1. Limited time only! Sign up for a free 30min personal 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!

Is this sufficiently proven? ( A set being a convex set)

  1. Jun 28, 2013 #1
    Hi, just a few details prior: I'm trying to study techniques for maths proofs in general after having completed A level maths as I feel it will be of benefit later when actually doing more advanced maths/physics. With this question what is important is the proof is correct which means I don't actually know what a convex set is or what it is useful for...

    P.S sorry if this is the wrong sub forum.

    1. The problem statement, all variables and given/known data
    Prove that if a and b are real numbers, then the set C= { real numbers x: ax ≤ b} is a convex set.


    2. Relevant equations

    tx + (1-t) y ( for convex sets)

    ax ≤ b

    3. The attempt at a solution

    To show that C = x ≤ [itex]\frac{b}{a}[/itex] is a convex set let x and x' be members of C.

    The inequality must hold:

    tx + (1-t)x' ≤ [itex]\frac{b}{a}[/itex]

    Case 1)

    Assume x ≤ x' ≤ [itex]\frac{b}{a}[/itex]

    Then

    t(x-x') ≤ 0

    and hence t(x-x') +x' ≤ [itex]\frac{b}{a}[/itex]

    Case 2)

    assume x' ≤ x ≤ [itex]\frac{b}{a}[/itex]

    Then

    (1-t)x' ≤ [itex]\frac{b}{a}[/itex] -tx ≤ (1-t)[itex]\frac{b}{a}[/itex]

    (1-t)x' ≤ (1-t)[itex]\frac{b}{a}[/itex]

    And hence the inequality holds for both cases Q.E.D
     
    Last edited by a moderator: Jun 28, 2013
  2. jcsd
  3. Jun 28, 2013 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    How do you get that middle term from (1-t)x?

    Why not just calculate ##a(tx + (1-t)x')## directly and show it is ##\le b##? No separate cases needed.
     
  4. Jun 28, 2013 #3
    I thought about that but not sure how to go about it. I mean wouldn't you essentially be doing the same thing?

    and sorry I'm don't know what term you are referring to. Do you mean the x' term instead of x?

    Thanks for reply btw.
     
  5. Jun 28, 2013 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Note: you need to be careful! The inequalities ax ≤ b and x ≤ b/a are equivalent only if a > 0. If a < 0 you need to reverse one of the inequality signs, and if a = 0, one of them does not make sense.
     
  6. Jun 28, 2013 #5
    whoops :/ .. well say I kept the inequality in the question and changed all the other inequalities accordingly e.g.

    at(x-x') + ax' ≤ b

    etc... does that then make the proof valid?
     
  7. Jun 28, 2013 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I'm referring to the expression in red.


    No. It's really simple. Expand it and use what you know about ax and ax'.
     
  8. Jun 28, 2013 #7
    rearraged tx + (1-t)x' ≤ b/a

    although now thinking about it.. looks like circular reasoning to me
     
  9. Jun 28, 2013 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    So try my suggestion...
     
  10. Jun 28, 2013 #9
    so tax + (1-t)ax' ≤ tb + (1-t)b = b ?
     
  11. Jun 28, 2013 #10

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    So simple it leaves you Dazed&Confused, eh? You should start with ##a(tx + (1-t)x')=## on the left to be complete.
     
  12. Jun 28, 2013 #11
    yh pretty simple... makes me feel like a simpleton.. thank you anyway
     
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: Is this sufficiently proven? ( A set being a convex set)
  1. Convexity of a set (Replies: 0)

  2. Is this a convex set? (Replies: 2)

Loading...