Prove algebraically that, if $x^2 \leq x$ then $0 \leq x \leq 1$ It's easy to just look at the graphs and see that $0 \leq x \leq 1$ satisfies $x^2 \leq x$, but how do I prove it using only the axioms from inequalities? (I mean: trichotomy and given two positive numbers, their sum and product is also positive).

$x \geq x^2$
If $x = x^2$,
$x - x^2$ = $0$ (Subtract x^2 from both sides)
$x^2 - x$ = $0$
$x(−x+1$) = $0$ (Factor left side of equation)
$x = 0$ or $−x+1 = 0$ (Set factors equal to 0)
$x = 0$ or $x = 1$
Check intervals in between critical points. (Test values in the intervals to see if they work.)
$x ≤ 0$ (Doesn't work in original inequality)
$0$ ≤ $x$ $≤1$ (Works in original inequality)
$x ≥ 1$ (Doesn't work in original inequality)
