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Spivak 4 (xiv): Inequality

  1. Feb 1, 2012 #1
    1. The problem statement, all variables and given/known data

    Find all x for which [itex]\frac{x-1}{x+1}>0 \qquad(1)[/itex]


    2. Relevant equations

    (2) AB > 0 if A,B >0 OR A,B < 0

    (3) 1/Z > 0 => Z > 0



    3. The attempt at a solution

    Since (1) holds if:

    [itex] (x-1) > 0 \text{ and } (x+1) > 0 \qquad x\ne -1[/itex]

    then we must have x>1 AND x>-1

    and since (1) also will hold if:

    [itex] (x-1) < 0 \text{ and } (x+1) < 0 \qquad x\ne -1[/itex]

    then we must have x<1 AND x<-1

    So that the solution is x on the interval [itex](-\infty,-1) \cup (1,\infty)[/itex].

    What is the proper way to write the solution using set builder notation?
     
  2. jcsd
  3. Feb 1, 2012 #2

    jbunniii

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    You could write, for example:

    [tex]\left\{x : \frac{x-1}{x+1} > 0\right\} = (-\infty, -1) \cup (1, \infty)[/tex]

    Or:

    [tex]\frac{x-1}{x+1} > 0 \iff x \in (-\infty, -1) \cup (1, \infty)[/tex]
     
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