Do binary operations need to be surjective functions?

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Let $star$ be a binary operation on the set $S=[0,1]$ defined to be $$star : [0,1] times [0,1] to [0,1] $$



$$textwhere a star b = textminleft(frac12 a , frac12 bright) $$



From observation we can see that the set $S$ is closed under $star$ and that each ordered pair $(a,b)$ is mapped to only one element in $S$.



For example, $1 star 0.3 = 0.15$



But we also don't have every element in the codomain being hit. There doesn't exist any $(a,b) in S^2$ such that $a star b = 0.75$, for example.



Does this cause a problem at all? Is $star$ still considered a binary operation on $S$? In class we were told all binary operations were surjective, but the textbook for the class states no such thing. And if it is not a problem, I am wondering if there are any more complicated or "elegant" examples. I am interested to see them if they are.



Thanks for any clarification on my confusion.










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  • 16




    Binary operations do not need to be surjective.
    – Mike Earnest
    Sep 5 at 16:02










  • Are you clear on the difference between codomain and range?
    – Eric Towers
    Sep 6 at 13:54










  • Yes, codomain is the set of elements that are being mapped to and the range or image is the subset of the codomain that actually has elements from the domain mapped to it
    – WaveX
    Sep 6 at 14:51














up vote
14
down vote

favorite
1












Let $star$ be a binary operation on the set $S=[0,1]$ defined to be $$star : [0,1] times [0,1] to [0,1] $$



$$textwhere a star b = textminleft(frac12 a , frac12 bright) $$



From observation we can see that the set $S$ is closed under $star$ and that each ordered pair $(a,b)$ is mapped to only one element in $S$.



For example, $1 star 0.3 = 0.15$



But we also don't have every element in the codomain being hit. There doesn't exist any $(a,b) in S^2$ such that $a star b = 0.75$, for example.



Does this cause a problem at all? Is $star$ still considered a binary operation on $S$? In class we were told all binary operations were surjective, but the textbook for the class states no such thing. And if it is not a problem, I am wondering if there are any more complicated or "elegant" examples. I am interested to see them if they are.



Thanks for any clarification on my confusion.










share|cite|improve this question



















  • 16




    Binary operations do not need to be surjective.
    – Mike Earnest
    Sep 5 at 16:02










  • Are you clear on the difference between codomain and range?
    – Eric Towers
    Sep 6 at 13:54










  • Yes, codomain is the set of elements that are being mapped to and the range or image is the subset of the codomain that actually has elements from the domain mapped to it
    – WaveX
    Sep 6 at 14:51












up vote
14
down vote

favorite
1









up vote
14
down vote

favorite
1






1





Let $star$ be a binary operation on the set $S=[0,1]$ defined to be $$star : [0,1] times [0,1] to [0,1] $$



$$textwhere a star b = textminleft(frac12 a , frac12 bright) $$



From observation we can see that the set $S$ is closed under $star$ and that each ordered pair $(a,b)$ is mapped to only one element in $S$.



For example, $1 star 0.3 = 0.15$



But we also don't have every element in the codomain being hit. There doesn't exist any $(a,b) in S^2$ such that $a star b = 0.75$, for example.



Does this cause a problem at all? Is $star$ still considered a binary operation on $S$? In class we were told all binary operations were surjective, but the textbook for the class states no such thing. And if it is not a problem, I am wondering if there are any more complicated or "elegant" examples. I am interested to see them if they are.



Thanks for any clarification on my confusion.










share|cite|improve this question















Let $star$ be a binary operation on the set $S=[0,1]$ defined to be $$star : [0,1] times [0,1] to [0,1] $$



$$textwhere a star b = textminleft(frac12 a , frac12 bright) $$



From observation we can see that the set $S$ is closed under $star$ and that each ordered pair $(a,b)$ is mapped to only one element in $S$.



For example, $1 star 0.3 = 0.15$



But we also don't have every element in the codomain being hit. There doesn't exist any $(a,b) in S^2$ such that $a star b = 0.75$, for example.



Does this cause a problem at all? Is $star$ still considered a binary operation on $S$? In class we were told all binary operations were surjective, but the textbook for the class states no such thing. And if it is not a problem, I am wondering if there are any more complicated or "elegant" examples. I am interested to see them if they are.



Thanks for any clarification on my confusion.







abstract-algebra binary-operations






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edited Sep 5 at 22:21

























asked Sep 5 at 16:01









WaveX

2,1252720




2,1252720







  • 16




    Binary operations do not need to be surjective.
    – Mike Earnest
    Sep 5 at 16:02










  • Are you clear on the difference between codomain and range?
    – Eric Towers
    Sep 6 at 13:54










  • Yes, codomain is the set of elements that are being mapped to and the range or image is the subset of the codomain that actually has elements from the domain mapped to it
    – WaveX
    Sep 6 at 14:51












  • 16




    Binary operations do not need to be surjective.
    – Mike Earnest
    Sep 5 at 16:02










  • Are you clear on the difference between codomain and range?
    – Eric Towers
    Sep 6 at 13:54










  • Yes, codomain is the set of elements that are being mapped to and the range or image is the subset of the codomain that actually has elements from the domain mapped to it
    – WaveX
    Sep 6 at 14:51







16




16




Binary operations do not need to be surjective.
– Mike Earnest
Sep 5 at 16:02




Binary operations do not need to be surjective.
– Mike Earnest
Sep 5 at 16:02












Are you clear on the difference between codomain and range?
– Eric Towers
Sep 6 at 13:54




Are you clear on the difference between codomain and range?
– Eric Towers
Sep 6 at 13:54












Yes, codomain is the set of elements that are being mapped to and the range or image is the subset of the codomain that actually has elements from the domain mapped to it
– WaveX
Sep 6 at 14:51




Yes, codomain is the set of elements that are being mapped to and the range or image is the subset of the codomain that actually has elements from the domain mapped to it
– WaveX
Sep 6 at 14:51










4 Answers
4






active

oldest

votes

















up vote
20
down vote













Binary operations do not need to be surjective. Here is a natural example:



Let $mathbb N = 1,2,3,dots $. Then $+: mathbb N times mathbb N to mathbb N$ is not surjective because $1$ is not in the image.



Here is another natural, more interesting example:



Let $mathbb N' = 2,3,dots $. Then $times: mathbb N' times mathbb N' to mathbb N'$ is not surjective because the prime numbers are not in the image.






share|cite|improve this answer


















  • 3




    Are these examples natural because they are not contrived, or because they involve $mathbb N$?
    – Misha Lavrov
    Sep 5 at 22:47






  • 5




    @MishaLavrov, natural because the operations are basic ones, not invented.
    – lhf
    Sep 5 at 22:53


















up vote
12
down vote













In general binary operations are not surjective.



Note that, say, for some set $S$ and a fixed $s in S$ the operation given by $a times b = s$ for all $a,b in S$ is a binary operation.



Just binary operation means really little. It's literally just a function from $Stimes S$ to $S$.



However, if the binary relation has an identity element (or just a left-identity or a right-identity would also suffice), then it can be directly seen that it is surjective. This may or may not be the reason for the discrepancy you observed. It also explains why not few of the most common binary operations are in fact surjective (they have an identity), and further shows a way how to construct some somewhat natural ones that don't.



Note that in the two examples in lhf's answer they judiciously avoided to have the respective natural neutral element in the set.



Let me add some more examples:



  • The reals greater than $0.5$ with addition. This also works for the reals greater than $t$ for any fixed positive $t$, yet not for the positive reals.


  • The reals in the interval $[-0.5,0.5]$ with multiplication (works for any closed interval, even non-symmetric ones in $(-1,1)$ yet not for $(-1,1)$ itself).


  • The reals or also the complex numbers with absolute value greater than $2$ with multiplication (works for greater than $t$ for any fixed $t > 1$, yet not for greater than $1$).


  • The $n times n$ (real) matrices with determinant greater than $2$ (works for any fixed $t>1$).






share|cite|improve this answer






















  • However, if the binary relation has a neutral element - What do you mean by neutral element?
    – tarit goswami
    Sep 5 at 17:22







  • 4




    @taritgoswami Another term for "identity" element.
    – Randall
    Sep 5 at 18:09






  • 1




    @taritgoswami What Randall said. But maybe your point was that I glossed over the right left issue. I clarified the answer and added a link.
    – quid♦
    Sep 5 at 19:06










  • @quid Thank you :)
    – tarit goswami
    Sep 5 at 20:13






  • 1




    +1 for identifying the possible source of the confusion.
    – Ilmari Karonen
    Sep 6 at 8:54

















up vote
6
down vote













I'll give a simple non-fancy example. Consider $f : mathbbR^2 to mathbbR$ defined by $$f(x, y) = 0$$ think of this function as assigning the value $0$ to every point on the plane. Certainly $f$ is a binary operation, but it's as non-surjective as such a function can get.






share|cite|improve this answer
















  • 4




    Nor is it injective. Great example of a boring old function that just disproves a statement.
    – WesleyGroupshaveFeelingsToo
    Sep 5 at 19:40

















up vote
0
down vote













If you define operation on a set $S$ as a function from $S times S$ on S, then it has no reason to be surjective, and you and others gave many example on such non surjective functions.



But common true operations are surjective, simply because real world operations are. Almost all (if not all) interesting operation have an invariant element: $i$ for which $ forall x in S, i otimes x = x$. Such an element is enough to guarantee that the operation will be surjective, and all operations with interesting properties have.






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    4 Answers
    4






    active

    oldest

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    4 Answers
    4






    active

    oldest

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    votes






    active

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    up vote
    20
    down vote













    Binary operations do not need to be surjective. Here is a natural example:



    Let $mathbb N = 1,2,3,dots $. Then $+: mathbb N times mathbb N to mathbb N$ is not surjective because $1$ is not in the image.



    Here is another natural, more interesting example:



    Let $mathbb N' = 2,3,dots $. Then $times: mathbb N' times mathbb N' to mathbb N'$ is not surjective because the prime numbers are not in the image.






    share|cite|improve this answer


















    • 3




      Are these examples natural because they are not contrived, or because they involve $mathbb N$?
      – Misha Lavrov
      Sep 5 at 22:47






    • 5




      @MishaLavrov, natural because the operations are basic ones, not invented.
      – lhf
      Sep 5 at 22:53















    up vote
    20
    down vote













    Binary operations do not need to be surjective. Here is a natural example:



    Let $mathbb N = 1,2,3,dots $. Then $+: mathbb N times mathbb N to mathbb N$ is not surjective because $1$ is not in the image.



    Here is another natural, more interesting example:



    Let $mathbb N' = 2,3,dots $. Then $times: mathbb N' times mathbb N' to mathbb N'$ is not surjective because the prime numbers are not in the image.






    share|cite|improve this answer


















    • 3




      Are these examples natural because they are not contrived, or because they involve $mathbb N$?
      – Misha Lavrov
      Sep 5 at 22:47






    • 5




      @MishaLavrov, natural because the operations are basic ones, not invented.
      – lhf
      Sep 5 at 22:53













    up vote
    20
    down vote










    up vote
    20
    down vote









    Binary operations do not need to be surjective. Here is a natural example:



    Let $mathbb N = 1,2,3,dots $. Then $+: mathbb N times mathbb N to mathbb N$ is not surjective because $1$ is not in the image.



    Here is another natural, more interesting example:



    Let $mathbb N' = 2,3,dots $. Then $times: mathbb N' times mathbb N' to mathbb N'$ is not surjective because the prime numbers are not in the image.






    share|cite|improve this answer














    Binary operations do not need to be surjective. Here is a natural example:



    Let $mathbb N = 1,2,3,dots $. Then $+: mathbb N times mathbb N to mathbb N$ is not surjective because $1$ is not in the image.



    Here is another natural, more interesting example:



    Let $mathbb N' = 2,3,dots $. Then $times: mathbb N' times mathbb N' to mathbb N'$ is not surjective because the prime numbers are not in the image.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Sep 5 at 16:11

























    answered Sep 5 at 16:04









    lhf

    157k9161374




    157k9161374







    • 3




      Are these examples natural because they are not contrived, or because they involve $mathbb N$?
      – Misha Lavrov
      Sep 5 at 22:47






    • 5




      @MishaLavrov, natural because the operations are basic ones, not invented.
      – lhf
      Sep 5 at 22:53













    • 3




      Are these examples natural because they are not contrived, or because they involve $mathbb N$?
      – Misha Lavrov
      Sep 5 at 22:47






    • 5




      @MishaLavrov, natural because the operations are basic ones, not invented.
      – lhf
      Sep 5 at 22:53








    3




    3




    Are these examples natural because they are not contrived, or because they involve $mathbb N$?
    – Misha Lavrov
    Sep 5 at 22:47




    Are these examples natural because they are not contrived, or because they involve $mathbb N$?
    – Misha Lavrov
    Sep 5 at 22:47




    5




    5




    @MishaLavrov, natural because the operations are basic ones, not invented.
    – lhf
    Sep 5 at 22:53





    @MishaLavrov, natural because the operations are basic ones, not invented.
    – lhf
    Sep 5 at 22:53











    up vote
    12
    down vote













    In general binary operations are not surjective.



    Note that, say, for some set $S$ and a fixed $s in S$ the operation given by $a times b = s$ for all $a,b in S$ is a binary operation.



    Just binary operation means really little. It's literally just a function from $Stimes S$ to $S$.



    However, if the binary relation has an identity element (or just a left-identity or a right-identity would also suffice), then it can be directly seen that it is surjective. This may or may not be the reason for the discrepancy you observed. It also explains why not few of the most common binary operations are in fact surjective (they have an identity), and further shows a way how to construct some somewhat natural ones that don't.



    Note that in the two examples in lhf's answer they judiciously avoided to have the respective natural neutral element in the set.



    Let me add some more examples:



    • The reals greater than $0.5$ with addition. This also works for the reals greater than $t$ for any fixed positive $t$, yet not for the positive reals.


    • The reals in the interval $[-0.5,0.5]$ with multiplication (works for any closed interval, even non-symmetric ones in $(-1,1)$ yet not for $(-1,1)$ itself).


    • The reals or also the complex numbers with absolute value greater than $2$ with multiplication (works for greater than $t$ for any fixed $t > 1$, yet not for greater than $1$).


    • The $n times n$ (real) matrices with determinant greater than $2$ (works for any fixed $t>1$).






    share|cite|improve this answer






















    • However, if the binary relation has a neutral element - What do you mean by neutral element?
      – tarit goswami
      Sep 5 at 17:22







    • 4




      @taritgoswami Another term for "identity" element.
      – Randall
      Sep 5 at 18:09






    • 1




      @taritgoswami What Randall said. But maybe your point was that I glossed over the right left issue. I clarified the answer and added a link.
      – quid♦
      Sep 5 at 19:06










    • @quid Thank you :)
      – tarit goswami
      Sep 5 at 20:13






    • 1




      +1 for identifying the possible source of the confusion.
      – Ilmari Karonen
      Sep 6 at 8:54














    up vote
    12
    down vote













    In general binary operations are not surjective.



    Note that, say, for some set $S$ and a fixed $s in S$ the operation given by $a times b = s$ for all $a,b in S$ is a binary operation.



    Just binary operation means really little. It's literally just a function from $Stimes S$ to $S$.



    However, if the binary relation has an identity element (or just a left-identity or a right-identity would also suffice), then it can be directly seen that it is surjective. This may or may not be the reason for the discrepancy you observed. It also explains why not few of the most common binary operations are in fact surjective (they have an identity), and further shows a way how to construct some somewhat natural ones that don't.



    Note that in the two examples in lhf's answer they judiciously avoided to have the respective natural neutral element in the set.



    Let me add some more examples:



    • The reals greater than $0.5$ with addition. This also works for the reals greater than $t$ for any fixed positive $t$, yet not for the positive reals.


    • The reals in the interval $[-0.5,0.5]$ with multiplication (works for any closed interval, even non-symmetric ones in $(-1,1)$ yet not for $(-1,1)$ itself).


    • The reals or also the complex numbers with absolute value greater than $2$ with multiplication (works for greater than $t$ for any fixed $t > 1$, yet not for greater than $1$).


    • The $n times n$ (real) matrices with determinant greater than $2$ (works for any fixed $t>1$).






    share|cite|improve this answer






















    • However, if the binary relation has a neutral element - What do you mean by neutral element?
      – tarit goswami
      Sep 5 at 17:22







    • 4




      @taritgoswami Another term for "identity" element.
      – Randall
      Sep 5 at 18:09






    • 1




      @taritgoswami What Randall said. But maybe your point was that I glossed over the right left issue. I clarified the answer and added a link.
      – quid♦
      Sep 5 at 19:06










    • @quid Thank you :)
      – tarit goswami
      Sep 5 at 20:13






    • 1




      +1 for identifying the possible source of the confusion.
      – Ilmari Karonen
      Sep 6 at 8:54












    up vote
    12
    down vote










    up vote
    12
    down vote









    In general binary operations are not surjective.



    Note that, say, for some set $S$ and a fixed $s in S$ the operation given by $a times b = s$ for all $a,b in S$ is a binary operation.



    Just binary operation means really little. It's literally just a function from $Stimes S$ to $S$.



    However, if the binary relation has an identity element (or just a left-identity or a right-identity would also suffice), then it can be directly seen that it is surjective. This may or may not be the reason for the discrepancy you observed. It also explains why not few of the most common binary operations are in fact surjective (they have an identity), and further shows a way how to construct some somewhat natural ones that don't.



    Note that in the two examples in lhf's answer they judiciously avoided to have the respective natural neutral element in the set.



    Let me add some more examples:



    • The reals greater than $0.5$ with addition. This also works for the reals greater than $t$ for any fixed positive $t$, yet not for the positive reals.


    • The reals in the interval $[-0.5,0.5]$ with multiplication (works for any closed interval, even non-symmetric ones in $(-1,1)$ yet not for $(-1,1)$ itself).


    • The reals or also the complex numbers with absolute value greater than $2$ with multiplication (works for greater than $t$ for any fixed $t > 1$, yet not for greater than $1$).


    • The $n times n$ (real) matrices with determinant greater than $2$ (works for any fixed $t>1$).






    share|cite|improve this answer














    In general binary operations are not surjective.



    Note that, say, for some set $S$ and a fixed $s in S$ the operation given by $a times b = s$ for all $a,b in S$ is a binary operation.



    Just binary operation means really little. It's literally just a function from $Stimes S$ to $S$.



    However, if the binary relation has an identity element (or just a left-identity or a right-identity would also suffice), then it can be directly seen that it is surjective. This may or may not be the reason for the discrepancy you observed. It also explains why not few of the most common binary operations are in fact surjective (they have an identity), and further shows a way how to construct some somewhat natural ones that don't.



    Note that in the two examples in lhf's answer they judiciously avoided to have the respective natural neutral element in the set.



    Let me add some more examples:



    • The reals greater than $0.5$ with addition. This also works for the reals greater than $t$ for any fixed positive $t$, yet not for the positive reals.


    • The reals in the interval $[-0.5,0.5]$ with multiplication (works for any closed interval, even non-symmetric ones in $(-1,1)$ yet not for $(-1,1)$ itself).


    • The reals or also the complex numbers with absolute value greater than $2$ with multiplication (works for greater than $t$ for any fixed $t > 1$, yet not for greater than $1$).


    • The $n times n$ (real) matrices with determinant greater than $2$ (works for any fixed $t>1$).







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Sep 5 at 19:05

























    answered Sep 5 at 16:14









    quid♦

    36.4k85091




    36.4k85091











    • However, if the binary relation has a neutral element - What do you mean by neutral element?
      – tarit goswami
      Sep 5 at 17:22







    • 4




      @taritgoswami Another term for "identity" element.
      – Randall
      Sep 5 at 18:09






    • 1




      @taritgoswami What Randall said. But maybe your point was that I glossed over the right left issue. I clarified the answer and added a link.
      – quid♦
      Sep 5 at 19:06










    • @quid Thank you :)
      – tarit goswami
      Sep 5 at 20:13






    • 1




      +1 for identifying the possible source of the confusion.
      – Ilmari Karonen
      Sep 6 at 8:54
















    • However, if the binary relation has a neutral element - What do you mean by neutral element?
      – tarit goswami
      Sep 5 at 17:22







    • 4




      @taritgoswami Another term for "identity" element.
      – Randall
      Sep 5 at 18:09






    • 1




      @taritgoswami What Randall said. But maybe your point was that I glossed over the right left issue. I clarified the answer and added a link.
      – quid♦
      Sep 5 at 19:06










    • @quid Thank you :)
      – tarit goswami
      Sep 5 at 20:13






    • 1




      +1 for identifying the possible source of the confusion.
      – Ilmari Karonen
      Sep 6 at 8:54















    However, if the binary relation has a neutral element - What do you mean by neutral element?
    – tarit goswami
    Sep 5 at 17:22





    However, if the binary relation has a neutral element - What do you mean by neutral element?
    – tarit goswami
    Sep 5 at 17:22





    4




    4




    @taritgoswami Another term for "identity" element.
    – Randall
    Sep 5 at 18:09




    @taritgoswami Another term for "identity" element.
    – Randall
    Sep 5 at 18:09




    1




    1




    @taritgoswami What Randall said. But maybe your point was that I glossed over the right left issue. I clarified the answer and added a link.
    – quid♦
    Sep 5 at 19:06




    @taritgoswami What Randall said. But maybe your point was that I glossed over the right left issue. I clarified the answer and added a link.
    – quid♦
    Sep 5 at 19:06












    @quid Thank you :)
    – tarit goswami
    Sep 5 at 20:13




    @quid Thank you :)
    – tarit goswami
    Sep 5 at 20:13




    1




    1




    +1 for identifying the possible source of the confusion.
    – Ilmari Karonen
    Sep 6 at 8:54




    +1 for identifying the possible source of the confusion.
    – Ilmari Karonen
    Sep 6 at 8:54










    up vote
    6
    down vote













    I'll give a simple non-fancy example. Consider $f : mathbbR^2 to mathbbR$ defined by $$f(x, y) = 0$$ think of this function as assigning the value $0$ to every point on the plane. Certainly $f$ is a binary operation, but it's as non-surjective as such a function can get.






    share|cite|improve this answer
















    • 4




      Nor is it injective. Great example of a boring old function that just disproves a statement.
      – WesleyGroupshaveFeelingsToo
      Sep 5 at 19:40














    up vote
    6
    down vote













    I'll give a simple non-fancy example. Consider $f : mathbbR^2 to mathbbR$ defined by $$f(x, y) = 0$$ think of this function as assigning the value $0$ to every point on the plane. Certainly $f$ is a binary operation, but it's as non-surjective as such a function can get.






    share|cite|improve this answer
















    • 4




      Nor is it injective. Great example of a boring old function that just disproves a statement.
      – WesleyGroupshaveFeelingsToo
      Sep 5 at 19:40












    up vote
    6
    down vote










    up vote
    6
    down vote









    I'll give a simple non-fancy example. Consider $f : mathbbR^2 to mathbbR$ defined by $$f(x, y) = 0$$ think of this function as assigning the value $0$ to every point on the plane. Certainly $f$ is a binary operation, but it's as non-surjective as such a function can get.






    share|cite|improve this answer












    I'll give a simple non-fancy example. Consider $f : mathbbR^2 to mathbbR$ defined by $$f(x, y) = 0$$ think of this function as assigning the value $0$ to every point on the plane. Certainly $f$ is a binary operation, but it's as non-surjective as such a function can get.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Sep 5 at 18:46









    Perturbative

    3,71711140




    3,71711140







    • 4




      Nor is it injective. Great example of a boring old function that just disproves a statement.
      – WesleyGroupshaveFeelingsToo
      Sep 5 at 19:40












    • 4




      Nor is it injective. Great example of a boring old function that just disproves a statement.
      – WesleyGroupshaveFeelingsToo
      Sep 5 at 19:40







    4




    4




    Nor is it injective. Great example of a boring old function that just disproves a statement.
    – WesleyGroupshaveFeelingsToo
    Sep 5 at 19:40




    Nor is it injective. Great example of a boring old function that just disproves a statement.
    – WesleyGroupshaveFeelingsToo
    Sep 5 at 19:40










    up vote
    0
    down vote













    If you define operation on a set $S$ as a function from $S times S$ on S, then it has no reason to be surjective, and you and others gave many example on such non surjective functions.



    But common true operations are surjective, simply because real world operations are. Almost all (if not all) interesting operation have an invariant element: $i$ for which $ forall x in S, i otimes x = x$. Such an element is enough to guarantee that the operation will be surjective, and all operations with interesting properties have.






    share|cite|improve this answer
























      up vote
      0
      down vote













      If you define operation on a set $S$ as a function from $S times S$ on S, then it has no reason to be surjective, and you and others gave many example on such non surjective functions.



      But common true operations are surjective, simply because real world operations are. Almost all (if not all) interesting operation have an invariant element: $i$ for which $ forall x in S, i otimes x = x$. Such an element is enough to guarantee that the operation will be surjective, and all operations with interesting properties have.






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        If you define operation on a set $S$ as a function from $S times S$ on S, then it has no reason to be surjective, and you and others gave many example on such non surjective functions.



        But common true operations are surjective, simply because real world operations are. Almost all (if not all) interesting operation have an invariant element: $i$ for which $ forall x in S, i otimes x = x$. Such an element is enough to guarantee that the operation will be surjective, and all operations with interesting properties have.






        share|cite|improve this answer












        If you define operation on a set $S$ as a function from $S times S$ on S, then it has no reason to be surjective, and you and others gave many example on such non surjective functions.



        But common true operations are surjective, simply because real world operations are. Almost all (if not all) interesting operation have an invariant element: $i$ for which $ forall x in S, i otimes x = x$. Such an element is enough to guarantee that the operation will be surjective, and all operations with interesting properties have.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 6 at 11:42









        Serge Ballesta

        39518




        39518



























             

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