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).
 A: Here's a hint:
$$
x - x^2  = x(1-x).
$$
A: We have$$x^2\le x\Rightarrow x^2-x\le0\Rightarrow x(x-1)\le0$$
For $x=0, x=1$ it holds as an equality. Let $x\neq 0,1$
$$x(x-1)\lt0$$
Let's consider the different cases:

$i)$ $x\gt0$ and $x-1\lt0$. It follows that $0\lt x\lt1$.
$ii)$ $x\lt0$ and $x-1\gt0$. They can't both hold.
So, it follows that $0\le x\le1$
A: Let $$x(1-x)\color{red}{<}0$$
Then $A\cdot B <0 \Leftrightarrow A<0, B>0 $ or $A>0, B<0$
A: If $x=0$ then $x^2 \le x$ is true.
If $x \neq 0$ there is the case $x < 0$: we multiply both sides of $x^2 \le x$ 
by the negative (in this case) number $\frac{1}{x}$ and we'd get $x \ge 1$, which cannot be when $x < 0$. So $x < 0$ is ruled out.
If $x > 0$, we multiply both sides of $x^2 \le x$ with the positive $\frac{1}{x}$ and we get $x \le 1$. This is consistent with $x > 0$ as well.
So $x^2 \le x$ iff ($0\le x$ and $x \le 1$), where we just showed the left to right, and the right to left follows by multiplication with the positive $x$, as you said.
A: $$x^2\le x \iff x^2-x\le0\iff4x^2-4x\le0\iff4x^2-4x+1\le1\iff(2x-1)^2\le1$$
$$\iff -1\le2x-1\le1\iff 0\le2x\le2\iff0\le x\le1$$
A: $x \geq x^2$
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)
