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The New Coffee Room

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  3. Puzzle time - algebra edition

Puzzle time - algebra edition

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  • KlausK Offline
    KlausK Offline
    Klaus
    wrote on last edited by
    #5

    Hey, what are you all doing in this thread? I was hoping that the title would scare off math-challenged posters ๐Ÿ˜‰

    1 Reply Last reply
    • jon-nycJ Online
      jon-nycJ Online
      jon-nyc
      wrote on last edited by
      #6

      Do the operator mappings have to be unique?

      Only non-witches get due process.

      • Cotton Mather, Salem Massachusetts, 1692
      KlausK 1 Reply Last reply
      • bachophileB Offline
        bachophileB Offline
        bachophile
        wrote on last edited by
        #7

        Jon chimes in to let us know he understands the question.

        Right....tell the truth Klaus, u just made up the word bijection.

        Sounds filthy to me.

        1 Reply Last reply
        • MikM Away
          MikM Away
          Mik
          wrote on last edited by
          #8

          Needs pronouns.

          โ€œI am fond of pigs. Dogs look up to us. Cats look down on us. Pigs treat us as equals.โ€ ~Winston S. Churchill

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          • jon-nycJ jon-nyc

            Do the operator mappings have to be unique?

            KlausK Offline
            KlausK Offline
            Klaus
            wrote on last edited by Klaus
            #9

            @jon-nyc said in Puzzle time - algebra edition:

            Do the operator mappings have to be unique?

            I'm not sure what you have in mind, but I mean that, for instance, "+" is interpreted as an operation on sets (for the first task) or logical propositions (for the second task). For instance, you could map "+" to "intersection" (that choice would satisfy equations 1 and 2, but it would be hard to make it work for the other equations, too).

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            • KlausK Offline
              KlausK Offline
              Klaus
              wrote on last edited by
              #10

              Just as a little teaser, this puzzle illustrates the probably most beautiful result of 20th century mathematics/logic. A result that inspired the works of Field medalists and created whole new subfields in mathematics.

              1 Reply Last reply
              • jon-nycJ Online
                jon-nycJ Online
                jon-nyc
                wrote on last edited by
                #11

                Could you map two symbols to the same operator?

                Only non-witches get due process.

                • Cotton Mather, Salem Massachusetts, 1692
                1 Reply Last reply
                • KlausK Offline
                  KlausK Offline
                  Klaus
                  wrote on last edited by
                  #12

                  Sure.

                  1 Reply Last reply
                  • KlausK Offline
                    KlausK Offline
                    Klaus
                    wrote on last edited by Klaus
                    #13

                    OK, maybe not the right audience here for this kind of puzzle ๐Ÿ™‚

                    Anyway, just for closure, here's a solution:

                    For sets:

                    the operator + is set union
                    ยท is set intersection
                    A^B is the set of functions from B to A
                    1 is any one-element set
                    0 is the empty set
                    2 is any two-element set

                    For logical propositions:

                    the operator + is disjunction
                    ยท is conjunction
                    A^B is the implication B -> A
                    1 is logical truth
                    0 is logical falsity
                    2 is also logical truth (because 2 = 1 + 1 = true OR true = true )

                    As usual for mathematics, I'll leave the verification of the equations from the original post as an exercise to the reader ๐Ÿ™‚ .

                    These mappings form the basis the so-called "Curry-Howard correspondence" and, more recently, "homotopy type theory", an influential modern attempt to "reboot" mathematics.

                    jon-nycJ 1 Reply Last reply
                    • KlausK Klaus

                      OK, maybe not the right audience here for this kind of puzzle ๐Ÿ™‚

                      Anyway, just for closure, here's a solution:

                      For sets:

                      the operator + is set union
                      ยท is set intersection
                      A^B is the set of functions from B to A
                      1 is any one-element set
                      0 is the empty set
                      2 is any two-element set

                      For logical propositions:

                      the operator + is disjunction
                      ยท is conjunction
                      A^B is the implication B -> A
                      1 is logical truth
                      0 is logical falsity
                      2 is also logical truth (because 2 = 1 + 1 = true OR true = true )

                      As usual for mathematics, I'll leave the verification of the equations from the original post as an exercise to the reader ๐Ÿ™‚ .

                      These mappings form the basis the so-called "Curry-Howard correspondence" and, more recently, "homotopy type theory", an influential modern attempt to "reboot" mathematics.

                      jon-nycJ Online
                      jon-nycJ Online
                      jon-nyc
                      wrote on last edited by
                      #14

                      @klaus said in Puzzle time - algebra edition:

                      A^B is the implication B -> A

                      I tried that and ruled it out (erroneously) thinking it failed 7

                      Only non-witches get due process.

                      • Cotton Mather, Salem Massachusetts, 1692
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