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

If , x < y. Show that if, x and y are rational, then an irrational number u, x < u < y.







I am not sure what to do now.

Contradiction, contrapositive??

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.

Originally Posted by dwsmith

If , x < y. Show that if, x and y are rational, then 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 .

So is between

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 ?