4 ms·

I invented two proofs for the Collatz Conjecture

sean_pedersen 4y ago
Who can find the mistakes?
- Someone 4y ago
  The text is not 100% clear to me (partly because it is a bogus proof), but you seem to redefine the arithmetic operators as i ‘+’ j := i + √j i ‘-‘ j := i - √j i ‘×’ j := i + j i ‘÷’ j := i - j That completely changes the structure of the ‘integers’. Essential properties such as i + j = j + i (i + j) + k = i + (j + k) i + i = 2 × i no longer hold, so you can’t use the usual rules of arithmetic. For example, the PDF does ‘3’ × X ‘+’ ‘4’ = (‘3’ × X) ‘+’ ‘4’ = (‘3’ + X) + ‘4’ = (‘3’ + X) + ‘√4’ and then uses the rule (‘3’ + X) + ‘√4’ = (X + ‘3’) + ‘√4’ but that rule doesn’t hold in the alternative arithmetic. Also, ‘even’ and ‘odd’ are completely different from the traditional even and odd. Fo example, for any even ‘n’, we have ‘n’ ÷ ‘2’ = n - 2 so division by two of an ‘even’ ‘number’ always keeps the number ‘even’, and for starting point ‘4’, we get an infinite sequence ‘2’, ‘0’, ‘-2’, etc. (also a direct counterexample to the claim that you always end up at ‘1’)
  - sean_pedersen 4y ago
    Ah ok thanks for the corrections, so I need to evaluate how my 2D (projected) arithmentic works for the proof to work out. Thank you so much for taking my work seriously.
  - sean_pedersen 4y ago
    You only criticized my first proof though. I shipped two proofs gladly! The second one just uses inverses of the collatz function and graph theory.
    - Someone 4y ago
      The second one is inscrutable to me, but seems to depend on the existence of the inverse of a single iteration of the Collatz sequence. That inverse doesn’t exist. An easy counterexample is that both 1 and 8 in a single step produce 4. In general, for any odd n, both n and 6n+2 in a single step lead to 3n+1.
      - sean_pedersen 4y ago
        "That inverse doesn’t exist." That is my new math contribution apparently then. I am happy to explain my proof. I can inverse f(x)=f(x+1) easily as g(x)=f(x)-1 since g(f(x)) = x do you understand my functional inversion approach now better?
        Khelavaster 4y ago
        What's the reverse of '4', since '1' and '8' both link to it?
        sean_pedersen 4y ago
        The reverse is a "Quantum" tree (DAG): 4->1,4->8
sean_pedersen 4y ago
I need an edorsment on arXiv to post it there in the math section... Can anyone help me? Here is my arXiv endorsment code: 9C8E9J

Who can find the mistakes?

Someone 4y ago
The text is not 100% clear to me (partly because it is a bogus proof), but you seem to redefine the arithmetic operators as i ‘+’ j := i + √j i ‘-‘ j := i - √j i ‘×’ j := i + j i ‘÷’ j := i - j That completely changes the structure of the ‘integers’. Essential properties such as i + j = j + i (i + j) + k = i + (j + k) i + i = 2 × i no longer hold, so you can’t use the usual rules of arithmetic. For example, the PDF does ‘3’ × X ‘+’ ‘4’ = (‘3’ × X) ‘+’ ‘4’ = (‘3’ + X) + ‘4’ = (‘3’ + X) + ‘√4’ and then uses the rule (‘3’ + X) + ‘√4’ = (X + ‘3’) + ‘√4’ but that rule doesn’t hold in the alternative arithmetic. Also, ‘even’ and ‘odd’ are completely different from the traditional even and odd. Fo example, for any even ‘n’, we have ‘n’ ÷ ‘2’ = n - 2 so division by two of an ‘even’ ‘number’ always keeps the number ‘even’, and for starting point ‘4’, we get an infinite sequence ‘2’, ‘0’, ‘-2’, etc. (also a direct counterexample to the claim that you always end up at ‘1’)
- sean_pedersen 4y ago
  Ah ok thanks for the corrections, so I need to evaluate how my 2D (projected) arithmentic works for the proof to work out. Thank you so much for taking my work seriously.
- sean_pedersen 4y ago
  You only criticized my first proof though. I shipped two proofs gladly! The second one just uses inverses of the collatz function and graph theory.
  - Someone 4y ago
    The second one is inscrutable to me, but seems to depend on the existence of the inverse of a single iteration of the Collatz sequence. That inverse doesn’t exist. An easy counterexample is that both 1 and 8 in a single step produce 4. In general, for any odd n, both n and 6n+2 in a single step lead to 3n+1.
    - sean_pedersen 4y ago
      "That inverse doesn’t exist." That is my new math contribution apparently then. I am happy to explain my proof. I can inverse f(x)=f(x+1) easily as g(x)=f(x)-1 since g(f(x)) = x do you understand my functional inversion approach now better?
      - Khelavaster 4y ago
        What's the reverse of '4', since '1' and '8' both link to it?
        sean_pedersen 4y ago
        The reverse is a "Quantum" tree (DAG): 4->1,4->8

I need an edorsment on arXiv to post it there in the math section... Can anyone help me? Here is my arXiv endorsment code: 9C8E9J