Prove that, for all real numbers $x$ and $y$, if $x$ and $y$ are irrational, then $x + y$ and
$x − y$ are not both rational. How to solve it?
What I have;
Suppose that $x+y$ is either irrational or rational. If irrational, then done.
If $x + y$ is rational, then I need to show that $x – y$ is irrational. To prove that, I use contradiction, and assume that x – y is rational. I set up the equation $x – y = a/b$, where $x, y$ are in $R/Q$ and $a, b \in\mathbb Z$. I add $2y$ to both sides, to get $x + y = a / b + 2y$. I use a lemma to state that $2y$ is irrational, and another one to state that $a/b + 2y$ is irrational. This arrives at a contradiction that $x + y$ is irrational, hence it must be that $x – y$ is irrational.
And since in both scenarios neither $x+y$ nor $x-y$ are both rational
Best Answer
A stronger statement can be proved:$$\text{if at least one of }x\text{ and }y\text{ is not rational then at least one of }x+y\text{ and }x-y\text{ is not rational}$$
or equivalently:$$\text{if }x+y\text{ and }x-y\text{ are both rational then }x\text{ and }y\text{ are both rational}$$
Observe that the premise implies that $2x=(x+y)+(x-y)$ and $2y=(x+y)-(x-y)$ are both rational, and consequently $x$ and $y$ are both rational.