How do I prove the following?
$[{x,y \epsilon \mathbb{R}: x < y}]$
Prove that there is an irrational number u between x and y,
s.t. x < u < y, where x and y are any two real numbers.
Apply the density of $[\mathbb{Q} to \frac{x}{\sqrt{2}} and \frac{y}{\sqrt{2}}]$
I believe this is similar to the Archimedean principle. So , square root 2 and 1 over square root 2 are both irrational . I believe, I have to incorporate the well ordering property of $\mathbb{N}$ , but I'm getting lost in showing this.
Let $\displaystyle t = \frac{x + y}{2}$.
If $t$ is irrational, set $t = u$, and you are done.
Therefore, without loss of generality, $~t~$ is rational.
If $x$ is irrational, then
set $~\displaystyle u = \frac{t + x}{2} \implies u~$ is irrational.
Further, then, $x < u < t < y.$
Therefore, without loss of generality, $x,t$ are both rational.
Let $s = t-x,~$ and set $\displaystyle ~u = t - \frac{s\sqrt{2}}{2} \implies u~$ is irrational.
Then $x = t - s < u < t < y.$