How can I show the incompleteness of the irrational numbers?

Question

To show the incompleteness of the rational numbers, we just had to find a set of rational numbers, that does not have an supremum / infimum which is element of $\mathbb{Q}$. For example, we could show that the set $\{x\in\mathbb{Q}|x²<2\}$ does not have a supremum in $\mathbb{Q}$, because $\sqrt{2}\notin\mathbb{Q}$. So we simply found a counter example and we were done, right?

Now I'm wondering, how can we show the same thing for $\mathbb{R}$ \ $\mathbb{Q}$ (let's call it $\mathbb{I}$ for simplicity reasons). It feels like, you could just do the same thing we did for $\mathbb{Q}$ also with $\mathbb{I}$, for example we know that $\frac{3}{2}\notin\mathbb{I}$, but how can I actually prove that? In $\mathbb{Q}$ we could atleast construct numbers, but in $\mathbb{I}$ this gets kinda hard. Does someone have a tip for me? Thanks in advance!

Answer 1

Let be $a_n=-\frac{\sqrt 2}{n}$ then all of them are irrational but $\sup a_n=0$.

Answer 2

Consider all irrational numbers smaller than $0$. This set doesn't have a supremum (in the irrational numbers).

If you want an explicit sequence in $\mathbb{I}$ which limits to a rational, try $a_n = -\sqrt{\frac{1}{n}}$ for $n$ non-square.

Answer 3

Since $\frac32\in\mathbb Q$ and since, by definition, $\mathbb I=\mathbb R\setminus\mathbb Q$, $\frac32\notin\mathbb I$. So, it is not hard to prove that $\left\{x\in\mathbb I\,\middle|\,x<\frac32\right\}$ does not have a supremum in $\mathbb I$.

Answer 4

What Stirling's formula asserts is a counterexample:

$$\lim_{n\to\infty} \frac{n!}{\sqrt{2\pi n}\,\Bigl(\dfrac{n}{\mathrm e}\Bigr)^{\!n}}=1. $$