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:$\displaystyle 0<x<y \ \therefore \ y-x >0 \Rightarrow 0<\tfrac{1}{z}<y-x, \ \mbox{where} \ \exists \ z>0, \ z\ \in \mathbb{R}, \ z \notin \mathbb{Q}$

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.