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Math Help - Quick Proof on real numbers

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    Quick Proof on real numbers

    For all positive real numbers x, (x2)-x≥0.

    I don't think this is true since for instance x can equal 2/3 and (2/3)2-(2/3)=-2/9 which is not greater than or equal to 0. Am i wrong?
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    Re: Quick Proof on real numbers

    Quote Originally Posted by Aquameatwad View Post
    For all positive real numbers x, (x2)-x≥0.

    I don't think this is true since for instance x can equal 2/3 and (2/3)2-(2/3)=-2/9 which is not greater than or equal to 0. Am i wrong?
    Hi Aquameatwad,

    You are correct. When x is less than one the statement is wrong. However,

    x^2-x\geq 0\mbox{ whenever x\geq 1} is a correct statement.
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    Re: Quick Proof on real numbers

    I see, so is the original proof "For all positive real numbers x, (x2)-x≥0. " is only correct if you set the parameter x≥1.
    So the original statement is false since 2/3 is a positive real number, and (2/3)2-(2/3)=-2/9 which is not greater than or equal to 0. Correct? Just want to make sure whats in my head is correct.
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    Re: Quick Proof on real numbers

    Quote Originally Posted by Aquameatwad View Post
    I see, so is the original proof "For all positive real numbers x, (x2)-x≥0. " is only correct if you set the parameter x≥1.
    So the original statement is false since 2/3 is a positive real number, and (2/3)2-(2/3)=-2/9 which is not greater than or equal to 0. Correct? Just want to make sure whats in my head is correct.
    Yes. Correct.
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    Re: Quick Proof on real numbers

    Quote Originally Posted by Aquameatwad View Post
    For all positive real numbers x, (x2)-x≥0.

    I don't think this is true since for instance x can equal 2/3 and (2/3)2-(2/3)=-2/9 which is not greater than or equal to 0. Am i wrong?
    If you want to prove this directly, you could complete the square...

    \displaystyle \begin{align*} x^2 - x &\geq 0 \\ x^2 - x + \left(-\frac{1}{2}\right)^2 &\geq \left(-\frac{1}{2}\right)^2 \\ \left(x - \frac{1}{2}\right)^2 &> \frac{1}{4} \\ \left|x - \frac{1}{2}\right| &> \frac{1}{2} \\ x - \frac{1}{2} \leq  -\frac{1}{2} \textrm{ or }x - \frac{1}{2} &\geq \frac{1}{2} \\ x \leq 0 \textrm{ or } x &\geq 1  \end{align*}

    So the inequality \displaystyle x^2 - x \geq 0 is only true when \displaystyle x \leq 0 or \displaystyle x \geq 1.
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    Re: Quick Proof on real numbers

    S0mewhat simpler, I think:

    x^2- x= x(x- 1)\ge 0 if and only if both factors on the left have the same sigh. If x\ge 1, x> 0 and x-1\ge 0, so that is true. If x\le 0, x< 1 so that is true. But if 0< x< 1, x is positive while x- 1 is negative so it is not true.
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