If the union of two subgroups of $G$ is the group $G$, does that mean one subgroup is $G$?
In a problem, it was proven to be true on this site. But take real numbers under addition as a group with Rational numbers as a subgroup and irrational numbers with ${0}$ as a subgroup. Then their union is a real number, but no subgroup is real numbers. So the statement is not true for all groups, right?
The result that one of the subgroups must be $G$ is correct. What goes wrong with your example is that the irrational numbers together with $0$ are not a subgroup of $\mathbb{R}$. For instance, $\pi$ and $1-\pi$ are both in this set, but their sum $\pi+(1-\pi)=1$ is not.
The irrational numbers with $0$ do not form a subgroup of $\mathbb{R}$. $(\sqrt{2}+1)-\sqrt{2}=1$, so the irrationals with $0$ aren't closed under subtraction.
You're correct that at least one of them must be $G$. This follows from the more general fact that if the union of two groups is a group, then one of the groups must be (isomorphic to) a subgroup of the other.
The decimal-part of an irrational number has a complement which together add to 1. Thus the integers are in any group of irrational numbers. SO your example doesn't work and you are false.
