What does this relation represents ?
$begingroup$
so i have a hard time understanding what would this relation looks like, we aren't given any precise function so it's hard to know what this would look like. We have to establish the relation and then say if it's an equivalence relation and give the differents equivalence classes.
My representation of this relation would be : ${ <0,0> , <1,1>, <2,2>, <3,3> }$ but im not quite sure if i'm right. Then i would say it is an equivalence relation but we cannot define the different classes since we aren't given any function.
Here's the function and the relation R that we have to create ( N represents natural numbers) :
f: N$,to,${0,1,2,3} a function. Construct the relation R included in $N^2$ the relation ( has the same image by f) which means : 
Thank you for your help !!
discrete-mathematics logic computer-science relations equivalence-relations
asked Dec 8 '18 at 15:24
Dany PépinDany Pépin
112
$endgroup$
add a comment |
$begingroup$
so i have a hard time understanding what would this relation looks like, we aren't given any precise function so it's hard to know what this would look like. We have to establish the relation and then say if it's an equivalence relation and give the differents equivalence classes.
My representation of this relation would be : ${ <0,0> , <1,1>, <2,2>, <3,3> }$ but im not quite sure if i'm right. Then i would say it is an equivalence relation but we cannot define the different classes since we aren't given any function.
Here's the function and the relation R that we have to create ( N represents natural numbers) :
f: N$,to,${0,1,2,3} a function. Construct the relation R included in $N^2$ the relation ( has the same image by f) which means :
Thank you for your help !!
discrete-mathematics logic computer-science relations equivalence-relations
asked Dec 8 '18 at 15:24
Dany PépinDany Pépin
112
$endgroup$
$begingroup$
This is a well-known equivalence relation whose elements (equivalence classes) are called fibers.
$endgroup$
– Wuestenfux
Dec 8 '18 at 15:26
$begingroup$
Is my understanding of it good ? The relation would be ${<0,0>,<1,1>,<2,2><3,3>}$ and its an equivalence relation but since we aren't given any specific function we cannot define the differents equivalence classes ? thank you
$endgroup$
– Dany Pépin
Dec 8 '18 at 15:29
add a comment |
$begingroup$
so i have a hard time understanding what would this relation looks like, we aren't given any precise function so it's hard to know what this would look like. We have to establish the relation and then say if it's an equivalence relation and give the differents equivalence classes.
My representation of this relation would be : ${ <0,0> , <1,1>, <2,2>, <3,3> }$ but im not quite sure if i'm right. Then i would say it is an equivalence relation but we cannot define the different classes since we aren't given any function.
Here's the function and the relation R that we have to create ( N represents natural numbers) :
f: N$,to,${0,1,2,3} a function. Construct the relation R included in $N^2$ the relation ( has the same image by f) which means :
Thank you for your help !!
discrete-mathematics logic computer-science relations equivalence-relations
asked Dec 8 '18 at 15:24
Dany PépinDany Pépin
112
$endgroup$
so i have a hard time understanding what would this relation looks like, we aren't given any precise function so it's hard to know what this would look like. We have to establish the relation and then say if it's an equivalence relation and give the differents equivalence classes.
My representation of this relation would be : ${ <0,0> , <1,1>, <2,2>, <3,3> }$ but im not quite sure if i'm right. Then i would say it is an equivalence relation but we cannot define the different classes since we aren't given any function.
Here's the function and the relation R that we have to create ( N represents natural numbers) :
f: N$,to,${0,1,2,3} a function. Construct the relation R included in $N^2$ the relation ( has the same image by f) which means :
Thank you for your help !!
discrete-mathematics logic computer-science relations equivalence-relations
discrete-mathematics logic computer-science relations equivalence-relations
asked Dec 8 '18 at 15:24
Dany PépinDany Pépin
112
asked Dec 8 '18 at 15:24
Dany PépinDany Pépin
112
asked Dec 8 '18 at 15:24
Dany PépinDany Pépin
112
asked Dec 8 '18 at 15:24
Dany PépinDany Pépin
112
asked Dec 8 '18 at 15:24
Dany PépinDany Pépin
112
112
$begingroup$
This is a well-known equivalence relation whose elements (equivalence classes) are called fibers.
$endgroup$
– Wuestenfux
Dec 8 '18 at 15:26
$begingroup$
Is my understanding of it good ? The relation would be ${<0,0>,<1,1>,<2,2><3,3>}$ and its an equivalence relation but since we aren't given any specific function we cannot define the differents equivalence classes ? thank you
$endgroup$
– Dany Pépin
Dec 8 '18 at 15:29
add a comment |
$begingroup$
This is a well-known equivalence relation whose elements (equivalence classes) are called fibers.
$endgroup$
– Wuestenfux
Dec 8 '18 at 15:26
$begingroup$
Is my understanding of it good ? The relation would be ${<0,0>,<1,1>,<2,2><3,3>}$ and its an equivalence relation but since we aren't given any specific function we cannot define the differents equivalence classes ? thank you
$endgroup$
– Dany Pépin
Dec 8 '18 at 15:29
$begingroup$
This is a well-known equivalence relation whose elements (equivalence classes) are called fibers.
$endgroup$
– Wuestenfux
Dec 8 '18 at 15:26
$begingroup$
This is a well-known equivalence relation whose elements (equivalence classes) are called fibers.
$endgroup$
– Wuestenfux
Dec 8 '18 at 15:26
$begingroup$
Is my understanding of it good ? The relation would be ${<0,0>,<1,1>,<2,2><3,3>}$ and its an equivalence relation but since we aren't given any specific function we cannot define the differents equivalence classes ? thank you
$endgroup$
– Dany Pépin
Dec 8 '18 at 15:29
$begingroup$
Is my understanding of it good ? The relation would be ${<0,0>,<1,1>,<2,2><3,3>}$ and its an equivalence relation but since we aren't given any specific function we cannot define the differents equivalence classes ? thank you
$endgroup$
– Dany Pépin
Dec 8 '18 at 15:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your answer makes no sense, because $R$ is an equivalence relation in $mathbb N$, not in ${0,1,2,3}$.
You should check that $R$ is an equivalence relation (that's easy). Then there are (at most) $4$ equivalence classes:
${ninmathbb{N},|,f(n)=0}$;
${ninmathbb{N},|,f(n)=1}$;
${ninmathbb{N},|,f(n)=2}$;
${ninmathbb{N},|,f(n)=3}$.
Of course, each of these sets is really an equivalence class when it is not empty. In particular, there are $4$ equivalence classes if and only if $f$ is surjective.
answered Dec 8 '18 at 15:33
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
$begingroup$
Ohhh makes sens now correct me if im wrong but i would write the 4 classes like that ?:- ${n,minmathbb{N},|,f(n)=f(m)=0}$; - ${n,minmathbb{N},|,f(n)=f(m)=1}$; - ${n,minmathbb{N},|,f(n)=f(m)=2}$; - ${n,minmathbb{N},|,f(n)=f(m)=3}$.
$endgroup$
– Dany Pépin
Dec 8 '18 at 16:00
add a comment |
$begingroup$
It concerns a relation on $mathbb N$ (so not on ${0,1,2,3}$ as you seem to think).
The equivalence classes form a partition of set $mathbb N$ and the class that is represented by element $ninmathbb N$ is the set:$$[n]={minmathbb Nmid f(m)=f(n)}$$
The number of equivalence classes corresponds with the cardinality of the image of $f$.
This because the classes take the shape: $${ninmathbb Nmid f(n)=i}$$ where $i$ ranges over the image of $f$.
More generally any equivalence relation $R$ on any set $X$ can be presented like this.
If $P$ denotes the set of equivalence classes of $R$ then we can prescribe function $nu:Xto P$ by $xmapsto[x]$ where $[x]$ denotes the equivalence class that is represented by $x$.
Then automatically we have:$$forall,yin X; [xRyiff nu(x)=nu(y)]$$
answered Dec 8 '18 at 15:38
drhabdrhab
101k544130
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your answer makes no sense, because $R$ is an equivalence relation in $mathbb N$, not in ${0,1,2,3}$.
You should check that $R$ is an equivalence relation (that's easy). Then there are (at most) $4$ equivalence classes:
${ninmathbb{N},|,f(n)=0}$;
${ninmathbb{N},|,f(n)=1}$;
${ninmathbb{N},|,f(n)=2}$;
${ninmathbb{N},|,f(n)=3}$.
Of course, each of these sets is really an equivalence class when it is not empty. In particular, there are $4$ equivalence classes if and only if $f$ is surjective.
answered Dec 8 '18 at 15:33
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
$begingroup$
Ohhh makes sens now correct me if im wrong but i would write the 4 classes like that ?:- ${n,minmathbb{N},|,f(n)=f(m)=0}$; - ${n,minmathbb{N},|,f(n)=f(m)=1}$; - ${n,minmathbb{N},|,f(n)=f(m)=2}$; - ${n,minmathbb{N},|,f(n)=f(m)=3}$.
$endgroup$
– Dany Pépin
Dec 8 '18 at 16:00
add a comment |
$begingroup$
Your answer makes no sense, because $R$ is an equivalence relation in $mathbb N$, not in ${0,1,2,3}$.
You should check that $R$ is an equivalence relation (that's easy). Then there are (at most) $4$ equivalence classes:
${ninmathbb{N},|,f(n)=0}$;
${ninmathbb{N},|,f(n)=1}$;
${ninmathbb{N},|,f(n)=2}$;
${ninmathbb{N},|,f(n)=3}$.
Of course, each of these sets is really an equivalence class when it is not empty. In particular, there are $4$ equivalence classes if and only if $f$ is surjective.
answered Dec 8 '18 at 15:33
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
$begingroup$
Ohhh makes sens now correct me if im wrong but i would write the 4 classes like that ?:- ${n,minmathbb{N},|,f(n)=f(m)=0}$; - ${n,minmathbb{N},|,f(n)=f(m)=1}$; - ${n,minmathbb{N},|,f(n)=f(m)=2}$; - ${n,minmathbb{N},|,f(n)=f(m)=3}$.
$endgroup$
– Dany Pépin
Dec 8 '18 at 16:00
add a comment |
$begingroup$
Your answer makes no sense, because $R$ is an equivalence relation in $mathbb N$, not in ${0,1,2,3}$.
You should check that $R$ is an equivalence relation (that's easy). Then there are (at most) $4$ equivalence classes:
${ninmathbb{N},|,f(n)=0}$;
${ninmathbb{N},|,f(n)=1}$;
${ninmathbb{N},|,f(n)=2}$;
${ninmathbb{N},|,f(n)=3}$.
Of course, each of these sets is really an equivalence class when it is not empty. In particular, there are $4$ equivalence classes if and only if $f$ is surjective.
answered Dec 8 '18 at 15:33
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
Your answer makes no sense, because $R$ is an equivalence relation in $mathbb N$, not in ${0,1,2,3}$.
You should check that $R$ is an equivalence relation (that's easy). Then there are (at most) $4$ equivalence classes:
${ninmathbb{N},|,f(n)=0}$;
${ninmathbb{N},|,f(n)=1}$;
${ninmathbb{N},|,f(n)=2}$;
${ninmathbb{N},|,f(n)=3}$.
Of course, each of these sets is really an equivalence class when it is not empty. In particular, there are $4$ equivalence classes if and only if $f$ is surjective.
answered Dec 8 '18 at 15:33
José Carlos SantosJosé Carlos Santos
160k22127232
answered Dec 8 '18 at 15:33
José Carlos SantosJosé Carlos Santos
160k22127232
answered Dec 8 '18 at 15:33
José Carlos SantosJosé Carlos Santos
160k22127232
answered Dec 8 '18 at 15:33
José Carlos SantosJosé Carlos Santos
160k22127232
160k22127232
$begingroup$
Ohhh makes sens now correct me if im wrong but i would write the 4 classes like that ?:- ${n,minmathbb{N},|,f(n)=f(m)=0}$; - ${n,minmathbb{N},|,f(n)=f(m)=1}$; - ${n,minmathbb{N},|,f(n)=f(m)=2}$; - ${n,minmathbb{N},|,f(n)=f(m)=3}$.
$endgroup$
– Dany Pépin
Dec 8 '18 at 16:00
add a comment |
$begingroup$
Ohhh makes sens now correct me if im wrong but i would write the 4 classes like that ?:- ${n,minmathbb{N},|,f(n)=f(m)=0}$; - ${n,minmathbb{N},|,f(n)=f(m)=1}$; - ${n,minmathbb{N},|,f(n)=f(m)=2}$; - ${n,minmathbb{N},|,f(n)=f(m)=3}$.
$endgroup$
– Dany Pépin
Dec 8 '18 at 16:00
$begingroup$
Ohhh makes sens now correct me if im wrong but i would write the 4 classes like that ?:- ${n,minmathbb{N},|,f(n)=f(m)=0}$; - ${n,minmathbb{N},|,f(n)=f(m)=1}$; - ${n,minmathbb{N},|,f(n)=f(m)=2}$; - ${n,minmathbb{N},|,f(n)=f(m)=3}$.
$endgroup$
– Dany Pépin
Dec 8 '18 at 16:00
$begingroup$
Ohhh makes sens now correct me if im wrong but i would write the 4 classes like that ?:- ${n,minmathbb{N},|,f(n)=f(m)=0}$; - ${n,minmathbb{N},|,f(n)=f(m)=1}$; - ${n,minmathbb{N},|,f(n)=f(m)=2}$; - ${n,minmathbb{N},|,f(n)=f(m)=3}$.
$endgroup$
– Dany Pépin
Dec 8 '18 at 16:00
add a comment |
$begingroup$
It concerns a relation on $mathbb N$ (so not on ${0,1,2,3}$ as you seem to think).
The equivalence classes form a partition of set $mathbb N$ and the class that is represented by element $ninmathbb N$ is the set:$$[n]={minmathbb Nmid f(m)=f(n)}$$
The number of equivalence classes corresponds with the cardinality of the image of $f$.
This because the classes take the shape: $${ninmathbb Nmid f(n)=i}$$ where $i$ ranges over the image of $f$.
More generally any equivalence relation $R$ on any set $X$ can be presented like this.
If $P$ denotes the set of equivalence classes of $R$ then we can prescribe function $nu:Xto P$ by $xmapsto[x]$ where $[x]$ denotes the equivalence class that is represented by $x$.
Then automatically we have:$$forall,yin X; [xRyiff nu(x)=nu(y)]$$
answered Dec 8 '18 at 15:38
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
It concerns a relation on $mathbb N$ (so not on ${0,1,2,3}$ as you seem to think).
The equivalence classes form a partition of set $mathbb N$ and the class that is represented by element $ninmathbb N$ is the set:$$[n]={minmathbb Nmid f(m)=f(n)}$$
The number of equivalence classes corresponds with the cardinality of the image of $f$.
This because the classes take the shape: $${ninmathbb Nmid f(n)=i}$$ where $i$ ranges over the image of $f$.
More generally any equivalence relation $R$ on any set $X$ can be presented like this.
If $P$ denotes the set of equivalence classes of $R$ then we can prescribe function $nu:Xto P$ by $xmapsto[x]$ where $[x]$ denotes the equivalence class that is represented by $x$.
Then automatically we have:$$forall,yin X; [xRyiff nu(x)=nu(y)]$$
answered Dec 8 '18 at 15:38
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
It concerns a relation on $mathbb N$ (so not on ${0,1,2,3}$ as you seem to think).
The equivalence classes form a partition of set $mathbb N$ and the class that is represented by element $ninmathbb N$ is the set:$$[n]={minmathbb Nmid f(m)=f(n)}$$
The number of equivalence classes corresponds with the cardinality of the image of $f$.
This because the classes take the shape: $${ninmathbb Nmid f(n)=i}$$ where $i$ ranges over the image of $f$.
More generally any equivalence relation $R$ on any set $X$ can be presented like this.
If $P$ denotes the set of equivalence classes of $R$ then we can prescribe function $nu:Xto P$ by $xmapsto[x]$ where $[x]$ denotes the equivalence class that is represented by $x$.
Then automatically we have:$$forall,yin X; [xRyiff nu(x)=nu(y)]$$
answered Dec 8 '18 at 15:38
drhabdrhab
101k544130
$endgroup$
It concerns a relation on $mathbb N$ (so not on ${0,1,2,3}$ as you seem to think).
The equivalence classes form a partition of set $mathbb N$ and the class that is represented by element $ninmathbb N$ is the set:$$[n]={minmathbb Nmid f(m)=f(n)}$$
The number of equivalence classes corresponds with the cardinality of the image of $f$.
This because the classes take the shape: $${ninmathbb Nmid f(n)=i}$$ where $i$ ranges over the image of $f$.
More generally any equivalence relation $R$ on any set $X$ can be presented like this.
If $P$ denotes the set of equivalence classes of $R$ then we can prescribe function $nu:Xto P$ by $xmapsto[x]$ where $[x]$ denotes the equivalence class that is represented by $x$.
Then automatically we have:$$forall,yin X; [xRyiff nu(x)=nu(y)]$$
answered Dec 8 '18 at 15:38
drhabdrhab
101k544130
edited Dec 8 '18 at 15:46
answered Dec 8 '18 at 15:38
drhabdrhab
101k544130
answered Dec 8 '18 at 15:38
drhabdrhab
101k544130
answered Dec 8 '18 at 15:38
drhabdrhab
101k544130
101k544130
add a comment |
add a comment |
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$begingroup$
This is a well-known equivalence relation whose elements (equivalence classes) are called fibers.
$endgroup$
– Wuestenfux
Dec 8 '18 at 15:26
$begingroup$
Is my understanding of it good ? The relation would be ${<0,0>,<1,1>,<2,2><3,3>}$ and its an equivalence relation but since we aren't given any specific function we cannot define the differents equivalence classes ? thank you
$endgroup$
– Dany Pépin
Dec 8 '18 at 15:29