# Totally Ordered Ring Zero Precedes Element or its Inverse

## Theorem

Let $\left({R, +, \circ, \preceq}\right)$ be an ordered ring.

From the definition of ordered ring, $\preceq$ is compatible with $+$.

Let $0_R$ be the zero element of the ring.

Let $x \ne 0_R$ be a non-zero element of the ring, and let $-x$ be its ring negative.

Then $0_R \prec x \lor 0_R \prec -x$, but not both.

## Proof

By the definition of total ordering, $\preceq$ is connected.

As $x \ne 0_R$, one of the following is true, but not both:

$(1): 0_R \prec x$
$(2): x \prec 0_R$

If $(2)$, because $\prec$ is compatible with $+$:

 $\displaystyle x$ $\prec$ $\displaystyle 0_R$ $\displaystyle \implies \ \$ $\displaystyle x + \left({- x}\right)$ $\prec$ $\displaystyle 0_R + \left({-x}\right)$ $\displaystyle \implies \ \$ $\displaystyle 0_R$ $\prec$ $\displaystyle -x$ definition of ring zero, definition of ring negative

$\blacksquare$