A dual number is given by an expression of the form $a + \epsilon b$, where $a$ and $b$ are real numbers and $\epsilon^2 = 0$ (but $\epsilon \ne 0$). The set of dual numbers is a topological vector space and a commutative algebra over the real numbers.

We can generalise (at least the algebraic aspects) from $\mathbb{R}$ to any commutative ring$R$.

Interpretations

This can be thought of as:

the vector space $\mathbb{R}^2$ made into an algebra by the rule

$(a, b) \cdot (c, d) = (a c, a d + b c) ;$

the subalgebra of those $2$-by-$2$ real matrices of the form

$\left(\array { a & b \\ 0 & a } \right);$

the polynomial ring $\mathbb{R}[\mathrm{x}]$ modulo $\mathrm{x}^2$;

if $\epsilon$ is regarded as being of degree 1 and $\mathbb{R} \oplus \epsilon \mathbb{R}$ is regarded accordingly as a superalgebra then this is the algebra of functions on the odd line$\mathbb{R}^{0|1}$.

the square-0-extension corresponding to the $\mathbb{R}$-module (see there) given by $\mathbb{R}$ itself.

We think of $\mathbb{R}$ as a subset of $\mathbb{D}$ by identifying $a$ with $a + 0 \epsilon$.

Properties

$\mathbb{D}$ is equipped with an involution that maps $\epsilon$ to $\bar{\epsilon} = -\epsilon$:

notice that the absolute value of a dual number is a non-negative real number, with

${|z|^2} = z \bar{z}.$

But this absolute value is degenerate, in that ${|z|} = 0$ need not imply that $z = 0$.

Some concepts in analysis can be extended from $\mathbb{R}$ to $\mathbb{D}$, but not as many as work for the complex numbers. Even algebraically, the dual numbers are not as nice as the real or complex numbers, as they do not form a field.

Revised on November 14, 2011 06:02:25
by Toby Bartels
(139.55.238.24)