- #1

- 768

- 4

Choose two reals A and B, B>A. There are two cases of B: B is irrational or B is rational. Assume B is irrational. Then [itex] B- \frac{1}{n} [/itex] (n is a natural number) is irrational. You can get as close as you like to B by taking n sufficienly large. Thus, between a real A and an irrational B, an irrational.

Next, assume B is rational. Then, [itex] B- \frac{\pi}{n} [/itex] is irrational. By making n sufficiently large you can get as close as you like to B. Thus, between a real A and a rational B, there is always an irrational.