Is there a ternary operator in math












3












$begingroup$


Is there a math equivalent of the programming ternary operator?



a = b + c > 0 ? 1 : 2


The above meaning that if c is greater than 0 then a = b + 1 else a = b + 2.






computer-science







share|cite|improve this question









$endgroup$












  • $begingroup$
    It probably depends on the form of the conditions and the results, but the example you gave can be expressed with the unit step $u(x)$ (with an appropriate definition at $x=1$): $a = u(b+c) + 1$
    $endgroup$
    – Alex
    1 hour ago












  • $begingroup$
    @Alex $a = b + 2 - u(c)$
    $endgroup$
    – eyeballfrog
    1 hour ago


















3












$begingroup$


Is there a math equivalent of the programming ternary operator?



a = b + c > 0 ? 1 : 2


The above meaning that if c is greater than 0 then a = b + 1 else a = b + 2.






computer-science







share|cite|improve this question









$endgroup$












  • $begingroup$
    It probably depends on the form of the conditions and the results, but the example you gave can be expressed with the unit step $u(x)$ (with an appropriate definition at $x=1$): $a = u(b+c) + 1$
    $endgroup$
    – Alex
    1 hour ago












  • $begingroup$
    @Alex $a = b + 2 - u(c)$
    $endgroup$
    – eyeballfrog
    1 hour ago
















3












3








3


1



$begingroup$


Is there a math equivalent of the programming ternary operator?



a = b + c > 0 ? 1 : 2


The above meaning that if c is greater than 0 then a = b + 1 else a = b + 2.






computer-science







share|cite|improve this question









$endgroup$




Is there a math equivalent of the programming ternary operator?



a = b + c > 0 ? 1 : 2


The above meaning that if c is greater than 0 then a = b + 1 else a = b + 2.






computer-science




computer-science






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









dataphiledataphile

415




415












  • $begingroup$
    It probably depends on the form of the conditions and the results, but the example you gave can be expressed with the unit step $u(x)$ (with an appropriate definition at $x=1$): $a = u(b+c) + 1$
    $endgroup$
    – Alex
    1 hour ago












  • $begingroup$
    @Alex $a = b + 2 - u(c)$
    $endgroup$
    – eyeballfrog
    1 hour ago




















  • $begingroup$
    It probably depends on the form of the conditions and the results, but the example you gave can be expressed with the unit step $u(x)$ (with an appropriate definition at $x=1$): $a = u(b+c) + 1$
    $endgroup$
    – Alex
    1 hour ago












  • $begingroup$
    @Alex $a = b + 2 - u(c)$
    $endgroup$
    – eyeballfrog
    1 hour ago


















$begingroup$
It probably depends on the form of the conditions and the results, but the example you gave can be expressed with the unit step $u(x)$ (with an appropriate definition at $x=1$): $a = u(b+c) + 1$
$endgroup$
– Alex
1 hour ago






$begingroup$
It probably depends on the form of the conditions and the results, but the example you gave can be expressed with the unit step $u(x)$ (with an appropriate definition at $x=1$): $a = u(b+c) + 1$
$endgroup$
– Alex
1 hour ago














$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
1 hour ago






$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
1 hour ago












4 Answers
4






active

oldest

votes


















6












$begingroup$

Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Indicator is definitely the way to go, since the conditional can define an arbitrary set.
    $endgroup$
    – eyeballfrog
    1 hour ago










  • $begingroup$
    Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
    $endgroup$
    – dataphile
    1 hour ago



















4












$begingroup$

In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".



There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    This is not a ternary operator, it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
    $$f(b,c)=begin {cases} b+1&c gt 0\
    b+2 & c le 0 end {cases}$$



    You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
      $$
      a = b + 2 - [c gt 0].
      $$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        aka en.wikipedia.org/wiki/Iverson_bracket
        $endgroup$
        – qwr
        14 mins ago










      • $begingroup$
        @qwr Thank you! I did not recall the name.
        $endgroup$
        – FredH
        12 mins ago










      • $begingroup$
        I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
        $endgroup$
        – Rahul
        4 mins ago













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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Indicator is definitely the way to go, since the conditional can define an arbitrary set.
        $endgroup$
        – eyeballfrog
        1 hour ago










      • $begingroup$
        Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
        $endgroup$
        – dataphile
        1 hour ago
















      6












      $begingroup$

      Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Indicator is definitely the way to go, since the conditional can define an arbitrary set.
        $endgroup$
        – eyeballfrog
        1 hour ago










      • $begingroup$
        Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
        $endgroup$
        – dataphile
        1 hour ago














      6












      6








      6





      $begingroup$

      Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$






      share|cite|improve this answer









      $endgroup$



      Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 1 hour ago









      Siong Thye GohSiong Thye Goh

      102k1466118




      102k1466118












      • $begingroup$
        Indicator is definitely the way to go, since the conditional can define an arbitrary set.
        $endgroup$
        – eyeballfrog
        1 hour ago










      • $begingroup$
        Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
        $endgroup$
        – dataphile
        1 hour ago


















      • $begingroup$
        Indicator is definitely the way to go, since the conditional can define an arbitrary set.
        $endgroup$
        – eyeballfrog
        1 hour ago










      • $begingroup$
        Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
        $endgroup$
        – dataphile
        1 hour ago
















      $begingroup$
      Indicator is definitely the way to go, since the conditional can define an arbitrary set.
      $endgroup$
      – eyeballfrog
      1 hour ago




      $begingroup$
      Indicator is definitely the way to go, since the conditional can define an arbitrary set.
      $endgroup$
      – eyeballfrog
      1 hour ago












      $begingroup$
      Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
      $endgroup$
      – dataphile
      1 hour ago




      $begingroup$
      Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
      $endgroup$
      – dataphile
      1 hour ago











      4












      $begingroup$

      In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".



      There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".



        There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".



          There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.






          share|cite|improve this answer









          $endgroup$



          In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".



          There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Paul ChildsPaul Childs

          3247




          3247























              2












              $begingroup$

              This is not a ternary operator, it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
              $$f(b,c)=begin {cases} b+1&c gt 0\
              b+2 & c le 0 end {cases}$$



              You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                This is not a ternary operator, it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
                $$f(b,c)=begin {cases} b+1&c gt 0\
                b+2 & c le 0 end {cases}$$



                You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  This is not a ternary operator, it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
                  $$f(b,c)=begin {cases} b+1&c gt 0\
                  b+2 & c le 0 end {cases}$$



                  You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.






                  share|cite|improve this answer









                  $endgroup$



                  This is not a ternary operator, it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
                  $$f(b,c)=begin {cases} b+1&c gt 0\
                  b+2 & c le 0 end {cases}$$



                  You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Ross MillikanRoss Millikan

                  298k23198371




                  298k23198371























                      1












                      $begingroup$

                      In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
                      $$
                      a = b + 2 - [c gt 0].
                      $$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        aka en.wikipedia.org/wiki/Iverson_bracket
                        $endgroup$
                        – qwr
                        14 mins ago










                      • $begingroup$
                        @qwr Thank you! I did not recall the name.
                        $endgroup$
                        – FredH
                        12 mins ago










                      • $begingroup$
                        I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
                        $endgroup$
                        – Rahul
                        4 mins ago


















                      1












                      $begingroup$

                      In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
                      $$
                      a = b + 2 - [c gt 0].
                      $$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        aka en.wikipedia.org/wiki/Iverson_bracket
                        $endgroup$
                        – qwr
                        14 mins ago










                      • $begingroup$
                        @qwr Thank you! I did not recall the name.
                        $endgroup$
                        – FredH
                        12 mins ago










                      • $begingroup$
                        I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
                        $endgroup$
                        – Rahul
                        4 mins ago
















                      1












                      1








                      1





                      $begingroup$

                      In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
                      $$
                      a = b + 2 - [c gt 0].
                      $$






                      share|cite|improve this answer











                      $endgroup$



                      In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
                      $$
                      a = b + 2 - [c gt 0].
                      $$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 9 mins ago

























                      answered 43 mins ago









                      FredHFredH

                      49629




                      49629












                      • $begingroup$
                        aka en.wikipedia.org/wiki/Iverson_bracket
                        $endgroup$
                        – qwr
                        14 mins ago










                      • $begingroup$
                        @qwr Thank you! I did not recall the name.
                        $endgroup$
                        – FredH
                        12 mins ago










                      • $begingroup$
                        I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
                        $endgroup$
                        – Rahul
                        4 mins ago




















                      • $begingroup$
                        aka en.wikipedia.org/wiki/Iverson_bracket
                        $endgroup$
                        – qwr
                        14 mins ago










                      • $begingroup$
                        @qwr Thank you! I did not recall the name.
                        $endgroup$
                        – FredH
                        12 mins ago










                      • $begingroup$
                        I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
                        $endgroup$
                        – Rahul
                        4 mins ago


















                      $begingroup$
                      aka en.wikipedia.org/wiki/Iverson_bracket
                      $endgroup$
                      – qwr
                      14 mins ago




                      $begingroup$
                      aka en.wikipedia.org/wiki/Iverson_bracket
                      $endgroup$
                      – qwr
                      14 mins ago












                      $begingroup$
                      @qwr Thank you! I did not recall the name.
                      $endgroup$
                      – FredH
                      12 mins ago




                      $begingroup$
                      @qwr Thank you! I did not recall the name.
                      $endgroup$
                      – FredH
                      12 mins ago












                      $begingroup$
                      I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
                      $endgroup$
                      – Rahul
                      4 mins ago






                      $begingroup$
                      I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
                      $endgroup$
                      – Rahul
                      4 mins ago




















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