# Irrationals Dense in R

• Oct 11th 2008, 05:53 PM
lllll
Irrationals Dense in R
I just want to know if me proof is correct:

Q: Show that the set of irrationals are dense in $\mathbb{R}$

Case 1: $x <0, \ y >0 \ \mbox{then} \ x<0

Case 2: $00, \ z\ \in \mathbb{R}, \ z \notin \mathbb{Q}: 0< \tfrac{1}{z}

Case 3: $00 \Rightarrow 0<\tfrac{1}{z}0, \ z\ \in \mathbb{R}, \ z \notin \mathbb{Q}$

Case 4: $x<0 \ \exists \ z<0, \ z\ \in \mathbb{R}, \ z \notin \mathbb{Q}: -x< - \tfrac{1}{z} <0$

case 5: $xy, \ \mbox{where} \ \exists \ z>0, \ z\ \in \mathbb{R}, \ z \notin \mathbb{Q}$

I based my solution on what I have in the text book for $\mathbb{Q}$, but they use the Archimedian Principle, and I'm not too sure if it applies to the set of irrationals.
• Oct 12th 2008, 01:03 AM
Moo
Hello,

The problem in your proof is, for example in case 1 : how can you say that there exists $z \in \mathbb{R} \backslash \mathbb{Q}$ between x and y ?
In fact, this is where the Archimedian property of $\mathbb{R}$ intervenes (Thinking)

Here is what I have for $\mathbb{R} \backslash \mathbb{Q}= \{\text{irrationals} \}$ dense in $\mathbb{R}$

We know that there exists an irrational in $\mathbb{R}$, let's call it $\delta$.
Let $x,y \in \mathbb{Q}$ such that $x

Since $\mathbb{R}$ is Archimedian, there exists $q \in \mathbb{N}^*$ such that $q(y-x)> \delta$ (both $y-x$ and $\delta$ belong to $\mathbb{R}$).

Let $z=x+\frac \delta q$.
We have $x :
$\square \quad q(y-x) > \delta \implies y-x > \frac \delta q \quad \square$

Since x and q are rationals and $\delta$ 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 $\mathbb{R} \backslash \mathbb{Q}$ is dense in $\mathbb{R}$
• Oct 12th 2008, 03:16 AM
Plato
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 $x < y\; \Rightarrow \;\frac{x}{{\sqrt 2 }} < \frac{y}
{\sqrt 2}\;\& \;\left( {\exists r \in \mathbb{Q}} \right)\left[ {\frac{x}
{{\sqrt 2 }} < r < \frac{y}{\sqrt 2}} \right]\; \Rightarrow \;x < r\sqrt 2 < y.$