Zero Strictly Precedes One

Theorem

Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.

Let $0$ be the zero of $S$.

Let $1$ be the one of $S$.

Then:

$0 \prec 1$

Proof

This follows directly from the definition of $\prec$.

First note that:

$\forall n \in S: 0 \preceq n$

from the definition of zero.

Next, from the definition of one:

$0 \ne 1$

Thus:

\(\ds \)

\(\)

\(\ds 0 \preceq 1 \land 0 \ne 1\)

\(\ds \)

\(\leadsto\)

\(\ds 0 \prec 1\)

Definition of Strictly Precede

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers