For all positive real numbers x, (xˆ2)-x≥0.
I don't think this is true since for instance x can equal 2/3 and (2/3)ˆ2-(2/3)=-2/9 which is not greater than or equal to 0. Am i wrong?
I see, so is the original proof "For all positive real numbers x, (xˆ2)-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)ˆ2-(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.
If you want to prove this directly, you could complete the square...
$\displaystyle \displaystyle \begin{align*} x^2 - x &\geq 0 \\ x^2 - x + \left(-\frac{1}{2}\right)^2 &\geq \left(-\frac{1}{2}\right)^2 \\ \left(x - \frac{1}{2}\right)^2 &> \frac{1}{4} \\ \left|x - \frac{1}{2}\right| &> \frac{1}{2} \\ x - \frac{1}{2} \leq -\frac{1}{2} \textrm{ or }x - \frac{1}{2} &\geq \frac{1}{2} \\ x \leq 0 \textrm{ or } x &\geq 1 \end{align*}$
So the inequality $\displaystyle \displaystyle x^2 - x \geq 0$ is only true when $\displaystyle \displaystyle x \leq 0$ or $\displaystyle \displaystyle x \geq 1$.
S0mewhat simpler, I think:
$\displaystyle x^2- x= x(x- 1)\ge 0$ if and only if both factors on the left have the same sigh. If $\displaystyle x\ge 1$, x> 0 and $\displaystyle x-1\ge 0$, so that is true. If $\displaystyle x\le 0$, x< 1 so that is true. But if $\displaystyle 0< x< 1$, x is positive while x- 1 is negative so it is not true.