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Analysis problems

  1. Nov 20, 2004 #1
    I am having some difficulty in understanding this problem.
    1. Prove that there is a real x, s.t x-|_ x _|=2 or that there isn't.
    (|_ x _|= m (element of) Z and m<=x<m+1.)

    Read as, "the floor function of x is equal to m, element of Z(integers), and m is less than or equal to x, which is less than m+1.

    How am i suppose to solve this problem, and what kind of definitions or facts can i use to solve this problem? Could someone tell me the steps that are required to solve this problem?

    And this is the second problem.
    2. Show that there is a real x s.t. f(x)=x^5+x+1=0. f(x) is continuous because it is a polynomial. Let s be a real s.t. f(-s)<0<f(s). Then by I.V.T.(intermediate value theorem) there is at least one x in (0-s,0+s) s.t. f(x)=0 if f(-s)<0<f(s).

    Could someone just clarify this for me, because i think i am able to do it, just dont understand on what to do first. Thanks in advance.
     
    Last edited: Nov 20, 2004
  2. jcsd
  3. Nov 20, 2004 #2
    LaTeX helps so much with these sorts of problems.

    1. Is part of the text proposing solutions as well? subtracting m from each part of the inequality you have fetches 0<= x-m < 1 which shows that x - floor(x) = 2 doesn't have a solution.
     
  4. Nov 21, 2004 #3
    Well not really, no solutions are given. I just dont understand how the floor functions work. And like you said for 1. it doesn't have a solution, then what do i have to say in the proof to prove that it has no solution.
     
  5. Nov 21, 2004 #4
    Anyone know how to do the second problem, urgent help needed.
     
  6. Nov 22, 2004 #5
    Hint for Q2: what's f(1) and f(-1)?
     
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