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Spivak's Calculus, 5(x) - Use (ix) backwards

  1. Oct 11, 2008 #1
    Spivak's Calculus, 5(x) -- "Use (ix) backwards..."

    1. The problem statement, all variables and given/known data
    Prove the following:
    (x) If [tex]a,b\geq0[/tex] and [tex]a^{2}<b^{2}[/tex], then [tex]a<b[/tex]. (Use (ix), backwards.)



    2. Relevant equations

    (ix) If [tex]0 \leq a<b[/tex], then [tex]a^{2}<b^{2}[/tex].


    3. The attempt at a solution

    Suppose [tex]a,b\geq0[/tex] and [tex]a^{2}<b^{2}[/tex].

    Here's my problem. What does "Use (ix) backwards" mean? I'll assume he means to use the converse of (ix). In that case...

    The converse of (ix):

    [tex]\neg(a^{2}<b^{2})\rightarrow\neg(0\leq a<b)[/tex]

    Hence [tex](a^{2}\geq b^{2})\rightarrow\neg(0\leq a\&\&a<b)[/tex];

    hence [tex](a^{2}\geq b^{2})\rightarrow(0>a)\Vert(a\geq b).[/tex] [tex](\star)[/tex]

    Since [tex]a^{2}<b^{2}[/tex], then [tex]a^{2} \leq b^{2}[/tex]. So [tex]b^{2} \geq a^{2}[/tex].

    Then by [tex](\star)[/tex], [tex](0>b)\Vert(b \geq a)[/tex].

    Since [tex]b \geq 0[/tex], then we know [tex]0>b[/tex] cannot be true.

    This means that [tex]b\geq[/tex] a must be true.

    But if [tex]b=a[/tex], then [tex]b^{2}=a^{2}[/tex]; this is a contradiction since we are given that [tex]a^{2}<b^{2}[/tex].

    Hence b>a must be true.

    Hence a<b.
     
  2. jcsd
  3. Oct 11, 2008 #2

    HallsofIvy

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    Re: Spivak's Calculus, 5(x) -- "Use (ix) backwards..."

    What you have done is perfectly good.
     
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