Find all non-trivial congruence relations on $(mathbbZ,+,0,-)$. The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What can we actually do with congruence relations, specifically?What does “induced operations” means in congruence operationsFinding all non trivial congruence relations on (N, +)What are the properties of congruent classes over symmetric matrices?Substitution in congruence relationsProof verification: Show that the set of congruence classes form a partition of $mathbbZ$.How do I find the least x that satisfies this congruence properties?Meaning of $phi / theta$ where $phi$,$theta$ are congruence relationsObjects of a congruence categoryFinding an equivalence relation that isn't a congruence.
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Find all non-trivial congruence relations on $(mathbbZ,+,0,-)$.
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What can we actually do with congruence relations, specifically?What does “induced operations” means in congruence operationsFinding all non trivial congruence relations on (N, +)What are the properties of congruent classes over symmetric matrices?Substitution in congruence relationsProof verification: Show that the set of congruence classes form a partition of $mathbbZ$.How do I find the least x that satisfies this congruence properties?Meaning of $phi / theta$ where $phi$,$theta$ are congruence relationsObjects of a congruence categoryFinding an equivalence relation that isn't a congruence.
$begingroup$
First of all, I have already proven that for every $dinmathbbN$ the relation
beginalign*
asim_d b:Leftrightarrowexists kinmathbbZ:,b-a=kd
endalign*
is a congruence relation on $(mathbbZ,+,0,-)$. Furthermore, the trivial ones $mathbbZtimesmathbbZ$ as well as $rm id_mathbbZ$ are congruence relations.
Are those all of them and how do I prove it?
abstract-algebra modular-arithmetic congruence-relations
asked Mar 31 at 13:51
Lemma 5Lemma 5
63
$endgroup$
add a comment |
$begingroup$
First of all, I have already proven that for every $dinmathbbN$ the relation
beginalign*
asim_d b:Leftrightarrowexists kinmathbbZ:,b-a=kd
endalign*
is a congruence relation on $(mathbbZ,+,0,-)$. Furthermore, the trivial ones $mathbbZtimesmathbbZ$ as well as $rm id_mathbbZ$ are congruence relations.
Are those all of them and how do I prove it?
abstract-algebra modular-arithmetic congruence-relations
asked Mar 31 at 13:51
Lemma 5Lemma 5
63
$endgroup$
$begingroup$
What do you mean exactly by a congruence relation on $(Bbb Z,+,0,-)$?
$endgroup$
– Saucy O'Path
Mar 31 at 14:25
$begingroup$
An equivalence relation that is compatible with all the operations on the algebraic structure (in this case $(mathbbZ,+,0,-)$). For the operation "$+$" this means $a_1sim b_1 land a_2sim b_2 Rightarrow a_1+a_2 sim b_1+b_2$. For "$-$" this means $asim bRightarrow -asim -b$. For "$0$" this just means $0sim 0$, which is trivial since $sim$ is reflexive.
$endgroup$
– Lemma 5
Mar 31 at 15:08
$begingroup$
Ok. For the record, the "trivial ones" you indicated are already included in the previous case by considering $d=1$ and $d=0$ respectively.
$endgroup$
– Saucy O'Path
Mar 31 at 15:39
$begingroup$
Indeed, I haven't noticed that in the first place :)
$endgroup$
– Lemma 5
Mar 31 at 16:01
add a comment |
$begingroup$
First of all, I have already proven that for every $dinmathbbN$ the relation
beginalign*
asim_d b:Leftrightarrowexists kinmathbbZ:,b-a=kd
endalign*
is a congruence relation on $(mathbbZ,+,0,-)$. Furthermore, the trivial ones $mathbbZtimesmathbbZ$ as well as $rm id_mathbbZ$ are congruence relations.
Are those all of them and how do I prove it?
abstract-algebra modular-arithmetic congruence-relations
asked Mar 31 at 13:51
Lemma 5Lemma 5
63
$endgroup$
First of all, I have already proven that for every $dinmathbbN$ the relation
beginalign*
asim_d b:Leftrightarrowexists kinmathbbZ:,b-a=kd
endalign*
is a congruence relation on $(mathbbZ,+,0,-)$. Furthermore, the trivial ones $mathbbZtimesmathbbZ$ as well as $rm id_mathbbZ$ are congruence relations.
Are those all of them and how do I prove it?
abstract-algebra modular-arithmetic congruence-relations
abstract-algebra modular-arithmetic congruence-relations
asked Mar 31 at 13:51
Lemma 5Lemma 5
63
asked Mar 31 at 13:51
Lemma 5Lemma 5
63
asked Mar 31 at 13:51
Lemma 5Lemma 5
63
asked Mar 31 at 13:51
Lemma 5Lemma 5
63
asked Mar 31 at 13:51
Lemma 5Lemma 5
63
63
$begingroup$
What do you mean exactly by a congruence relation on $(Bbb Z,+,0,-)$?
$endgroup$
– Saucy O'Path
Mar 31 at 14:25
$begingroup$
An equivalence relation that is compatible with all the operations on the algebraic structure (in this case $(mathbbZ,+,0,-)$). For the operation "$+$" this means $a_1sim b_1 land a_2sim b_2 Rightarrow a_1+a_2 sim b_1+b_2$. For "$-$" this means $asim bRightarrow -asim -b$. For "$0$" this just means $0sim 0$, which is trivial since $sim$ is reflexive.
$endgroup$
– Lemma 5
Mar 31 at 15:08
$begingroup$
Ok. For the record, the "trivial ones" you indicated are already included in the previous case by considering $d=1$ and $d=0$ respectively.
$endgroup$
– Saucy O'Path
Mar 31 at 15:39
$begingroup$
Indeed, I haven't noticed that in the first place :)
$endgroup$
– Lemma 5
Mar 31 at 16:01
add a comment |
$begingroup$
What do you mean exactly by a congruence relation on $(Bbb Z,+,0,-)$?
$endgroup$
– Saucy O'Path
Mar 31 at 14:25
$begingroup$
An equivalence relation that is compatible with all the operations on the algebraic structure (in this case $(mathbbZ,+,0,-)$). For the operation "$+$" this means $a_1sim b_1 land a_2sim b_2 Rightarrow a_1+a_2 sim b_1+b_2$. For "$-$" this means $asim bRightarrow -asim -b$. For "$0$" this just means $0sim 0$, which is trivial since $sim$ is reflexive.
$endgroup$
– Lemma 5
Mar 31 at 15:08
$begingroup$
Ok. For the record, the "trivial ones" you indicated are already included in the previous case by considering $d=1$ and $d=0$ respectively.
$endgroup$
– Saucy O'Path
Mar 31 at 15:39
$begingroup$
Indeed, I haven't noticed that in the first place :)
$endgroup$
– Lemma 5
Mar 31 at 16:01
$begingroup$
What do you mean exactly by a congruence relation on $(Bbb Z,+,0,-)$?
$endgroup$
– Saucy O'Path
Mar 31 at 14:25
$begingroup$
What do you mean exactly by a congruence relation on $(Bbb Z,+,0,-)$?
$endgroup$
– Saucy O'Path
Mar 31 at 14:25
$begingroup$
An equivalence relation that is compatible with all the operations on the algebraic structure (in this case $(mathbbZ,+,0,-)$). For the operation "$+$" this means $a_1sim b_1 land a_2sim b_2 Rightarrow a_1+a_2 sim b_1+b_2$. For "$-$" this means $asim bRightarrow -asim -b$. For "$0$" this just means $0sim 0$, which is trivial since $sim$ is reflexive.
$endgroup$
– Lemma 5
Mar 31 at 15:08
$begingroup$
An equivalence relation that is compatible with all the operations on the algebraic structure (in this case $(mathbbZ,+,0,-)$). For the operation "$+$" this means $a_1sim b_1 land a_2sim b_2 Rightarrow a_1+a_2 sim b_1+b_2$. For "$-$" this means $asim bRightarrow -asim -b$. For "$0$" this just means $0sim 0$, which is trivial since $sim$ is reflexive.
$endgroup$
– Lemma 5
Mar 31 at 15:08
$begingroup$
Ok. For the record, the "trivial ones" you indicated are already included in the previous case by considering $d=1$ and $d=0$ respectively.
$endgroup$
– Saucy O'Path
Mar 31 at 15:39
$begingroup$
Ok. For the record, the "trivial ones" you indicated are already included in the previous case by considering $d=1$ and $d=0$ respectively.
$endgroup$
– Saucy O'Path
Mar 31 at 15:39
$begingroup$
Indeed, I haven't noticed that in the first place :)
$endgroup$
– Lemma 5
Mar 31 at 16:01
$begingroup$
Indeed, I haven't noticed that in the first place :)
$endgroup$
– Lemma 5
Mar 31 at 16:01
add a comment |
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$begingroup$
What do you mean exactly by a congruence relation on $(Bbb Z,+,0,-)$?
$endgroup$
– Saucy O'Path
Mar 31 at 14:25
$begingroup$
An equivalence relation that is compatible with all the operations on the algebraic structure (in this case $(mathbbZ,+,0,-)$). For the operation "$+$" this means $a_1sim b_1 land a_2sim b_2 Rightarrow a_1+a_2 sim b_1+b_2$. For "$-$" this means $asim bRightarrow -asim -b$. For "$0$" this just means $0sim 0$, which is trivial since $sim$ is reflexive.
$endgroup$
– Lemma 5
Mar 31 at 15:08
$begingroup$
Ok. For the record, the "trivial ones" you indicated are already included in the previous case by considering $d=1$ and $d=0$ respectively.
$endgroup$
– Saucy O'Path
Mar 31 at 15:39
$begingroup$
Indeed, I haven't noticed that in the first place :)
$endgroup$
– Lemma 5
Mar 31 at 16:01