- #1
ArcanaNoir
- 779
- 4
I'd like to know if this indeed proves that between any 2 reals is an irrational.
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.
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.