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Intervals Proof

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that I+J is the smallest interval containing all x+y, for x [tex]\in[/tex] I and y [tex]\in[/tex] J.

    2. Relevant equations
    I+J=[r+u,s+v]


    3. The attempt at a solution
    Let I=[r,s] and J=[u,v]
    Then I+J=[r+u,s+v] for all x [tex]\in[/tex] I and y [tex]\in[/tex] J
    x [tex]\in[/tex] I means r [tex]\leq[/tex] x [tex]\leq[/tex] s
    y [tex]\in[/tex] J means u [tex]\leq[/tex] y[tex]\leq[/tex] v
    Then r+u [tex]\leq[/tex] x+y [tex]\leq[/tex] s+v
    So x+y [tex]\in[/tex] I+J

    Is this proof sufficient? I feel like I should say something at the end but don't quite know what to say? Did I miss anything in the proof?
     
  2. jcsd
  3. Sep 16, 2009 #2

    Mark44

    Staff: Mentor

    I don't believe that you have shown this to be the smallest interval. The operation of addition of intervals is new to me, so I'm uncertain whether how you have defined I + J applies to all cases. E.g., does this apply to intervals that are separated, or that overlap, or where one interval is contained within another?

    To prove that the interval you show is the smallest, one approach is to assume that I + J is not the smallest interval and work towards a contradiction.
     
  4. Sep 16, 2009 #3
    Is this a better proof?

    Let I=[r,s] and J=[u,v]
    Then I+J=[r+u,s+v] for all x [tex]\in[/tex] I and y [tex]\in[/tex] J
    x [tex]\in[/tex] I means r [tex]\leq[/tex] x [tex]\leq[/tex] s
    So the most x can be is s and the least x can be is r
    y [tex]\in[/tex] J means u [tex]\leq[/tex] y[tex]\leq[/tex] v
    So the most y can be is u and the least y can be is v
    Then r+u [tex]\leq[/tex] x+y [tex]\leq[/tex] s+v
    So the most x+y can be is s+v and the least x+y can be is r+u
    So x+y [tex]\in[/tex] I+J and [r+u,s+v] is the smallest interval that contains all x+y for x [tex]\in[/tex] I and y [tex]\in[/tex] J
     
  5. Sep 16, 2009 #4

    Mark44

    Staff: Mentor

    Looks OK to me. Maybe someone else will weigh in if not.
     
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