I posted this problem some time ago, and it didn't elicit any feedback. So, in compliance with MHF Rules, I am now reposing the problem with some additional info. The result is bound to be long-winded; apologies.

Recall the topological notion of a nowhere dense set. There are a number of equivalent definitions of this, but I shall be using the following: "A set $\displaystyle S$ in $\displaystyle \mathbb{R}$ is said to benowhere denseprovided that its complement $\displaystyle S^\text{C}$ contains a dense, open set." If a set in $\displaystyle \mathbb{R}$ can be expressed as a union of countably many nowhere dense sets, it is said to be offirst (Baire) category,and if it cannot be so expressed, it is said to be ofsecond (Baire) category.The complement of a set of first category is often calledresidual.It is easy to show that, in $\displaystyle \mathbb{R},$ any residual set is of second category.

I need to prove the following:

Let $\displaystyle A$ and $\displaystyle B$ be any two residual sets in $\displaystyle \mathbb{R}.$ For fixed real numbers $\displaystyle x$ and $\displaystyle y$, the sets $\displaystyle A_x=\{x-a : a\in A\}$ and $\displaystyle B_y = \{y/b : b\in B, b\neq 0\}$ are both residual in $\displaystyle \mathbb{R}$; that is, they are both complements of sets of first category.

I can see how this is true intuitively, but I have never been able to find a rigorous proof. Any guidance anyone can provide would be of great help.