prove rational $+$ irrational is irrational without contradiction. Similar questions can be found on the web, mostly by contradiction, but I came up with this one, not sure if it sounds.
Let $x\in\mathbb{Q}$ be fixed. Prove without contradiction: if $y\in\mathbb{R}\setminus\mathbb{Q}\,$, then $x+y\in\mathbb{R}\setminus\mathbb{Q}$
attempt
Equivatenly, if $x+y\in\mathbb{Q}$, then $x\in\mathbb{R}\setminus\mathbb{Q}\,$ or $\,y\in\mathbb{Q}$.
we know $x\in\mathbb{Q}$ so it can't be in $\mathbb{R}\setminus\mathbb{Q}$, left to prove $y\in\mathbb{Q}$.
$$x+y=\frac{m}{n},x=\frac{p}{q}\implies y=\frac{mq-pn}{mq}\in\mathbb{Q}$$
Is this proof sound?

Appreciate anyone who has thought could leave the method on direct proof
 A: Your proof seems right as far as I can see, but probably there are ways to express it more formally. I will try it.
As far as I can see the statement to be proved is
$$
\forall x,y(x\in \mathbb{Q}\,\land\, y\in \mathbb{R}\setminus \mathbb{Q} \implies x+y\in \mathbb{R}\setminus \mathbb{Q})\tag1
$$
Now, we have the following equivalences
$$
A\implies B\quad \text{ is logically equivalent to }\quad \lnot B \implies \lnot A\\
A\implies B\,\lor\, C\quad \text{ is logically equivalent to }\quad A\,\land\, \lnot B \implies C
$$
Using the first equivalence above gives a contrapositive statement, and the second equivalence above (unnamed as far as I know) is usually used without noticing it, its a metatheorem that can be proven using the inference rules of natural deduction.
Then applying the first equivalence to (1) gives
$$
\forall x,y(x+y\in \mathbb{Q}\implies x\in \mathbb{R}\setminus \mathbb{Q} \,\lor\, y\in \mathbb{Q})\tag2
$$
and as I dont see a clear way to give a direct proof from (2) I will use the second equivalence in (2) what gives
$$
\forall x,y(x+y\in \mathbb{Q}\,\land\, x\in \mathbb{Q}\implies y\in \mathbb{Q})\tag3
$$
Now (3) is very easy to prove using representations $x+y=\frac{r_1}{r_2}$ and $x=\frac{s_1}{s_2}$ for $s_1,r_1\in \mathbb{Z}$ and $r_2,s_2\in \mathbb{N}\setminus \{0\}$, then we arrive to
$$
y=(x+y)-x=\frac{r_1}{r_2}-\frac{s_1}{s_2}=\frac{r_1s_2-s_1r_2}{s_2r_2}
$$
so we conclude that $y\in \mathbb{Q}$ as both, numerator and denominator of the last expression are integers and $s_2r_2>0$.∎
