As normed division algebras, when we go from the complex numbers to the quaternions, we lose commutativity. Moving on to the octonions, we lose associativity. Is there some analogous property that we lose moving from the reals to the complex numbers?
Complex Numbers – What is Lost Passing from Reals to Complex Numbers?
complex numbersdivision-algebrasoctonionsquaternions
Related Solutions
Essentially one first proves that any real division algebra $D$ is a Clifford algebra (i.e. it's generated by elements of some inner product vector space I subject to relations $v^2=\langle v, v\rangle$): first one splits $D$ as $\mathbb R\oplus D_0$ where $D_0$ is the space of elements with $Tr=0$ and then one observes that minimal polynomial of a traceless element has the form $x^2-a=0$ (it's quadratic because it's irreducible and the coefficient of $x$ is zero because it is the trace). Now it remains to find out which Clifford algebras are division algebras which is pretty straightforward (well, and it follows from the classification of Clifford algebras).
This proof is written in Wikipedia.
I'd never heard of a bicomplex number, but I like algebra, so let's go for it. From what I understand, reading Wikipedia and playing around:
We assume that, in addition to $i$, our favorite square root of $-1$, we have another, $h$, so that $h^2 = -1 = i^2$. Further, we force $i$ and $h$ to commute, so that $ih = hi$.
You can see for yourself that $(ih)^2 = 1$, since $h$ and $i$ commute. So, $ih$ is a square root of $1$. We know of some other square roots of $1$, let's see what happens when we multiply something by $ih$. At this point, we'll assume our numbers are associative (just like we assumed $h$ and $i$ commute).
$(ih)h = ih^2 = i(-1) = -i$. Thus, multiplication by $ih$ sent $h$ to $-i$. This means $ih$ can't be $1$ or $-1$, the square roots of $1$ we already know and love, since $i$ and $h$ are distinct (that is, $h \neq -i$).
It turns out the set $\{1, i, h, ih\}$ forms a basis for these numbers. Thus, a generic "bicomplex number" looks like $a + bi + ch + dih$.
But, these bicomplex numbers are certainly don't act like quaternions. For one, they form a commutative algebra; the order of multiplication never matters! But more seriously, we have zero-divisors:
$$(1 + ih)(-1 + ih) = -1 + ih - ih +(ih)^2 = -1 +ih - ih - 1 = 0.$$
This is a pretty serious departure from the quaternions.
It turns out that the biquaternions are isomorphic to the direct sum $\mathbb{C} \oplus \mathbb{C}$ of the field of complex numbers with itself (here, addition and multiplication are performed component-wise, like vector addition). This is not obvious to me, but it's believable.
Direct sums are fairly uninteresting: Just slap two copies of something together, and there you go! It explains why we have zero-divisors, and commutativity. We just do everything we normally would, but do it twice; business as usual.
The quaternions on the other hand - now that's an algebra! Associative, but not commutative; no zero divisors; we can define a norm. Generally, life with the quaternions is great.
To build the complex numbers, quaternions, or octonians from the reals, the situation is a lot more subtle; the construction is an example of a "twisted group algebra." The complex numbers, quaternions, and octonions are all examples of twisted group algebras over the real numbers. In particular, we get the quaternions by introducing a "twisted" way of multiplying elements of the group algebra $\mathbb{R}[\mathbb{Z}_2^2]$.
Unfortunately, I'm still processing all the details myself, so it's up to you to figure out exactly what "twisting" something entails. Essentially, it's like a direct sum, but with the terms interacting multiplicatively more than just "component-wise," we need some extra magic thrown in.
Sources:
John Baez' write up on octonions, where I first saw the term "twisted group algebra." Lacking on details of the construction, lots of attention to the big picture.
John Bales' write up on "Properly Twisted Group Algebras", which is way more detail than I'd hoped for, but could fill in some of the gaps.
Best Answer
The most obvious property that we lose is the linear (or total) ordering of the real line.