This is admittedly a very simple proof by contradiction from Rudin, but I just want to be sure that my thinking on this is correct.
Proof: We have some rational, $r \neq 0$, and an irrational number, $x \in \mathbb{I}$, and need to establish that $r + x \in \mathbb{I}$. For a contradiction, suppose $r + x \in \mathbb{Q}$. So, by closure of rationals under multiplication, $-r \in \mathbb{Q}$, and by closure under addition, $\left(r + x\right) + \left(-r \right) \in \mathbb{Q}$. But, $\left(r + x \right) + \left(-r \right) = \left(x + r\right) + \left(-r \right) = x + \left(r + \left(-r\right)\right) = x + 0 = x$, so $x \in \mathbb{Q}$, a contradiction, so $r + x \in \mathbb{I}$.
My question on this proof is as follows:
First, it seems that we're defining irrational numbers so that $y \in \mathbb{I} \equiv y \not \in \mathbb{Q}$. But, $\mathbb{I} = \mathbb{R} \setminus \mathbb{Q}$. So, in proceeding by contradiction to establish that $y \in \mathbb{I}$, we would assume $y \not \in \mathbb{I}$. But, this doesn't necessitate that $y \in \mathbb{Q}$: if $y$ is a natural, integer, or real, but is not irrational, then surely $y \in \mathbb{Q}$, but it could be the case that $y \in \mathbb{C}$ with a non-zero imaginary component.
So, are we assuming before writing this proof that we are working within the domain of the reals? (In other words, there's an unspoken line at the top of the proof to the effect of 'let $r$ and $x$ be real numbers.') Or, is $y \in \mathbb{I}$ interchangeable with $y \not \in \mathbb{Q}$, in which case we can go back and forth between these descriptions, as I've done above?
Thanks.
First of all, your proof is completely correct.
To address your question: $\mathbb{I}$ is defined as $\Bbb R \setminus \Bbb Q$. So when we say "let $x \in \mathbb{I}$", we are also saying that $x \in \Bbb R$, because $\Bbb I \subset \Bbb R$. So when you reach the step in your proof where you get $x \in \Bbb Q$, this is a contradiction because $\mathbb{I}$ and $\mathbb{Q}$ are disjoint sets by definition. Thus the initial assumption $r + x \in \Bbb Q$ is false. Since $r,x \in \Bbb R$ (they are in $\Bbb Q$ and $\Bbb I$, both of which are subsets of the reals), this implies that $r + x \in \Bbb R \setminus \Bbb Q = \Bbb I$. Does this clear things up?