Clarification of example of Clifford Algebra I just started studying Clifford algebras and I am puzzled by the following example.
Let $X$ be a Hilbert space with $\mathrm{dim}\ X = 1$. Let $\{e_1\}$ be the basis for $X$. Then the  Clifford algebra $\mathcal{C}(X)$ consists of all elements of the form $\alpha + \beta e_1$ where $\alpha,\beta \in \mathbb{R}$. Hence, $\mathrm{dim}\ \mathcal{C}(X) = 2$.
However, when I try taking $X = \mathbb{R}$ then $\mathrm{dim}\ X = 1$ and all elements of $\mathcal{C}(X)$ are of the form $\alpha + \beta e_1 \in \mathbb{R}$. But then $\mathrm{dim}\ \mathcal{C}(X) = 1$ ?
 A: This is a bit tricky to answer without knowing your background, so I'll try to err on the side of a more accessible answer.
Basically, "$\mathbb R$ as a vector space" is a different thing from $\mathbb R$ (where numbers can be multiplied), and if you keep them separate you won't have this kind of issue. One choice, but not the only one, would be to distinguish them in notation by writing "$\mathbb R$ as a vector space" as "$\mathbb R^1$" (as discussed in Is $\mathbb{R}^1$ really the same as $\mathbb{R}$?). Let's say that when we have elements of "$\mathbb R$ as a vector space" (or $\mathbb{R}^1$), we write them in angle brackets so that $-2\langle-1\rangle=\langle2\rangle$ and $\langle-1\rangle+\langle2\rangle=\langle1\rangle$, but the product $\langle3\rangle\langle4\rangle$ is not defined since you can't just "multiply" vectors. There is also a natural inner product where $\langle2\rangle\cdot\langle3\rangle=6\ne\langle6\rangle$. With this ($\mathbb{R}^1$) as our Hilbert space $X$, elements of $\mathcal{C}(X)$ would be like $2+3\langle1\rangle\notin\mathbb R$.

It may be easier to understand the the $\mathcal{C}(X)$ example with $X$ being a one-dimensional vector space that is more clearly different from $\mathbb R$, like let $X$ be "the vector space of horizontal arrows, with addition given by attaching arrows tail-to-tip, and scalar multiplication given in the standard way where multiplication by $2$ doubles the length, multiplication by $-1$ flips the direction, etc." We can define $|\mathbf{v}|$ to be the length of $\mathbf v$. Then $X$ is a Hilbert space where $\mathbf v\cdot\mathbf w$ is given by $\mathbf v\cdot\mathbf w=\pm|\mathbf{v}||\mathbf{w}|$ with the minus sign when $\mathbf v$ and $\mathbf w$ point in opposite directions.
Any arrow of nonzero length could be a basis vector for $X$, say $e_1=``\to"$ (so that $-e_1=``\leftarrow"$, etc.). Then $\mathcal{C}(X)$ consists of all expressions of the form $\alpha +\beta e_1$, like $``3+2\to"=``3+\longrightarrow"$.
