1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof: between 2 reals is an irrational

  1. Jul 26, 2011 #1
    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.
     
  2. jcsd
  3. Jul 26, 2011 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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].
     
  4. Jul 26, 2011 #3

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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)
     
  5. Jul 26, 2011 #4
    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?
     
  6. Jul 26, 2011 #5
    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...
     
  7. Jul 26, 2011 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook