# Prove $x \in \mathbb{Q}$ and $y \notin \mathbb{Q} \implies (x + y) \notin \mathbb{Q}$

Looking for tips to prove the homework

$\forall x,y \in \mathbb{R}, x \in \mathbb{Q} \land y \notin \mathbb{Q} \implies (x + y) \notin \mathbb{Q}$

Can I assume the hypothesis and to yield a contradiction assume that $(x + y)$ is rational, or rather, how should I approach it? So far I know the definitions for an integer $n$ being odd, even, divisible by an integer $m$, has a remainder $r$, and the definition of a rational having integers $p, q$ s.t. $q \neq 0$ and $\frac{p}{q} = Q$

Is there some other tool I might find useful for solving this particular problem? How might I use it? Thanks.

• If $x+y$ and $x$ rational, then $-x$ is rational. so $y=(x+y)-x$ is also rational. – Hanul Jeon Feb 5 '13 at 5:12
• For this, seems like contradiction is the only way. I don't see any way to prove it directly. – marty cohen Dec 26 '15 at 1:58

## 1 Answer

An approach by contradiction is a good idea. Suppose $x+y\in\mathbb{Q}$. What happens when you subtract $x$? Recall or prove that $\mathbb{Q}$ is closed under addition (and hence subtraction).

• I guess if a rational plus a rational is a rational, then assuming x + y is rational means (x + y) - x is also rational. Therefore leading to the conclusion that y is also rational, contradicting the hypothesis. – Leonardo Feb 5 '13 at 5:16
• @Leonardo Exactly. – Ben West Feb 5 '13 at 5:17