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!

Trivial and Vacuous Proofs

  1. Feb 18, 2014 #1
    1. The problem statement, all variables and given/known data
    Let ##n\in N ##. Prove that if ##|n-1|+|n+1|\leq 1,## then ## |n^{2}-1|\leq 4##


    2. Relevant equations



    3. The attempt at a solution
    I am trying to show by a counter example that this statement is not true.
    Consider this statement:
    ##|n-1|+|n+1| \leq 1##
    Assume :
    |n-1|=n-1 >0 and |n+1|=n+1>0
    Then, by our former statement, ## n-1+n+1 \leq 1,##
    which gives
    ## 2n \leq 1 , \text{where n} \in N##
    Now, divide by 2, and ##n \leq \dfrac{1}{2}##, which is not possible since ##n \in N##

    Call ##|n-1|+|n+1| \leq 1## , P.
    In logic, P→Q. If P is false, then Q does not matter, the implication is always true.

    Is this non-elegant proof correct?
     
  2. jcsd
  3. Feb 19, 2014 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You said you were "trying to prove by a counter example that this statement is not true." First, this is not a counter example. A "counter example" would be a specific n, satisfying the hypothesis, such that the conclusion is not true.

    Further, you said yourself that "In logic, P→Q. If P is false, then Q does not matter, the implication is always true." So, by showing that there is NO n in N such that [tex]|n- 1|+ |n+1|\le 1[/tex], you have shown that the hypothesis is always false and so have shown that the statement itself is true.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Trivial and Vacuous Proofs
  1. A proof (Replies: 3)

  2. "Arithmetic?" proof (Replies: 3)

Loading...