Proof: between 2 reals is an irrational

  • Thread starter ArcanaNoir
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  • #1
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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.
 

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  • #2
HallsofIvy
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Yes, that's a valid proof. You might want to clarify it a little by letting [itex]\delta= B- A[/itex] and saying that, for sufficiently large n, 1/n and [itex]\pi/n[/itex] is less than [itex]\delta[/itex].
 
  • #3
micromass
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This looks quite good, ArcanaNoir! :smile:

(of course the proof isn't elementary since it requires you to know that [itex]\pi[/itex] is irrational, which can be difficult to prove)
 
  • #4
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This looks quite good, ArcanaNoir! :smile:

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

I think this could be made elementary by using [itex]\sqrt{2}[/itex] in place of [itex]\pi[/itex], since the irrationality of [itex]\sqrt{2}[/itex] is quite easy to prove by contradiction. :biggrin:
But then I suppose using [itex]\pi[/itex] for this proof also works to prove that there are transcendental numbers (a stronger attribute than irrational) between any two reals, as well (since [itex]\pi[/itex] is also transcendental), am I right?
 
  • #5
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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...
 
  • #6
micromass
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Also note that your proof actually shows something stronger. It shows that between every two reals is a transcendental!! :smile:

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

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