Finding an equivalence relation that isn't a congruence.
Let $B=S times T$ be a rectangular band such that $|S|=|T|=3$.
I've got to find an equivalence relation which is not a congruence in order to prove that at least one exists.
I've tried many different equivalence relations, but they have all turned out to be a congruence.
Any help on how I would find one would be great.
Thanks.
EDIT: The operation on $B$ is: $(s,t)(s',t')=(s,t')$
Congruence is defined as : for $((s,t),(s',t')),((x,y),(x',y'))$ contained in the the equivalence relation $R$, $((s,t)(x,y),(s',t')(x',y'))$ is also contained in $R$.
modular-arithmetic equivalence-relations semigroups
edited Nov 28 '18 at 17:40
J.-E. Pin
18.3k21754
asked Nov 28 '18 at 16:31
ASynosure
254
add a comment |
Let $B=S times T$ be a rectangular band such that $|S|=|T|=3$.
I've got to find an equivalence relation which is not a congruence in order to prove that at least one exists.
I've tried many different equivalence relations, but they have all turned out to be a congruence.
Any help on how I would find one would be great.
Thanks.
EDIT: The operation on $B$ is: $(s,t)(s',t')=(s,t')$
Congruence is defined as : for $((s,t),(s',t')),((x,y),(x',y'))$ contained in the the equivalence relation $R$, $((s,t)(x,y),(s',t')(x',y'))$ is also contained in $R$.
modular-arithmetic equivalence-relations semigroups
edited Nov 28 '18 at 17:40
J.-E. Pin
18.3k21754
asked Nov 28 '18 at 16:31
ASynosure
254
On which set are you working?
– Bernard
Nov 28 '18 at 16:33
1
What is the definition of congruence in this context?
– Hagen von Eitzen
Nov 28 '18 at 16:34
1
Which operations?
– Wuestenfux
Nov 28 '18 at 16:34
The operation on B is: (s,t)(s',t')=(s,t') Congruence is defined as : for ((s,t),(s',t')),((x,y),(x',y')) contained in the the equivalence relation r, ((s,t)(x,y),(s',t')(x',y')) is also contained in r.
– ASynosure
Nov 28 '18 at 16:40
add a comment |
Let $B=S times T$ be a rectangular band such that $|S|=|T|=3$.
I've got to find an equivalence relation which is not a congruence in order to prove that at least one exists.
I've tried many different equivalence relations, but they have all turned out to be a congruence.
Any help on how I would find one would be great.
Thanks.
EDIT: The operation on $B$ is: $(s,t)(s',t')=(s,t')$
Congruence is defined as : for $((s,t),(s',t')),((x,y),(x',y'))$ contained in the the equivalence relation $R$, $((s,t)(x,y),(s',t')(x',y'))$ is also contained in $R$.
modular-arithmetic equivalence-relations semigroups
edited Nov 28 '18 at 17:40
J.-E. Pin
18.3k21754
asked Nov 28 '18 at 16:31
ASynosure
254
Let $B=S times T$ be a rectangular band such that $|S|=|T|=3$.
I've got to find an equivalence relation which is not a congruence in order to prove that at least one exists.
I've tried many different equivalence relations, but they have all turned out to be a congruence.
Any help on how I would find one would be great.
Thanks.
EDIT: The operation on $B$ is: $(s,t)(s',t')=(s,t')$
Congruence is defined as : for $((s,t),(s',t')),((x,y),(x',y'))$ contained in the the equivalence relation $R$, $((s,t)(x,y),(s',t')(x',y'))$ is also contained in $R$.
modular-arithmetic equivalence-relations semigroups
modular-arithmetic equivalence-relations semigroups
edited Nov 28 '18 at 17:40
J.-E. Pin
18.3k21754
asked Nov 28 '18 at 16:31
ASynosure
254
edited Nov 28 '18 at 17:40
J.-E. Pin
18.3k21754
asked Nov 28 '18 at 16:31
ASynosure
254
edited Nov 28 '18 at 17:40
J.-E. Pin
18.3k21754
edited Nov 28 '18 at 17:40
J.-E. Pin
18.3k21754
edited Nov 28 '18 at 17:40
J.-E. Pin
18.3k21754
18.3k21754
asked Nov 28 '18 at 16:31
ASynosure
254
asked Nov 28 '18 at 16:31
ASynosure
254
asked Nov 28 '18 at 16:31
ASynosure
254
254
On which set are you working?
– Bernard
Nov 28 '18 at 16:33
1
What is the definition of congruence in this context?
– Hagen von Eitzen
Nov 28 '18 at 16:34
1
Which operations?
– Wuestenfux
Nov 28 '18 at 16:34
The operation on B is: (s,t)(s',t')=(s,t') Congruence is defined as : for ((s,t),(s',t')),((x,y),(x',y')) contained in the the equivalence relation r, ((s,t)(x,y),(s',t')(x',y')) is also contained in r.
– ASynosure
Nov 28 '18 at 16:40
add a comment |
On which set are you working?
– Bernard
Nov 28 '18 at 16:33
1
What is the definition of congruence in this context?
– Hagen von Eitzen
Nov 28 '18 at 16:34
1
Which operations?
– Wuestenfux
Nov 28 '18 at 16:34
The operation on B is: (s,t)(s',t')=(s,t') Congruence is defined as : for ((s,t),(s',t')),((x,y),(x',y')) contained in the the equivalence relation r, ((s,t)(x,y),(s',t')(x',y')) is also contained in r.
– ASynosure
Nov 28 '18 at 16:40
On which set are you working?
– Bernard
Nov 28 '18 at 16:33
On which set are you working?
– Bernard
Nov 28 '18 at 16:33
1
1
What is the definition of congruence in this context?
– Hagen von Eitzen
Nov 28 '18 at 16:34
What is the definition of congruence in this context?
– Hagen von Eitzen
Nov 28 '18 at 16:34
1
1
Which operations?
– Wuestenfux
Nov 28 '18 at 16:34
Which operations?
– Wuestenfux
Nov 28 '18 at 16:34
The operation on B is: (s,t)(s',t')=(s,t') Congruence is defined as : for ((s,t),(s',t')),((x,y),(x',y')) contained in the the equivalence relation r, ((s,t)(x,y),(s',t')(x',y')) is also contained in r.
– ASynosure
Nov 28 '18 at 16:40
The operation on B is: (s,t)(s',t')=(s,t') Congruence is defined as : for ((s,t),(s',t')),((x,y),(x',y')) contained in the the equivalence relation r, ((s,t)(x,y),(s',t')(x',y')) is also contained in r.
– ASynosure
Nov 28 '18 at 16:40
add a comment |
1 Answer
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Hint. Try to find a counterexample when $|S| = |T| = 2$, this is easier. An equivalence relation on a set defines a partition of the set. Just find a partition of $S times T$ into two sets which does not define a congruence on $S times T$.
answered Nov 28 '18 at 17:46
J.-E. Pin
18.3k21754
add a comment |
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Hint. Try to find a counterexample when $|S| = |T| = 2$, this is easier. An equivalence relation on a set defines a partition of the set. Just find a partition of $S times T$ into two sets which does not define a congruence on $S times T$.
answered Nov 28 '18 at 17:46
J.-E. Pin
18.3k21754
add a comment |
Hint. Try to find a counterexample when $|S| = |T| = 2$, this is easier. An equivalence relation on a set defines a partition of the set. Just find a partition of $S times T$ into two sets which does not define a congruence on $S times T$.
answered Nov 28 '18 at 17:46
J.-E. Pin
18.3k21754
add a comment |
Hint. Try to find a counterexample when $|S| = |T| = 2$, this is easier. An equivalence relation on a set defines a partition of the set. Just find a partition of $S times T$ into two sets which does not define a congruence on $S times T$.
answered Nov 28 '18 at 17:46
J.-E. Pin
18.3k21754
Hint. Try to find a counterexample when $|S| = |T| = 2$, this is easier. An equivalence relation on a set defines a partition of the set. Just find a partition of $S times T$ into two sets which does not define a congruence on $S times T$.
answered Nov 28 '18 at 17:46
J.-E. Pin
18.3k21754
answered Nov 28 '18 at 17:46
J.-E. Pin
18.3k21754
answered Nov 28 '18 at 17:46
J.-E. Pin
18.3k21754
answered Nov 28 '18 at 17:46
J.-E. Pin
18.3k21754
18.3k21754
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On which set are you working?
– Bernard
Nov 28 '18 at 16:33
1
What is the definition of congruence in this context?
– Hagen von Eitzen
Nov 28 '18 at 16:34
1
Which operations?
– Wuestenfux
Nov 28 '18 at 16:34
The operation on B is: (s,t)(s',t')=(s,t') Congruence is defined as : for ((s,t),(s',t')),((x,y),(x',y')) contained in the the equivalence relation r, ((s,t)(x,y),(s',t')(x',y')) is also contained in r.
– ASynosure
Nov 28 '18 at 16:40