1. Not finding help here? Sign up for a free 30min 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!

Real Analysis Proof.

  1. Jan 11, 2006 #1
    Prove: For every real number x>1, there exists two distinct positive real numbers y and z such that

    x = (y^2 +9)/(6y) = (z^2 +9)/(6z)

    Okay.. this has real got me beat. Firstly (this sounds stupid and obvious), when they give us a proof, is it really true? Do we just naturally believe that its true and with this in mind, prove it? Or could it be wrong and we eventually find that out if we do the proof?

    I first tried to isolate y and z in terms of x and set them equal to each other, but that led me to a dead end.

    Then i tried the contrapositive, but that didn't make it any easier.

    So.. I tried contradiction.

    For every real number x>1, there DOES NOT exists two distinct positive real numbers y and z such that

    x = (y^2 +9)/(6y) = (z^2 +9)/(6z)

    and now im out of ideas?

    Can somebody please help me out?

  2. jcsd
  3. Jan 11, 2006 #2


    User Avatar
    Homework Helper

    Consider solving the equation

    [tex]x=\frac{w^2+9}{6w},[/tex] where x>1 is a constant.

    If [itex]w\neq 0[/itex], then


    so by the quadratic equation, we have

    [tex]w=\frac{6x\pm\sqrt{36x^2-36}}{2}=3x\pm 3\sqrt{x^2-1}[/tex]

    which are both real (since x>1) and thus we note that:


    are certianly distinct real numbers satisfing said conditions.
  4. Jan 12, 2006 #3

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    There is of course another way of showing existence without actually finding the answers y aand z in terms of x.

    Consider f(w)=(w^2+9)/6w - x

    when w tends to zero this is positive, when w tends to infinity this is positive. If there exists any point where it is negative then the solution follows. We can finid the minimum: f'(w)= 1/6 - 3/(4w^2), and the minimal value there will be negative (for x>1) and we are done.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Real Analysis Proof.
  1. Real Analysis Proof (Replies: 7)