Ordering of Squares in Reals

Square Always Positive
Let $x \in \R$.

Then $0 \le x^2$.

Square of Less Than One
Let $x \in \R$.

Let $0 < x < 1$.

Then $0 < x^2 < x$.

Square of Greater Than One
Let $x > 1$.

Then $x^2 > x$.

Square Always Positive
From the Trichotomy Law for Real Numbers, there are three possibilities: $x < 0$, $x = 0$ and $x > 0$.


 * Let $x = 0$. Then $x^2 = 0$ and thus $0 \le x^2$.


 * Let $x > 0$.

Then:

Thus $x^2 > 0$ and so $0 \le x^2$.


 * Finally, let $x < 0$.

Then:

Thus $0 < x^2$ and so $0 \le x^2$.

Square of Less Than One
We are given that $0 < x < 1$.

By direct application of Ordering is Compatible with Multiplication, it follows that $0 \times x < x \times x < 1 \times x$ and the result follows.

Square of Greater Than One
As $x > 1$ it follows that $x > 0$.

Thus by Ordering is Compatible with Multiplication, $x \times x > 1 \times x$ and the result follows.