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
asked Jan 1 at 21:03
user604574user604574
146
$endgroup$
add a comment |
-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
asked Jan 1 at 21:03
user604574user604574
146
$endgroup$
$begingroup$
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
add a comment |
-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
asked Jan 1 at 21:03
user604574user604574
146
$endgroup$
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
asked Jan 1 at 21:03
user604574user604574
146
asked Jan 1 at 21:03
user604574user604574
146
edited Jan 1 at 21:21
user604574
asked Jan 1 at 21:03
user604574user604574
146
asked Jan 1 at 21:03
user604574user604574
146
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
add a comment |
$begingroup$
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
add a comment |
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.
answered Jan 1 at 21:05
user3482749user3482749
4,3211119
$endgroup$
$begingroup$
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
|
show 1 more comment
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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.
answered Jan 1 at 21:05
user3482749user3482749
4,3211119
$endgroup$
$begingroup$
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
|
show 1 more comment
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.
answered Jan 1 at 21:05
user3482749user3482749
4,3211119
$endgroup$
$begingroup$
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
|
show 1 more comment
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.
answered Jan 1 at 21:05
user3482749user3482749
4,3211119
$endgroup$
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.
answered Jan 1 at 21:05
user3482749user3482749
4,3211119
answered Jan 1 at 21:05
user3482749user3482749
4,3211119
answered Jan 1 at 21:05
user3482749user3482749
4,3211119
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
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$begingroup$
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