Let $x$ be a real number. Prove that if $x^2\lt x$, then $x\lt1$ Let $x$ be a real number. Prove that if $x^2\lt x$, then $x\lt1$
I do not know how to proceed. I assume its proof is by contrapositive?

$$x^2\lt x$$
$$x^2-x\lt 0$$
$$x(x-1)\lt 0$$
Thus, either $x\lt 0$ and $x\gt 1$ 
or 
$x\gt0$ and $x\lt1$.
Clearly, the former is not possible.
Hence,
$$0\lt x\lt1$$
