Are there real algebras that don't have rational structure constants? Take a finite dimensional associative algebra $A$ over the reals. Fix a basis $\{x_1, x_2, \ldots x_n\}$. The multiplication is completely specified by specifying structure constants $c^{ij}_k$ defined by the following equation:
$$x_i \cdot x_j = \sum_k c^{ij}_k x_k \quad\forall i, j$$
Of course, the structure constants depend on the choice of basis.
My question is:
Are there algebras such that no choice of basis leads to structure constants that are all rational?
My conjectured example would be the algebra spanned by two variables $a$ and $b$, and relations $a^2 = a, ab = ba = b, b^2 = \sqrt{2}a$, but I couldn't prove it yet.
Note: One could also ask "integer" instead of "rational", this is equivalent.
 A: Nice question! Here is a coarse heuristic argument suggesting that the answer should be yes: the structure constants of a multiplication on a finite-dimensional real vector space $V$ of dimension $n$ form an $n \times n \times n$ table of real numbers, so the space of these should have dimension $n^3$ (ignoring associativity, since it's hard to tell how many independent constraints this adds). The action of change of basis on these structure constants is the action of $\text{GL}(V)$, which has dimension $n^2$, hence the quotient of structure constants by change of basis should have dimension at most $n^3 - n^2$, which in particular is positive as soon as $n \ge 2$.
On the other hand, there are only countably many choices of rational structure constants even before quotienting by change of basis. So as soon as $n \ge 2$ the "generic" algebra should be a counterexample (although again it's hard to tell how strong a constraint associativity is; perhaps you need to go a few dimensions higher).
Edit: Including units, the dimension count is as follows. If we always include the unit as the first element of our basis, then a multiplication (still not assumed to be associative) on an $(n+1)$-dimensional real vector space $V$ is a table of $n \times n \times (n+1)$ numbers, and the action of change of basis is a group of dimension $n^2 + n$ (we also want to be able to add the unit to other elements of our basis). So now the dimension count is
$$n^2 (n + 1) - n^2 - n = n^3 - n$$
which is positive as soon as $n \ge 2$, hence as soon as $n+1 \ge 3$, so this is the dimension where we should start seeing generic counterexamples. 
