Prove $0\leq x \leq 1 \implies 0 \leq x^2 \leq x \leq 1 $

Question

I try to prove that $0\leq x \leq 1 \implies 0 \leq x^2 \leq x \leq 1 $ only with the Axioms of the real numbers. The statement is clear and i am able to prove similar things, but in this case i don't see it.

I proved that $\forall x \in \mathbb{R}: x^2 \geq 0 $

Answer 1

$0\le x \le 1$

$\Leftrightarrow x\ge0$ and $x \le 1$

$\Leftrightarrow x\ge0$ and $x-1\le0$

$\Leftrightarrow x(x-1)\le0$ $\Leftrightarrow x^2-x\le0$

$\Leftrightarrow x^2\le x$

The other inequalities are obviously true.

Answer 2

$0\le x\le 1$ can be expressed as $0 \le \cfrac 1n \le 1$, where $n$ is some number $\ge1$.

$\left(\cfrac 1n\right)^2=\cfrac 1{n^2}$. If $n\ge1$, then $n^2\ge n$.

$\cfrac {n^2}{n^3}\ge \cfrac n{n^3}$

$\cfrac 1n \ge \cfrac 1{n^2}$

Therefore, $\cfrac 1{n^2}\le\cfrac 1n$, and thus $0\le x^2 \le x\le1$.

EDIT: As per Martín's comment, the only case where the above does not apply is when $x=0$, but then $x^2=0$, which satisfies $0\le x^2\le x \le1$.

Answer 3

Multiply by $x\ge 0$: $$0\cdot x \le x\cdot x \le 1\cdot x.$$ Hence the result.