# x,y\in\mathbb{R} x<y, there exist an irrational between x and y

• May 12th 2011, 12:23 PM
dwsmith
x,y\in\mathbb{R} x<y, there exist an irrational between x and y
If $\displaystyle x,y\in\mathbb{R}$, x < y. Show that if, x and y are rational, then $\displaystyle \exists$ an irrational number u, x < u < y.

$\displaystyle m,n,r,s\in\mathbb{Z}, \ n\neq 0, \ s\neq 0$

$\displaystyle x=\frac{m}{n}, \ y=\frac{r}{s}$

$\displaystyle \frac{m}{n}<u<\frac{r}{s}$

I am not sure what to do now.

• May 12th 2011, 12:33 PM
DrSteve
Here's a simple solution using basic set theory:

The interval (x,y) is uncountable.
Since there are only countably many rational numbers, there must be an irrational number in the interval.

Note: this proof doesn't require x and y to be rational, only that x<y.
• May 12th 2011, 12:46 PM
Plato
Quote:

Originally Posted by dwsmith
If $\displaystyle x,y\in\mathbb{R}$, x < y. Show that if, x and y are rational, then $\displaystyle \exists$ an irrational number u, x < u < y.

This is standard proof in any basic analysis course.
But it depends upon this theorem: Between any two real numbers there is a rational number.

In this case $\displaystyle \left( {\exists r \in \mathbb{Q}} \right)\left[ {x\sqrt 2 < r < y\sqrt 2 } \right]$.

So $\displaystyle \frac{r}{\sqrt{2}}$ is between $\displaystyle x~\&~y.$
• May 12th 2011, 12:47 PM
Tinyboss
While DrSteve's answer is very elegant and is how I'd answer it if asked, you may not have those concepts available to you yet. Since the sum of a rational and an irrational is irrational, it suffices to find an irrational number between 0 and y-x. Look at the "Archimedean property" of the reals. Is there an n such that $\displaystyle \sqrt2/n<y-x$?