Results 1 to 6 of 6

Math Help - Analysis

  1. #1
    Junior Member
    Joined
    Sep 2006
    Posts
    32

    Analysis

    Have a question:

    Suppose a, b can be any real number and for every e greater than 0, a is less than or equal to b +e. Prove that a is less than or equal to b.

    It makes sense, just thinking about it, but how do you actually prove it?

    MK
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,712
    Thanks
    1641
    Awards
    1
    Say it is false, then a>b.
    Then a-b>0. So let e=a-b.
    See what happens when you use the e in the given.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2006
    Posts
    32
    I still don't see where the contradiction occurs.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    You have,
    a<b+e
    then,
    a-b<e
    Make a contradiction,
    If,
    a>b then, a-b>0
    Say,
    c=a-b
    then,
    c<e
    c>0
    thus
    0<c<e
    but for any positive integer c
    we can choose a smaller e, i.e. e=c/2
    thus,
    0<c<e is not true for all e.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,965
    Thanks
    350
    Awards
    1
    Quote Originally Posted by MKLyon View Post
    I still don't see where the contradiction occurs.
    Let e = a - b > 0 (the last step is by assumption of a > b.)

    Thus b + e = b + (a - b) = a. (Which holds for the condition in the problem statement.)

    Now, e is positive so there is some real number e' such that 0 < e' < e.

    The problem statement says that a =< b + e' also, since any positive real number will do.

    But e' < e implies b + e' < b + e.

    Thus b + e' < a, contrary to the problem statement.

    Thus a is not greater than b.

    What can you do to show that a is not equal to b?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,712
    Thanks
    1641
    Awards
    1
    We are given that for each e>0, a<b+e.
    If it were the case that a>b then let e=a-b>0.
    By the given a<b+e=b+(a-b)=a.
    The above says a<a. That is the clear contradiction.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with Analysis please :)
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 16th 2009, 04:38 AM
  2. analysis help
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: October 8th 2009, 03:11 PM
  3. analysis help
    Posted in the Calculus Forum
    Replies: 0
    Last Post: January 19th 2009, 02:38 PM
  4. analysis
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 15th 2008, 03:07 PM
  5. Analysis Help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 8th 2008, 05:59 PM

Search Tags


/mathhelpforum @mathhelpforum