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

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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.

$\blacksquare$