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Thread: Prove or disprove help

  1. #1
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    Prove or disprove help

    I can't firgure out how to show that this prove or disprove problem is false:

    Prove or disprove that there exists a real number x such that x^2 + x + 1 ≤ 0.

    As i said i know the problem is false for every number. I just don't know how to show it. I thought maybe by cases with x ≤ 0 and x ≥ 0.
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  2. #2
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    Quote Originally Posted by jconfer View Post
    I can't firgure out how to show that this prove or disprove problem is false:

    Prove or disprove that there exists a real number x such that x^2 + x + 1 ≤ 0.

    As i said i know the problem is false for every number. I just don't know how to show it. I thought maybe by cases with x ≤ 0 and x ≥ 0.
    Assume there exists a $\displaystyle y \in \mathbb{R} $ such that

    $\displaystyle (y^2+y) \leq -1 $

    Then either:
    (i) $\displaystyle y\geq 0$
    (ii) $\displaystyle -1<y<0 $
    (ii) $\displaystyle y \leq -1 $

    Consider (i), i.e. $\displaystyle y \geq 0 $
    Then since y is positive, y+1 is certainly positive $\displaystyle \Rightarrow y(y+1) > 0 $ Contradiction. Hence y<0

    Consider (ii), i.e. $\displaystyle -1<y<0 $
    Then $\displaystyle y^2 + y \leq -1 $
    $\displaystyle y(y+1) \leq -1 $
    But since y > -1, we certainly have that y+1 is positive and $\displaystyle 0<y+1<1 $$\displaystyle \Rightarrow y(y+1) > -1 $ Contradiction. Hence y must be less than or equal to 1.

    Consider (iii), i.e. $\displaystyle y \leq -1 $
    Then $\displaystyle y \leq -1 \Rightarrow y+1 <0 \Rightarrow y(y+1) > 0 $. Contradiction.

    Hence, there exists no such $\displaystyle y \in \mathbb{R} $.

    EDIT: I don't know why I used y instead of x. Sorry.
    Last edited by WWTL@WHL; Mar 27th 2008 at 07:11 PM.
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  3. #3
    Junior Member teuthid's Avatar
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    Does this work?

    There are three cases that need to be considered: $\displaystyle |x| \leq 1$ and $\displaystyle x \geq 1$ and $\displaystyle x \leq -1$

    Case 1:
    $\displaystyle |x| < 1$
    $\displaystyle \rightarrow x+1 > 0$
    since $\displaystyle x^2 \geq 0$ for all real numbers, we can conclude that $\displaystyle x^2+x+1 > 0$

    Case 2:
    $\displaystyle x \geq 1$
    $\displaystyle \rightarrow x^2 +x \geq 1^2+1 > 0$
    $\displaystyle \rightarrow x^2+x+1>0$

    Case 3:
    $\displaystyle x \leq -1$
    $\displaystyle \rightarrow x^2 \geq 1$
    $\displaystyle \rightarrow x^2 +x \geq 1 + (-1) \geq 0$
    $\displaystyle \rightarrow x^2+x+1>0$

    Our three cases cover all possible real values of x, and in each case $\displaystyle x^2+x+1>0$. So we have exhaustively demonstrated that there is no real number x such that $\displaystyle x^2+x+1 \leq 0$
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  4. #4
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    thank you both for your help.
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