What is the general definition of the conjugate of a multiple-component number? I know that the conjugate of a binomial is the negation of the second part.  So the conjugate of (a + b) would be (a - b).
I know that the conjugate of a complex number (a + bi), similarly, is (a - bi).
I also know that the conjugate of a quaternion (w + xi + yj + zk) is (w - xi - yj - zk).
So I'm thinking that the general definition of the conjugate of a multiple-component number is the negation of all non-real parts.
If true, then the conjugate of a dual number (a + be), where 'e' is shorthand for the dual operator epsilon, would be (a - be).
Complication: But I read in A Beginners Guide to Dual Quaternions (http://wscg.zcu.cz/wscg2012/short/a29-full.pdf, page 6) that the conjugate of a dual quaternion is calculated by taking the conjugate of both the real and dual parts.  That is, for a dual quaternion dq = q_real + (q_dual)e, the conjugate dq* = q_real* + (q_dual*)e.  The scalar in the dual component quaternion is not negated, which I expected.
So what is definition of the conjugate of a generic multiple-component number?  Also, is the paper wrong in its definition of the conjugate of a dual quaternion?
 A: Given the dual quaternion defined by $z = (a+a_0\epsilon)+(b+b_0\epsilon)i+(c+c_0\epsilon)j+(d+d_0\epsilon)k$ $= (a+bi+cj+dk)+(a_0+b_0i+c_0j+d_0k)\epsilon$, for $a,a_0,b,b_0,c,c_0,d,d_0 \in \Bbb R$, $i,j,k$ defined by $i^2=j^2=k^2=ijk=-1$, and $\epsilon$ defined by $\epsilon \ne 0$ and $\epsilon^2 = 0$, the conjugate $\overline{z}$ is defined by :
$$\overline{z} = \overline{(a+a_0\epsilon)+(b+b_0\epsilon)i+(c+c_0\epsilon)j+(d+d_0\epsilon)k}$$ $$= (a+a_0\epsilon)-(b+b_0\epsilon)i-(c+c_0\epsilon)j-(d+d_0\epsilon)k$$ $$= (a-bi-cj-dk)+(a_0-b_0i-c_0j-d_0k)\epsilon$$ $$= \overline{(a+bi+cj+dk)}+\overline{(a_0+b_0i+c_0j+d_0k)}\epsilon$$ $$= \overline{(a+bi+cj+dk)+(a_0+b_0i+c_0j+d_0k)\epsilon}$$
Notice in this case we have (after some algebra) $\|z\|^2 = z\overline{z} = (a+a_0\epsilon)^2 + (b+b_0\epsilon)^2 + (c+c_0\epsilon)^2 + (d+d_0\epsilon)^2$.  From this last expression, you can see we no longer have any $i$'s, $j$'s, or $k$'s and thus the norm of a dual quaternion is a dual number.
For general multi-component numbers, it depends on how you construct your number system, so I don't think I can give you a good answer.  For instance, notice in this example that the conjugation is with respect to the quaternion parts as opposed to the dual parts of these numbers.  This is because the dual quaternions are constructed as quaternions over the ring of dual numbers and thus the norm should be an element of that ring -- a dual number.  However you construct your number system and however you define your conjugation, though, you want the bilinear form $\langle A, B\rangle = \frac {\overline A B + A\overline B}{2}$ to produce an element of our ring.  We then define $\|A\|^2 = \langle A, A\rangle$.  Note, for constructions where $A\overline A =\overline AA$ (which is most of them, including the complex numbers and dual numbers), this reduces to $\|A\|^2 = A\overline A$.
For more information, there are several pdf's floating around the internet on the subject.  A cursory search yielded this pdf which expands much further on dual quaternions.  The real problem is that there is no standard nomenclature for multicomponent numbers.  See here for $16$ different terms for the cousins of "dual numbers" (which also have several synomyns) which are called by some authors "split-complex numbers".

Edit:
The method of conjugation and the type of number that the norm is will always depend on how your number system was constructed.  The standard method of conjugation is just to negate all of the imaginary parts of our construction -- but not of the ring over which we constructed our numbers.  For example, we constructed the dual quaternions as quaternions over the commutative ring of dual numbers, therefore with negated the quaternion parts while NOT conjugating the dual parts.  Let's look at $2$ other ways of constructing objects which look like the dual quaternions.
$\#1$
Let's construct a set of objects from dual numbers, but instead of real number coefficients, we'll instead have coefficients from the ring of quaternions.  So our system will be of the type $z=a+b\epsilon$, where $a=a_0 + a_1i+ a_2j + a_3k$ and $b=b_0 + b_1i+b_2j+b_3k$ are quaternions.  Then $\langle A, B \rangle$ (as defined above) should always result in a quaternion.  Let's check if the "standard" conjugation $\overline z = a-b\epsilon$ will accomplish this.  Let's see:
$$\langle a+b\epsilon, c+d\epsilon\rangle = \frac 12((a-b\epsilon)(c+d\epsilon)+(a+b\epsilon)(c-d\epsilon)) = \frac 12((ac+(ad-bc)\epsilon)+(ac+(bc-ad)\epsilon)) = \frac 12(2ac)=ac$$
Where $ac$, as the product of quaternions, is a quaternion -- success!  This conjugation results in a quaternionic bilinear form, and thus a quaternionic norm.  Notice here that we didn't even need to consider the structure of quaternions (i.e. we didn't even need to plug in $a_0+a_1i+a_2j+a_k$ for $a$) to get our result.  It just fell out of the definition.
$\#2$
Now let's construct a different number system.  This time let's construct our number system over the reals (which is not only a commutative ring, but a field).  That means that this time $i,j,k, \epsilon, i\epsilon, j\epsilon,$ and $k\epsilon$ are all different objects.  If you're familiar with vector spaces, you can see that, together with $1$, these $8$ objects form a basis for our space.
We'll define those objects via the following table:
$$\text{row $\times$ column}
\begin{array}{c|lcr}
 & 1 & i & j & k & \epsilon & i\epsilon & j\epsilon & k\epsilon \\
\hline
1 & 1 & i & j & k & \epsilon & i\epsilon & j\epsilon & k\epsilon \\
i & i & -1 & k & -j & i\epsilon & -\epsilon & k\epsilon & -j\epsilon \\
j & j & -k & -1 & i & j\epsilon & -k\epsilon &  -\epsilon & i\epsilon \\
k & k & j & -i & -1 & k\epsilon & j\epsilon & -i\epsilon & -\epsilon \\
\epsilon & \epsilon & i\epsilon & j\epsilon & k\epsilon & 0 & 0 & 0 & 0 \\
i\epsilon & i\epsilon & -\epsilon & k\epsilon & -j\epsilon & 0 & 0 & 0 & 0 \\
j\epsilon & j\epsilon & -k\epsilon & -\epsilon & i\epsilon & 0 & 0 & 0 & 0 \\
k\epsilon & k\epsilon & j\epsilon & -i\epsilon & -\epsilon & 0 & 0 & 0 & 0
\end{array}$$
So if $z=z_0 + z_1i + z_2j + z_3k + z_4\epsilon + z_5i\epsilon + z_6j\epsilon + z_7k\epsilon$, we define the standard conjugate as $\overline z = z_0 - z_1i - z_2j - z_3k - z_4\epsilon - z_5i\epsilon - z_6j\epsilon - z_7k\epsilon$.  Multiplying two arbitary elements would be really annoying so I just tried to find the modulus of this system: $\langle z, \overline z\rangle$.  If I didn't make a mistake, this will be $z\overline z = z_0^2+z_1^2+z_2^2+z_3^2$. Again, we've got an element of our ring -- a real number -- thus this conjugation is a good one for this system of "numbers".
NOTE: It would be a good exercise to multiply out two arbitrary elements of this number system via our bilinear form and verify that we always get a real number.  We should, but if not, let me know.
Conclusion:
So as you can see, the conjugate of a multicomponent system is completely dependent on how we construct our number system.  There is a "standard" method which is to negate the "imaginary" parts of your system, and it seems to work for constructions made by embedding one of other usual number systems within another (constructing a algebra over a ring).
In actuality, I never actually constructed (in the algebraic sense) the number systems above, and so I'm not completely sure that they are self-consistent and behave well (form an algebra) under the usual notions of multiplication and addition.  Nevertheless, we seemingly have a good definition of conjugation.
