Principled construction of the quaternions Is there a construction of the quaternions that doesn't proceed through generators and relations, and which makes the connection with 3D rotations clear?
I'm not happy with Clifford Algebra as an approach, as that also uses generators and relations.
I have one in mind myself, which is constructing the quaternions as the unique coordinate system for 3D rotations with the following properties:


*

*The coordinates form a set equal to $\mathbb R^n$ for some $n$.

*All coordinates are arbitrarily close to some other coordinate that represents a rotation. In particular, this allows for zero divisors to maybe exist.

*If $v$ represents a 3D rotation, then any $\lambda v$ where $\lambda$ is a non-zero scalar, represents the same rotation. The motivation for this axiom is that we want to find a projective coordinate system for 3D rotations.

*The composition of rotations corresponds to a bilinear product between coordinates. In particular, such a product consists of only the operations $+, \times, -$.


One can show using a long calculation that the quaternions are the only coordinate system that satisfies the above axioms.
 A: Let's take a modified version of the axioms above. We want a coordinate system for 3D rotations with the following features:


*

*The coordinates form a set equal to $\mathbb R^n$ for some $n$.

*All non-zero coordinates represent some rotation.

*Every rotation is represented by some coordinate.

*If $v$ represents a 3D rotation, then any $\lambda v$ where $\lambda$ is a non-zero scalar, represents the same rotation. The motivation for this axiom is that we want to find a projective coordinate system for 3D rotations.

*The composition of rotations corresponds to a bilinear product between coordinates. In particular, such a product consists of only the operations $+, \times, -$.


Let $\mathbf 1$ be the coordinate for a null rotation, $i$ be the coordinate for a half-turn around the $\vec i$ vector, and $j$ and $k$ be likewise for the $\vec j$ and $\vec k$ vectors. This all follows from axiom 3.
Let's consider the point $i$. Immediately, it's clear that $i^2 = \lambda \mathbf 1$ for some scalar $\lambda$. This scalar could be anything at this point except $0$.
By axiom (1) and (2), every non-zero linear combination of $\mathbf 1$ and $i$ represents some rotation. It is weird to take linear combinations of projective coordinate, but it is certainly possible, and it allows us to narrow down the space of possibilities. We see that these linear combinations generate a 2D algebra. It turns out there are only three 2D algebras up to isomorphism: The complex numbers, the dual numbers, and the split-complex numbers. We see that up to isomorphism, the algebra we've generated must be the complex numbers as the other ones contain zero divisors, and all rotations are invertible. Alternatively, we may argue using the orders of the elements: Rotations contain elements of any order.
The subalgebra generated by $\{\mathbf 1, i\}$ must correspond to some continuous subgroup of the 3D rotations. The only such subgroups that exist are rotations about some axes. Since $i$ represents a rotation about the $\vec i$ axis by a half turn, the group we've generated must be the set of all rotations about $\vec i$.
A similar argument applies to $j$ and $k$.
Finally, what is $ij$? Well, for geometric reasons it must be $\lambda_{ij} k$ for some scalar $\lambda_{ij}$. Moreover, we have $(iji^{-1})^2 = ij^2i^{-1}=-1$; so $\lambda_{ij}=\pm 1$. By geometric considerations also, $ji = \lambda_{ji}k$, and we can likewise conclude that ${ji} = \pm k$. Finally, we conclude that $i$ and $j$ anti-commute by starting with $(ij)^2 = -1$ and rearranging to get $ij = -(ij)^{-1}=-j^{-1}i^{-1}=-ji$.
What's the motivation?
Essentially, I posit that a transformation group can often be endowed with a coordinate system or representation. This coordinate system is either cartesian or projective.
A Cartesian coordinate system for a transformation group is one where (except for a negligible subset) all elements of $\mathbb R^n$ for some $n$ are in one-to-one correspondence with that transformation group. We insist that the product be bilinear. The rotation-dilation group can be represented by the Cartesian coordinate system $\mathbb C$. The complex number $i$ now represents a quarter-turn.
In fact, the classic examples of Cartesian systems for transformation groups are rings of square matrices.
Now consider as an alternative projective coordinate systems for transformation groups. Here, a scalar multiple of a coordinate represents the same transformation. Again, we insist that almost all elements of $\mathbb R^n$ stand for some element of the group, and that the product be bilinear.
The classic example is the matrix representation of the group of affine transformations. Here, the matrices are specified up to a ratio. We are generalising this because we may want our coordinate system to be more specific for computational efficiency and accuracy reasons.
The complex numbers $\mathbb C$ are also a projective coordinate system for rotations (not dilations). But then $i$ represents a half-turn, not a quarter turn.
[Removed a subjective remark that's perhaps non-mathematical. If someone who's seen it thinks it's appropriate to include it here, then I might do that].
A: $\def\RR{\mathbb{R}}$This is an answer to flesh out what I wrote in comments. Let $V$ be a real vector space with positive definite inner product. For a unit vector $\vec{u} \in V$, let $s_{\vec{u}}$ be the reflection over $\vec{v}$. Here is something we might hope for:

Vague hope: Can we build a ring $R$ whose additive structure comes from the vector space $V$ and whose multiplicative structure comes from multiplying the reflections $s_{\vec{u}}$ in the orthogonal group $O(V)$?

Note that, in particular, if we want multiplication to come from $O(V)$, we should have $\vec{u}^2 = 1$.
More precisely, 

Precise idea: Can we find an $\RR$-algebra $R$ which contains $V$ as a vector subspace, such that $\vec{u}^2=1$ for every unit vector $\vec{u} \in V$?

Well, if we ask for a linear map from $V$ to $R$ rather than insisting on a subspace, we can clearly do this. We simply define $R$ by generators and relations. This is why I don't understand the OP's objection to generators and relations: They are the correct tool whenever you want to build an algebraic object that has certain elements in it with certain properties.

Definition: Let $R$ be the $\RR$-algebra generated by $V$, subject to the relations $\vec{u}^2 = 1$ for every unit vector $\vec{u}$ in $V$.

The ring $R$ is the Clifford algebra, and it clearly has an $\RR$-linear map $V \to R$. It isn't obvious that this map is injective, but that is a lemma which you can find in any textbook on Clifford algebras, so I'll omit it. (The more precise statement is that, if $e_1$, $e_2$, ..., $e_n$ is any basis for $V$ then the $2^n$ products $e_{i_1} e_{i_2} \cdots e_{i_k}$ are a basis of $R$.)
Let's connect this to the standard definition. For any vector $\vec{v} \in V$, we can write $\vec{v} = c \vec{u}$ where $\vec{u}$ is a unit vector and $c \in \RR_{\geq 0}$. Since we asked $R$ to be an $\RR$-algebra, we have $\vec{v}^2 = (c \vec{u})^2 = c^2 = |\vec{v}|^2$. For any $\vec{v}$ and $\vec{w}$, we then have
$$\vec{v}^2 = \vec{v} \cdot \vec{v},\ \vec{w}^2 = \vec{w} \cdot \vec{w},\ \mbox{and}\ (\vec{v}+\vec{w})^2 = (\vec{v}+\vec{w}) \cdot (\vec{v}+\vec{w}).$$
Subtracting the first and second equation from the third,
$$\vec{v} \vec{w} + \vec{w} \vec{v} = 2 \vec{v} \cdot \vec{w}. \qquad (\ast)$$
Here the left hand side is multiplication in the ring $R$ and the right hand side is dot product. That's the more usual definition of the Clifford algebra. (Some people choose the opposite sign convention, $\vec{v} \vec{w} + \vec{w} \vec{v} = - 2 \vec{v} \cdot \vec{w}$. I don't have a firm opinion, but I started writing the answer this way, so I'll stick to it.)
Then a miracle happens! For any unit vector $\vec{u}$ and any vector $\vec{v}$, write $\vec{v} = \vec{v}^{\parallel} + \vec{v}^{\perp}$, where $\vec{v}^{\parallel}$ is parallel to $\vec{u}$ and $\vec{v}^{\perp}$ is orthogonal. We have $\vec{u} \vec{v}^{\parallel} = \vec{u} \cdot \vec{v} = \vec{v}^{\parallel} \vec{u}$, because $\vec{v}^{\parallel} = (\vec{u} \cdot \vec{v}) \vec{u}$ and $\vec{u}^2 = 1$. Equation $(\ast)$ also gives $\vec{u} \vec{v}^{\perp} = - \vec{v}^{\perp} \vec{u}$. So we have
$$\vec{u} \vec{v} = \vec{u} \left( \vec{v}^{\parallel} + \vec{v}^{\perp} \right) = \left( \vec{v}^{\parallel} - \vec{v}^{\perp} \right) \vec{u} = - s_{\vec{u}}(\vec{v}) \vec{u}. \qquad (\dagger)$$
Let $\vec{u}_1$, $\vec{u}_2$, ..., $\vec{u}_k$ be any sequence of reflections and let $g$ be the product $s_{\vec{u}_1} s_{\vec{u}_2} \cdots s_{\vec{u}_k}$ in the orthogonal group $O(V)$. Then using $(\dagger)$ repeatedly gives
$$\vec{u}_1 \vec{u}_2 \cdots \vec{u}_k \vec{v} = (-1)^k g(\vec{v}) \vec{u}_1 \vec{u}_2 \cdots \vec{u}_k. \qquad (!)$$
Let $\mathrm{Pin}(V)$ be the subgroup of the unit group of $R$ generated by the unit vectors $\vec{u}$. It is easy to see that there is a group map $\texttt{sign} : \mathrm{Pin}(V) \to \{ \pm 1 \}$ with $\texttt{sign}(\vec{u}) = -1$ for every unit vector. Then equation $(!)$ tells us that there is a map $\pi: \mathrm{Pin}(V) \to O(V)$ with $\pi(x)(\vec{v}) = \texttt{sign}(x) x v x^{-1}$. 
In this sense, we have fulfilled our hope: $V$ is a vector subspace of $R$, and the fact that $\pi$ is a map of groups tells us that any multiplicative identities that hold between the $\vec{u}$ in $R^{\times}$ also hold between the $s_{\vec{u}}$ in $O(V)$.
The ring $R$ decomposes as $R_+ \oplus R_-$ where $R_+$ is spanned by products of an even number of vectors and $R_-$ uses an odd number. The subgroup $\mathrm{Spin}(V) = \mathrm{Pin}(V) \cap R_+$ maps to $SO(V)$, so this is where to look for rotations.

So far, $V$ can have any dimension. Now we specialize to $\RR^3$, with orthogonal basis $e_1$, $e_2$, $e_3$. It is easy to check that the elements $e_1 e_2$, $e_2 e_3$ and $e_1 e_3$ multiply by the same relations as $i$, $j$ and $k$ in the quaternions and, in fact, $R_+ \cong \mathbb{H}$.
The special magic thing that happens in the quaterions is that $\mathrm{Spin}(\RR^3)$ is the unit sphere in $\mathbb{H}$. In general, $\mathrm{Spin}(\RR^n)$ is an $\binom{n}{2}$-dimensional submanifold of a $2^{n-1}-1$ dimensional unit sphere in $R_+$. It is only when $n=1$, $2$ or $3$ that $\binom{n}{2} = 2^{n-1} - 1$.
