1
$\begingroup$

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. A set of positive elements $x$ of a cyclically ordered group such that $2x$ is positive is the first quadrant of the group.

Definition 3.2. A set of positive elements $x$ of a cyclically ordered group such that $2x$ is negative is the second quadrant of the group.

Definition 3.3. A set of negative elements $x$ of a cyclically ordered group such that $2x$ is positive is the third quadrant of the group.

Definition 3.4. A set of 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:

  1. $[0, a, \frac\pi2] \implies a$ 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.
  2. $a$ 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.
  3. $a$ 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.
  4. $a$ 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:

  1. $a$ 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: $[0, a, \pi] \land [0, \frac\pi2, a] \implies [0, \frac\pi2, a, \pi]$;
    • $[0, \frac\pi2, a, \pi] \implies [\frac\pi2, a, \pi]$.
  2. $[\frac\pi2, a, \pi] \implies a$ 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:

  1. $a + \frac\pi2$ 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.
  2. $a - \frac\pi2$ 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:

  1. $a + \frac\pi2$ 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.
  2. $a - \frac\pi2$ 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:

  1. Case $a = b$: $[0, a, -a]$ and $[0, 2a, -2a]$ by the conditions;
  2. 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.
  3. 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:

  1. Case $a = b$: $[0, a, -a]$ and $[0, -2a, 2a]$ by the conditions;
  2. 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.
  3. 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:

  1. Assuming $a$ and $b$ are in the second quadrant:
    • Lemma 3.7: $a + b$ is negative, contradiction;
  2. Assuming $a$ is in the second quadrant and $b = \frac\pi2$:
    • Lemma 3.5: $a + \frac\pi2$ is negative, contradiction;
  3. 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:

  1. Assuming $a$ and $b$ are in the first quadrant:
    • Lemma 3.6: $a + b$ is positive, contradiction;
  2. Assuming $a$ is in the first quadrant and $b = \frac\pi2$:
    • Lemma 3.4: $a + \frac\pi2$ is positive, contradiction;
  3. 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.

  1. 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.
  2. 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:
    1. 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;
        1. Case $b$ is in the second quadrant:
          • Lemma 3.10: $b \ne a \implies (a - b)$ or $(b - a)$ is in the first quadrant;
        2. Case $b = \frac\pi2$:
          • Lemma 3.5: $(a - \frac\pi2)$ is in the first quadrant.
    2. 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 $x \ne \frac\pi2$;
      • $(a + x) + (y + z) = 0 \implies (y + z) = -(a + x)$;
        1. 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.
        2. 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.
  • Lemma 3.11: the first and the second quadrants are not empty, therefore the third and the fourth quadrants are not empty.
$\endgroup$
  • 1
    $\begingroup$ I don't think that this is the correct place for questions such as this. "Verify my working" is fine if its really short, but when its pages and pages like this you are unlikely to get an answer (which I guess is why your two similar questions which you've just edited have not been answered, even after 2 years...). $\endgroup$ – user1729 May 20 at 13:08
  • $\begingroup$ Should I remove my work leaving just the original question? Where can I verify my work? $\endgroup$ – Alex C May 20 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.