How single elements belong to partition though they can't satisfy equivalence relation? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Understanding equivalence class, equivalence relation, partitionWhy is the term “composition” used to mean a certain binary operation on the set of relations on a given set?Partition and equivalence relationFinding an Equivalence Relation from a Partition?Symmetricity of composition of equivalence relationsAre equivalence relations total, i.e. are all elements of an equivalence relation related one to each other?How many classes does the equivalence relation partition the set?Show that $xRy$ if and only if $x, y$ are in the same part of the partition defines an equivalence relation.Equivalence relation gives unique partition?How to think intuitively quotient of quotient set in equivalence relations
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How single elements belong to partition though they can't satisfy equivalence relation?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Understanding equivalence class, equivalence relation, partitionWhy is the term “composition” used to mean a certain binary operation on the set of relations on a given set?Partition and equivalence relationFinding an Equivalence Relation from a Partition?Symmetricity of composition of equivalence relationsAre equivalence relations total, i.e. are all elements of an equivalence relation related one to each other?How many classes does the equivalence relation partition the set?Show that $xRy$ if and only if $x, y$ are in the same part of the partition defines an equivalence relation.Equivalence relation gives unique partition?How to think intuitively quotient of quotient set in equivalence relations
$begingroup$
Let $A = 2, 3, 5, 15$ and $G$ is the equivalence relation of elements divisible by 3 and $H$ is the equivalence relation divisible by 5.
Now the quotient set $A/G = 3, 15, 5, 2,$
quotient set $A/H = 5, 15, 3, 2$.
1. 3$G$15 is valid relation, but how come 5$G$5 and 2$G$2 belongs to quotient set or partition though they are not divisible by 3 ? and what is the intuition benhind including elements those doesn't satisfy the relation ?
2.If we have to calculate the composition of relations,i.e, $(Gcirc H)$ do we have to consider single elements that doesn't satisfy the relation?
To simplify, what will be the value of $(Gcirc H)$ ?
a). $Gcirc H = 5, 15, 5, 3$
or,
b). $Gcirc H = 2, 2, 3, 3, 5, 5, 15, 15, 15, 3, 5, 15, 5, 3, 15, 5 $
3. What is the quotient set $A/(Gcirc H)$ ?
elementary-set-theory relations equivalence-relations quotient-group
asked Mar 31 at 20:01
krishnakrishna
12
$endgroup$
add a comment |
$begingroup$
Let $A = 2, 3, 5, 15$ and $G$ is the equivalence relation of elements divisible by 3 and $H$ is the equivalence relation divisible by 5.
Now the quotient set $A/G = 3, 15, 5, 2,$
quotient set $A/H = 5, 15, 3, 2$.
1. 3$G$15 is valid relation, but how come 5$G$5 and 2$G$2 belongs to quotient set or partition though they are not divisible by 3 ? and what is the intuition benhind including elements those doesn't satisfy the relation ?
2.If we have to calculate the composition of relations,i.e, $(Gcirc H)$ do we have to consider single elements that doesn't satisfy the relation?
To simplify, what will be the value of $(Gcirc H)$ ?
a). $Gcirc H = 5, 15, 5, 3$
or,
b). $Gcirc H = 2, 2, 3, 3, 5, 5, 15, 15, 15, 3, 5, 15, 5, 3, 15, 5 $
3. What is the quotient set $A/(Gcirc H)$ ?
elementary-set-theory relations equivalence-relations quotient-group
asked Mar 31 at 20:01
krishnakrishna
12
$endgroup$
5
$begingroup$
"elements divisible by 3" does not define an equivalence relation. What is probably intended is that the difference between two elements is divisible by $3$. Then $2G2$ is true, since $2-2$ is divisible by $3$.
$endgroup$
– FredH
Mar 31 at 20:09
1
$begingroup$
The comment by @FredH is correct about the first sentence of the question being wrong. I think his guess about what's intended is also correct, but if it is then $2$ and $5$ are equivalent and $A/G$ should be $3,15,2,5$.
$endgroup$
– Andreas Blass
Mar 31 at 20:12
add a comment |
$begingroup$
Let $A = 2, 3, 5, 15$ and $G$ is the equivalence relation of elements divisible by 3 and $H$ is the equivalence relation divisible by 5.
Now the quotient set $A/G = 3, 15, 5, 2,$
quotient set $A/H = 5, 15, 3, 2$.
1. 3$G$15 is valid relation, but how come 5$G$5 and 2$G$2 belongs to quotient set or partition though they are not divisible by 3 ? and what is the intuition benhind including elements those doesn't satisfy the relation ?
2.If we have to calculate the composition of relations,i.e, $(Gcirc H)$ do we have to consider single elements that doesn't satisfy the relation?
To simplify, what will be the value of $(Gcirc H)$ ?
a). $Gcirc H = 5, 15, 5, 3$
or,
b). $Gcirc H = 2, 2, 3, 3, 5, 5, 15, 15, 15, 3, 5, 15, 5, 3, 15, 5 $
3. What is the quotient set $A/(Gcirc H)$ ?
elementary-set-theory relations equivalence-relations quotient-group
asked Mar 31 at 20:01
krishnakrishna
12
$endgroup$
Let $A = 2, 3, 5, 15$ and $G$ is the equivalence relation of elements divisible by 3 and $H$ is the equivalence relation divisible by 5.
Now the quotient set $A/G = 3, 15, 5, 2,$
quotient set $A/H = 5, 15, 3, 2$.
1. 3$G$15 is valid relation, but how come 5$G$5 and 2$G$2 belongs to quotient set or partition though they are not divisible by 3 ? and what is the intuition benhind including elements those doesn't satisfy the relation ?
2.If we have to calculate the composition of relations,i.e, $(Gcirc H)$ do we have to consider single elements that doesn't satisfy the relation?
To simplify, what will be the value of $(Gcirc H)$ ?
a). $Gcirc H = 5, 15, 5, 3$
or,
b). $Gcirc H = 2, 2, 3, 3, 5, 5, 15, 15, 15, 3, 5, 15, 5, 3, 15, 5 $
3. What is the quotient set $A/(Gcirc H)$ ?
elementary-set-theory relations equivalence-relations quotient-group
elementary-set-theory relations equivalence-relations quotient-group
asked Mar 31 at 20:01
krishnakrishna
12
asked Mar 31 at 20:01
krishnakrishna
12
asked Mar 31 at 20:01
krishnakrishna
12
asked Mar 31 at 20:01
krishnakrishna
12
asked Mar 31 at 20:01
krishnakrishna
12
12
5
$begingroup$
"elements divisible by 3" does not define an equivalence relation. What is probably intended is that the difference between two elements is divisible by $3$. Then $2G2$ is true, since $2-2$ is divisible by $3$.
$endgroup$
– FredH
Mar 31 at 20:09
1
$begingroup$
The comment by @FredH is correct about the first sentence of the question being wrong. I think his guess about what's intended is also correct, but if it is then $2$ and $5$ are equivalent and $A/G$ should be $3,15,2,5$.
$endgroup$
– Andreas Blass
Mar 31 at 20:12
add a comment |
5
$begingroup$
"elements divisible by 3" does not define an equivalence relation. What is probably intended is that the difference between two elements is divisible by $3$. Then $2G2$ is true, since $2-2$ is divisible by $3$.
$endgroup$
– FredH
Mar 31 at 20:09
1
$begingroup$
The comment by @FredH is correct about the first sentence of the question being wrong. I think his guess about what's intended is also correct, but if it is then $2$ and $5$ are equivalent and $A/G$ should be $3,15,2,5$.
$endgroup$
– Andreas Blass
Mar 31 at 20:12
5
5
$begingroup$
"elements divisible by 3" does not define an equivalence relation. What is probably intended is that the difference between two elements is divisible by $3$. Then $2G2$ is true, since $2-2$ is divisible by $3$.
$endgroup$
– FredH
Mar 31 at 20:09
$begingroup$
"elements divisible by 3" does not define an equivalence relation. What is probably intended is that the difference between two elements is divisible by $3$. Then $2G2$ is true, since $2-2$ is divisible by $3$.
$endgroup$
– FredH
Mar 31 at 20:09
1
1
$begingroup$
The comment by @FredH is correct about the first sentence of the question being wrong. I think his guess about what's intended is also correct, but if it is then $2$ and $5$ are equivalent and $A/G$ should be $3,15,2,5$.
$endgroup$
– Andreas Blass
Mar 31 at 20:12
$begingroup$
The comment by @FredH is correct about the first sentence of the question being wrong. I think his guess about what's intended is also correct, but if it is then $2$ and $5$ are equivalent and $A/G$ should be $3,15,2,5$.
$endgroup$
– Andreas Blass
Mar 31 at 20:12
add a comment |
0
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5
$begingroup$
"elements divisible by 3" does not define an equivalence relation. What is probably intended is that the difference between two elements is divisible by $3$. Then $2G2$ is true, since $2-2$ is divisible by $3$.
$endgroup$
– FredH
Mar 31 at 20:09
1
$begingroup$
The comment by @FredH is correct about the first sentence of the question being wrong. I think his guess about what's intended is also correct, but if it is then $2$ and $5$ are equivalent and $A/G$ should be $3,15,2,5$.
$endgroup$
– Andreas Blass
Mar 31 at 20:12