TITLE: Density of irrationals
QUESTION [36 upvotes]: 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?

REPLY [4 votes]: If the rationals are dense then there are at least two distinct rationals $a$ and $b$ between $x$ and $y$.  
Then $t=a \dfrac{\sqrt{2}}{2} + b \dfrac{2-\sqrt{2}}{2}$ is irrational and also between $x$ and $y$.