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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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  • 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.

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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˜‰

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      • 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 Offline
            MikM Offline
            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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