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Out of curiosity, I was seeing about hypercomplex numbers. In that article, the definition says that, "Where possible, it is conventional to choose the basis so that $i_k^2 \in \{ -1, 0, +1 \}$. A technical approach to hypercomplex numbers directs attention first to those of dimension two.", where $i_k$ is the basis of hypercomplex number.

Later on, "Clifford algebraic construction" and "Cayley–Dickson construction" (along with "modified Cayley–Dickson construction") of hypercomplex numbers have been given. But both of these does not consider $i_k^2$ being $0$. Is there something that makes studying about dual numbers (separately) not much useful?

So, is there any other methods of construction which algorithmically produces dual numbers and its successors? Also, are there any methods in general that produce hypercomplex numbers like these?

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  • $\begingroup$ There are Grassmann numbers - not commutative generalization of dual numbers $\endgroup$
    – Anixx
    Dec 11, 2022 at 3:10

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methods of construction

Dual numbers have a matrix representation (in fact more than one), or can be identified with $\Bbb R[X]/X^2$. Either shows duals are equiconsistent with reals. See here for a treatment of more general extensions of $\Bbb R$.

Is there something that makes studying about dual numbers (separately) not much useful?

Dual numbers and their generalizations have at least four applications beside their original motive, but you can't be blamed for not seeing them often. In principle, one can use square roots of $0,\,1,\,-1$ in an arbitrary sequence as one doubles the dimension, starting from $\Bbb R$. One can even vary anti/commutativity rules; for example, one can use $-1$ twice in more than one way, but they haven't been useful for equally many things.

In principle, an example of mathematics is useful insofar as it describes something we encounter, so why some examples are more useful than others amounts to asking why we encounter certain things, or even why the world is a certain way.

Philosophy aside, complex numbers have the advantage over split-complex numbers that they avoid zero divisors so are invertible, while dual numbers have the disadvantage relative to both that, because they're nilpotent, sums/differences of squares don't need them, because $(a\epsilon)^2=0$ just disappears. Mind you, sometimes that's useful.

Clifford algebraic construction and Cayley–Dickson construction (along with modified Cayley–Dickson construction) of hypercomplex numbers have been given. But both of these does not consider $i_k^2$ being $0$.

One modification of CD is to use a square root of $0$ rather than $-1$ to double the dimension. Although dual numbers can be related to a suitable Clifford algebra, most applications of such algebras warrant their being related instead to complex or split-complex numbers.

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  • $\begingroup$ "one can use square roots of 0,1,−1 in an arbitrary sequence as one doubles the dimension, starting from R". Should I only double the dimension? Algebra with dimensions 3, 5, 6, … doesn't exist? $\endgroup$
    – Sourav Kannantha B
    Mar 13, 2022 at 16:40
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    $\begingroup$ @SouravKannanthaB Oh, they do. For example, there's a $3$-dimensional cousin of the dual numbers, namely $\Bbb R[X]/X^3$. (They're useful to differentiate a function twice instead of once.) But you brought up values of $i_k^2$, so I addressed those. With linear transformations over $\Bbb R$, only quadratics reduce to such a simple condition; the equivalent with a vanishing cubic in some new number would set $i_k^3+\alpha i_k\in\{0,\,1,\,-1\}$ for some $\alpha\in\Bbb R$. $\endgroup$
    – J.G.
    Mar 13, 2022 at 16:44
  • $\begingroup$ Also, are quaternions 3-dimensional or 4? It has basis of ${0, i, j, k}$. But $k$ can be expressed as $ij$, so only i, j are independent! $\endgroup$
    – Sourav Kannantha B
    Mar 13, 2022 at 18:57
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    $\begingroup$ @SouravKannanthaB Of course, by $0$ you meant $1$. $\Bbb H$ is $1$-dimensional over $\Bbb H$, $2$-dimensional over $\Bbb C$ and $4$-dimensional over $\Bbb R$, and in each case the "dimension" is that of a vector space, so counts linearly independent numbers; you can make $k$ from $1,\,i,\,j$ by multiplication, but not by addition, not even with arbitrary real coefficients. Of course, $4=2^2$ because $\Bbb H$ can be thought of as $\Bbb R[i][j]$. $\endgroup$
    – J.G.
    Mar 13, 2022 at 19:03

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