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Homework Help: Proving a Statement

  1. Sep 26, 2010 #1

    silvermane

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    1. The problem statement:
    Suppose that a, b, x, and y are real numbers satisfying
    a < x < b and a < y < b.
    Show that |x - y| < b - a.


    3. The attempt at a solution
    We may start with the fact that:
    i.) a < x < b
    ii.) a < y < b

    Subtracting a from all sides, we have that

    i.) 0 = a - a < x - a < b - a
    ii.) 0 = a - a < y - a < b - a

    and

    0 < x - a < b - a
    0 < y - a < b - a

    but I'm stuck here in justifying my steps. I thought about subtracting the inequalities from each other but that would just yield

    0 < x - y < 0

    which doesn't help my case :(
    Any help and/or tips would be lovely!
     
  2. jcsd
  3. Sep 26, 2010 #2
    (1) a < x < b
    (2) a < y < b

    well take -(2) which is
    -b < -y < -a call that (2*)

    now take (1)+(2*) which gives

    a-b < x-y < b-a
     
    Last edited: Sep 26, 2010
  4. Sep 26, 2010 #3

    silvermane

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    Isn't it true that when we multiply by -1, we reverse the equality?

    so -(2) would be -b > -y > -a?
     
    Last edited: Sep 26, 2010
  5. Sep 26, 2010 #4

    silvermane

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    Just out of curiosity, what if we wanted to prove
    |x-y| < b - a

    I'm thinking we break it down into two cases, one where (x-y) is positive, and one where it's negative. What do you think?
     
  6. Sep 26, 2010 #5
    but i did reverse the equality then i mult. by-1
     
  7. Sep 26, 2010 #6
    This is a correct solution. What more do you want ?;P
     
  8. Sep 27, 2010 #7

    silvermane

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    lol I don't want anymore, I was just checking. This is just for my understanding. I guess one could call it a blonde moment ;)

    I just asked about |x-y| to help deepen my understanding. I seemed to have come off a little sour and I'm sorry for that! o:)

    Either way, thank you greatly for your help. I really do appreciate it :blushing:
     
  9. Sep 27, 2010 #8
    i
    You did not come off a little sour; it was I who came off sour. Sorry about that.

    You want to do the following, right ?


    Case 1:

    x-y < 0

    a < b By definition so

    This implies

    b-a > 0
    x-y < b-a

    Thus

    a-b< x-y < b-a [ Some steps are needed to get here and i didn't want to do it twice; so read case 2 . This inequality will follow from there.]

    Case 2

    x-y >0
    x>y

    y-x <0

    y-x < b-a

    x-y> a-b

    definiton and assumption of case 2
    b>x>y >a

    b-y > x-y

    y>a

    -y < -a

    b-y < b-a

    b-a > x-y
    ( the. last of case 1 can be derive from the above 5 steps)

    Thus

    a-b < x-y < b-a

    So case 1 and case 2 is equivalent to saying .

    |x-y| < b-a



    The first solution is a lot easier than the cases.

    Because you have to prove an upper and lower bound of x-y when x-y < 0 and x-y >0. That is like proving 4 things instead of 1.
     
    Last edited: Sep 27, 2010
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