Jones–Wenzl projectors exist up to the first zero quantum integer

Since $q^2$ has exact order $r$, for $1\le j<r$ one has $q^{2j}\mathrm{\ne }1$. Also $q-q^{-1}\mathrm{\ne }0$, because $q^2\mathrm{\ne }1$. Thus $q^j-q^{-j}\mathrm{\ne }0$ and $[j]\mathrm{\ne }0$ for every $1\le j<r$. At $j=r$, $q^{2r}=1$, so $q^r=q^{-r}$ and $[r]=0$.

For $p\ge 6$, one has $r\ge 3$, so every denominator $[2],\dots ,[r-1]$ is nonzero and therefore invertible in $\mathbb{C}$. By [F2] and [F3], the recursion defines nonzero idempotents through $p_{r-1}$ and gives ${\mathrm{Tr}}_{r-1}(p_{r-1})=[r]=0$. For $p=4$, $r=2$ and $p_{r-1}=p_1=1_1$ is the initial projector; its closure is $\delta =[2]=0$, so the same conclusion holds without using a rank-two denominator.

For any $g\in {\mathrm{T}\mathrm{L}}_{r-1}$, [F3] gives $p_{r-1}g{p}_{r-1}=\lambda p_{r-1}$ for some $\lambda \in \mathbb{C}$. Cyclicity [F4] gives $\mathrm{Tr}(g{p}_{r-1})=\mathrm{Tr}(p_{r-1}g{p}_{r-1})=\lambda \mathrm{Tr}(p_{r-1})=0$. Thus $p_{r-1}$ is negligible by [F5].

The recursive formula for $p_r$ requires the inverse of $[r]$, which does not exist because $[r]=0$. Therefore that formula has no value at this step; in particular, assigning a value by cancelling a zero denominator is not part of the recursion. This leaves open the separate existence question for normalized idempotents constructed by other methods.