algebra, by contrast, carries four conjugations and six eigenspace-type subspaces; here the lattice degenerates to two submodules and the single relation $R_{\mathbb{D}'} \cap \varepsilon R_{\mathbb{D}'} = 0$.

The Nilpotents and the Maximal Ideal

Squaring the Maximal Ideal

Example (worked). Every element of $\mathrm{M} = \varepsilon\mathbb{R}$ is nilpotent of index two:

$$ (3\varepsilon)^2 = 9\varepsilon^2 = 0, \qquad (-5\varepsilon)^2 = 25\varepsilon^2 = 0, \qquad (7\varepsilon)^2 = 0. $$

So $\mathrm{M}^2 = 0$ and $\mathrm{M} \neq 0$: the maximal ideal is nilpotent of index exactly two.

Example (worked). A mixed element is never nilpotent: $(2 + 3\varepsilon)^2 = 4 + 12\varepsilon \neq 0$, and more generally $(a + \varepsilon a')^n = a^n + na^{n-1}\varepsilon a'$, which vanishes only when $a = 0$.

The Powers of an Element

Proposition. For every integer $n \geq 1$, $(a + \varepsilon a')^n = a^n + n\,a^{n-1}\varepsilon a'$, interpreted for $n = 1$ as the element itself.

Proof. Induction on $n$ using $(a + \varepsilon a')^n(a + \varepsilon a') = a^{n+1} + (n a^{n-1}a'\,a + a^n a')\varepsilon = a^{n+1} + (n+1)a^n \varepsilon a'$.

Example (worked). $(1 + \varepsilon)^{5} = 1 + 5\varepsilon$, and $(2 + \varepsilon)^3 = 8 + 12\varepsilon$.

Explicit Zero-Divisor Pairs

The General Pair

By Dual-Numbers Zero Divisors the zero divisors are exactly the nonzero elements of $\mathrm{M}$, that is, the elements $\varepsilon a'$ with $a' \neq 0$, and any two of them multiply to zero:

$$ (\varepsilon a')(\varepsilon b') = a' b'\,\varepsilon^2 = 0 \qquad \text{for all } a', b'. $$

A Table

Each row is a pair of nonzero zero divisors whose product is zero; the product is computed in the last column.

$A$	$B$	$AB$
$\varepsilon$	$\varepsilon$	$0$
$2\varepsilon$	$5\varepsilon$	$0$
$-3\varepsilon$	$\varepsilon$	$0$
$7\varepsilon$	$-4\varepsilon$	$0$

The point is that the annihilation is not special to any pair: it holds for every pair of nonzero elements of the line $\mathrm{M}$. Equivalently, $\operatorname{Ann}(\varepsilon a') = \mathrm{M}$ for every $a' \neq 0$, so every nonzero zero divisor annihilates the entire punctured line.

Example (worked, mixed pair). The element $2 + 3\varepsilon$ is not a zero divisor and annihilates nothing nonzero:

$$ (2 + 3\varepsilon)(b + \varepsilon b') = 2 b + (2 b' + 3 b)\varepsilon, $$

which vanishes only when $b = 0$ and $2b' + 3b = 0$, that is $b = b' = 0$.

Units and Inverses, Worked

Units

Example (worked). $A = 2 + 3\varepsilon$ has real part $2 \neq 0$, so it is a unit. Its inverse is

$$ A^{-1} = \frac{1}{2} - \frac{3}{4}\,\varepsilon, $$

and the check is

$$ (2 + 3\varepsilon)\Bigl(\tfrac{1}{2} - \tfrac{3}{4}\varepsilon\Bigr) = 1 + \Bigl(-\tfrac{3}{2} + \tfrac{3}{2}\Bigr)\varepsilon = 1. $$

Non-Units

Example (worked). $A = 0 + 3\varepsilon$ has real part $0$, so it is not a unit. Any attempted inverse fails: if $(3\varepsilon)(b + \varepsilon b') = 3b\,\varepsilon = 1$, then $3b = 1$ and $0 = 1$, a contradiction.

Example (worked). $A = -4 + \varepsilon$ has real part $-4 \neq 0$, so it is a unit, since the criterion is $a \neq 0$ and not $a > 0$:

$$ (-4 + \varepsilon)^{-1} = -\frac{1}{4} - \frac{1}{16}\varepsilon. $$

The Inverse Worked, in General

Example (worked). For $A = a + \varepsilon a'$ with $a \neq 0$, the inverse formula gives

$$ A^{-1} = \frac{1}{a} - \frac{a'}{a^2}\,\varepsilon, $$

and the general check is

$$ (a + \varepsilon a')\Bigl(\frac{1}{a} - \frac{a'}{a^2}\varepsilon\Bigr) = 1 + \Bigl(-\frac{a'}{a} + \frac{a'}{a}\Bigr)\varepsilon = 1. $$

The Matrix Model and the Shear

The Matrix of a Worked Product

By Dual-Numbers Matrix Element Representation the algebra embeds faithfully in $M_2(\mathbb{R})$ through

$$ \Phi(a + \varepsilon a') = \begin{pmatrix} a & a' \\ 0 & a \end{pmatrix}. $$

Worked, $\Phi(2 + 3\varepsilon) = \begin{pmatrix} 2 & 3 \\ 0 & 2 \end{pmatrix}$ and $\Phi(4 + 5\varepsilon) = \begin{pmatrix} 4 & 5 \\ 0 & 4 \end{pmatrix}$, and

$$ \Phi(2 + 3\varepsilon)\,\Phi(4 + 5\varepsilon) = \begin{pmatrix} 8 & 2\cdot 5 + 3\cdot 4 \\ 0 & 8 \end{pmatrix} = \begin{pmatrix} 8 & 22 \\ 0 & 8 \end{pmatrix} = \Phi(8 + 22\varepsilon), $$

matching the coordinate product worked above. The matrix is upper triangular with equal diagonal entries, and $\Phi(1 + \varepsilon) = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$.

Multiplication by a Unit Is a Shear

Let $A = a + \varepsilon a'$ be a unit, so $a \neq 0$, and write $s = a'/a$, giving $A = a(1 + s\varepsilon)$. Acting on $B = b + \varepsilon b'$,

$$ A B = a(1 + s\varepsilon)(b + \varepsilon b') = a b + (a b' + a s b)\varepsilon = a\bigl(b + (b' + s b)\varepsilon\bigr). $$

In the coordinates $(b, b')$ of $B$ this is the linear map

$$ (b, b') \longmapsto \bigl(a b,\ a(b' + s b)\bigr), $$

a dilation by $a$ followed by the shear $(b, b') \mapsto (b, b' + s b)$; the shear alone is multiplication by $1 + s\varepsilon$, which fixes the real axis pointwise.

Example (worked). For $A = 1 + \varepsilon$ and $B = 2 + 5\varepsilon$,

$$ (1 + \varepsilon)(2 + 5\varepsilon) = 2 + (5 + 2)\varepsilon = 2 + 7\varepsilon, $$

so the point $(2, 5)$ goes to $(2, 7)$, raised by the shear amount $s b = 1 \cdot 2 = 2$.

Example (worked). For $A = 2 + 3\varepsilon$ one has $a = 2$, $s = 3/2$, and for $B = 2 + 5\varepsilon$,

$$ (2 + 3\varepsilon)(2 + 5\varepsilon) = 4 + (10 + 6)\varepsilon = 4 + 16\varepsilon = 2\Bigl(2 + \bigl(5 + \tfrac{3}{2}\cdot 2\bigr)\varepsilon\Bigr), $$

a dilation by $2$ of the sheared point $(2, 8)$.

Worked Hyperbolic Elements

The Two Factors of a Unit

Every unit factors as $A = a(1 + s\varepsilon)$ with $a \neq 0$ and $s = a'/a$: the real factor $a$ is the scale and $1 + s\varepsilon$ the parabolic factor.

Example (worked). $2 + 3\varepsilon = 2\bigl(1 + \tfrac{3}{2}\varepsilon\bigr)$ and $-4 + \varepsilon = -4\bigl(1 - \tfrac{1}{4}\varepsilon\bigr)$.

Powers of a Worked Element

Example (worked). For $A = 2 + 3\varepsilon$,

$$ (2 + 3\varepsilon)^3 = 2^3 + 3\cdot 2^2 \cdot 3\,\varepsilon = 8 + 36\varepsilon, $$

and in the factorised form $2^3\bigl(1 + \tfrac{3}{2}\varepsilon\bigr)^3 = 8\bigl(1 + 3\cdot\tfrac{3}{2}\varepsilon\bigr) = 8 + 36\varepsilon$, the two computations agreeing.

The Hyperbolic Case Degenerates

In the split-complex algebra the units $a + jb$ satisfy $a^2 - b^2 = 1$ and form a hyperbola, a one-parameter multiplicative family with a genuine hyperbolic parameter; in the complex algebra the units form a circle. In $\mathbb{D}'$ the corresponding family of units is the pair of parallel lines $a = \pm 1$, and each line is carried by the parabolic family $1 + s\varepsilon$ with the additive parameter $s$. The hyperbola and the circle have both degenerated to lines, the hyperbolic parameter being replaced by the shear parameter; this is the arithmetic content of $\varepsilon^2 = 0$.

Relation Between the Decompositions

The two decompositions of the dual-number algebra coincide, and the worked examples make the coincidence explicit.

The conjugate decomposition is $\mathbb{D}' = R_{\mathbb{D}'} \oplus \varepsilon R_{\mathbb{D}'}$, from the eigenspaces of $\bar{\cdot}$; worked for $A = 2 + 3\varepsilon$ as $A = 2 + 3\varepsilon$.
The real–infinitesimal decomposition is the same identity written as $A = \operatorname{Re}(A) + \operatorname{Inf}(A)\varepsilon$.

There is no third decomposition, and no nontrivial Peirce decomposition: the only idempotents are $0$ and $1$, so the corners of Dual-Numbers Ideals and the Maximal Ideal are trivial. The single relation $R_{\mathbb{D}'} \cap \varepsilon R_{\mathbb{D}'} = \{0\}$ is the only intersection of the two submodules, and the worked example $2 + 3\varepsilon = A$ is its arithmetic content.

Summary

The worked computations in the dual-number algebra are these. The two involutions are the identity and dual conjugation; the identity has fixed set all of $\mathbb{D}'$, while dual conjugation has $+1$-eigenspace $R_{\mathbb{D}'} = \mathbb{R}$ and $-1$-eigenspace $\varepsilon R_{\mathbb{D}'} = \varepsilon\mathbb{R}$, meeting only in $0$. Every element of the maximal ideal is nilpotent of index two, $(\varepsilon a')^2 = 0$, while a mixed element has $(a + \varepsilon a')^n = a^n + n a^{n-1}\varepsilon a'$ and is never nilpotent unless $a = 0$. The zero divisors are the nonzero elements $\varepsilon a'$ of $\mathrm{M}$, and any two of them multiply to zero; for instance $(2\varepsilon)(5\varepsilon) = 0$. The criterion for a unit is $a \neq 0$, and the inverse is $a^{-1} - a^{-2}\varepsilon a'$; the check $(2 + 3\varepsilon)(\tfrac{1}{2} - \tfrac{3}{4}\varepsilon) = 1$ is the worked instance. Under the matrix model $\Phi(a + \varepsilon a') = \begin{pmatrix} a & a' \\ 0 & a \end{pmatrix}$ the worked product $\Phi(2 + 3\varepsilon)\Phi(4 + 5\varepsilon) = \Phi(8 + 22\varepsilon)$ reproduces the coordinate product, and multiplication by the unit $a(1 + s\varepsilon)$ acts as the dilation by $a$ followed by the shear $(b, b') \mapsto (b, b' + s b)$. The two decompositions of the algebra coincide, and no nontrivial Peirce decomposition exists.

Summary of Notation

Symbol	Meaning
$\mathbb{D}' = \mathbb{R}[\varepsilon]/(\varepsilon^2)$	Dual-number algebra; $\varepsilon^2 = 0$
$A = a + \varepsilon a'$	General dual number
$a = \operatorname{Re} A$, $a' = \operatorname{Inf} A$	Real and infinitesimal parts
$\bar{A} = a - \varepsilon a'$	Dual conjugation, the nontrivial involution
$R_{\mathbb{D}'}$, $\varepsilon R_{\mathbb{D}'}$	Real and infinitesimal submodules
$\mathrm{M} = (\varepsilon)$	Maximal ideal, $\mathrm{M}^2 = 0$
$\Phi(a + \varepsilon a') = \begin{pmatrix} a & a' \\ 0 & a \end{pmatrix}$	The faithful matrix model
$a(1 + s\varepsilon)$	Factorisation of a unit, $s = a'/a$
$(b, b') \mapsto (b, b' + s b)$	The shear of multiplication by $1 + s\varepsilon$