How are injections, surjections and bijections described for functions with an ordered pair as an...












-1












$begingroup$


When given a function with an ordered pair as argument i.e.



Z × Z → Z given by f(m, n) = m − n − 1



How should injections, surjections and bijections be described?
Should there be a domain of tuples ie. (1,1) (1,2) (2,1) (2,2)? Or possibly an extra domain or codomain?



Any help would be greatly appreciated.






graph-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you perhaps extend your question with some more concrete examples of what it is that confuses you? For example give an example of a function where you're not sure whether it is injective (or surjective), together with an explanation of how you think both answers might make sense depending on how you understand the definitions?
    $endgroup$
    – Henning Makholm
    Jan 1 at 21:34










  • $begingroup$
    A function from $A$ to $B$ is e.g. injective if for all $x,y in A$ we have $f(x)=f(y)$ implies $x=y$. The same definition applies whether $f: mathbb{Z} times mathbb{Z} to B$ or whatever $A$ is, as user3482749 points out. What about this confuses you?
    $endgroup$
    – LoveTooNap29
    Jan 1 at 22:36
















-1












$begingroup$


When given a function with an ordered pair as argument i.e.



Z × Z → Z given by f(m, n) = m − n − 1



How should injections, surjections and bijections be described?
Should there be a domain of tuples ie. (1,1) (1,2) (2,1) (2,2)? Or possibly an extra domain or codomain?



Any help would be greatly appreciated.






graph-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you perhaps extend your question with some more concrete examples of what it is that confuses you? For example give an example of a function where you're not sure whether it is injective (or surjective), together with an explanation of how you think both answers might make sense depending on how you understand the definitions?
    $endgroup$
    – Henning Makholm
    Jan 1 at 21:34










  • $begingroup$
    A function from $A$ to $B$ is e.g. injective if for all $x,y in A$ we have $f(x)=f(y)$ implies $x=y$. The same definition applies whether $f: mathbb{Z} times mathbb{Z} to B$ or whatever $A$ is, as user3482749 points out. What about this confuses you?
    $endgroup$
    – LoveTooNap29
    Jan 1 at 22:36














-1












-1








-1





$begingroup$


When given a function with an ordered pair as argument i.e.



Z × Z → Z given by f(m, n) = m − n − 1



How should injections, surjections and bijections be described?
Should there be a domain of tuples ie. (1,1) (1,2) (2,1) (2,2)? Or possibly an extra domain or codomain?



Any help would be greatly appreciated.






graph-theory







share|cite|improve this question











$endgroup$




When given a function with an ordered pair as argument i.e.



Z × Z → Z given by f(m, n) = m − n − 1



How should injections, surjections and bijections be described?
Should there be a domain of tuples ie. (1,1) (1,2) (2,1) (2,2)? Or possibly an extra domain or codomain?



Any help would be greatly appreciated.






graph-theory




graph-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 21:21







user604574

















asked Jan 1 at 21:03









user604574user604574

146




146












  • $begingroup$
    Can you perhaps extend your question with some more concrete examples of what it is that confuses you? For example give an example of a function where you're not sure whether it is injective (or surjective), together with an explanation of how you think both answers might make sense depending on how you understand the definitions?
    $endgroup$
    – Henning Makholm
    Jan 1 at 21:34










  • $begingroup$
    A function from $A$ to $B$ is e.g. injective if for all $x,y in A$ we have $f(x)=f(y)$ implies $x=y$. The same definition applies whether $f: mathbb{Z} times mathbb{Z} to B$ or whatever $A$ is, as user3482749 points out. What about this confuses you?
    $endgroup$
    – LoveTooNap29
    Jan 1 at 22:36


















  • $begingroup$
    Can you perhaps extend your question with some more concrete examples of what it is that confuses you? For example give an example of a function where you're not sure whether it is injective (or surjective), together with an explanation of how you think both answers might make sense depending on how you understand the definitions?
    $endgroup$
    – Henning Makholm
    Jan 1 at 21:34










  • $begingroup$
    A function from $A$ to $B$ is e.g. injective if for all $x,y in A$ we have $f(x)=f(y)$ implies $x=y$. The same definition applies whether $f: mathbb{Z} times mathbb{Z} to B$ or whatever $A$ is, as user3482749 points out. What about this confuses you?
    $endgroup$
    – LoveTooNap29
    Jan 1 at 22:36
















$begingroup$
Can you perhaps extend your question with some more concrete examples of what it is that confuses you? For example give an example of a function where you're not sure whether it is injective (or surjective), together with an explanation of how you think both answers might make sense depending on how you understand the definitions?
$endgroup$
– Henning Makholm
Jan 1 at 21:34




$begingroup$
Can you perhaps extend your question with some more concrete examples of what it is that confuses you? For example give an example of a function where you're not sure whether it is injective (or surjective), together with an explanation of how you think both answers might make sense depending on how you understand the definitions?
$endgroup$
– Henning Makholm
Jan 1 at 21:34












$begingroup$
A function from $A$ to $B$ is e.g. injective if for all $x,y in A$ we have $f(x)=f(y)$ implies $x=y$. The same definition applies whether $f: mathbb{Z} times mathbb{Z} to B$ or whatever $A$ is, as user3482749 points out. What about this confuses you?
$endgroup$
– LoveTooNap29
Jan 1 at 22:36




$begingroup$
A function from $A$ to $B$ is e.g. injective if for all $x,y in A$ we have $f(x)=f(y)$ implies $x=y$. The same definition applies whether $f: mathbb{Z} times mathbb{Z} to B$ or whatever $A$ is, as user3482749 points out. What about this confuses you?
$endgroup$
– LoveTooNap29
Jan 1 at 22:36










1 Answer
1






active

oldest

votes


















3












$begingroup$

Those aren't functions with multiple arguments. Those are functions taking a single argument in the domain $mathbb{Z}timesmathbb{Z}$. The definitions are exactly as for any other function.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Aren't 'm' and 'n' separate arguments to the function? How are injections, surjections and bijections described for this?
    $endgroup$
    – user604574
    Jan 1 at 21:08






  • 2




    $begingroup$
    @user604574 Describing two separate arguments $m,n$ is the same as describing a single argument which is an ordered pair: $(m,n)$.
    $endgroup$
    – Wojowu
    Jan 1 at 21:08










  • $begingroup$
    How would injections, surjections and bijections described for this?
    $endgroup$
    – user604574
    Jan 1 at 21:11










  • $begingroup$
    This is not an answer to the question, please remove so that someone can provide help with this topic
    $endgroup$
    – user604574
    Jan 1 at 21:15






  • 3




    $begingroup$
    This absolutely is an answer to the question: the definitions are precisely the standard ones, with no changes whatsoever, because the definitions do not specify anything about the domain and codomain other than that they are sets.
    $endgroup$
    – user3482749
    Jan 1 at 21:28












Your Answer





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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Those aren't functions with multiple arguments. Those are functions taking a single argument in the domain $mathbb{Z}timesmathbb{Z}$. The definitions are exactly as for any other function.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Aren't 'm' and 'n' separate arguments to the function? How are injections, surjections and bijections described for this?
    $endgroup$
    – user604574
    Jan 1 at 21:08






  • 2




    $begingroup$
    @user604574 Describing two separate arguments $m,n$ is the same as describing a single argument which is an ordered pair: $(m,n)$.
    $endgroup$
    – Wojowu
    Jan 1 at 21:08










  • $begingroup$
    How would injections, surjections and bijections described for this?
    $endgroup$
    – user604574
    Jan 1 at 21:11










  • $begingroup$
    This is not an answer to the question, please remove so that someone can provide help with this topic
    $endgroup$
    – user604574
    Jan 1 at 21:15






  • 3




    $begingroup$
    This absolutely is an answer to the question: the definitions are precisely the standard ones, with no changes whatsoever, because the definitions do not specify anything about the domain and codomain other than that they are sets.
    $endgroup$
    – user3482749
    Jan 1 at 21:28
















3












$begingroup$

Those aren't functions with multiple arguments. Those are functions taking a single argument in the domain $mathbb{Z}timesmathbb{Z}$. The definitions are exactly as for any other function.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Aren't 'm' and 'n' separate arguments to the function? How are injections, surjections and bijections described for this?
    $endgroup$
    – user604574
    Jan 1 at 21:08






  • 2




    $begingroup$
    @user604574 Describing two separate arguments $m,n$ is the same as describing a single argument which is an ordered pair: $(m,n)$.
    $endgroup$
    – Wojowu
    Jan 1 at 21:08










  • $begingroup$
    How would injections, surjections and bijections described for this?
    $endgroup$
    – user604574
    Jan 1 at 21:11










  • $begingroup$
    This is not an answer to the question, please remove so that someone can provide help with this topic
    $endgroup$
    – user604574
    Jan 1 at 21:15






  • 3




    $begingroup$
    This absolutely is an answer to the question: the definitions are precisely the standard ones, with no changes whatsoever, because the definitions do not specify anything about the domain and codomain other than that they are sets.
    $endgroup$
    – user3482749
    Jan 1 at 21:28














3












3








3





$begingroup$

Those aren't functions with multiple arguments. Those are functions taking a single argument in the domain $mathbb{Z}timesmathbb{Z}$. The definitions are exactly as for any other function.






share|cite|improve this answer









$endgroup$



Those aren't functions with multiple arguments. Those are functions taking a single argument in the domain $mathbb{Z}timesmathbb{Z}$. The definitions are exactly as for any other function.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 1 at 21:05









user3482749user3482749

4,3211119




4,3211119












  • $begingroup$
    Aren't 'm' and 'n' separate arguments to the function? How are injections, surjections and bijections described for this?
    $endgroup$
    – user604574
    Jan 1 at 21:08






  • 2




    $begingroup$
    @user604574 Describing two separate arguments $m,n$ is the same as describing a single argument which is an ordered pair: $(m,n)$.
    $endgroup$
    – Wojowu
    Jan 1 at 21:08










  • $begingroup$
    How would injections, surjections and bijections described for this?
    $endgroup$
    – user604574
    Jan 1 at 21:11










  • $begingroup$
    This is not an answer to the question, please remove so that someone can provide help with this topic
    $endgroup$
    – user604574
    Jan 1 at 21:15






  • 3




    $begingroup$
    This absolutely is an answer to the question: the definitions are precisely the standard ones, with no changes whatsoever, because the definitions do not specify anything about the domain and codomain other than that they are sets.
    $endgroup$
    – user3482749
    Jan 1 at 21:28


















  • $begingroup$
    Aren't 'm' and 'n' separate arguments to the function? How are injections, surjections and bijections described for this?
    $endgroup$
    – user604574
    Jan 1 at 21:08






  • 2




    $begingroup$
    @user604574 Describing two separate arguments $m,n$ is the same as describing a single argument which is an ordered pair: $(m,n)$.
    $endgroup$
    – Wojowu
    Jan 1 at 21:08










  • $begingroup$
    How would injections, surjections and bijections described for this?
    $endgroup$
    – user604574
    Jan 1 at 21:11










  • $begingroup$
    This is not an answer to the question, please remove so that someone can provide help with this topic
    $endgroup$
    – user604574
    Jan 1 at 21:15






  • 3




    $begingroup$
    This absolutely is an answer to the question: the definitions are precisely the standard ones, with no changes whatsoever, because the definitions do not specify anything about the domain and codomain other than that they are sets.
    $endgroup$
    – user3482749
    Jan 1 at 21:28
















$begingroup$
Aren't 'm' and 'n' separate arguments to the function? How are injections, surjections and bijections described for this?
$endgroup$
– user604574
Jan 1 at 21:08




$begingroup$
Aren't 'm' and 'n' separate arguments to the function? How are injections, surjections and bijections described for this?
$endgroup$
– user604574
Jan 1 at 21:08




2




2




$begingroup$
@user604574 Describing two separate arguments $m,n$ is the same as describing a single argument which is an ordered pair: $(m,n)$.
$endgroup$
– Wojowu
Jan 1 at 21:08




$begingroup$
@user604574 Describing two separate arguments $m,n$ is the same as describing a single argument which is an ordered pair: $(m,n)$.
$endgroup$
– Wojowu
Jan 1 at 21:08












$begingroup$
How would injections, surjections and bijections described for this?
$endgroup$
– user604574
Jan 1 at 21:11




$begingroup$
How would injections, surjections and bijections described for this?
$endgroup$
– user604574
Jan 1 at 21:11












$begingroup$
This is not an answer to the question, please remove so that someone can provide help with this topic
$endgroup$
– user604574
Jan 1 at 21:15




$begingroup$
This is not an answer to the question, please remove so that someone can provide help with this topic
$endgroup$
– user604574
Jan 1 at 21:15




3




3




$begingroup$
This absolutely is an answer to the question: the definitions are precisely the standard ones, with no changes whatsoever, because the definitions do not specify anything about the domain and codomain other than that they are sets.
$endgroup$
– user3482749
Jan 1 at 21:28




$begingroup$
This absolutely is an answer to the question: the definitions are precisely the standard ones, with no changes whatsoever, because the definitions do not specify anything about the domain and codomain other than that they are sets.
$endgroup$
– user3482749
Jan 1 at 21:28


















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