• JeeBaiChow@lemmy.world
      link
      fedilink
      English
      arrow-up
      7
      ·
      2 months ago

      It’s fucking obvious!

      Seriously, I once had to prove that mulplying a value by a number between 0 and 1 decreased it’s original value, i.e. effectively defining the unary, which should be an axiom.

      • Sop
        link
        fedilink
        English
        arrow-up
        5
        ·
        2 months ago

        Mathematicians like to have as little axioms as possible because any axiom is essentially an assumption that can be wrong.

        Also proving elementary results like your example with as little tools as possible is a great exercise to learn mathematical deduction and to understand the relation between certain elementary mathematical properties.

      • friendlymessage@feddit.org
        link
        fedilink
        English
        arrow-up
        3
        ·
        edit-2
        2 months ago

        So you need to proof x•c < x for 0<=c<1?

        Isn’t that just:

        xc < x | ÷x

        c < x/x (for x=/=0)

        c < 1 q.e.d.

        What am I missing?

        • bleistift2@sopuli.xyz
          link
          fedilink
          English
          arrow-up
          5
          ·
          2 months ago

          My math teacher would be angry because you started from the conclusion and derived the premise, rather than the other way around. Note also that you assumed that division is defined. That may not have been the case in the original problem.

          • friendlymessage@feddit.org
            link
            fedilink
            English
            arrow-up
            1
            ·
            edit-2
            2 months ago

            Your math teacher is weird. But you can just turn it around:

            c < 1

            c < x/x | •x

            xc < x q.e.d.

            This also shows, that c≥0 is not actually a requirement, but x>0 is

            I guess if your math teacher is completely insufferable, you need to add the definitions of the arithmetic operations but at that point you should also need to introduce Latin letters and Arabic numerals.

      • Superb
        link
        fedilink
        English
        arrow-up
        3
        ·
        2 months ago

        It can’t be an axiom if it can be defined by other axioms. An axiom can not be formally proven

      • erin (she/her)
        link
        fedilink
        English
        arrow-up
        6
        ·
        2 months ago

        This isn’t a rigorous mathematic proof that would prove that it holds true in every case. You aren’t wrong, but this is a colloquial definition of proof, not a mathematical proof.

        • humblebun@sh.itjust.works
          link
          fedilink
          English
          arrow-up
          1
          ·
          2 months ago

          Sorry, I’ve spent too much of my earthly time on reading and writing formal proofs. I’m not gonna write it now, but I will insist that it’s easy

          • erin (she/her)
            link
            fedilink
            English
            arrow-up
            1
            ·
            2 months ago

            Oh trust me, I believe you. Especially using modern set theory and not the Principia Mathematica.

      • davidagain@lemmy.world
        link
        fedilink
        English
        arrow-up
        1
        ·
        edit-2
        2 months ago

        Only works for a smooth curve with a neighbourhood around it. I think you need the transverse regular theorem or something.