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!

Check my three proofs

  1. Jul 31, 2012 #1
    I am doing exercises form Velleman's How to Prove It

    1. The problem statement, all variables and given/known data

    1. Suppose a and b are real numbers. Prove that if a < b < 0 then a^2 > b^2.
    2. Suppose a and b are real numbers. Prove that if 0 < a < b then 1/b < 1/a.
    3. Suppose a and b are real numbers. Prove that if a < b then (a+b)/2 < b.

    2. The attempt at a solution

    1. Since a < b < 0 , this means that both a and b are negative numbers. Then i multiply
    a < b by a and get a^2 > ab and then again multiply a < b by b and get ab > b^2.
    Then since, a^2 > ab and ab > b^2 therefore a^2 > b^2.

    2. I multiply a < b by ab and get 1/b < 1/a.

    3. I add b to a < b and get a + b < 2b and divide this by 2 and get (a+b)/2 < b
     
  2. jcsd
  3. Jul 31, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that's excellent.

    Well, you meandivide by ab, not multiply. And you should specifically say that "because 0< a< b, ab> 0".

    Yes. Notice that the same kind of argument shows that a< (a+b)/2 so that the "mean" of two distinct numbers always lies between them.
     
  4. Jul 31, 2012 #3
    Thank you for response.

    2. Yes i meant divide not multiply.
    3. I haven't noticed that but it's a nice little insight.

    So, is the key in proofs to manipulate equation to get for A to B without putting in numbers since that isn't proof?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Check my three proofs
  1. Check my proof (Replies: 5)

Loading...