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 Real Number Ordering is Compatible with Multiplication, it follows that $0 \times x < x \times x < 1 \times x$ and the result follows.

Alternatively we can use the fact that Real Numbers form Ordered Integral Domain and apply Square of Element Less than Unity in Ordered Integral Domain directly.

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

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