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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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  • bachophileB Offline
    bachophileB Offline
    bachophile
    wrote on last edited by
    #2

    Link to video

    taiwan_girlT 1 Reply Last reply
    • bachophileB bachophile

      Link to video

      taiwan_girlT Offline
      taiwan_girlT Offline
      taiwan_girl
      wrote on last edited by
      #3

      @bachophile ๐Ÿ˜† I am glad I am not the only one who did not understand. 555

      George KG 1 Reply Last reply
      • taiwan_girlT taiwan_girl

        @bachophile ๐Ÿ˜† I am glad I am not the only one who did not understand. 555

        George KG Offline
        George KG Offline
        George K
        wrote on last edited by
        #4

        @taiwan_girl said in Puzzle time - algebra edition:

        @bachophile ๐Ÿ˜† I am glad I am not the only one who did not understand. 555

        That makes at least three of us.

        "Now look here, you Baltic gas passer... " - Mik, 6/14/08

        The saying, "Lite is just one damn thing after another," is a gross understatement. The damn things overlap.

        1 Reply Last reply
        • 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?

            Person. Woman. Man. Camera. TV.

            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 Offline
                MikM Offline
                Mik
                wrote on last edited by
                #8

                Needs pronouns.

                "You cannot subsidize irresponsibility and expect people to become more responsible." โ€” Thomas Sowell

                1 Reply Last reply
                • 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).

                  1 Reply Last reply
                  • 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?

                      Person. Woman. Man. Camera. TV.

                      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

                            Person. Woman. Man. Camera. TV.

                            1 Reply Last reply

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