Ordering of Squares in Reals

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): $\S 1.8 \ (1)$