Real Number between Zero and One is Greater than Square/Proof 1

Theorem
Let $x \in \R$.

Let $0 < x < 1$.

Then:
 * $0 < x^2 < x$

Proof
We are given that $0 < x < 1$.

By direct application of Real Number Ordering is Compatible with Multiplication, it follows that:
 * $0 \times x < x \times x < 1 \times x$

and the result follows.