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Inequality implication?

  1. Apr 28, 2008 #1
    if

    [tex]s^{'}_{n} = s_n + b[/tex]

    and

    [tex]S^{'}_{n} = S_n + b [/tex]

    and

    [tex]s_n < A < S_n [/tex]

    does that imply that

    [tex]s^{'}_{n} < A+b < S^{'}_{n}[/tex]

    ?

    ahh why can't i delete, i think this is probably obvious

    the actual point of contention is if A is the only number between [itex]s_n[/itex] and [itex]S_n[/itex]
    is A+b the only number between [itex]s^{'}_{n}[/itex] and [itex]S^{'}_{n}[/itex] ? Surely it is but does it obviously follow?
     
    Last edited: Apr 28, 2008
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  3. Apr 28, 2008 #2

    rock.freak667

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    If you add b to everything you will get the last inequality. So it doesn't really imply it but you can get it by addition of b.
     
  4. Apr 28, 2008 #3

    Hurkyl

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    Could you be more precise in what you're asking?
     
  5. Apr 28, 2008 #4
    [itex] s_n [/itex] and [itex]S_n[/itex] are the inner and outer rectangles of an integral of f(x) and [itex] s^{'}_{n} [/itex] and [itex]S^{'}_{n}[/itex] are similarly for f(x)+b. I want to use the results of the proof of the first integral for the second one.
     
  6. Apr 29, 2008 #5

    HallsofIvy

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    Yes, as rockfreak667 said, if you start with [itex]s_n< A< S_n[/itex] and add b to each part, [itex]s_n+ b< A+ b< S_n+ b[/itex] so [itex]s'_n< A+ b< S'_n[/itex].

    If you want to be more specific, separate [itex]s_n< A< S_n[/itex] into [itex]s_n< A[/itex] and [itex]A< S_n[/itex]. Adding b to each side of those, [itex]s_n+ b< A+ b[/itex] and [itex]A+ b< S_n+ b[/itex] so [itex]s'_n< A+ b[/itex] and [itex]A+ b< S'_n[/itex] which combine to [itex]s'_n< A+ b< S'_n[/itex].
     
  7. Apr 29, 2008 #6

    Hurkyl

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    How are you ordering rectangles? Or did you mean s_n and S_n are their height? So I presume A, s_n, and S_n are all supposed to be real numbers? Then (no matter what their values are!) it cannot possibly be true that A is the only number between s_n and S_n.
     
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