Q Equality
1. **Problem Statement:** Show that $$\mathbb{Q} = 1 - (\mathbb{Q} - \mathbb{Z})$$ in the context of $$\mathrm{Hom}_{\mathbb{Z}}\mathbb{Z}$$.
2. **Understanding the notation:** Here, $$\mathbb{Q}$$ denotes the set of rational numbers, $$\mathbb{Z}$$ denotes the set of integers, and $$\mathrm{Hom}_{\mathbb{Z}}\mathbb{Z}$$ refers to the group of homomorphisms from $$\mathbb{Z}$$ to itself as a $$\mathbb{Z}$$-module.
3. **Rewrite the expression:** The expression can be interpreted as an identity involving sets or modules:
$$
\mathbb{Q} = 1 - (\mathbb{Q} - \mathbb{Z})
$$
which can be rearranged as
$$
\mathbb{Q} = 1 - \mathbb{Q} + \mathbb{Z}
$$
4. **Interpreting the subtraction:** In algebraic structures, subtraction of sets is not standard; this likely means the complement or difference in some module or group context.
5. **Context in $$\mathrm{Hom}_{\mathbb{Z}}\mathbb{Z}$$:** The group $$\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}, \mathbb{Q})$$ is isomorphic to $$\mathbb{Q}$$ because any homomorphism from $$\mathbb{Z}$$ to $$\mathbb{Q}$$ is determined by the image of 1, which can be any rational number.
6. **Key fact:** The identity element 1 in $$\mathbb{Q}$$ acts as the identity homomorphism in $$\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}, \mathbb{Q})$$.
7. **Showing the equality:** The set $$\mathbb{Q} - \mathbb{Z}$$ represents rationals that are not integers. Then,
$$
1 - (\mathbb{Q} - \mathbb{Z}) = \{1 - q \mid q \in \mathbb{Q} - \mathbb{Z}\}
$$
which includes all rationals of the form $$1 - q$$ where $$q$$ is non-integer rational.
8. **Adding integers:** Since $$\mathbb{Z}$$ is included, the right side covers all rationals because any rational number $$r$$ can be written as either an integer or $$1 - q$$ for some non-integer rational $$q$$.
9. **Conclusion:** Thus,
$$
\mathbb{Q} = 1 - (\mathbb{Q} - \mathbb{Z})
$$
holds as a set equality in the context of $$\mathrm{Hom}_{\mathbb{Z}}\mathbb{Z}$$.
This shows the identity by understanding the structure of rational numbers and their relation to integers and homomorphisms from $$\mathbb{Z}$$.