We have a definite quaternion algebra $K$, we have to show that it is ramified at $\infty$.
For any prime $p$ including $\infty$, $\mathbb{Q}_p$ be it's $p$-adic completion if $p$ is finite and $\mathbb{Q}_{\infty} = \mathbb{R}$. We say a quaternion algebra $K$ splits at $p$ if $$ K \otimes_{\mathbb{Q}} \mathbb{Q}_p \cong M_2(\mathbb{Q}_p )$$ Here, $\alpha^{2} , \beta^{2} \in \mathbb{Q}$ and $<0$, $\alpha. \beta = - \beta. \alpha$ otherwise we say $K$ is ramified at $p$.
(This isomorphism is algebra isomorphism)
Now, here for $K= \mathbb{Q} + \mathbb{Q} \alpha + \mathbb{Q} \beta + \mathbb{Q} \alpha. \beta $, and $\mathbb{Q}_{\infty} = \mathbb{R}$ and
$$ K \otimes \mathbb{R} = (\mathbb{Q} + \mathbb{Q} \alpha + \mathbb{Q} \beta + \mathbb{Q} \alpha. \beta) \otimes \mathbb{R}$$
$$= (\mathbb{Q} \otimes \mathbb{R}) + (\mathbb{Q} \alpha \otimes \mathbb{R}) + (\mathbb{Q} \beta \otimes \mathbb{R}) + (\mathbb{Q} \alpha. \beta \otimes \mathbb{R})$$
$$= \mathbb{R} + \alpha (\mathbb{Q} \otimes \mathbb{R}) + \beta (\mathbb{Q} \otimes \mathbb{R}) + \alpha. \beta (\mathbb{Q} \otimes \mathbb{R})$$
$$\cong \mathbb{R} + \alpha \mathbb{R} +\alpha \mathbb{R} + \alpha. \beta \mathbb{R} $$
And $M_2(\mathbb{R})$ on the right side. I know I have to show that these two algebras are not isomorphic. The hint says that $M_2(\mathbb{R})$ has zero divisors and since $K \otimes \mathbb{R}$ is a division algebra, it does not. Hence, they cannot be algebra isomorphic.
Is my solution correct?
Any kind of help is appreciated.
Thanks!