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Eat your words with professor?

  1. Jul 23, 2011 #1
    So I was hanging out in my professor's office on Friday, playing maths, and I asked, "Couldn't I have a small problem to work on over the weekend?" So he thinks for a minute, and then he says, "If [itex]f[/itex] is a continuous function such that [itex] f: [0,1] \to [0,1] [/itex], then there exists a fixed [itex] x \in [0,1] : f(x)=x [/itex]. Prove this" (I'm in his elementary proofs class.) So I take it in for a brief moment (too brief) and I smile and say, "easy, it's one of those basic calc function theorems, they're all in my book.....Rolle's? no.......ah yes! Intermediate value theorem will do the trick! I just have to look up the definition to be sure, then it will be simple." And he looks at me and says, "I'd like to give you a hint but it would then be obvious." And I shoot back straight away, "Hint? What you have said is already more than sufficient! This will take five minutes. I got this."

    I. Got. This.

    I so don't got this.

    I first had a twinge of doubt when I opened my calc book and under the intermediate value theorem where I expect to see a proof I see instead: "This proof is given in more advanced books on calculus." Uh oh. Okay, I think, this is a rather abridged book, I've completed the entire thing, I can probably do this. They just don't want to put it here because it's like, chapter 1. .... It's only a matter of minutes before I realize this isn't the theorem that I need at all. I discover that I need a "fixed point theorem". Such a thing is definitely not in my calc book. So here I am on Saturday morning thinking, "omg, this is going to be very hard for someone at my level." But I better do it, because I said it was a cakewalk!

    Anyone else said something to a prof they later regretted?
  2. jcsd
  3. Jul 23, 2011 #2
    I think this is a valuable lesson: never tell a professor that something was easy, especially when you did not prove it yet :biggrin:


    Just look at the function [itex]g(x)=f(x)-x[/itex] and apply the intermediate value theorem.
  4. Jul 23, 2011 #3
    Thank you for the spoiler alert! I look forward to reading what I imagine is your proof when I am done with mine :)
  5. Jul 23, 2011 #4


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    Draw a picture!
  6. Jul 23, 2011 #5
    lol that's what my friend did and my professor said that was sort of philosophical and less than a proof.
  7. Jul 23, 2011 #6


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    You're supposed to turn the picture into a proof, not submit it as a proof. :tongue:
  8. Jul 23, 2011 #7
    Anyway, your professor is pretty smart. Because even if you could answer his question on the spot, he could immediately ask to prove the same thing for [itex]g:[0,1]^2\rightarrow [0,1]^2[/itex]. Or, god forbid, for [itex]g:[0,1]^n\rightarrow [0,1]^n[/itex]. These things are a bit... harder. :biggrin:

    So, whatever you did, you would always have lost. :tongue2:
  9. Jul 24, 2011 #8
    Would you mind if I post my attempt at a proof? :smile:
  10. Jul 24, 2011 #9
    I think I proved it (at least I've finished the piece of paper I'm going to hand him and hope he finds adequate), so anyone who wants to post a proof can feel free now.
  11. Jul 24, 2011 #10
    Okay, here's my attempt. Hopefully it's correct. Tell me if it's similar to what you have. :smile:

    Attached Files:

  12. Jul 24, 2011 #11
    Kinda. I took cases, the most important being that if f(0) not 0 and f(1) not one then f(x)> x and f(x)< x and then since its continuous it has to "switch" somewhere and converges to a point where |f(x)-x| goes to zero so f(x)=x on account of the limit as it approaches from each side.
  13. Jul 25, 2011 #12
    I also thought this would take me 5 minutes, in fact I remember working on this problem several years ago and making heavy weather of it! Well I was thinking about it while I was driving to the nursery and it took me about 20 minutes to get the proof and sort out all the details. It's basically the fact that f has a maximum and the intermediate value theorem, and looking at cases.
  14. Jul 25, 2011 #13
    I go to the pub at weekends.
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