I wanted to check whether I have done this proof right. I have not fully convinced myself.
Proof:
Let $S=\{x \in \mathbb{R}\mid x^2 < x\}$. Since $x \in \mathbb{R}$, we know that $x^2 > 0$, by a result of the positivity axioms, it follows that $x > 0$. Now if $x^2 < x$, then $x < 1$. Then by definition of $S$, $1$ is an upper bound for $S$. Since $S$ is bounded above, then by the Completeness Axiom, there exists a least upper bound, $\sup S$, such that $\sup S \leq 1$. So $x < \sup S \leq 1$, then $x^2 < x < \sup S \leq 1$. Then $1 < \sup S \leq 1$. Therefore $\sup S=1$.
Best Answer
You know that $1^2=1$, and you know that $f(x)=x$ grows slower than $g(x)=x^2$ for $x>1$, since $f'(x)=1$ and $g'(x)=2x$, and $2x>1$ for $x>1$. It follows that $x^2-x<0 $ for $x>1$. Now use too, the fact that $x(x-1)<0$ for $0<x<1$