I came across the following problem:
Show that if $x$ and $y$ are real numbers with $x <y$, then there exists an irrational number $t$ such that $x < t < y$.
We know that $y-x>0$.
By the Archimedean property, there exists a positive integer $n$ such that $n(y-x)>1$ or $1/n < y-x$. There exists an integer $m$ such that $m \leq nx < m+1$ or $\displaystyle \frac{m}{n} \leq x \leq \frac{m+1}{n} < y$.
This is essentially the proof for the denseness of the rationals. Instead of $\large \frac{m+1}{n}$ I need something of the form $\large\frac{\text{irrational}}{n}$. How would I get the numerator?
Best Answer
Suggestion: I expect that you can use the fact that $\sqrt{2}$ is irrational.
From the denseness of the rationals, you know that there is a non-zero rational $r$ such that $$\frac{x}{\sqrt{2}} <r <\frac{y}{\sqrt{2}}.$$
Now it's essentially over. (I almost forgot to insist that $r$ be non-zero!)