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!

Proofs help

  1. Jun 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that for every two distinct real numbers a and b, either (a+b)/2>a or (a+b)/2>b


    2. Relevant equations



    3. The attempt at a solution
    Proof:
    if two distinct numbers a and b then (a+b)/2>a
    Since a≠b and a,bεR, (a+b)/2>a=a+b>2a=b>a. Therefore (a+b)/2>a if b>a.
    and
    if two distinct numbers a and b then (a+b)/2>b
    Since a≠b and a,bεR, (a+b)/2>b=a+b>2b=a>b.
    Therefore (a+b)/2>b if a>b.

    would this suffice as a proof or no?
     
    Last edited: Jun 14, 2012
  2. jcsd
  3. Jun 14, 2012 #2
    Something is missing. I think you ought to state that a≠b implies either a>b or b>a
    and then show that
    a>b is an equivalent statement to (a+b)/2>b
    and
    b>a => (a+b)/2>a

    Would you have to prove that a≠b => a>b or b>a?
     
  4. Jun 14, 2012 #3

    Mark44

    Staff: Mentor

    No. You're assuming part of what you need to prove.
    Also, this makes no sense: (a+b)/2>a=a+b>2a=b>a
    You're saying that (a + b)/2 > a, which equals a + b, which is greater than 2a, which equals b, which is greater than a. The problem is that you are apparently connecting inequalities (such as (a+b)/2>a and a + b > 2a), with =. That's not the right symbol. Equations and inequalities aren't equal to anything; they might be equivalent, or one might imply another, but they're not equal.
     
    Last edited: Jun 14, 2012
  5. Jun 14, 2012 #4
    yeah i knew this was wrong. I think the problem is that im not approaching the conclusion of the result correctly. let me try this again
     
  6. Jun 15, 2012 #5

    Mark44

    Staff: Mentor

    Break it up into two cases, along the lines of what cryora suggests.

    Case 1: Suppose a < b.
    Show that (a + b)/2 > a.

    Case 2: Suppose that b < a.
    Show that (a + b)/2 > b.
     
  7. Jun 15, 2012 #6
    proof: Since a and b are distinct numbers this implies that either a>b or b>a

    Case 1: Let a>b. Then a/2>b/2. a/2+b/2>2b/2. (a+b)/2>b. Since a>b then (a+b)/2>b.

    Something like this. Damn proving is tough. Has anyone used a transition to advanced math by chartrand? Im trying to learn how to do proofs from this book. I can do about 70%-80% of the problems but some of them are tricky like this one. Anyone have other books that are good at showing different kinds of proofs that I can use to supplement this book?
     
  8. Jun 15, 2012 #7

    Mark44

    Staff: Mentor

    Here are a couple that I think would be helpful.
    How to Read and Do Proofs (http://books.google.com/books/about/How_to_read_and_do_proofs.html?id=K3itQwAACAAJ)
    The Nuts and Bolts of Proofs (https://www.amazon.com/Nuts-Bolts-Proofs-Fourth-Edition/dp/0123822173)
     
  9. Jun 15, 2012 #8
    ok thanks. I'll look into these books.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proofs help
  1. Help with proofs (Replies: 6)

  2. Help with a proof. (Replies: 10)

  3. Help with a proof! (Replies: 8)

  4. Help with a Proof (Replies: 1)

  5. Proof help! (Replies: 3)

Loading...