What are the mathematical properties of ⊥ in wheel theory?











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I'm talking about the value of 0/0 in wheel theory, often denoted as "⊥." What are the behaviours of operations like addition, multiplication, exponentiation, trigonometric functions, their inverses, etc when ⊥ is involved? As far as I can guess, the outputs of all operations involving ⊥ should be ⊥, but I can't prove that for all operations.










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    I'm talking about the value of 0/0 in wheel theory, often denoted as "⊥." What are the behaviours of operations like addition, multiplication, exponentiation, trigonometric functions, their inverses, etc when ⊥ is involved? As far as I can guess, the outputs of all operations involving ⊥ should be ⊥, but I can't prove that for all operations.










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I'm talking about the value of 0/0 in wheel theory, often denoted as "⊥." What are the behaviours of operations like addition, multiplication, exponentiation, trigonometric functions, their inverses, etc when ⊥ is involved? As far as I can guess, the outputs of all operations involving ⊥ should be ⊥, but I can't prove that for all operations.










      share|cite|improve this question















      I'm talking about the value of 0/0 in wheel theory, often denoted as "⊥." What are the behaviours of operations like addition, multiplication, exponentiation, trigonometric functions, their inverses, etc when ⊥ is involved? As far as I can guess, the outputs of all operations involving ⊥ should be ⊥, but I can't prove that for all operations.







      ring-theory riemann-sphere






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      edited Nov 18 at 22:11









      user26857

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      asked Nov 18 at 15:49









      ozigzagor

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          Yes, the nullity element will absorb everything in all operations, similar to "NaN" in IEEE 754 floating point (except that it compares equal to itself because it is a "number", or at least equal citizen as an element of the wheel.).



          For addition, it is axiomatic that



          $$x + bot = bot$$



          so no proof here is required. For multiplication, we note that



          $$bot = 0/0 = 0 cdot /0$$



          Now consider $xbot$. This equals $x (0/0)$ which by associativity equals $(x0)/0$. But from here, we cannot simply go directly here because $0x = 0$ is an identity that does not hold in wheels - after all we need $0/0 = 0cdot /0 = 0x = bot$ for $x = /0$. Instead, we have the other axiom that



          $$(x + yz)/y = x/y + z + 0y$$



          and we take $x = 0$ and $y = 0$ so the left side becomes



          $$(0 + 0z)/0 = (0z)/0 = 0/0 + z + 0(0) = 0/0 + z = z + 0/0 = 0/0$$



          Thus, of course, since it doesn't matter what we call it, we have $(x0)/0 = 0/0$ so $xbot = bot$.



          That $/bot = bot$ is, of course, trivial. Thus we have proven the desired result since the full set of operations on a wheel is $+$, $cdot$, and $/$.



          One way you can think of this intuitively was discussed in an earlier thread of mine here about how one could go about imagining a "topological wheel" where we put a spatial structure upon the "extended projective real wheel":



          "Wheel Theory", Extended Reals, Limits, and "Nullity": Can DNE limits be made to equal the element "$0/0$"?



          In particular, $bot$ (or $Phi$) could be considered here to be a sort of internal metaphor for the entire wheel itself. Thus the result of doing any operation to it is, essentially, "anything" and thus itself. Note that this also rather nicely gels with the intuitive notion of calculus of $frac{0}{0}$ as an "indeterminate form" that can take on any, and every, value. Geometrically, this also makes some sense, as you could say that "technically", the graph of the "function"



          $$f(x) := frac{0}{x}$$



          when $x$ is real, "should" have a vertical line at $x = 0$ just as it has a horizontal one at $y = 0$, looking at the limit as the numerator shrinks, but of course we can't allow that as it's a function so we leave the value at $x = 0$ undefined. Nonetheless, if you do go with this, it would agree conceptually also with this understanding of $Phi$, which would be what we'd get if we consider it as a function in the wheel, and moreover, geometrically such a shape is a degenerate hyperbola, which again makes sense given this is the limiting case of non-degenerate hyperbolae (namely of $f_a(x) := frac{a}{x}$ as $a rightarrow 0$). (But this also makezz mvee gvoo Pveezzgh!)






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            up vote
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            accepted










            Yes, the nullity element will absorb everything in all operations, similar to "NaN" in IEEE 754 floating point (except that it compares equal to itself because it is a "number", or at least equal citizen as an element of the wheel.).



            For addition, it is axiomatic that



            $$x + bot = bot$$



            so no proof here is required. For multiplication, we note that



            $$bot = 0/0 = 0 cdot /0$$



            Now consider $xbot$. This equals $x (0/0)$ which by associativity equals $(x0)/0$. But from here, we cannot simply go directly here because $0x = 0$ is an identity that does not hold in wheels - after all we need $0/0 = 0cdot /0 = 0x = bot$ for $x = /0$. Instead, we have the other axiom that



            $$(x + yz)/y = x/y + z + 0y$$



            and we take $x = 0$ and $y = 0$ so the left side becomes



            $$(0 + 0z)/0 = (0z)/0 = 0/0 + z + 0(0) = 0/0 + z = z + 0/0 = 0/0$$



            Thus, of course, since it doesn't matter what we call it, we have $(x0)/0 = 0/0$ so $xbot = bot$.



            That $/bot = bot$ is, of course, trivial. Thus we have proven the desired result since the full set of operations on a wheel is $+$, $cdot$, and $/$.



            One way you can think of this intuitively was discussed in an earlier thread of mine here about how one could go about imagining a "topological wheel" where we put a spatial structure upon the "extended projective real wheel":



            "Wheel Theory", Extended Reals, Limits, and "Nullity": Can DNE limits be made to equal the element "$0/0$"?



            In particular, $bot$ (or $Phi$) could be considered here to be a sort of internal metaphor for the entire wheel itself. Thus the result of doing any operation to it is, essentially, "anything" and thus itself. Note that this also rather nicely gels with the intuitive notion of calculus of $frac{0}{0}$ as an "indeterminate form" that can take on any, and every, value. Geometrically, this also makes some sense, as you could say that "technically", the graph of the "function"



            $$f(x) := frac{0}{x}$$



            when $x$ is real, "should" have a vertical line at $x = 0$ just as it has a horizontal one at $y = 0$, looking at the limit as the numerator shrinks, but of course we can't allow that as it's a function so we leave the value at $x = 0$ undefined. Nonetheless, if you do go with this, it would agree conceptually also with this understanding of $Phi$, which would be what we'd get if we consider it as a function in the wheel, and moreover, geometrically such a shape is a degenerate hyperbola, which again makes sense given this is the limiting case of non-degenerate hyperbolae (namely of $f_a(x) := frac{a}{x}$ as $a rightarrow 0$). (But this also makezz mvee gvoo Pveezzgh!)






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              Yes, the nullity element will absorb everything in all operations, similar to "NaN" in IEEE 754 floating point (except that it compares equal to itself because it is a "number", or at least equal citizen as an element of the wheel.).



              For addition, it is axiomatic that



              $$x + bot = bot$$



              so no proof here is required. For multiplication, we note that



              $$bot = 0/0 = 0 cdot /0$$



              Now consider $xbot$. This equals $x (0/0)$ which by associativity equals $(x0)/0$. But from here, we cannot simply go directly here because $0x = 0$ is an identity that does not hold in wheels - after all we need $0/0 = 0cdot /0 = 0x = bot$ for $x = /0$. Instead, we have the other axiom that



              $$(x + yz)/y = x/y + z + 0y$$



              and we take $x = 0$ and $y = 0$ so the left side becomes



              $$(0 + 0z)/0 = (0z)/0 = 0/0 + z + 0(0) = 0/0 + z = z + 0/0 = 0/0$$



              Thus, of course, since it doesn't matter what we call it, we have $(x0)/0 = 0/0$ so $xbot = bot$.



              That $/bot = bot$ is, of course, trivial. Thus we have proven the desired result since the full set of operations on a wheel is $+$, $cdot$, and $/$.



              One way you can think of this intuitively was discussed in an earlier thread of mine here about how one could go about imagining a "topological wheel" where we put a spatial structure upon the "extended projective real wheel":



              "Wheel Theory", Extended Reals, Limits, and "Nullity": Can DNE limits be made to equal the element "$0/0$"?



              In particular, $bot$ (or $Phi$) could be considered here to be a sort of internal metaphor for the entire wheel itself. Thus the result of doing any operation to it is, essentially, "anything" and thus itself. Note that this also rather nicely gels with the intuitive notion of calculus of $frac{0}{0}$ as an "indeterminate form" that can take on any, and every, value. Geometrically, this also makes some sense, as you could say that "technically", the graph of the "function"



              $$f(x) := frac{0}{x}$$



              when $x$ is real, "should" have a vertical line at $x = 0$ just as it has a horizontal one at $y = 0$, looking at the limit as the numerator shrinks, but of course we can't allow that as it's a function so we leave the value at $x = 0$ undefined. Nonetheless, if you do go with this, it would agree conceptually also with this understanding of $Phi$, which would be what we'd get if we consider it as a function in the wheel, and moreover, geometrically such a shape is a degenerate hyperbola, which again makes sense given this is the limiting case of non-degenerate hyperbolae (namely of $f_a(x) := frac{a}{x}$ as $a rightarrow 0$). (But this also makezz mvee gvoo Pveezzgh!)






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Yes, the nullity element will absorb everything in all operations, similar to "NaN" in IEEE 754 floating point (except that it compares equal to itself because it is a "number", or at least equal citizen as an element of the wheel.).



                For addition, it is axiomatic that



                $$x + bot = bot$$



                so no proof here is required. For multiplication, we note that



                $$bot = 0/0 = 0 cdot /0$$



                Now consider $xbot$. This equals $x (0/0)$ which by associativity equals $(x0)/0$. But from here, we cannot simply go directly here because $0x = 0$ is an identity that does not hold in wheels - after all we need $0/0 = 0cdot /0 = 0x = bot$ for $x = /0$. Instead, we have the other axiom that



                $$(x + yz)/y = x/y + z + 0y$$



                and we take $x = 0$ and $y = 0$ so the left side becomes



                $$(0 + 0z)/0 = (0z)/0 = 0/0 + z + 0(0) = 0/0 + z = z + 0/0 = 0/0$$



                Thus, of course, since it doesn't matter what we call it, we have $(x0)/0 = 0/0$ so $xbot = bot$.



                That $/bot = bot$ is, of course, trivial. Thus we have proven the desired result since the full set of operations on a wheel is $+$, $cdot$, and $/$.



                One way you can think of this intuitively was discussed in an earlier thread of mine here about how one could go about imagining a "topological wheel" where we put a spatial structure upon the "extended projective real wheel":



                "Wheel Theory", Extended Reals, Limits, and "Nullity": Can DNE limits be made to equal the element "$0/0$"?



                In particular, $bot$ (or $Phi$) could be considered here to be a sort of internal metaphor for the entire wheel itself. Thus the result of doing any operation to it is, essentially, "anything" and thus itself. Note that this also rather nicely gels with the intuitive notion of calculus of $frac{0}{0}$ as an "indeterminate form" that can take on any, and every, value. Geometrically, this also makes some sense, as you could say that "technically", the graph of the "function"



                $$f(x) := frac{0}{x}$$



                when $x$ is real, "should" have a vertical line at $x = 0$ just as it has a horizontal one at $y = 0$, looking at the limit as the numerator shrinks, but of course we can't allow that as it's a function so we leave the value at $x = 0$ undefined. Nonetheless, if you do go with this, it would agree conceptually also with this understanding of $Phi$, which would be what we'd get if we consider it as a function in the wheel, and moreover, geometrically such a shape is a degenerate hyperbola, which again makes sense given this is the limiting case of non-degenerate hyperbolae (namely of $f_a(x) := frac{a}{x}$ as $a rightarrow 0$). (But this also makezz mvee gvoo Pveezzgh!)






                share|cite|improve this answer












                Yes, the nullity element will absorb everything in all operations, similar to "NaN" in IEEE 754 floating point (except that it compares equal to itself because it is a "number", or at least equal citizen as an element of the wheel.).



                For addition, it is axiomatic that



                $$x + bot = bot$$



                so no proof here is required. For multiplication, we note that



                $$bot = 0/0 = 0 cdot /0$$



                Now consider $xbot$. This equals $x (0/0)$ which by associativity equals $(x0)/0$. But from here, we cannot simply go directly here because $0x = 0$ is an identity that does not hold in wheels - after all we need $0/0 = 0cdot /0 = 0x = bot$ for $x = /0$. Instead, we have the other axiom that



                $$(x + yz)/y = x/y + z + 0y$$



                and we take $x = 0$ and $y = 0$ so the left side becomes



                $$(0 + 0z)/0 = (0z)/0 = 0/0 + z + 0(0) = 0/0 + z = z + 0/0 = 0/0$$



                Thus, of course, since it doesn't matter what we call it, we have $(x0)/0 = 0/0$ so $xbot = bot$.



                That $/bot = bot$ is, of course, trivial. Thus we have proven the desired result since the full set of operations on a wheel is $+$, $cdot$, and $/$.



                One way you can think of this intuitively was discussed in an earlier thread of mine here about how one could go about imagining a "topological wheel" where we put a spatial structure upon the "extended projective real wheel":



                "Wheel Theory", Extended Reals, Limits, and "Nullity": Can DNE limits be made to equal the element "$0/0$"?



                In particular, $bot$ (or $Phi$) could be considered here to be a sort of internal metaphor for the entire wheel itself. Thus the result of doing any operation to it is, essentially, "anything" and thus itself. Note that this also rather nicely gels with the intuitive notion of calculus of $frac{0}{0}$ as an "indeterminate form" that can take on any, and every, value. Geometrically, this also makes some sense, as you could say that "technically", the graph of the "function"



                $$f(x) := frac{0}{x}$$



                when $x$ is real, "should" have a vertical line at $x = 0$ just as it has a horizontal one at $y = 0$, looking at the limit as the numerator shrinks, but of course we can't allow that as it's a function so we leave the value at $x = 0$ undefined. Nonetheless, if you do go with this, it would agree conceptually also with this understanding of $Phi$, which would be what we'd get if we consider it as a function in the wheel, and moreover, geometrically such a shape is a degenerate hyperbola, which again makes sense given this is the limiting case of non-degenerate hyperbolae (namely of $f_a(x) := frac{a}{x}$ as $a rightarrow 0$). (But this also makezz mvee gvoo Pveezzgh!)







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                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 19 at 0:33









                The_Sympathizer

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                7,0972243






























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