TITLE: Existence of irrationals in arbitrary intervals
QUESTION [6 upvotes]: I was studying for my analysis mid-term paper and was going over the properties of real numbers. I was wondering how to prove the following statement: (Not a textbook problem, it just popped into my head.)

Given rational numbers $p$ and $q$ such that $p < q$, show that there exists an irrational number $r$ such that $p < r < q$.

I know some ways of proving it, like picking a known irrational and shifting it into the open interval $(p,q)$. I was wondering whether there is a way to prove it without referencing to any previously known irrationals. Specifically I am trying to construct a sequence of rational numbers which converges to a irrational in the interval $(p,q)$. Is there any way to do that?

REPLY [3 votes]: I want to mention how the notions of countable and uncountable can be used to give a solution. Because $\mathbb Q$ is countable, so is $(p,q)\cap \mathbb Q$.  But $(p,q)$ is uncountable, because $$\displaystyle{f(x)=\frac{x-\frac{p+q}{2}}{(x-p)(q-x)}}$$ is a bijective map from $(p,q)$ to $\mathbb R$, and $\mathbb R$ is uncountable, or because $g(x)=\frac{x-p}{q-p}$ is a bijective map from $(p,q)$ to $(0,1)$, and $(0,1)$ is uncountable.  Therefore $(p,q)\cap \mathbb Q\neq(p,q)$, meaning that there are irrational numbers in $(p,q)$.