# Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive

## Theorem

Let $\struct {D, +, \times, \le}$ be an ordered integral domain whose zero is $0_D$.

Then:

- $\forall x \in D: x \ne 0_D \iff \map P {x \times x}$

where $\map P {x \times x}$ denotes that $x \times x$ has the (strict) positivity property.

That is, the square of any element of an ordered integral domain is (strictly) positive if and only if that element is non-zero.

## Proof

Let $x = 0_D$.

Then $x \times x = 0_D \times 0_D = 0_D$ by the properties of the ring zero.

Thus by definition of strict positivity property:

- $\neg \map P {0_D \times 0_D}$

$\Box$

Now suppose $x \ne 0_D$.

One of two cases applies:

- $\map P x$

- $\neg \map P x$

Let $\map P x$.

Then by definition of (strict) positivity:

- $\map P {x \times x}$

Now suppose $\neg \map P x$.

Then by the trichotomy law of ordered integral domains:

- $\map P {-x}$

Then again by definition:

- $\map P {\paren {-x} \times \paren {-x} }$.

But by Product of Ring Negatives:

- $\paren {-x} \times \paren {-x} = x \times x$

So again:

- $\map P {x \times x}$

Hence the result.

$\blacksquare$

## Sources

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $7 \ \text{(i)}$