Can the overall order of a reaction ever be strictly less than $1$?

In A-level chemistry we are taught that:



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> To every reaction: $$\sum\_{i=1}^n c\_iA\_i\to\sum\_{j=1}^m d\_jB\_j$$There is a rate equation: $$r=\kappa[A\_1]^{\alpha\_1}[A\_2]^{\alpha\_2}\cdots[A\_n]^{\alpha\_n}$$With the $(\alpha\_i)$ called the partial orders and $\sum\_{i=1}^n\alpha\_i$ the overall order, $\kappa$ some constant.
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Since the case of multiple reactants is confusing (e.g. I have not been able to get a clear answer of how 'rate' is defined in general) I'll stick to a reaction with only one reactant, $A\to bB+cC+dD+\cdots$.


We can then explicitly solve the differential equations, for $[A]\_t$ treated as a function in time $t\ge0$, with $[A]\_0>0$: $$\frac{\mathrm{d}[A]\_t}{\mathrm{d}t}=\kappa\cdot[A]\_t^\alpha\implies\kappa t+C=\begin{cases}\frac{1}{1-\alpha}[A]\_t^{1-\alpha}&\alpha\neq0\\\ln[A]\_t&\alpha=0\end{cases}$$After some shuffling, you get: $$[A]\_t=\begin{cases}\{(1-\alpha)(\kappa\cdot t+C)\}^{\frac{1}{1-\alpha}}&\alpha\neq1\\Ce^{\kappa t}&\alpha=1\end{cases}$$Where $C$ is some constant in both cases. If $\alpha=1$, we get $C=[A]\_t$: otherwise, $C=\frac{1}{1-\alpha}[A]\_0^{1-\alpha}$.


I played with this a little, plotting concentration-time reaction curves on Desmos, and I quickly noticed a significant difference between the cases $\alpha<1$ and $\alpha\ge1$: if the order $\alpha$ is less than $1$, the model predicts the reaction terminates in finite time, at $t=-C/\kappa$. Else, the reaction never terminates!


Intuitively, I don't think a chemical reaction can ever *fully* terminate, so I'm inclined to believe the following conjecture:



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> In a reaction with only one reactant, the order of reaction must be greater than or equal to $1$.
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More generally, I suppose:



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> In any reaction for which the rate equation holds, the overall order must be greater than or equal to $1$.
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Does that make sense? I imagine I'm quite wrong, since I'm doing this on the back of relatively little chemical knowledge. In particular, it's hard to picture what solving the differential equations looks like for multiple reactants.


