Critique my proof of: For every real number x, if $x^2 \geq x$, then either $x \leq 0 \lor x \geq 1$.

Solution 1:

Perhaps give more information why is it so.

Since $x>0$, multiplying to an inequality doesn't change the sign.

Hence we can multiply positive $x$ to $x<1$ and conclude that $x^2 < x$.

Solution 2:

Your proof sounds good.

If you are interested in another way to tackle this problem, you can do it as follows: \begin{align*} x^{2} \geq x & \Longleftrightarrow x^{2} - x \geq 0\\\\ & \Longleftrightarrow \left(x^{2} - x + \frac{1}{4}\right) - \frac{1}{4} \geq 0\\\\ & \Longleftrightarrow \left(x - \frac{1}{2}\right)^{2} \geq \frac{1}{4}\\\\ & \Longleftrightarrow \left|x - \frac{1}{2}\right| \geq \frac{1}{2}\\\\ & \Longleftrightarrow (x\leq 0)\vee(x \geq 1) \end{align*}

Hopefully this helps !