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Proof about irrationals between reals.

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Given any 2 reals a<b there exists an irrational number t such that
    a<t<b . It tells us to use a theorem that states there is a rational number between any 2 reals.
    3. The attempt at a solution
    so If we use this and pick reals of the form p=a-√2 and q=b-√2
    so now we have a-√2<t<b-√2 , and t is a rational number. then we add √2
    to everything so we get a<t+√2<b
    so now t+√2 is irrational , so now we have an irrational between any 2 reals.
     
  2. jcsd
  3. Feb 2, 2012 #2

    Dick

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    Again, no arguments with your reasoning. Do you have any doubts?
     
  4. Feb 2, 2012 #3
    no, im just new to writing proofs, so I just wanted more experienced people to check my work.
     
  5. Feb 2, 2012 #4

    jbunniii

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    deleted
     
  6. Feb 3, 2012 #5
    From the highlighted above, you are stating that t is rational and that it is irrational.

    These contradict each other, yes?
     
  7. Feb 3, 2012 #6
    t is rational , t+√2 is irrational
     
  8. Feb 3, 2012 #7

    HallsofIvy

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    Yes, checkitagain's point was that you state the problem as "Given any 2 reals a<b there exists an irrational number t such that a<t<b" then later use the same letter, t, to represent a rational number.
     
  9. Feb 3, 2012 #8
    ok so I just need to be careful with how I define my variables.
     
  10. Feb 3, 2012 #9
    check it again, checkitagain
     
  11. Feb 3, 2012 #10
    yes but honestly it looks fine to me
     
  12. Feb 3, 2012 #11
    just replace this

    so now we have a-√2<t<b-√2 , and t is a rational number. then we add √2

    with this

    so now we have a-√2<s<b-√2 , and s is a rational number. then we add √2

    and the other lines as necessary
     
  13. Feb 5, 2012 #12
    There is no need to "check it again." That OP's post is as wrong now
    because of the wrong variable as it was then.



    Originally Posted by cragar:
    ------------------------------------------------------------
    ok so I just need to be careful with how I define my variables.
    ------------------------------------------------------------

    Stating "yes," and then it looks "fine" are contradictory.
    That's why my post and HallsofIvy's accurate posts
    are all that was needed.


    @ cragar, once you have altered your proof with what you admitted
    to in post # 8, then it will be "fine."
     
    Last edited: Feb 5, 2012
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