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.