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Homework Help: Easy Analysis Problem

  1. Jan 7, 2010 #1
    1. The problem statement, all variables and given/known data
    This is from Wade's Intro to Analysis book; problem 5, Chapter 1.

    Prove that if [tex]0 < a < 1[/tex] and [tex] b = 1 - \sqrt{1 - a} [/tex] then [tex]0 < b < a[/tex]

    2. Relevant equations

    The book hints to use this result: [tex]0 < a < 1[/tex] implies [tex]0 < a^{2} < a[/tex]

    3. The attempt at a solution
    First I substitute b into the equation to get:
    [tex]0 < 1 - \sqrt{1 - a} < a < 1[/tex]

    By the additive property I add [tex]-1[/tex] and get:
    [tex]-1 < -\sqrt{1 - a} < a - 1 < 0[/tex]

    Using the multiplicative property I multiply by [tex]-1[/tex] and get:
    [tex]0 < 1 - a < \sqrt{1 - a} < 1[/tex]

    That above bears resemblance to the hint equation but I don't know exactly where to go (and my inkling of an idea doesn't seem to prove the initial statement).
  2. jcsd
  3. Jan 7, 2010 #2


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    Hi iamalexalright! :smile:

    (have a square-root: √ :wink:)
    But that's the answer!

    Start with the question: b = 1 - √(1 - a).

    Now use the hint. :wink:
  4. Jan 7, 2010 #3
    hrm, and you'll have to forgive me but I am new at this :)

    [tex]b = 1 - \sqrt{1 - a}[/tex]
    [tex]b^{2} = 2 - 2\sqrt{1-a} - a = 2b - a[/tex]

    This certainly feels like the wrong direction (or I just can't find the next logical step). If I go from here I have trouble comparing a to b ...

    If I could get a bigger nudge in the right direction I'd be much appreciative (but nothing more than a nudge please)
  5. Jan 7, 2010 #4


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    Alright, let's first suppose that the proposition that [itex]0 < b < a[/itex] is false. Then clearly we must have that [itex]0 < a \leq b = 1 - \sqrt{(1-a)}[/itex]. From this, we know that [itex]0 < \sqrt{(1-a)} < 1 - a[/itex]. Now, can you use the hint to arrive at a contradiction? What does this contradiction suggest about the actual inequality?
  6. Jan 7, 2010 #5
    ah cool, I see it now. For some reason I don't like proofs by contradiction (I really can't say that since I'm just starting to write them). For some reason it's tougher for me to wrap my head around them. Anyway, thanks for the help you two!
  7. Jan 8, 2010 #6


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    (just got up :zzz: …)
    Nor do I! :wink:

    But you don't need it here …

    just start with 1 - b = √(1 - a), and use the hint. :smile:
  8. Jan 8, 2010 #7
    Yes trying a proof by contradiction for basic inequalities in analysis is ugly, but it can sometimes get the job done well. Once you get to least upper bounds or basic metric topology, you might find contradiction approaches more congenial or elegant.
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