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  4. Perfectly balanced, as all things should be.

Perfectly balanced, as all things should be.

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  • C cm0002@lemmy.world
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    leate_wonceslace@lemmy.dbzer0.comL This user is from outside of this forum
    leate_wonceslace@lemmy.dbzer0.comL This user is from outside of this forum
    leate_wonceslace@lemmy.dbzer0.com
    wrote on last edited by
    #2

    Okay, so I had a personal project for a long time that addressed the potential for an algebra that allowed for the multipicitive inverse of the additive identity.

    In the context of the resulting non-associative algebra, 0/0=1, rather than 0.

    For anyone wondering, the foundation goes as such: Ω0=1, Ωx=ΩΩ=Ω, x+Ω=Ω, Ω-Ω=Ω+Ω=0.

    A fun consequence of this is the exponential function exp(x)=Σ((x^n)/n!) diverges at exp(Ω). Specifically you can reduce it to Σ(Ω), which when you try to evaluate it, you find that it evaluates to either 0 or Ω. This is particularly fitting, because e^x has a divergent limit at infinity. Specially, it approaches infinity when going towards the positive end and it approaches 0 when approaching the negative.

    There's more cool things you can do with that, but I'll leave it there for now.

    kogasa@programming.devK 1 Reply Last reply
    0
    • leate_wonceslace@lemmy.dbzer0.comL leate_wonceslace@lemmy.dbzer0.com

      Okay, so I had a personal project for a long time that addressed the potential for an algebra that allowed for the multipicitive inverse of the additive identity.

      In the context of the resulting non-associative algebra, 0/0=1, rather than 0.

      For anyone wondering, the foundation goes as such: Ω0=1, Ωx=ΩΩ=Ω, x+Ω=Ω, Ω-Ω=Ω+Ω=0.

      A fun consequence of this is the exponential function exp(x)=Σ((x^n)/n!) diverges at exp(Ω). Specifically you can reduce it to Σ(Ω), which when you try to evaluate it, you find that it evaluates to either 0 or Ω. This is particularly fitting, because e^x has a divergent limit at infinity. Specially, it approaches infinity when going towards the positive end and it approaches 0 when approaching the negative.

      There's more cool things you can do with that, but I'll leave it there for now.

      kogasa@programming.devK This user is from outside of this forum
      kogasa@programming.devK This user is from outside of this forum
      kogasa@programming.dev
      wrote on last edited by
      #3

      There's not much coherent algebraic structure left with these "definitions." If Ωx=ΩΩ=Ω then there is no multiplicative identity, hence no such thing as a multiplicative inverse.

      leate_wonceslace@lemmy.dbzer0.comL 1 Reply Last reply
      0
      • kogasa@programming.devK kogasa@programming.dev

        There's not much coherent algebraic structure left with these "definitions." If Ωx=ΩΩ=Ω then there is no multiplicative identity, hence no such thing as a multiplicative inverse.

        leate_wonceslace@lemmy.dbzer0.comL This user is from outside of this forum
        leate_wonceslace@lemmy.dbzer0.comL This user is from outside of this forum
        leate_wonceslace@lemmy.dbzer0.com
        wrote last edited by
        #4

        No; 1 is the multiplicative identity.

        1Ω=Ω, and for all x in C 1x=x. Thus, 1 fulfills the definition of an identity.

        kogasa@programming.devK 1 Reply Last reply
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        • leate_wonceslace@lemmy.dbzer0.comL leate_wonceslace@lemmy.dbzer0.com

          No; 1 is the multiplicative identity.

          1Ω=Ω, and for all x in C 1x=x. Thus, 1 fulfills the definition of an identity.

          kogasa@programming.devK This user is from outside of this forum
          kogasa@programming.devK This user is from outside of this forum
          kogasa@programming.dev
          wrote last edited by
          #5

          1 = Ω0 = Ω(Ω + Ω) = ΩΩ + ΩΩ = Ω + Ω = 0

          so distributivity is out or else 1 = 0

          leate_wonceslace@lemmy.dbzer0.comL 1 Reply Last reply
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          • kogasa@programming.devK kogasa@programming.dev

            1 = Ω0 = Ω(Ω + Ω) = ΩΩ + ΩΩ = Ω + Ω = 0

            so distributivity is out or else 1 = 0

            leate_wonceslace@lemmy.dbzer0.comL This user is from outside of this forum
            leate_wonceslace@lemmy.dbzer0.comL This user is from outside of this forum
            leate_wonceslace@lemmy.dbzer0.com
            wrote last edited by
            #6

            Correct; multiplying by Ω doesn't distribute over addition.

            kogasa@programming.devK 1 Reply Last reply
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            • leate_wonceslace@lemmy.dbzer0.comL leate_wonceslace@lemmy.dbzer0.com

              Correct; multiplying by Ω doesn't distribute over addition.

              kogasa@programming.devK This user is from outside of this forum
              kogasa@programming.devK This user is from outside of this forum
              kogasa@programming.dev
              wrote last edited by
              #7

              Distributivity is a requirement for non associative algebras. So whatever structure is left is not one of those

              leate_wonceslace@lemmy.dbzer0.comL 1 Reply Last reply
              0
              • kogasa@programming.devK kogasa@programming.dev

                Distributivity is a requirement for non associative algebras. So whatever structure is left is not one of those

                leate_wonceslace@lemmy.dbzer0.comL This user is from outside of this forum
                leate_wonceslace@lemmy.dbzer0.comL This user is from outside of this forum
                leate_wonceslace@lemmy.dbzer0.com
                wrote last edited by
                #8

                What the fuck are you talking about? That's incorrect as a matter of simple fact.

                Associativity is a property possessed by a single operation, whereas distribution is a property possessed by pairs of operations. Non-associative algebras aren't even generally ones that posses multiple operations, so how the hell do you think one implies the other?

                Edit: actually, while we're on it, your first comment was nonsense too; you don't know what an identity is and you think that there's no notion of inverses without an identity. While that's generally the case there are exceptions like in Latin Squares, which describe the Cayley Tables of finite algebras for which every element can be operated with some other element to produce any one target element. In this way we can formulate a notion of "division" without using an identity.

                kogasa@programming.devK 1 Reply Last reply
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                • leate_wonceslace@lemmy.dbzer0.comL leate_wonceslace@lemmy.dbzer0.com

                  What the fuck are you talking about? That's incorrect as a matter of simple fact.

                  Associativity is a property possessed by a single operation, whereas distribution is a property possessed by pairs of operations. Non-associative algebras aren't even generally ones that posses multiple operations, so how the hell do you think one implies the other?

                  Edit: actually, while we're on it, your first comment was nonsense too; you don't know what an identity is and you think that there's no notion of inverses without an identity. While that's generally the case there are exceptions like in Latin Squares, which describe the Cayley Tables of finite algebras for which every element can be operated with some other element to produce any one target element. In this way we can formulate a notion of "division" without using an identity.

                  kogasa@programming.devK This user is from outside of this forum
                  kogasa@programming.devK This user is from outside of this forum
                  kogasa@programming.dev
                  wrote last edited by
                  #9

                  Algebras have two operations by definition and the one thing they have in common is that the multiplication distributes over addition.

                  Yes, there is no notion of inverses without an identity, the definition of an inverse is in terms of an identity.

                  Stop posting.

                  leate_wonceslace@lemmy.dbzer0.comL 1 Reply Last reply
                  0
                  • kogasa@programming.devK kogasa@programming.dev

                    Algebras have two operations by definition and the one thing they have in common is that the multiplication distributes over addition.

                    Yes, there is no notion of inverses without an identity, the definition of an inverse is in terms of an identity.

                    Stop posting.

                    leate_wonceslace@lemmy.dbzer0.comL This user is from outside of this forum
                    leate_wonceslace@lemmy.dbzer0.comL This user is from outside of this forum
                    leate_wonceslace@lemmy.dbzer0.com
                    wrote last edited by
                    #10

                    Do you think a group isn't an algebra? What, by your definitions make an "Algebra" different from a "Ring"?

                    C 1 Reply Last reply
                    0
                    • leate_wonceslace@lemmy.dbzer0.comL leate_wonceslace@lemmy.dbzer0.com

                      Do you think a group isn't an algebra? What, by your definitions make an "Algebra" different from a "Ring"?

                      C This user is from outside of this forum
                      C This user is from outside of this forum
                      compassred@discuss.tchncs.de
                      wrote last edited by
                      #11

                      A group is not an algebra. A group consists of a single associative binary operation with an identity element and inverses for each element.

                      A ring is an abelian (commutative) group under addition, along with an additional associative binary operation (multiplication) that distributes over addition. The additive identity is called zero.

                      A field is a ring in which every nonzero element has a multiplicative inverse.

                      A vector space over a field consists of an abelian group (the vectors) together with scalar multiplication by elements of the field, satisfying distributivity and compatibility conditions.

                      A non-associative algebra is a vector space equipped with a bilinear multiplication operation that distributes over vector addition and is compatible with scalar multiplication.

                      An (associative) algebra is a non-associative algebra whose multiplication operation is associative.

                      You can read more about these definitions online and in textbooks - these are standard definitions. If you are using different definitions, then it would help your case to provide them so we can better understand your claims.

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