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Proof by contradiction

  1. Apr 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Please check that the proof is correct or not.
    Let ℝ+ = {x[itex]\inℝ[/itex]: x>0} and T = {x[itex]\inℝ[/itex]: 0<x<1}.

    Let x∈ℝ+ and t∈T

    Prove: If x[itex]\leq [/itex]xt then x[itex]\leq [/itex]1.

    * You may assume any common properties of log(x) as well as : if 0<a[itex]\leq b[/itex] then log(a) ≤ log(b)

    Any help is appreciated.

    2. Relevant equations



    3. The attempt at a solution
    First, I assume the theorem is false, so negation of If x[itex]\leq [/itex]xt then x[itex]\leq [/itex]1 is true.

    The negation of the theorem is: x[itex]\leq [/itex]xt [itex]\wedge[/itex] x>1
    x[itex]\leq [/itex]xt [itex]\wedge[/itex] x>1 Premis
    x[itex]\leq [/itex]xt Inference rule for conjunction
    log(x) ≤ log(xt) log both side
    log(x) ≤ t*log(x) properties of log
    1 ≤ t
    which is a contradiction with the domain of t since 0<t<1

    Therefore, the negation of If x[itex]\leq [/itex]xt then x[itex]\leq [/itex]1 is false
    Thus, If x[itex]\leq [/itex]xt then x[itex]\leq [/itex]1
     
    Last edited: Apr 13, 2013
  2. jcsd
  3. Apr 13, 2013 #2
    What is the sign of log(x)?
     
  4. Apr 13, 2013 #3
    It is positive.
     
  5. Apr 14, 2013 #4
    The negation of the statement should be x≤xt => x>1 instead of x≤xt ∧ x>1.
    Other than that, I believe your proof is valid.
     
  6. Apr 15, 2013 #5
    Thank you aleph-aleph for helping me on this!
     
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