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Textbook's corollary of [itex]S := \{y \in \mathbb{R} : 0 \leq y^2 \leq 2 \}[/itex]

  1. Jul 28, 2012 #1
    I can't understand a statement in a proof in a textbook.

    I'm going to terminate the proof at the line that I don't understand.

    1. The problem statement, all variables and given/known data
    Prove that there exists an [itex]x \in \mathbb{R}[/itex] such that x2=2.


    2. Their proof until line I don't understand
    For this, we define [tex] S:= \{y \in \mathbb{R} : 0 \leq y^2 \leq 2\} [/tex] Since [itex]0 \in S[/itex], S is a nonempty. Further, S does not contain real numbers y≥2 since the last inequality implies that [itex]y^2 - 2 = (y-2)(y+2)+2 \geq 2[/itex] ............

    3. Reason why I don't understand
    In the definition of S, they say that [itex]0 \leq y^2 \leq 2[/itex]. Subtracting 2 from everything, [itex]-2 \leq y^2 - 2 \leq 0[/itex], which leads to [itex]y^2 - 2 \geq -2[/itex] ... But they've claimed that [itex]y^2 - 2 \geq 2[/itex] - why?
     
  2. jcsd
  3. Jul 28, 2012 #2
    Re: Textbook's corollary of [itex]S := \{y \in \mathbb{R} : 0 \leq y^2 \leq 2 \}[/ite

    It is most likely a typo. If y^2 - 2 >= 2, then y^2 >= 4, which is clearly wrong.
     
  4. Jul 28, 2012 #3
    Re: Textbook's corollary of [itex]S := \{y \in \mathbb{R} : 0 \leq y^2 \leq 2 \}[/ite

    I think it means this:

    If [itex]y \in S[/itex] then [itex]y^2 \leq 2[/itex] and so [itex]y^2 - 2 \leq 0[/itex].

    However, if [itex]y \geq 2[/itex] then [itex]y-2 \geq 0[/itex] and so [itex]y^2 - 2 = (y-2)(y+2)+2 \geq 2[/itex]. That is [itex]y^2 - 2 > 0[/itex], which is a contradiction.
     
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