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!

Homework Help: Can someone tell me why this inequality proof is wrong?

  1. Oct 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Let x and y be real numbers such that x<y. There exists z that is a real number such that x<z<y.

    2. Relevant equations

    3. The attempt at a solution
    I wrote the following: We are given that x<y. We know from previous proof that (1/2)<1.
    Consider y such that y=x+1
    x+1<x and we know (1/2)<1. Therefore x+(1/2)<x+1
    Also, x<x+(1/2). Therefore x<x+(1/2)<x+1 which shows the existence of a z such that x<z<y.

    For this problem my professor marked me wrong at the part where I said let y=x+1. He just draws a red line and leaves it up to us to figure what we did wrong. I do not see how this argument is invalid though. Can someone explain?
  2. jcsd
  3. Oct 24, 2012 #2
    Correction in line 3 under my solution. I meant to type x+1>x
  4. Oct 24, 2012 #3


    User Avatar
    Science Advisor
    Homework Helper

    The argument is valid. It just only works if y=x+1. That really a special case of what you want to show. Suppose y isn't equal to x+1?
  5. Oct 24, 2012 #4
    Hi nate,
    well your teacher is right, x and y are already "given"
    why would you say let y=x+1 ?
    y can be anything, it is just > x, but with an arbitrarily small or big difference (it could be x+1000000000, or x+0.000000001 or whatever, you just know y>x).
    since you are talking about real numbers, you know that (x+y)/2 exists and is right between x and y

  6. Oct 24, 2012 #5
    I had my suspicions that that was why Dick, but I wanted to get another opinion, so thank you. Same to Oli4. In reference to Oli's post would this work:
    You have a given x,y such that x<y. Adding x to both sides you have
    x+x<y+x which implies 2x<y+x. Now divide by 2 and you have x<(y+x)/2.
    Similarly, taking x<y and adding y you gave x+y<2y and again dividing by 2 yields (x+y)/2<y. Therefore you have x<(x+y)/2<y. Other then the non-formal way in which I typed everything, this is what Oli was referring to if I am not mistaken?
  7. Oct 24, 2012 #6


    User Avatar
    Science Advisor
    Homework Helper

    Now that works for all y, not just y=x+1. Good job!
  8. Oct 24, 2012 #7
    Yup, you got it, congrats :)
  9. Oct 24, 2012 #8
    Awesome! Thanks :)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook