I am trying to prove that the statement "Let $x,y \in \mathbb{R}$. If $x + y$ is not rational, then $x$ or $y$ must be irrational." is true using proof by contradiction.
The following is my proof.
Suppose that the negation of the statement is true.
It follows that $x+y$ is irrational and both $x$ and $y$ are rational.
Thus,
\begin{equation}
x+y = \frac{p}{q} + \frac{r}{s} \text{ where $p,q,r,s \in \mathbb{Z}$ and $q,s \ne 0$}
\end{equation}
Re-expressing the $RHS$ of the above equation, one obtains
\begin{equation}
x+y = \frac{ps + rq}{qs}\text{ where $p,q,r,s \in \mathbb{Z}$ and $q,s \ne 0$}
\end{equation}
Since the $LHS$ of the equation is irrational and the $RHS$ is rational, one obtains a contradiction.
Therefore, if $x+y$ is irrational then $x$ or $y$ must be irrational.
Is the proof correct?
Best Answer
Your proof seems correct to me.
Alternatively, we can make the proof shorter:
Suppose that, $x+y \in\mathbb R\setminus \mathbb Q.$
Then if $x,y \in\mathbb Q$, we get $x+y\in\mathbb Q$, which gives a contradiction. Thus, we can at least conclude that $x\in\mathbb R\setminus \mathbb Q$ or $y\in\mathbb R\setminus \mathbb Q.$