
Irrationals Dense in R
I just want to know if me proof is correct:
Q: Show that the set of irrationals are dense in $\displaystyle \mathbb{R}$
Case 1: $\displaystyle x <0, \ y >0 \ \mbox{then} \ x<0<y \ \exists \ z: x<z<y \ \mbox{where} \ z \in \mathbb{R}, \ z \notin \mathbb{Q} $
Case 2: $\displaystyle 0<x \ \exists \ z>0, \ z\ \in \mathbb{R}, \ z \notin \mathbb{Q}: 0< \tfrac{1}{z} <x$
Case 3: $\displaystyle 0<x<y \ \therefore \ yx >0 \Rightarrow 0<\tfrac{1}{z}<yx, \ \mbox{where} \ \exists \ z>0, \ z\ \in \mathbb{R}, \ z \notin \mathbb{Q}$
Case 4: $\displaystyle x<0 \ \exists \ z<0, \ z\ \in \mathbb{R}, \ z \notin \mathbb{Q}: x<  \tfrac{1}{z} <0$
case 5: $\displaystyle x<y<0 \ \therefore \ x<  \tfrac{1}{z} >y, \ \mbox{where} \ \exists \ z>0, \ z\ \in \mathbb{R}, \ z \notin \mathbb{Q}$
I based my solution on what I have in the text book for $\displaystyle \mathbb{Q}$, but they use the Archimedian Principle, and I'm not too sure if it applies to the set of irrationals.

Hello,
The problem in your proof is, for example in case 1 : how can you say that there exists $\displaystyle z \in \mathbb{R} \backslash \mathbb{Q}$ between x and y ?
In fact, this is where the Archimedian property of $\displaystyle \mathbb{R}$ intervenes (Thinking)
Here is what I have for $\displaystyle \mathbb{R} \backslash \mathbb{Q}= \{\text{irrationals} \}$ dense in $\displaystyle \mathbb{R}$
We know that there exists an irrational in $\displaystyle \mathbb{R}$, let's call it $\displaystyle \delta$.
Let $\displaystyle x,y \in \mathbb{Q}$ such that $\displaystyle x<y$
Since $\displaystyle \mathbb{R}$ is Archimedian, there exists $\displaystyle q \in \mathbb{N}^*$ such that $\displaystyle q(yx)> \delta$ (both $\displaystyle yx$ and $\displaystyle \delta$ belong to $\displaystyle \mathbb{R}$).
Let $\displaystyle z=x+\frac \delta q$.
We have $\displaystyle x<z<y$ :
$\displaystyle \square \quad q(yx) > \delta \implies yx > \frac \delta q \quad \square$
Since x and q are rationals and $\displaystyle \delta$ is irrational, z is irrational (a proof by contradiction is suitable. And I know I replied to this a while ago on MHF)
Therefore, we can find an irrational between any 2 rationals (which are real numbers).
Hence $\displaystyle \mathbb{R} \backslash \mathbb{Q}$ is dense in $\displaystyle \mathbb{R}$

If you have already proved that the rational numbers are dense the here is a really simple proof that the irrational numbers are also.
If $\displaystyle x < y\; \Rightarrow \;\frac{x}{{\sqrt 2 }} < \frac{y}
{\sqrt 2}\;\& \;\left( {\exists r \in \mathbb{Q}} \right)\left[ {\frac{x}
{{\sqrt 2 }} < r < \frac{y}{\sqrt 2}} \right]\; \Rightarrow \;x < r\sqrt 2 < y.$