The question is: Let $r$ be rational and $x$ be irrational. Prove that $r+x$ is irrational.
I realize that this question has been asked before, but I have a doubt about an approach different from assuming $r = \frac{p}{q}$.
Assume $r+x$ is rational. Since $r \in \mathbb{Q}, \,\,\exists (-r) \in \mathbb{Q}$ such that $r + (-r) = 0$.
By the closure property of the field of rational numbers, if $ (-r), (r+x) \in \mathbb{Q}$, then
$$ x = 0 + x = (-r + r) + x = (-r) + (r+x) \in \mathbb{Q}$$
which is a contradiction, therefore $r+x$ must be irrational.
However, I have a doubt about the first equality. Since $x$ is irrational, would we be able to conclude that $0+x=x$ as this axiom of additive identity only holds for the rational numbers, and we haven't been introduced to the field of real numbers yet.
Besides Anne Bauval's pertinent remark in the comments, in the first place you need to have already assumed or constructed addition on ℝ, otherwise you are simply forbidden from writing "$r+x$" in the first place given reals $r,x$. And certainly you would have the basic arithmetic fact $0+x = x$, either assumed or proven, otherwise that addition would be meaningless and you would not be using it.