I am trying to prove that if there are two different positive elements in a non-linearly cyclically ordered group, then every quadrant of the group is not empty. Is this a correct statement?
Definition 3.1. The set of all positive elements $x$ of a cyclically ordered group such that $2x$ is positive is the first quadrant of the group.
Definition 3.2. The set of all positive elements $x$ of a cyclically ordered group such that $2x$ is negative is the second quadrant of the group.
Definition 3.3. The set of all negative elements $x$ of a cyclically ordered group such that $2x$ is positive is the third quadrant of the group.
Definition 3.4. The set of all negative elements $x$ of a cyclically ordered group such that $2x$ is negative is the fourth quadrant of the group.
(Positive and negative elements of a cyclically ordered group)
My attempt to prove the main statement:
The following lemmas are applicable to any cyclically ordered group:
Lemma 3.1. There cannot be more than one positive element $\frac\pi2$ such that $\frac\pi2 + \frac\pi2 = \pi$, where $π$ is the apex of the group (Apex of a cyclically ordered group).
Proof:
- Assuming there is another positive element $\frac\pi2'$ such that $\frac\pi2' + \frac\pi2' = \pi$;
- Lemma 1.3: $\frac\pi2' + \frac\pi2' = \frac\pi2 + \frac\pi2 \implies \frac\pi2' = \frac\pi2$.
Lemma 3.2. If there is element $\frac\pi2$, then for any element $a$:
$a$ is in the first quadrant $\iff [0, a, \frac\pi2]$;
$a$ is in the second quadrant $\iff a$ is positive and $[0, \frac\pi2, a]$.
Proof:
- $[0, a, \frac\pi2] \implies a$ is in the first quadrant:
- Lemma 1.4: $[0, a, \frac\pi2] \implies a$ is positive;
- Corollary 1.9: $[0, a, \frac\pi2] \implies [0, 2a, \pi]$;
- Lemma 2.2: $[0, 2a, \pi] \iff 2a$ is positive.
- $a$ is positive and $[0, \frac\pi2, a] \implies a$ is in the second quadrant:
- Corollary 1.9: $[0, \frac\pi2, a] \implies [0, \pi, 2a]$;
- Lemma 2.2: $[0, \pi, 2a] \iff 2a$ is negative.
- $a$ is in the first quadrant $\implies [0, a, \frac\pi2]$:
- Assuming $[0, \frac\pi2, a]$;
- Applying 2: $[0, \frac\pi2, a] \implies a$ is in the second quadrant; contradiction.
- $a$ is in the second quadrant $\implies [0, \frac\pi2, a]$:
- Assuming $[0, a, \frac\pi2]$;
- Applying 1: $[0, a, \frac\pi2] \implies a$ is in the first quadrant; contradiction.
Lemma 3.3. If there is element $\frac\pi2$, then for any element $a$:
$a$ is in the second quadrant $\iff [\frac\pi2, a, \pi]$.
Proof:
- $a$ is in the second quadrant $\implies [\frac\pi2, a, \pi]$:
- Lemma 2.2: $a$ is positive $\iff [0, a, \pi]$;
- Lemma 3.2: $a$ is in the second quadrant $\implies [0, \frac\pi2, a]$;
- 4-Cycle: $[0, a, \pi] \land [0, \frac\pi2, a] \implies [0, \frac\pi2, a, \pi]$;
- $[0, \frac\pi2, a, \pi] \implies [\frac\pi2, a, \pi]$.
- $[\frac\pi2, a, \pi] \implies a$ is in the second quadrant:
- Lemma 2.2: $\frac\pi2$ is positive $\iff [0, \frac\pi2, \pi]$;
- 4-Cycle: $[0, \frac\pi2, \pi] \land [\frac\pi2, a, \pi] \iff [0, \frac\pi2, a, \pi]$;
- $[0, \frac\pi2, a, \pi] \implies [0, a, \pi]$ and $[0, \frac\pi2, a]$;
- Lemma 2.2: $[0, a, \pi] \implies a$ is positive;
- Lemma 3.2: $a$ is positive and $[0, \frac\pi2, a] \implies a$ is in the second quadrant.
Lemma 3.4. If there is element $\frac\pi2$, and an element $a$ is in the first quadrant, then $a + \frac\pi2$ is in the second quadrant, and $a - \frac\pi2$ is negative.
Proof:
- $a + \frac\pi2$ is in the second quadrant:
- Lemma 3.2: $[0, a, \frac\pi2]$;
- Compatibility with $\frac\pi2$: $[0, a, \frac\pi2 ] \implies [\frac\pi2, a + \frac\pi2, \pi]$;
- Lemma 3.3: $[\frac\pi2, a + \frac\pi2, \pi] \iff a + \frac\pi2$ is in the second quadrant.
- $a - \frac\pi2$ is negative:
- Lemma 3.2: $[0, a, \frac\pi2]$;
- Compatibility with $-\frac\pi2$: $[-\frac\pi2, a - \frac\pi2, 0]$;
- Cyclicity: $[-\frac\pi2, a - \frac\pi2, 0] \iff [0, -\frac\pi2, a - \frac\pi2]$;
- Negation: $[0, -\frac\pi2, a - \frac\pi2] \iff [0, -(a - \frac\pi2), \frac\pi2]$;
- Lemma 1.4: $[0, -(a - \frac\pi2), \frac\pi2] \implies -(a - \frac\pi2)$ is positive;
- Lemma 1.1: $a - \frac\pi2$ is negative.
Lemma 3.5. If there is element $\frac\pi2$, and an element $a$ is in the second quadrant, then $a + \frac\pi2$ is negative, and $a - \frac\pi2$ is in the first quadrant.
Proof:
- $a + \frac\pi2$ is negative:
- Lemma 3.3: $[\frac\pi2, a, \pi]$;
- Compatibility with $\frac\pi2$: $[\frac\pi2, a, \pi] \implies [\pi, a + \frac\pi2, \pi + \frac\pi2]$;
- Lemma 2.4: $\pi + \frac\pi2$ is negative;
- Lemma 2.2: $[0, \pi, \pi + \frac\pi2]$;
- 4-Cycle: $[\pi, a + \frac\pi2, \pi + \frac\pi2] \land [0, \pi, \pi + \frac\pi2] \iff [0, \pi, a + \frac\pi2, \pi + \frac\pi2]$;
- $[0, \pi, a + \frac\pi2, \pi + \frac\pi2] \implies [0, \pi, a + \frac\pi2]$;
- Lemma 2.2: $a + \frac\pi2$ is negative.
- $a - \frac\pi2$ is in the first quadrant:
- Lemma 3.2: $[0, \frac\pi2, a]$;
- Lemma 1.7: $a - \frac\pi2$ is positive;
- If $a - \frac\pi2 = \frac\pi2$, then $a = \pi$, contradiction;
- Assuming $a - \frac\pi2$ is in the second quadrant;
- Applying 1: $(a - \frac\pi2) + \frac\pi2 = a$ is negative, contradiction.
Lemma 3.6. If two elements $a, b$ are in the first quadrant, then $a + b$ is positive.
Proof:
- Case $a = b$: $[0, a, -a]$ and $[0, 2a, -2a]$ by the conditions;
- Case $[0, a, b]$:
- Corollary 1.8: $[0, a + b, 2b]$;
- Lemma 1.4: $2b$ is positive, $[0, a + b, 2b] \implies a + b$ is positive.
- Case $[0, b, a]$: same as Case 2.
Lemma 3.7. If two elements $a, b$ are in the second quadrant, then $a + b$ is negative.
Proof:
- Case $a = b$: $[0, a, -a]$ and $[0, -2a, 2a]$ by the conditions;
- Case $[0, a, b]$:
- Corollary 1.8: $[0, 2a, a + b]$;
- Negation: $[0, 2a, a + b] \iff [0, -(a + b), -2a]$;
- Lemma 1.4: $-2a$ is positive, $[0, -(a + b), -2a] \implies -(a + b)$ is positive;
- Lemma 1.1: $a + b$ is negative.
- Case $[0, b, a]$: same as Case 2.
Lemma 3.8. If $a + b$ is positive for some positive $a, b$, then at least one of $a, b$ is in the first quadrant.
Proof:
- Assuming $a$ and $b$ are in the second quadrant:
- Lemma 3.7: $a + b$ is negative, contradiction;
- Assuming $a$ is in the second quadrant and $b = \frac\pi2$:
- Lemma 3.5: $a + \frac\pi2$ is negative, contradiction;
- Assuming $a = \frac\pi2$ and $b = \frac\pi2$:
- $a + b = \pi$, contradiction.
Lemma 3.9. If $a + b$ is negative for some positive $a, b$, then at least one of $a, b$ is in the second quadrant.
Proof:
- Assuming $a$ and $b$ are in the first quadrant:
- Lemma 3.6: $a + b$ is positive, contradiction;
- Assuming $a$ is in the first quadrant and $b = \frac\pi2$:
- Lemma 3.4: $a + \frac\pi2$ is positive, contradiction;
- Assuming $a = \frac\pi2$ and $b = \frac\pi2$:
- $a + b = \pi$, contradiction.
Lemma 3.10. If two positive elements $a, b$ are in the same quadrant, and $[0, b, a]$, then $a - b$ is in the first quadrant.
Case $a$ and $b$ are in the first quadrant:
- Lemma 1.7: $(a - b)$ is positive;
- Lemma 1.7: $[0, a - b, a]$;
- Corollary 1.9: $[0, a - b, a] \implies [0, 2(a - b), 2a]$;
- Lemma 1.4: $2a$ is positive, $[0, 2(a - b), 2a] \implies 2(a - b)$ is positive.
Case $a$ and $b$ are in the second quadrant:
- Lemma 1.7: $(a - b)$ is positive;
- $(a - b) + b = a$;
- Lemma 3.9. $a$ is positive $\implies$ one of $(a - b), b$ is in the first quadrant;
- $b$ is in the second quadrant $\implies (a - b)$ is in the first quadrant.
Lemma 3.11. An element $a$ is in the first quadrant (in the second quadrant) $\iff -a$ is in the fourth quadrant (in the third quadrant).
Proof:
- Lemma 1.1: $a$ is positive $\iff -a$ is negative;
- Lemma 1.1: $2a$ is positive $\iff -2a$ is negative;
- Lemma 1.1: $2a$ is negative $\iff -2a$ is positive.
Theorem. If there are two different positive elements in a non-linearly cyclically ordered group, then each quadrant of the group is not empty.
Proof:
- For a positive element $x$:
- Corollary 1.2: $2x \ne 0$;
- Corollary 2.1: $2x$ is positive, or $2x$ is negative, or $2x$ is the apex;
- Lemma 3.1: if $2x$ is the apex, then $x$ is unique;
- Assuming there are two positive elements $a \ne b$;
- If one of $2a$ or $2b$ is the apex, then another one is positive or negative;
- Taking $2a$ is not the apex, $a \ne \frac\pi2$;
- The rule of three steps for a cyclically ordered group:
- Case there are two positive elements $x, y$ such that $a + x + y = 0$:
- $a + x = -y$;
- Lemma 1.1: $(a + x)$ is negative;
- Lemma 3.9: one of $a$, $x$ is in the second quadrant;
- Assuming $a$ is in the second quadrant;
- Case $b$ is in the second quadrant:
- Lemma 3.10: $b \ne a \implies (a - b)$ or $(b - a)$ is in the first quadrant;
- Case $b = \frac\pi2$:
- Lemma 3.5: $(a - \frac\pi2)$ is in the first quadrant.
- Case $b$ is in the second quadrant:
- Case there are three positive elements $x, y, z$ such that $a + x + y + z = 0$:
- Noticing if $x = y = z = \frac\pi2$, then $a = \frac\pi2$, contradiction;
- At least one of $x, y, z$ is not $\frac\pi2$;
- Taking $x \ne \frac\pi2$;
- $(a + x) + (y + z) = 0 \implies (y + z) = -(a + x)$;
- Case $a$ and $x$ are in the first quadrant:
- Lemma 3.6: $(a + x)$ is positive;
- Lemma 1.1: $(y + z)$ is negative;
- Lemma 3.9: one of $y, z$ is in the second quadrant.
- Case $a$ and $x$ are in the second quadrant:
- Lemma 3.7: $a + x$ is negative;
- Lemma 1.1: $y + z$ is positive;
- Lemma 3.8: one of $y, z$ is in the first quadrant.
- Case $a$ and $x$ are in the first quadrant:
- Case there are two positive elements $x, y$ such that $a + x + y = 0$:
- Lemma 3.11: the first and the second quadrants are not empty, therefore the third and the fourth quadrants are not empty.