# Proof: between 2 reals is an irrational

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 $B- \frac{1}{n}$ (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, $B- \frac{\pi}{n}$ 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.

## Answers and Replies

HallsofIvy
Homework Helper
Yes, that's a valid proof. You might want to clarify it a little by letting $\delta= B- A$ and saying that, for sufficiently large n, 1/n and $\pi/n$ is less than $\delta$.

micromass
Staff Emeritus
Homework Helper
This looks quite good, ArcanaNoir!

(of course the proof isn't elementary since it requires you to know that $\pi$ is irrational, which can be difficult to prove)

This looks quite good, ArcanaNoir!

(of course the proof isn't elementary since it requires you to know that $\pi$ is irrational, which can be difficult to prove)

I think this could be made elementary by using $\sqrt{2}$ in place of $\pi$, since the irrationality of $\sqrt{2}$ is quite easy to prove by contradiction.
But then I suppose using $\pi$ for this proof also works to prove that there are transcendental numbers (a stronger attribute than irrational) between any two reals, as well (since $\pi$ is also transcendental), am I right?

Thanks for the feedback. @HallsofIvy: yes, I was thinking that too, about using a delta or maybe an epsilon.

@micromass + BrianMath: pi was just the first irrational I thought of, and since the irrationality of accepted irrationals was not in question, I went with it. I agree the sq root of 2 might be a safer choice though.

Incidentally, so there's not one "blanket proof" of whether a number is irrational? Slippery little buggers they are...

micromass
Staff Emeritus
And with some modifications, you can even show the following: between every two reals is an element of the form $\pi+q$ with q a rational. So $\pi+\mathbb{Q}$ is dense.