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Prove Existence of Real Number

  1. May 18, 2014 #1
    1. The problem statement, all variables and given/known data
    a.) Prove ## \exists x, x \in \mathbb{R} | x^3 -x^2 = 5##
    I know that x = 2.1163, but how do I find this without a calculator?

    b.) Prove that ## \not \exists x, x \in \mathbb{R} | x^4 - 2x^2 +2 =0##


    3. The attempt at a solution
    x^2 ( x-1 )=5 for part a)?


    =====================
    b.)
    I find it easier. If x =0, then it's false.
    ## x^2 ( x^2 -2 )+2 = 0 \\ x^2 \geq 0 \\ x =1 \vee -1 \\ 1(-1) +2 =1 \neq 0 ##
    Then since x is only greater than 1, it is false that x exists that satisfies the equation.

    What do you think?
     
    Last edited: May 18, 2014
  2. jcsd
  3. May 18, 2014 #2

    haruspex

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    You're not asked to find its value, only to prove such a number exists.
    Have you typed the question correctly? Should it say [itex]\nexists[/itex]?
    I cannot even begin to follow your reasoning. Why must x be 1 or -1?
    Try applying the usual formula for solving quadratics.
     
  4. May 18, 2014 #3
    I said 1 or -1 because I was trying the base case after 0, but yeah let me try the quadratic formula.
     
  5. May 18, 2014 #4
    Can I do this to the quadratic equation? :
    ## x^4 - 2x^2 +2 = 0 \\ (x^2)^2 - 2(x^2) +2 =0 \\ ##
    And, I solve for ## x^2 = -1 +- \sqrt{-1}##
     
  6. May 18, 2014 #5

    haruspex

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    So is there a real x satisfying that? If not, why not?
     
  7. May 18, 2014 #6
    ## \not\exists x | x \in \mathbb{R}, ## since ## x = \sqrt{x^2} = \sqrt{ -1 +- \sqrt{-1} }, x \in \mathbb{C}##
     
  8. May 18, 2014 #7

    haruspex

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    ##x \in \mathbb{C}## does not prove it is not in ##\mathbb{R}##.
    Work backwards. Suppose x is real. What does that tell you about x2?
     
  9. May 18, 2014 #8

    LCKurtz

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    Try using the intermediate value theorem.

    Try completing the square with ##x^2## as the variable.
     
  10. May 19, 2014 #9
    Thanks, Kurtz!

    ## (x^2 - 1)^2 = -1 ##, since the square of any number is always non-negative, it is impossible that x is real.
     
  11. May 19, 2014 #10

    Ray Vickson

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    What tools are you allowed to use? Calculus makes these easy.

    In (a), let ##f(x) = x^3-x^2##. We have ##f(0) = 0 < 5## but ##f(x) \to \infty## for ##x \to \infty##, so ##f(x_0) > 5## for some large-enough ##x_0##. By continuity there is an ##x \in(0,x_0)## giving ##f(x) = 5##.

    For (b), look at ##f(x) = x^4 - 2 x^2 = x^2(x^2-2)##. This a quadratic in ##y = x^2##, whose minimum is at ##y = 1##, giving ##f_{\min} = f(1) = -1>-2##. Therefore, it is impossible to find any ##x## giving ##f(x) = -2##.
     
  12. May 19, 2014 #11

    Curious3141

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    Are you familiar with Descartes' Rule of Signs? It gives a very quick solution here.

    Treat that as a quadratic equation in ##x^2## and consider the discriminant.
     
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