• anton@lemmy.blahaj.zone
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    18 hours ago

    That axiomisation is incomplete as it doesn’t preclude stuff like loops, a predecessor to zero or a second number line.

    • Eq0@literature.cafe
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      9 hours ago

      I think you are missing some properties of successors (uniqueness and s(n) different than any m<= n)

      That would avoid “branching” of two different successors to n and loops in which a successor is a smaller number than n

    • Kogasa@programming.dev
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      13 hours ago

      There are non-standard models of arithmetic. They follow the original first-order Peano axioms and any theorem about the naturals is true for them, but they have some wacky extra stuff in them like you mention.

    • TeddE@lemmy.world
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      16 hours ago

      Not sure what you mean by ‘loops’ - except perhaps modular arithmetic, but there are no natural numbers that are negative - you may be thinking of integers, which is constructed from the natural numbers. Similarly, rational numbers, real numbers, and complex numbers are also constructed from the naturals. Complex numbers are often expressed as though they’re two dimensional, since the imaginary part cannot be properly reduced, e.g. 3+2i.

      I recommend this playlist by mathematician another roof: https://www.youtube.com/playlist?list=PLsdeQ7TnWVm_EQG1rmb34ZBYe5ohrkL3t

      They build the whole modern number system ‘from scratch’

      • anton@lemmy.blahaj.zone
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        9 hours ago

        I know how how natural numbers work, but the axioms in the comment i replied to are not enough to define them.

        Not sure what you mean by ‘loops’

        There could be a number n such that m=s(n) and n=s(m). This would be precluded by taking the axiom of induction or the trichotomy axiom.

        If we only take the latter we can still make a second number line, that runs “parallel” to the “propper number line” like:

        n,s(n),s(s(n)),s(s(s(n))),...
        0,s(0),s(s(0)),s(s(s(0))),...
        

        there are no natural numbers that are negative

        I know, but the given axioms don’t preclude it. Under the peano axioms it’s explicitly spelled out:
        0 is not the successor of any natural number