Why are the only division algebras over the

Explanation:
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 ${\displaystyle v^2=⟨v,v⟩}$: first one splits D as ${\displaystyle \mathbb{R}\oplus D_0}$ where ${\displaystyle D_0}$ is the space of elements with ${\displaystyle Tr=0}$ and then one observes that minimal polynomial of a traceless element has the form ${\displaystyle x^2-a=0}$ (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).

Step 1
One should consider this theorem to be two theorems: (1) ${\displaystyle \mathbb{C}}$ is the only C-central division algebra and (2) ${\displaystyle \mathbb{R}}$ and H are the only R-central division algebras. The reason there are so few choices is that C is alg. closed and R is nearly so. Division algebras with center equal to a particular field can be created using cyclic Galois extensions* and since Q has such extensions of arb. high degree there are Q-central division alg. of arb. high dimension.
Step 2
*There are further technical conditions to be satisfied on the cyclic extension in order for the construction of a division algebra to work, e.g., a finite field has a cyclic extension of each degree but there are no central div. alg. of dim>1 over a finite field. The relevant technical conditions are satisfied when the base field is the rational numbers.