# Ordering of Squares in Reals

## Contents

## 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

- K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*(1977)... (previous)... (next): $\S 1.8 \ (1)$