Can anyone help me understand the notation of binary operator in the definition

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The definition goes,




Let $S$ be a non-empty set. Then a function $f : S times S to S$ is
a binary operator.




If I understand correctly, then it means $(a,b) in S times S Rightarrow f(a,b)=atext*b in S$ then $text*$ is a binary operator. Alright fine. But that doesn't explain that why is $f$ a binary operator?



Let me explain my doubt more clearly, I'm having trouble with notations here,



If $f(a,b) = a text* b$ where $text*$ is a binary operator. And if I say that $f$ is the same binary operator as well, then $f= text*$.
So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is not the image of $a$ under $f$. The correct expression (according to me) should be $(a,b)f(atext*b)$



So, $f$ can't be $text*$. I don't understand how can a relation (function in this case) be equal to a binary operator







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  • I hope I'm making myself clear.
    – William
    Aug 23 at 10:37














up vote
2
down vote

favorite












The definition goes,




Let $S$ be a non-empty set. Then a function $f : S times S to S$ is
a binary operator.




If I understand correctly, then it means $(a,b) in S times S Rightarrow f(a,b)=atext*b in S$ then $text*$ is a binary operator. Alright fine. But that doesn't explain that why is $f$ a binary operator?



Let me explain my doubt more clearly, I'm having trouble with notations here,



If $f(a,b) = a text* b$ where $text*$ is a binary operator. And if I say that $f$ is the same binary operator as well, then $f= text*$.
So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is not the image of $a$ under $f$. The correct expression (according to me) should be $(a,b)f(atext*b)$



So, $f$ can't be $text*$. I don't understand how can a relation (function in this case) be equal to a binary operator







share|cite|improve this question




















  • I hope I'm making myself clear.
    – William
    Aug 23 at 10:37












up vote
2
down vote

favorite









up vote
2
down vote

favorite











The definition goes,




Let $S$ be a non-empty set. Then a function $f : S times S to S$ is
a binary operator.




If I understand correctly, then it means $(a,b) in S times S Rightarrow f(a,b)=atext*b in S$ then $text*$ is a binary operator. Alright fine. But that doesn't explain that why is $f$ a binary operator?



Let me explain my doubt more clearly, I'm having trouble with notations here,



If $f(a,b) = a text* b$ where $text*$ is a binary operator. And if I say that $f$ is the same binary operator as well, then $f= text*$.
So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is not the image of $a$ under $f$. The correct expression (according to me) should be $(a,b)f(atext*b)$



So, $f$ can't be $text*$. I don't understand how can a relation (function in this case) be equal to a binary operator







share|cite|improve this question












The definition goes,




Let $S$ be a non-empty set. Then a function $f : S times S to S$ is
a binary operator.




If I understand correctly, then it means $(a,b) in S times S Rightarrow f(a,b)=atext*b in S$ then $text*$ is a binary operator. Alright fine. But that doesn't explain that why is $f$ a binary operator?



Let me explain my doubt more clearly, I'm having trouble with notations here,



If $f(a,b) = a text* b$ where $text*$ is a binary operator. And if I say that $f$ is the same binary operator as well, then $f= text*$.
So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is not the image of $a$ under $f$. The correct expression (according to me) should be $(a,b)f(atext*b)$



So, $f$ can't be $text*$. I don't understand how can a relation (function in this case) be equal to a binary operator









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asked Aug 23 at 10:29









William

903314




903314











  • I hope I'm making myself clear.
    – William
    Aug 23 at 10:37
















  • I hope I'm making myself clear.
    – William
    Aug 23 at 10:37















I hope I'm making myself clear.
– William
Aug 23 at 10:37




I hope I'm making myself clear.
– William
Aug 23 at 10:37










4 Answers
4






active

oldest

votes

















up vote
2
down vote













$*$ and $f$ are just different notations for the same underlying function. So



$a*b=c$



$f(a,b)=c$



$(a,b) xrightarrowf c$



all have the same meaning. A binary operator simply maps each ordered pair of elements from $S$ (i.e. each member of $S times S$) to an element of $S$.






share|cite|improve this answer



























    up vote
    1
    down vote













    It is called binary, because $f$ has two arguments. It acts on $2$-tuples.




    So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is
    not the image of $a$ under $f$.




    $a f b$ is called infix notation. ($f a b$ is called prefix notation and $a b f$ postfix notation). All mean $f(a, b)$.



    As usual, the interpretation of how you parse a given term depends on your context.






    share|cite|improve this answer





























      up vote
      1
      down vote













      Perhaps it is better to use the viewpoint that some definitions can be relaxed, in some sense, and that notation changes depending on how we want to use an object.



      Say that $ast$ is a binary operator. Then we can take two elements, say $a, b in S$, and combine them using $ast$ to get $aast bin S$. So a binary operator is something that takes two elements and combines them to give one.



      Now define $fcolon Stimes S to S,enspace f(x, y) = xast y$. Then $f$ takes two elements and combines them to give one, and hence $f$ is a binary operator by using the binary operation $ast$, but we will use the notation $f(x, y)$ instead of $xast y$ to emphasise that we're using some function $f$, rather than just the binary operation $ast$.



      This is like being able to write $f(a) = b$ and $a stackrelfmapsto b$ interchangeably. Sometimes it's better to write the former, and sometimes it's better to write the latter, but they both mean the same thing.






      share|cite|improve this answer



























        up vote
        1
        down vote













        When you write $afb$ then you seem to confuse the infix notation of a relation (of which a function is a special kind) and the infix notation of a binary operator (like $+$).



        What the definition tries to say is that a binary operator on a set $S$ is a function $S times S to S.$ It doesn't however say anything about notation. But as you know, we seldom write $+(3,5) = 8,$ but instead $3+5 = 8.$ That does not mean that $+$ relate $3$ and $5$ like $3 < 5$ does.



        If you want to use the infix notation of the function $+$ as a relation, then the correct way to write it is $(3,5)+8.$ But that's a really confusing way to write the identity that is normally written $3+5=8.$






        share|cite|improve this answer




















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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote













          $*$ and $f$ are just different notations for the same underlying function. So



          $a*b=c$



          $f(a,b)=c$



          $(a,b) xrightarrowf c$



          all have the same meaning. A binary operator simply maps each ordered pair of elements from $S$ (i.e. each member of $S times S$) to an element of $S$.






          share|cite|improve this answer
























            up vote
            2
            down vote













            $*$ and $f$ are just different notations for the same underlying function. So



            $a*b=c$



            $f(a,b)=c$



            $(a,b) xrightarrowf c$



            all have the same meaning. A binary operator simply maps each ordered pair of elements from $S$ (i.e. each member of $S times S$) to an element of $S$.






            share|cite|improve this answer






















              up vote
              2
              down vote










              up vote
              2
              down vote









              $*$ and $f$ are just different notations for the same underlying function. So



              $a*b=c$



              $f(a,b)=c$



              $(a,b) xrightarrowf c$



              all have the same meaning. A binary operator simply maps each ordered pair of elements from $S$ (i.e. each member of $S times S$) to an element of $S$.






              share|cite|improve this answer












              $*$ and $f$ are just different notations for the same underlying function. So



              $a*b=c$



              $f(a,b)=c$



              $(a,b) xrightarrowf c$



              all have the same meaning. A binary operator simply maps each ordered pair of elements from $S$ (i.e. each member of $S times S$) to an element of $S$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Aug 23 at 10:45









              gandalf61

              5,966522




              5,966522




















                  up vote
                  1
                  down vote













                  It is called binary, because $f$ has two arguments. It acts on $2$-tuples.




                  So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is
                  not the image of $a$ under $f$.




                  $a f b$ is called infix notation. ($f a b$ is called prefix notation and $a b f$ postfix notation). All mean $f(a, b)$.



                  As usual, the interpretation of how you parse a given term depends on your context.






                  share|cite|improve this answer


























                    up vote
                    1
                    down vote













                    It is called binary, because $f$ has two arguments. It acts on $2$-tuples.




                    So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is
                    not the image of $a$ under $f$.




                    $a f b$ is called infix notation. ($f a b$ is called prefix notation and $a b f$ postfix notation). All mean $f(a, b)$.



                    As usual, the interpretation of how you parse a given term depends on your context.






                    share|cite|improve this answer
























                      up vote
                      1
                      down vote










                      up vote
                      1
                      down vote









                      It is called binary, because $f$ has two arguments. It acts on $2$-tuples.




                      So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is
                      not the image of $a$ under $f$.




                      $a f b$ is called infix notation. ($f a b$ is called prefix notation and $a b f$ postfix notation). All mean $f(a, b)$.



                      As usual, the interpretation of how you parse a given term depends on your context.






                      share|cite|improve this answer














                      It is called binary, because $f$ has two arguments. It acts on $2$-tuples.




                      So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is
                      not the image of $a$ under $f$.




                      $a f b$ is called infix notation. ($f a b$ is called prefix notation and $a b f$ postfix notation). All mean $f(a, b)$.



                      As usual, the interpretation of how you parse a given term depends on your context.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Aug 23 at 10:40

























                      answered Aug 23 at 10:35









                      mvw

                      30.8k22252




                      30.8k22252




















                          up vote
                          1
                          down vote













                          Perhaps it is better to use the viewpoint that some definitions can be relaxed, in some sense, and that notation changes depending on how we want to use an object.



                          Say that $ast$ is a binary operator. Then we can take two elements, say $a, b in S$, and combine them using $ast$ to get $aast bin S$. So a binary operator is something that takes two elements and combines them to give one.



                          Now define $fcolon Stimes S to S,enspace f(x, y) = xast y$. Then $f$ takes two elements and combines them to give one, and hence $f$ is a binary operator by using the binary operation $ast$, but we will use the notation $f(x, y)$ instead of $xast y$ to emphasise that we're using some function $f$, rather than just the binary operation $ast$.



                          This is like being able to write $f(a) = b$ and $a stackrelfmapsto b$ interchangeably. Sometimes it's better to write the former, and sometimes it's better to write the latter, but they both mean the same thing.






                          share|cite|improve this answer
























                            up vote
                            1
                            down vote













                            Perhaps it is better to use the viewpoint that some definitions can be relaxed, in some sense, and that notation changes depending on how we want to use an object.



                            Say that $ast$ is a binary operator. Then we can take two elements, say $a, b in S$, and combine them using $ast$ to get $aast bin S$. So a binary operator is something that takes two elements and combines them to give one.



                            Now define $fcolon Stimes S to S,enspace f(x, y) = xast y$. Then $f$ takes two elements and combines them to give one, and hence $f$ is a binary operator by using the binary operation $ast$, but we will use the notation $f(x, y)$ instead of $xast y$ to emphasise that we're using some function $f$, rather than just the binary operation $ast$.



                            This is like being able to write $f(a) = b$ and $a stackrelfmapsto b$ interchangeably. Sometimes it's better to write the former, and sometimes it's better to write the latter, but they both mean the same thing.






                            share|cite|improve this answer






















                              up vote
                              1
                              down vote










                              up vote
                              1
                              down vote









                              Perhaps it is better to use the viewpoint that some definitions can be relaxed, in some sense, and that notation changes depending on how we want to use an object.



                              Say that $ast$ is a binary operator. Then we can take two elements, say $a, b in S$, and combine them using $ast$ to get $aast bin S$. So a binary operator is something that takes two elements and combines them to give one.



                              Now define $fcolon Stimes S to S,enspace f(x, y) = xast y$. Then $f$ takes two elements and combines them to give one, and hence $f$ is a binary operator by using the binary operation $ast$, but we will use the notation $f(x, y)$ instead of $xast y$ to emphasise that we're using some function $f$, rather than just the binary operation $ast$.



                              This is like being able to write $f(a) = b$ and $a stackrelfmapsto b$ interchangeably. Sometimes it's better to write the former, and sometimes it's better to write the latter, but they both mean the same thing.






                              share|cite|improve this answer












                              Perhaps it is better to use the viewpoint that some definitions can be relaxed, in some sense, and that notation changes depending on how we want to use an object.



                              Say that $ast$ is a binary operator. Then we can take two elements, say $a, b in S$, and combine them using $ast$ to get $aast bin S$. So a binary operator is something that takes two elements and combines them to give one.



                              Now define $fcolon Stimes S to S,enspace f(x, y) = xast y$. Then $f$ takes two elements and combines them to give one, and hence $f$ is a binary operator by using the binary operation $ast$, but we will use the notation $f(x, y)$ instead of $xast y$ to emphasise that we're using some function $f$, rather than just the binary operation $ast$.



                              This is like being able to write $f(a) = b$ and $a stackrelfmapsto b$ interchangeably. Sometimes it's better to write the former, and sometimes it's better to write the latter, but they both mean the same thing.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Aug 23 at 10:43









                              Bill Wallis

                              2,2361826




                              2,2361826




















                                  up vote
                                  1
                                  down vote













                                  When you write $afb$ then you seem to confuse the infix notation of a relation (of which a function is a special kind) and the infix notation of a binary operator (like $+$).



                                  What the definition tries to say is that a binary operator on a set $S$ is a function $S times S to S.$ It doesn't however say anything about notation. But as you know, we seldom write $+(3,5) = 8,$ but instead $3+5 = 8.$ That does not mean that $+$ relate $3$ and $5$ like $3 < 5$ does.



                                  If you want to use the infix notation of the function $+$ as a relation, then the correct way to write it is $(3,5)+8.$ But that's a really confusing way to write the identity that is normally written $3+5=8.$






                                  share|cite|improve this answer
























                                    up vote
                                    1
                                    down vote













                                    When you write $afb$ then you seem to confuse the infix notation of a relation (of which a function is a special kind) and the infix notation of a binary operator (like $+$).



                                    What the definition tries to say is that a binary operator on a set $S$ is a function $S times S to S.$ It doesn't however say anything about notation. But as you know, we seldom write $+(3,5) = 8,$ but instead $3+5 = 8.$ That does not mean that $+$ relate $3$ and $5$ like $3 < 5$ does.



                                    If you want to use the infix notation of the function $+$ as a relation, then the correct way to write it is $(3,5)+8.$ But that's a really confusing way to write the identity that is normally written $3+5=8.$






                                    share|cite|improve this answer






















                                      up vote
                                      1
                                      down vote










                                      up vote
                                      1
                                      down vote









                                      When you write $afb$ then you seem to confuse the infix notation of a relation (of which a function is a special kind) and the infix notation of a binary operator (like $+$).



                                      What the definition tries to say is that a binary operator on a set $S$ is a function $S times S to S.$ It doesn't however say anything about notation. But as you know, we seldom write $+(3,5) = 8,$ but instead $3+5 = 8.$ That does not mean that $+$ relate $3$ and $5$ like $3 < 5$ does.



                                      If you want to use the infix notation of the function $+$ as a relation, then the correct way to write it is $(3,5)+8.$ But that's a really confusing way to write the identity that is normally written $3+5=8.$






                                      share|cite|improve this answer












                                      When you write $afb$ then you seem to confuse the infix notation of a relation (of which a function is a special kind) and the infix notation of a binary operator (like $+$).



                                      What the definition tries to say is that a binary operator on a set $S$ is a function $S times S to S.$ It doesn't however say anything about notation. But as you know, we seldom write $+(3,5) = 8,$ but instead $3+5 = 8.$ That does not mean that $+$ relate $3$ and $5$ like $3 < 5$ does.



                                      If you want to use the infix notation of the function $+$ as a relation, then the correct way to write it is $(3,5)+8.$ But that's a really confusing way to write the identity that is normally written $3+5=8.$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Aug 23 at 12:31









                                      md2perpe

                                      6,40811023




                                      6,40811023



























                                           

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