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Homework Help: Proofs regarding inequalities and number line stuff

  1. Sep 27, 2010 #1
    1. The problem statement, all variables and given/known data
    "There is a very useful way of describing the points of the closed interval [a,b] (where we assume, as usual, that a < b)

    a. Consider the interval [0,b]. Prove that if x is in [0,b] then x = tb for some t with 0 <= t <= 1. What is the significance of the number t? What is the mid-point of the interval [0,b]

    b. Now prove that if x is in [a,b], then x = (1-t)a + tb for some t with 0 <= t <= 1. Hint: Can also be written as a + t(b-a). What is the midpoint of the interval [a,b]? What is the point 1/3 of the way from a to b?

    c. Prove, conversely, that if 0 <= t <= 1, then (1-t)a + tb is in [a,b]

    2. Relevant equations
    N/A

    3. The attempt at a solution

    I think I'm really close and I'm just missing something extremely stupid. I solved part a like so
    0 <= x <= b
    0/b <= x/b <= b/b
    0 <= x/b <= 1
    Let x/b = t
    0 <= t <= 1

    I'm stuck on part b however. No matter how many times I did the algebra I just never could come up with the right equation. I assume to start out like this:
    a <= x <= b
    a/b <= x/b <= b/b
    Let x/b = t
    a/b <= t <= 1
    a <= tb <= b
    From here I'm somewhat stuck. I see that "tb" is part of the equation I"m looking for, I just need a (1-t)a from there.

    I also think I have solved c:
    0 <= t <= 1
    0 * (b-a) <= t * (b-a) <= 1 (b-a)
    0 <= t(b-a) <= b -a
    a <= a + t(b-a) <= b

    Thanks for all of your help in advance!
     
    Last edited: Sep 27, 2010
  2. jcsd
  3. Sep 27, 2010 #2

    HallsofIvy

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    For (b), given interval [a, b], consider instead the interval [0, b-a]. You have, in part (a) shown that any point, x, in that interval, can be written in the form x= t(b- a) for some t between 0 and 1. Now "shift" that to put x between a and b.
    (You are looking at b/a. Look at b- a instead.)
     
  4. Sep 27, 2010 #3
    Thanks HallsofIvy!

    Now I am on part d.) which states: The points of the OPEN interval (a,b) are those of the form (1-t)a + tb for 0 < t < 1.

    Is there anything that is different about approaching this problem because of an open interval?
     
  5. Sep 28, 2010 #4

    HallsofIvy

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    Science Advisor

    No difference at all. Except, of course, since t= 0 gives a and t= 1 gives b, you have to drop those values.
     
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