Is there a name for this relation: for all $x$ there is $y$ such that $xRy$, and for all $x,y,z$, if $xRy$...
up vote
2
down vote
favorite
Suppose for all $x$ there is $y$ such that $xRy$, and for all $x,y,z$, if $xRy$ and $xRz$, then $y=z$.
Does there exist such a binary relation $R$ on some set such that the above properties are satisfied by $R$?
discrete-mathematics relations
edited Dec 3 at 4:29
Brahadeesh
5,97942059
asked Dec 3 at 4:01
mdryizk
153
add a comment |
up vote
2
down vote
favorite
Suppose for all $x$ there is $y$ such that $xRy$, and for all $x,y,z$, if $xRy$ and $xRz$, then $y=z$.
Does there exist such a binary relation $R$ on some set such that the above properties are satisfied by $R$?
discrete-mathematics relations
edited Dec 3 at 4:29
Brahadeesh
5,97942059
asked Dec 3 at 4:01
mdryizk
153
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose for all $x$ there is $y$ such that $xRy$, and for all $x,y,z$, if $xRy$ and $xRz$, then $y=z$.
Does there exist such a binary relation $R$ on some set such that the above properties are satisfied by $R$?
discrete-mathematics relations
edited Dec 3 at 4:29
Brahadeesh
5,97942059
asked Dec 3 at 4:01
mdryizk
153
Suppose for all $x$ there is $y$ such that $xRy$, and for all $x,y,z$, if $xRy$ and $xRz$, then $y=z$.
Does there exist such a binary relation $R$ on some set such that the above properties are satisfied by $R$?
discrete-mathematics relations
discrete-mathematics relations
edited Dec 3 at 4:29
Brahadeesh
5,97942059
asked Dec 3 at 4:01
mdryizk
153
edited Dec 3 at 4:29
Brahadeesh
5,97942059
asked Dec 3 at 4:01
mdryizk
153
edited Dec 3 at 4:29
Brahadeesh
5,97942059
edited Dec 3 at 4:29
Brahadeesh
5,97942059
edited Dec 3 at 4:29
Brahadeesh
5,97942059
5,97942059
asked Dec 3 at 4:01
mdryizk
153
asked Dec 3 at 4:01
mdryizk
153
asked Dec 3 at 4:01
mdryizk
153
153
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
One set of examples is functions from $A$ to $B$. For this let $xRy$ mean that $(x,y) in f$ [using the "function as ordered pairs" formulation]. Then your first requirement expresses that $f$ produces an output $f(x)$ for each $x$ in $A,$ while your second expresses that $f$ is a function.
There may be more examples.
Edit: In usual math terminology, the term "function" implies it is "single valued". That is, a single input doesn't map to more than one output. That's what your second condition expresses. The first condition really says each element of the domain $A$ maps to at leastone thing in $B$ [the "codomain"].
There is a version of a so-called "partial function" for which not every element of domain needs to map to something in codomain. [I've seen that more used by logicians]
answered Dec 3 at 4:19
coffeemath
2,1501413
I see, so the second condition says that R is well-defined?
– mdryizk
Dec 3 at 4:25
@mdryizk I added something to clarify this issue. I think well-defined is more often used when one works with a definition where it isn't clear one gets a unique output because one takes a representative in the definition. Like when defining the sum of two residue cases mod $n$ by selecting one from each, then adding as usual, then taking class of result.
– coffeemath
Dec 3 at 4:49
Aren't the two requirements precisely the definition of a function, so there cannot be any more examples?
– Rahul
Dec 3 at 8:13
@Rahul On thinking about it, I agree no more examples.
– coffeemath
Dec 3 at 19:54
add a comment |
up vote
0
down vote
Define a relation $operatorname{R}$ on $Bbb{Z}$ such that $xoperatorname{R}y$ if and only if $x+y=0$,where '$+$' denotes the usual addition.
answered Dec 3 at 8:10
Thomas Shelby
1,190116
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
One set of examples is functions from $A$ to $B$. For this let $xRy$ mean that $(x,y) in f$ [using the "function as ordered pairs" formulation]. Then your first requirement expresses that $f$ produces an output $f(x)$ for each $x$ in $A,$ while your second expresses that $f$ is a function.
There may be more examples.
Edit: In usual math terminology, the term "function" implies it is "single valued". That is, a single input doesn't map to more than one output. That's what your second condition expresses. The first condition really says each element of the domain $A$ maps to at leastone thing in $B$ [the "codomain"].
There is a version of a so-called "partial function" for which not every element of domain needs to map to something in codomain. [I've seen that more used by logicians]
answered Dec 3 at 4:19
coffeemath
2,1501413
I see, so the second condition says that R is well-defined?
– mdryizk
Dec 3 at 4:25
@mdryizk I added something to clarify this issue. I think well-defined is more often used when one works with a definition where it isn't clear one gets a unique output because one takes a representative in the definition. Like when defining the sum of two residue cases mod $n$ by selecting one from each, then adding as usual, then taking class of result.
– coffeemath
Dec 3 at 4:49
Aren't the two requirements precisely the definition of a function, so there cannot be any more examples?
– Rahul
Dec 3 at 8:13
@Rahul On thinking about it, I agree no more examples.
– coffeemath
Dec 3 at 19:54
add a comment |
up vote
3
down vote
accepted
One set of examples is functions from $A$ to $B$. For this let $xRy$ mean that $(x,y) in f$ [using the "function as ordered pairs" formulation]. Then your first requirement expresses that $f$ produces an output $f(x)$ for each $x$ in $A,$ while your second expresses that $f$ is a function.
There may be more examples.
Edit: In usual math terminology, the term "function" implies it is "single valued". That is, a single input doesn't map to more than one output. That's what your second condition expresses. The first condition really says each element of the domain $A$ maps to at leastone thing in $B$ [the "codomain"].
There is a version of a so-called "partial function" for which not every element of domain needs to map to something in codomain. [I've seen that more used by logicians]
answered Dec 3 at 4:19
coffeemath
2,1501413
I see, so the second condition says that R is well-defined?
– mdryizk
Dec 3 at 4:25
@mdryizk I added something to clarify this issue. I think well-defined is more often used when one works with a definition where it isn't clear one gets a unique output because one takes a representative in the definition. Like when defining the sum of two residue cases mod $n$ by selecting one from each, then adding as usual, then taking class of result.
– coffeemath
Dec 3 at 4:49
Aren't the two requirements precisely the definition of a function, so there cannot be any more examples?
– Rahul
Dec 3 at 8:13
@Rahul On thinking about it, I agree no more examples.
– coffeemath
Dec 3 at 19:54
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
One set of examples is functions from $A$ to $B$. For this let $xRy$ mean that $(x,y) in f$ [using the "function as ordered pairs" formulation]. Then your first requirement expresses that $f$ produces an output $f(x)$ for each $x$ in $A,$ while your second expresses that $f$ is a function.
There may be more examples.
Edit: In usual math terminology, the term "function" implies it is "single valued". That is, a single input doesn't map to more than one output. That's what your second condition expresses. The first condition really says each element of the domain $A$ maps to at leastone thing in $B$ [the "codomain"].
There is a version of a so-called "partial function" for which not every element of domain needs to map to something in codomain. [I've seen that more used by logicians]
answered Dec 3 at 4:19
coffeemath
2,1501413
One set of examples is functions from $A$ to $B$. For this let $xRy$ mean that $(x,y) in f$ [using the "function as ordered pairs" formulation]. Then your first requirement expresses that $f$ produces an output $f(x)$ for each $x$ in $A,$ while your second expresses that $f$ is a function.
There may be more examples.
Edit: In usual math terminology, the term "function" implies it is "single valued". That is, a single input doesn't map to more than one output. That's what your second condition expresses. The first condition really says each element of the domain $A$ maps to at leastone thing in $B$ [the "codomain"].
There is a version of a so-called "partial function" for which not every element of domain needs to map to something in codomain. [I've seen that more used by logicians]
answered Dec 3 at 4:19
coffeemath
2,1501413
edited Dec 3 at 4:45
answered Dec 3 at 4:19
coffeemath
2,1501413
answered Dec 3 at 4:19
coffeemath
2,1501413
answered Dec 3 at 4:19
coffeemath
2,1501413
2,1501413
I see, so the second condition says that R is well-defined?
– mdryizk
Dec 3 at 4:25
@mdryizk I added something to clarify this issue. I think well-defined is more often used when one works with a definition where it isn't clear one gets a unique output because one takes a representative in the definition. Like when defining the sum of two residue cases mod $n$ by selecting one from each, then adding as usual, then taking class of result.
– coffeemath
Dec 3 at 4:49
Aren't the two requirements precisely the definition of a function, so there cannot be any more examples?
– Rahul
Dec 3 at 8:13
@Rahul On thinking about it, I agree no more examples.
– coffeemath
Dec 3 at 19:54
add a comment |
I see, so the second condition says that R is well-defined?
– mdryizk
Dec 3 at 4:25
@mdryizk I added something to clarify this issue. I think well-defined is more often used when one works with a definition where it isn't clear one gets a unique output because one takes a representative in the definition. Like when defining the sum of two residue cases mod $n$ by selecting one from each, then adding as usual, then taking class of result.
– coffeemath
Dec 3 at 4:49
Aren't the two requirements precisely the definition of a function, so there cannot be any more examples?
– Rahul
Dec 3 at 8:13
@Rahul On thinking about it, I agree no more examples.
– coffeemath
Dec 3 at 19:54
I see, so the second condition says that R is well-defined?
– mdryizk
Dec 3 at 4:25
I see, so the second condition says that R is well-defined?
– mdryizk
Dec 3 at 4:25
@mdryizk I added something to clarify this issue. I think well-defined is more often used when one works with a definition where it isn't clear one gets a unique output because one takes a representative in the definition. Like when defining the sum of two residue cases mod $n$ by selecting one from each, then adding as usual, then taking class of result.
– coffeemath
Dec 3 at 4:49
@mdryizk I added something to clarify this issue. I think well-defined is more often used when one works with a definition where it isn't clear one gets a unique output because one takes a representative in the definition. Like when defining the sum of two residue cases mod $n$ by selecting one from each, then adding as usual, then taking class of result.
– coffeemath
Dec 3 at 4:49
Aren't the two requirements precisely the definition of a function, so there cannot be any more examples?
– Rahul
Dec 3 at 8:13
Aren't the two requirements precisely the definition of a function, so there cannot be any more examples?
– Rahul
Dec 3 at 8:13
@Rahul On thinking about it, I agree no more examples.
– coffeemath
Dec 3 at 19:54
@Rahul On thinking about it, I agree no more examples.
– coffeemath
Dec 3 at 19:54
add a comment |
up vote
0
down vote
Define a relation $operatorname{R}$ on $Bbb{Z}$ such that $xoperatorname{R}y$ if and only if $x+y=0$,where '$+$' denotes the usual addition.
answered Dec 3 at 8:10
Thomas Shelby
1,190116
add a comment |
up vote
0
down vote
Define a relation $operatorname{R}$ on $Bbb{Z}$ such that $xoperatorname{R}y$ if and only if $x+y=0$,where '$+$' denotes the usual addition.
answered Dec 3 at 8:10
Thomas Shelby
1,190116
add a comment |
up vote
0
down vote
up vote
0
down vote
Define a relation $operatorname{R}$ on $Bbb{Z}$ such that $xoperatorname{R}y$ if and only if $x+y=0$,where '$+$' denotes the usual addition.
answered Dec 3 at 8:10
Thomas Shelby
1,190116
Define a relation $operatorname{R}$ on $Bbb{Z}$ such that $xoperatorname{R}y$ if and only if $x+y=0$,where '$+$' denotes the usual addition.
answered Dec 3 at 8:10
Thomas Shelby
1,190116
answered Dec 3 at 8:10
Thomas Shelby
1,190116
answered Dec 3 at 8:10
Thomas Shelby
1,190116
answered Dec 3 at 8:10
Thomas Shelby
1,190116
1,190116
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023603%2fis-there-a-name-for-this-relation-for-all-x-there-is-y-such-that-xry-and%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
