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I searched the backlog of this website for something along the lines of my question to no avail. Not to mention a myriad of PDFs online which just drop the definition and allege more details can be found in Clifford's paper published in the 1873 Proceedings of the London Mathematical Society which is conveniently hidden behind a paywall.

It is easy to see what motivates the development of complex numbers $\mathbb{C}$, the inability of real numbers $\mathbb{R}$ to provide a solution in $x^2 + 1 = 0$ which leads to defining the indeterminate $i^2 = -1$. Naturally, extending or changing algebraic rules doesn't necessarily have to be motivated by the inability of the current abstraction to provide a solve to a particular, previously unconsidered equation.

What troubles me is that I am not sure how Clifford or Study (or anyone else in between) struck upon defining $\epsilon^2 = \epsilon^n = 0$ beyond just taking complex numbers, stating $i^n = p$ and throwing $p$s at the wall until one sticks. I understand the results, such as the way dual numbers lend themselves to various useful constructs such as encoding the first derivative in the "dual part" after evaluating a function extended to the realm of dual numbers... Or the notion of dual quaternions which extend quaternions to handle translations in addition to rotations (applicable to rigid body kinematics). But not what led Clifford to them, discounting divine providence.

Fundamentally, did Clifford just poke at the work of Hamilton and those before him, seeing what sticks or did he have a problem in mind which led him to the notion of dual numbers when he was working on the "preliminary sketch of biquaternions"?

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    $\begingroup$ For those who have neither heard of dual numbers before, nor want to penetrate behind that paywall: Here is the Wikipedia entry that explains what this question is all about: en.wikipedia.org/wiki/Dual_number . $\endgroup$ – Christian Blatter Aug 25 '18 at 18:25
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If $p\in\mathbb{R}[X]$ is irreducible over $\mathbb{R}$, by the fundamental theorem of algebra $\operatorname{deg}p\le 2$, and any non-real root of $p$ we invent is complex, split-complex or dual. It makes sense to study all three and see what they help with. In the case of dual numbers, the most obvious application is automatic differentiation.

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  • $\begingroup$ While I agree dual numbers are natural to consider for differentiation, this does not seem to be Clifford's motivation (see my answer). $\endgroup$ – Kimball Aug 26 '18 at 23:07
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A vector, as thought of by Hamilton, can be thought of as a direction and a magnitude but no position. One can add these and get another vector. One can also look at transformations taking vectors to vectors (scalings and rotations), which is Hamilton's geometric meaning of his quaternions. These one can add and multiply.

Clifford, in his 1873 paper Preliminary Sketch of Biquaternions, wanted to study what he called rotors (short for rotators), which are vectors with initial positions. I.e., these are quantities representing direction, magnitude and position. Now the sum of rotors is not a rotor, but Clifford calls a formal sum of rotors a motor.

The reason for this interest is, if one thinks of rotations in 3-space, then Hamilton's quaternions only describe rotations about lines passing through a given fixed point (usually taken to be the origin). However, in 3-space we can rotate about other lines and would like to describe, say, what happens when you do one rotation about the line $x=y=0$ and then one about the line $x=y=1$. Clifford proposes to study these kinds of rigid motions by studying transformations of motors to motors (which is essentially what he means by the biquaternion in the title, different than Hamilton's original usage).

In this study, Clifford is led to the introduction of an operation $\omega$ which takes motors to vectors, and finds that $\omega^2 = 0$ though $\omega$ itself is not 0. Then Clifford expresses his motors in the form $a + \omega b$, which one can interpret now as a set of dual numbers.

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Actually, complex numbers were not considered because we were so eager to have a square root of $-1$. But they turned out to be useful to solve cubic equations, and they gave meaningful and correct answers that were real numbers (see Wikipedia).

So I guess the same might be true with dual numbers. They turned out to be useful, that is why they are teached and used.

I should note that I'm no expert on dual numbers, but maybe you can think about $\epsilon$ like the algebraic version of $\epsilon$ in analysis. It is so small, that if you multiply it with itself it yields zero.

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  • $\begingroup$ Yes, it makes sense to use dual numbers in analysis, but this was not Clifford's motivation (see my answer). $\endgroup$ – Kimball Aug 26 '18 at 23:08

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