Proof: between 2 reals is an irrational

[ArcanaNoir]

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.

 

[HallsofIvy]

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]

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)

 

[BrianMath]

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?

 

[ArcanaNoir]

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]

Also note that your proof actually shows something stronger. It shows that between every two reals is a transcendental!

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.