"Write it down as an inequality: number $\displaystyle x$ is (on a number line, I guess) less distant from $\displaystyle 1$ (one), than it is from $\displaystyle x^2$."

$\displaystyle |x-1|<|x-x^2|=|x(x-1)|$

For $\displaystyle x\geq1$: $\displaystyle x-1<x-x^2 \Rightarrow x^2<1$FALSE($\displaystyle x$ is greater than $\displaystyle 1$)

For $\displaystyle 0\leq x<1$: $\displaystyle -(x-1)<-(x-x^2) \Rightarrow x^2>1$FALSE($\displaystyle x$ is less than $\displaystyle 1$ and greater than $\displaystyle 0$)

For $\displaystyle x<0$: $\displaystyle -(x-1)<x-x^2 \rightarrow x^2-2x+1<0 \Rightarrow (x-1)^2<0$FALSE(always positive)

$\displaystyle \Rightarrow$ inequality hasno solutions

Is this absolutely(!) correct?