# Prove that if $x$ and $y$ are irrational numbers, there exists an irrational number $z$ such that $y < z < x$

My teacher proposed this question a few days ago along with the similar case for rational numbers. I've already figured out the proof for rational numbers (just prove that their arithmetic mean is rational), but I'm not really sure where to start with this proof. I guess I'm stuck because it isn't as easy to represent irrational numbers as it is for rational numbers. Also, the arithmetic mean of two irrational numbers isn't necessarily irrational, so that approach doesn't work either. Could someone help me get started?

• Can you use the fact that irrational numbers are non-repeating decimals? – Joe Johnson 126 Sep 1 '14 at 2:22

If $\frac{x+y}{2}$ is irrational, there you go. If not, then $q=\frac{x+y}{2}$ is rational and what can we say about $\frac{q+x}{2}$?
$x-y>0$ so $(x-y)^2>0$. By the Archimedean property, there is a positive integer $m$ such that $\frac{1}{m}<(x-y)^2$. By increasing $m$ if necessary, you can assume that $m=2n^2$ so that $\frac{1}{2n^2}<(x-y)^2$. This means $$\frac{1}{\sqrt{2}n}<x-y\implies 1+\sqrt{2}{n}y<\sqrt{2}nx.$$ Now let $k$ be the largest integer that is less than or equal to $\sqrt{2}ny$. This implies $$1+k\leq1+\sqrt{2}ny<\sqrt{2}nx.$$ But we also have $\sqrt{2}ny<k+1$, so $$\sqrt{2}ny<k+1<\sqrt{2}nx\implies y<\frac{k+1}{\sqrt{2}n}<x.$$
Hint: We can first transform the problem by subtracting $y$; we get $0 < z' < x'$ where $z' = z - y$ and $x' = x - y$. Note, $y$ and $x$ are given and we are trying to choose $z$, so in the altered version, $x'$ is given and we are choosing $z'$ (which will give a choice of $z$).
We have $z = z' + y$. Given that $y$ is irrational, if $z'$ were irrational, could we guarantee that $z$ was irrational? What about if $z'$ were rational? The answer to one of these questions is yes. We just need to show that we can find $z'$ of the corresponding type such that $0 < z' < x'$. Do you know why this is the case?