Help with a proof my professor gave my class

Question

Let $x,y \in \mathbb{R}$ with $x<y$. There exists an irrational number $z$ such that $x<z<y$.

My proof so far:

Let $x,y \in \mathbb{R}$ and assume $x<y$. Then, by Theorem 11.8 (in our textbook), we can find $a,b \in \mathbb{Q}$ such that $x<a<b<y$.

I assume that all I have left to do is show that $a + \frac{b-a}{\sqrt{2}}$ is irrational and between x and y, but I don't have any idea how to show this. Any help showing this would be greatly appreciated.

Answer 1

Strategy:

1. Prove that $a+\frac{b-a}{\sqrt{2}}>a$.
2. Prove that $a+\frac{b-a}{\sqrt{2}}<b$.
3. Prove that if $a+\frac{b-a}{\sqrt{2}}$ is rational, then so is $\sqrt{2}$. Since this is known to be false, $a+\frac{b-a}{\sqrt{2}}$ must be irrational.

In step 3, you will need to have $a,b$ be rational, which I assume Thm 11.8 gives you.

If you have trouble with any of these three steps, please don't hesitate to ask for further help.

Answer 2

There are many possible constructions for an irrational between $x$ and $y$.

For example, if their mean, $\displaystyle\frac{x+y}2$ is rational, then choose $n$ big enough such that $(1/2)^n<\displaystyle\frac{y-x}2$, so that the following number is in the given range, and it is irrational: $$\frac{x+y}2+\left(\frac12\right)^n\cdot\frac1{\sqrt2}$$