I am working on (Baby) Rudin's Problem 1.6d and really struggling, so I looked up a solution given by some professor at Harvard. In the solution, the assertion is made that "Moreover, if $z \in \mathbb{Q}, z \leq x+y$ , then there exist rational $s',t'$ such that $z−y \leq s' ≤ x$ and $z−x \leq t' \leq y$" ($x$ and $y$ are understood to be arbitrary reals).
The problem appears to me to occur in the case that $z=x+y$ (if the inequality were strict we would have a simple appeal to the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$). We would conclude, for instance, that $z-x=y$, and the only possible $t'$ would be $t'=y$, which is not in general rational. More explicitly, $x=\sqrt2, y=-\sqrt2$ would seem to furnish a trivial counterexample to the statement given.
Where am I going wrong here?
