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Non-calc problem taken out from Spivak's Calculus, I reach a dead end

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data

    The problem states that:

    [tex] y_0 \neq 0 [/tex]
    [tex] |y - y_0| < \frac{|y_0|}{2} [/tex]
    [tex] |y - y_0| < \frac{\epsilon|y_0|^2}{2} [/tex]

    And I am supposed to use these to prove that:

    [tex] y \neq 0 [/tex]
    [tex] |\frac{1}{y} - \frac{1}{y_0}| < \epsilon [/tex]

    2. Relevant equations

    [tex] |a| - |b| \leq |a - b| [/tex]
    (Reverse Triangle Inequality - my professor hinted me to use it)

    3. The attempt at a solution

    Ok proving the first part was easy, I simply used the reverse triangle inequality like this:

    1.I change [tex] |y - y_0| [/tex] to [tex] |y_0 - y| [/tex]
    2.I use the inequality: [tex] |y_0 - y| \geq |y_0| - |y| < \frac{|y_0|}{2} => -|y| < \frac{|y_0|}{2} - |y_0| => -|y| < -\frac{|y_0|}{2} => |y| > \frac{|y_0|}{2} [/tex] and since we know [tex] y_0 [/tex] is greater than 0, it's square and it's fractions are also greater than 0, so this means [tex] y [/tex] is greater than 0.

    Now for part 2 I played around with [tex] |\frac{1}{y} - \frac{1}{y_0}| [/tex] and even after transforming it by completing the fraction addition: [tex] |\frac{y_0 - y}{yy_0}| [/tex] I couldn't figure out how to proceed.Could someone help for this second part?
     
  2. jcsd
  3. Nov 18, 2013 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    For ##v > 0## the inequality ##|u| < v## is the same as ##-v < u < v##. Now just consider the two cases ##y_0 \geq 0## and ##y_0 < 0##.
     
  4. Nov 18, 2013 #3
    I thought [tex] y_o [/tex] could not be equal to 0.You mean [tex] y_0 > 0 [/tex] instead of [tex] y_0 \geq 0 [/tex] right?
     
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