What about a module of rank $\frac{1}{2}$? Let $R$ be a commutative ring. The possible ranks of free $R$-modules are $0,1,2,\dotsc$. But what about a generalized notion of an $R$-module where ranks may be rational numbers such as $\frac{1}{2}$?
More specifically, is there any (natural) cocomplete symmetric monoidal category which contains $\mathsf{Mod}(R)$ and a distinguished object $T$ satisfying $T \oplus T \cong R$?
Remark that for $R \neq 0$ there is no $R$-module $T$ which satisfies $T \oplus T \cong R$, because otherwise $T$ would be finitely generated projective, hence locally free of finite rank $d$ with $2d=1$, a contradiction.
Background: I try to develope some kind of "functional calculus" for monoidal categories. This example here deals with $n$th roots, for which we have to adjoin $\frac{1}{n}$ first. More precisely, if some object $T$ has the property $n \cdot T \cong 1$, where $1$ is the unit for $\otimes$, we may think of $T$ as a categorification of $\frac{1}{n}$ and define an object $(1 \oplus M)^{\frac{1}{n}} := \bigoplus_{p=0}^{\infty} \Lambda^p(T) \otimes M^{\otimes p}$, which in fact is a $n$th root of $1 \oplus M$ with respect to $\otimes$. This is a categorification of the binomial theorem $(1+z)^{\frac{1}{n}}=\sum_{p=0}^{\infty} \binom{1/n}{p} z^p$ for $z \in \mathbb{C},\,|z|<1$.
 A: Maybe you can adapt the idea of groupoid cardinality.
Given a skeletal groupoid, we define its cardinality to be the
$$ |G| = \sum_X \frac{1}{|\hom(X, X)|} $$
where $X$ ranges over its objects.
So, for example, the cyclic group of two elements, viewed as a category, is a groupoid with cardinality $1/2$.
So, maybe you can make sense of having fractional if you consider modules with endomorphisms acting on them. Maybe an example of a "rank $1/2$" module should be a module $M$ together with a choice of involution acting on it?
More generally, maybe it's worth looking at combinatorial species, which can be used to do something that sounds similar to what you're trying to do.
A: Unfortunately the answer is no, at least when $2 \in R^*$. If $C$ is a cocomplete $R$-linear symmetric monoidal category with an object $T$ satisfying $1 \cong T \oplus T$, then $C=0$.
Proof: We have exterior powers in $C$ with the usual properties. Hence, $0 = \Lambda^2 (T \oplus T) \cong (\Lambda^2 T) \oplus (\Lambda^1 T \otimes \Lambda^1 T) \oplus (\Lambda^2 T)$, which means $\Lambda^2 T = 0$ and $T \otimes T = 0$. But then $T \cong 1 \otimes T \cong (T \oplus T) \otimes T \cong (T \otimes T) \oplus (T \otimes T) = 0$, hence $1 = 0$. $\square$
