Ordering of Squares in Reals

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Theorem

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 $x^2 > x$.


Proof

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:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle x^2\) \(=\) \(\displaystyle \) \(\displaystyle x x\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(>\) \(\displaystyle \) \(\displaystyle 0 x\) \(\displaystyle \) \(\displaystyle \)          Real Number Ordering is Compatible with Multiplication          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle 0\) \(\displaystyle \) \(\displaystyle \)                    


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


  • Finally, let $x < 0$.

Then:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle 0\) \(=\) \(\displaystyle \) \(\displaystyle 0 x\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(<\) \(\displaystyle \) \(\displaystyle x x\) \(\displaystyle \) \(\displaystyle \)          Real Number Ordering is Compatible with Multiplication          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle x^2\) \(\displaystyle \) \(\displaystyle \)                    


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

$\blacksquare$


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.

$\blacksquare$


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.

$\blacksquare$


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.

$\blacksquare$


Sources