# Math Help - Irrationals Dense in R

1. ## 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.

2. 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

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 $\mathbb{R}$

3. 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 $x < y\; \Rightarrow \;\frac{x}{{\sqrt 2 }} < \frac{y}
{\sqrt 2}\;\& \;\left( {\exists r \in \mathbb{Q}} \right)\left[ {\frac{x}
{{\sqrt 2 }} < r < \frac{y}{\sqrt 2}} \right]\; \Rightarrow \;x < r\sqrt 2 < y.$