In this paper by John Baez: http://math.ucr.edu/home/baez/octonions/node6.html he says that assuming a Clifford algebra with $vw + wv = -2<v, w>$, you can see that Cliff(0) = $\mathbb{R}$, Cliff(1) = $\mathbb{C}$ and Cliff(2) = $\mathbb{H}$...
How is that possible? The minus sign would spoil everything... Thanks in advance.
Let $(V,\langle\cdot,\cdot\rangle)$ be an inner product space over $\mathbb{K}$. First, recall that the tensor algebra generated by $V$ is the vector space $$T(V)=\mathbb{K}\oplus V\oplus(V\otimes V)\oplus(V\otimes V\otimes V)\oplus\dots$$ In the text, Baez defines the Clifford algebra as an associative algebra generated by $V$ modulo the relations $$vv=-\|v\|,$$ or equivalently $$vw+wv=-2\langle v,w\rangle.$$ When he says "generated by $V$ modulo", it means that the product given by juxtaposition of elements $v,w\in V$ in the equations is defined as a product satisfying those equations. More precisely, since the product must be bilinear, the Clifford algebra will be the quotient of the tensor algebra $T(V)$ by an equivalence relation that will make those relations true.
Aiming to be concise, let $V=\mathbb{R}^n$ and $\mathbb{K}=\mathbb{R}$. Now, if we set $n=0$, we would have $V=\{0\}$. Notice that since the juxtaposition product is bilinear, the equations above are automatically satisfied. Therefore, the Clifford algebra will be exactly the tensor algebra. However, $T(V)=\mathbb{K}\oplus\{0\}\oplus(\{0\}\otimes\{0\})\oplus\dots=\mathbb{K}=\mathbb{R}$. It yields $$\text{Cliff}(0)=\mathbb{R}.$$
Remark: the symbols $v$ and $w$ in the equations are elements of $V$, not from $\mathbb{K}$. Therefore, the example given in the comments of the question does not provide a counter-example.