Since there are complex numbers (2 dimensions) and quaternions (4 dimensions), it follows intuitively that there ought to be something in between for 3 dimensions ("triernions").
Yet no one uses these. Why is this?
Since there are complex numbers (2 dimensions) and quaternions (4 dimensions), it follows intuitively that there ought to be something in between for 3 dimensions ("triernions").
Yet no one uses these. Why is this?
It's because there isn't one! (Indeed, Hamilton was originally searching for such a thing, and found the quaternions instead; it was only later that people understood why he hadn't been successful, initially.)
The quaternions - along with the real numbers and the complex numbers - have a number of nice properties: specifically, they form a real division algebra. This is a mouthful, but basically amounts to:
Addition/multiplication of quaternions satisfy the ring axioms.
We can divide by quaternions.
We can multiply a quaternion by a real (and this "scalar multiplication" satisfies the basic properties it should).
It turns out the only finite-dimensional real division algebras are $\mathbb{R}$, $\mathbb{C}$, and the quaternions; see this. (I include associativity in the definition of algebra: if we allow non-associative algebras, then the octonions also count.)
By the way, there is a way to (sort of) keep going past the quaternions: the Cayley-Dickson construction. This produces things like the octonions and the sedenions, and other delightfully weird algebraic structures. However, it has a couple drawbacks:
Each time you apply Cayley-Dickson, the dimension of the starting algebra doubles. So this won't help us get to $3$.
Also, you keep losing nice properties. Passing from the reals to the complex numbers, we lose order; going from the complexes to the quaternions, we lose commutativity of multiplication. If we keep going, we lose associativity of multiplication, in increasing degrees: the sedenions are even less associative than the octonions, etc.
Assume $A$ is a three-dimensional (associative) algebra over $\mathbb{R}$. We can assume $\mathbb{R}$ is embedded in $A$. If $a\in A$, $a\notin\mathbb{R}$ the map $l_a\colon A\to A$, $l_a(x)=ax$, is an endomorphism of $A$ as a vector space over $\mathbb{R}$.
Let $\lambda$ be a real eigenvalue of $l_a$, with eigenvector $b\ne0$, so $ab=\lambda b$. Such an eigenvalue exists, because the characteristic polynomial of $l_a$ has degree $3$. Then $(a-\lambda)b=0$. Note that $a-\lambda\ne0$, so $A$ has zero divisors, in particular $A$ is not a division algebra.
It's a bit more complicated showing that a finite-dimensional division algebra over $\mathbb{R}$ can only have dimension $1$, $2$ or $4$ and it is isomorphic to $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$ (the quaternions); this is known as Frobenius' theorem.
On the other hand, three-dimensional algebras over $\mathbb{R}$ exist (but they have zero divisors, as shown above). A simple example is $\mathbb{R}[X]/(X^3-1)$, but they can be non-commutative as well.
There is an algebra in dimension 6, halfway between 4 and 8, that is not a division algebra but correctly interpolates various constructions.
http://arxiv.org/abs/math/0411428
Nothing like that is known for dimension 3 sitting between 2 and 4.
The closest thing to triterniums would be the structure $[\mathbb R^3, +, \times]$ where $``\times"$ represents the cross product
$$(a_1 \mathbf i + b_1 \mathbf j + c_1 \mathbf k) \times (a_2 \mathbf i + b_2 \mathbf j + c_2 \mathbf k) = \left| \begin{matrix} \mathbf i & \mathbf j & \mathbf k \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ \end{matrix} \right|$$
It distributes over $``+"$,
is anti commutative,
and isn't associative, but $[a \times (b \times c)] + [b \times (c \times a)] + [c \times (a \times b)] = 0$
Have you heard of the Frobenius theorem?
https://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras)
Triernions would not be an associative division algebra.
Among the vectorial spaces $\mathbb R^n$, only $\mathbb R$ and $\mathbb R^2\space ( \approx \mathbb C)$ admit a multiplication that gives them a structure of field. For the other values, only for $\mathbb R^4$ we can have a multiplication without divisors of zero and associative seeming a field; actually,with this multiplication, $\mathbb R^4$ becomes a division ring or, also called, skew field and is named quaternions.
According to a celebrated theorem of Wedderburn all finite division rings are necessarily commutative so quaternions are the first example of a non-commutative skew field. French mathematicians used the terminology “corps” for both “fields and skew fields” so there are for them commutative and non-commutative corps. By the theorem of Wedderburn, quaternions give the first example of a non-commutative “corp”.
All of these other answers go way over my head, but this is the way I think about it.
The complex numbers can be represented on a 2-dimensional plane, but they are an extension of the one-dimensional real number line. They are not really 2-dimensional, it's just a convenient way for us to represent them. When we expand the reals from 1 dimension into 2 dimensions, the corresponding complex numbers must double their dimension as well, going from 2 to 4. That is why they appear to "skip" 3.
Perhaps I am wrong, or this argument was folded into the other explanations, and I just didn't see it.