• friendlymessage@feddit.org
    link
    fedilink
    English
    arrow-up
    3
    ·
    edit-2
    1 month 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
      link
      fedilink
      English
      arrow-up
      5
      ·
      1 month 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.

      • lseif
        link
        fedilink
        English
        arrow-up
        2
        ·
        1 month ago

        isnt that how methods like proof by contrapositive work ??

        • bleistift2
          link
          fedilink
          English
          arrow-up
          3
          ·
          1 month ago

          Proof by contrapositive would be c<0 ∨ c≥1 ⇒ … ⇒ xc≥x. That is not just starting from the conclusion and deriving the premise.

          • lseif
            link
            fedilink
            English
            arrow-up
            1
            arrow-down
            3
            ·
            1 month ago

            i really dont care

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