Prove that any real number $r$ can be expressed as the sum of two irrational numbers $x$ and $y$.
Progress: I have a specific example for any rational number $r$: $x = r-\pi$ and $y = \pi$ (or replace $\pi$ with any irrational number.) However, I can't seem to find a way to prove this in general for irrational $r$.
Any help is greatly appreciated.
Best Answer
$\forall x\in\Bbb R$, the law of excluded middle tells us that $x$ must either be irrational or rational. This should be obvious since, after all, $x\in\Bbb Q$ or $x\in\Bbb R \setminus\Bbb Q$ by definition of rational and irrational.
If $x$ is irrational then $x = \frac{x}{2} + \frac{x}{2}$, where $\frac{x}{2}$ is irrational. This is clear since a rational, $\frac12$, times an irrational, $x$, is an irrational.
If $x$ is rational then $x = (\sqrt{2}) + (x-\sqrt{2})$, where both $\sqrt{2}$ and $x-\sqrt{2}$ are irrational. This is clear since a rational, $x$, added to an irrational $-\sqrt{2}$, is still irrational.
So all $x\in\Bbb R$ can be expressed as a sum of irrationals. $\;\;\;\blacksquare$
Try the general equality $x = (\frac12 x + z) + (\frac12 x -z)$, $\forall x \in\Bbb R$, where we choose an arbitrary $z\in\Bbb R$ such that $z\in\Bbb Q$ when $x\in\Bbb R\setminus \Bbb Q$ and $z\in\Bbb R\setminus\Bbb Q$ when $x\in\Bbb Q$. In this way both $(\frac12 x + z)$ and $(\frac12 x - z)$ are guaranteed irrational, and the sum is always $x$, regardless of which subset it comes from. If we further ensure that $z\ne 0$ then the two number forms are guaranteed distinct.