Why would that be a problem? The caller has to provide a function for 
and "multiplying" an N, and as long as I define what it means to multiply 
add strings it shouldn't matter that I'm using a dual number where both
components are strings.

But I think this is a case of the rectangle-square problem: dual scalars 
and a
dual vectors a both a subset of dual quaternions that contain less 
than their supertype. I guess what I'm really looking for is a way to
"magically" promote objects.

A quaternion is an object of

  H = {a + bi + cj + dk | a,b,c,d ∈ R},

a vector is an object of

  V = {ai + bj + ck | a,b,c ∈ R},

but we can also view H as

  H = R × V = {(a, v) | a ∈ R, v ∈ V}.

The first definition of H is how it is usually defined and written out, but 
second definition makes it easier to compute the product:

  (p_r, p_v) (q_r, q_v) = (p_r q_r - p_v ⋅ q_v, p_r q_v + q_r p_v + p_v × 

So far this is a simple hierarchy. But the set of scalars and the set of
vectors can be embedded in the set of quaternions:

  R → H, a ↦ (a, 0)  and V → H, v ↦ (0, v)

We can also define things like the "quaternion cross product" and 
dot product" for quaternions where the scalar part is zero:

  (0, p) × (0, q) := (p, p × q)
  (0, p) ⋅ (0, q) := (p ⋅ q, 0)

I'm starting to think this is becoming a pointless exercise. Maybe I should
just limit myself to "dual quaternions are a pair of quaternions" and
"quaternions are pairs of a scalar and a vector" and forget about the magic

On Tuesday, February 6, 2018 at 12:01:42 AM UTC+1, Sam Tobin-Hochstadt 
> I'm not sure how the "If" got there. 
> But to say more, consider your function: 
>   (: dual-* (∀ (N) (→ (Dual-Number N) (Dual-Number N) (→ N N N) (→ N N 
> N) (Dual-Number N)))) 
>   (define (dual-* d1 d2 * +) 
>     (cond 
>       [(D? d1) 
>        (D 
>          (D-real d1) 
>          (D-dual d1))] 
>       [else (D d1 d1)])) 
> Now you imagine instantiating `N` with things like `(Vector3 Real)`, 
> but if we instantiated it instead with `(Dual-Number String)`, then 
> you'd have a problem. 
> Sam 

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