Density of irrationals

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?

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!)

Pick your favorite positive irrational, which is $\sqrt{2}$. By the Archimedean property, there exists $n$ such that $\frac{\sqrt{2}}{n}\lt \frac{y-x}{2}$. Again by the Archimedean property, we know there exists an integer $m$ such that $m\left(\frac{\sqrt{2}}{n}\right)\gt x$. Pick $M$ to be the least such $m$. Can you show that $M\left(\frac{\sqrt{2}}{n}\right)$ is strictly between $x$ and $y$?

Choose any real numbers $a$ and $b$ with $a<b$. The interval $(a,b)$ is not denumerable. However, the rationals inside of it are so $(a,b) - \Bbb Q$ is nonvoid; it has an element. Hence every open interval contains an irrational. It follows immediately the irrationals are dense in the line.

One way to show this would be to use the fact that the rationals are countable, whereas the interval $(x,y)$ is uncountable (these facts must be proven, though), and therefore $(x,y)$ must contain some irrational number $t$, which will satisfy $x<t<y$.