Equivalence relation! Make formula saying that R is not an equivalence relation [closed]
$begingroup$
I don't know how to do this particular thing.
Using quantifiers and logical links as
$$and, or, =>, <=>$$
and expressions like $$xin A, xnotin A, R(x, y), lnot R(x, y)$$
Make formula saying that R is not an equivalence relation on set A.
discrete-mathematics logic relations
asked Dec 16 '18 at 12:50
KarolKarol
205
$endgroup$
closed as off-topic by José Carlos Santos, Paul Frost, Lord_Farin, amWhy, Cesareo Dec 17 '18 at 0:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Paul Frost, Lord_Farin, amWhy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I don't know how to do this particular thing.
Using quantifiers and logical links as
$$and, or, =>, <=>$$
and expressions like $$xin A, xnotin A, R(x, y), lnot R(x, y)$$
Make formula saying that R is not an equivalence relation on set A.
discrete-mathematics logic relations
asked Dec 16 '18 at 12:50
KarolKarol
205
$endgroup$
closed as off-topic by José Carlos Santos, Paul Frost, Lord_Farin, amWhy, Cesareo Dec 17 '18 at 0:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Paul Frost, Lord_Farin, amWhy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Do you know what is an equivalence relation ?
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 12:56
2
$begingroup$
A binary relation $R$ on a set $A$ is an equivalence relation iff three properties of $R$ do hold. Thus, two steps : (i) express the three properties with three formulas with quantifiers and connectives; (ii) express the fact that the conjunction of the three fromula is false.
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 12:59
$begingroup$
@MauroALLEGRANZA, hi, I know that I should show that the conjunction is false but with given things for me to use I don't know how.
$endgroup$
– Karol
Dec 16 '18 at 13:22
add a comment |
$begingroup$
I don't know how to do this particular thing.
Using quantifiers and logical links as
$$and, or, =>, <=>$$
and expressions like $$xin A, xnotin A, R(x, y), lnot R(x, y)$$
Make formula saying that R is not an equivalence relation on set A.
discrete-mathematics logic relations
asked Dec 16 '18 at 12:50
KarolKarol
205
$endgroup$
I don't know how to do this particular thing.
Using quantifiers and logical links as
$$and, or, =>, <=>$$
and expressions like $$xin A, xnotin A, R(x, y), lnot R(x, y)$$
Make formula saying that R is not an equivalence relation on set A.
discrete-mathematics logic relations
discrete-mathematics logic relations
asked Dec 16 '18 at 12:50
KarolKarol
205
asked Dec 16 '18 at 12:50
KarolKarol
205
asked Dec 16 '18 at 12:50
KarolKarol
205
asked Dec 16 '18 at 12:50
KarolKarol
205
asked Dec 16 '18 at 12:50
KarolKarol
205
205
closed as off-topic by José Carlos Santos, Paul Frost, Lord_Farin, amWhy, Cesareo Dec 17 '18 at 0:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Paul Frost, Lord_Farin, amWhy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Paul Frost, Lord_Farin, amWhy, Cesareo Dec 17 '18 at 0:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Paul Frost, Lord_Farin, amWhy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Do you know what is an equivalence relation ?
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 12:56
2
$begingroup$
A binary relation $R$ on a set $A$ is an equivalence relation iff three properties of $R$ do hold. Thus, two steps : (i) express the three properties with three formulas with quantifiers and connectives; (ii) express the fact that the conjunction of the three fromula is false.
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 12:59
$begingroup$
@MauroALLEGRANZA, hi, I know that I should show that the conjunction is false but with given things for me to use I don't know how.
$endgroup$
– Karol
Dec 16 '18 at 13:22
add a comment |
1
$begingroup$
Do you know what is an equivalence relation ?
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 12:56
2
$begingroup$
A binary relation $R$ on a set $A$ is an equivalence relation iff three properties of $R$ do hold. Thus, two steps : (i) express the three properties with three formulas with quantifiers and connectives; (ii) express the fact that the conjunction of the three fromula is false.
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 12:59
$begingroup$
@MauroALLEGRANZA, hi, I know that I should show that the conjunction is false but with given things for me to use I don't know how.
$endgroup$
– Karol
Dec 16 '18 at 13:22
1
1
$begingroup$
Do you know what is an equivalence relation ?
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 12:56
$begingroup$
Do you know what is an equivalence relation ?
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 12:56
2
2
$begingroup$
A binary relation $R$ on a set $A$ is an equivalence relation iff three properties of $R$ do hold. Thus, two steps : (i) express the three properties with three formulas with quantifiers and connectives; (ii) express the fact that the conjunction of the three fromula is false.
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 12:59
$begingroup$
A binary relation $R$ on a set $A$ is an equivalence relation iff three properties of $R$ do hold. Thus, two steps : (i) express the three properties with three formulas with quantifiers and connectives; (ii) express the fact that the conjunction of the three fromula is false.
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 12:59
$begingroup$
@MauroALLEGRANZA, hi, I know that I should show that the conjunction is false but with given things for me to use I don't know how.
$endgroup$
– Karol
Dec 16 '18 at 13:22
$begingroup$
@MauroALLEGRANZA, hi, I know that I should show that the conjunction is false but with given things for me to use I don't know how.
$endgroup$
– Karol
Dec 16 '18 at 13:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
Recall the definition of an equivalence relation $R$ on a set $A$. $R$ will be a subset of $A times A$. It also satisfies three properties:
- Reflexivity: Each element is related to itself, i.e. $(x,x) in R$.
- Symmetry: If two elements are related if and only if they're related going "both ways." (Not the best words for it but the formula clarifies it.) Explicitly, $(x,y) in R Leftrightarrow (y,x) in R$.
- Transitivity: Self-explanatory. $(x,y) in R$ and $(y,z) in R Rightarrow (x,z) in R$.
Begin by defining a subset $R subseteq A times A$ and translate the necessary properties into propositional logic and it should, within reason, all come together pretty smoothly, at least insofar as being an equivalence relation.
To be not an equivalence relation, one or more of the above properties must be false.
answered Dec 16 '18 at 12:58
Eevee TrainerEevee Trainer
6,53811237
$endgroup$
$begingroup$
Hi, I know the definition and I understand it, but even though I tried I don't know how to make such formula just using only given things that I can use
$endgroup$
– Karol
Dec 16 '18 at 13:09
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Recall the definition of an equivalence relation $R$ on a set $A$. $R$ will be a subset of $A times A$. It also satisfies three properties:
- Reflexivity: Each element is related to itself, i.e. $(x,x) in R$.
- Symmetry: If two elements are related if and only if they're related going "both ways." (Not the best words for it but the formula clarifies it.) Explicitly, $(x,y) in R Leftrightarrow (y,x) in R$.
- Transitivity: Self-explanatory. $(x,y) in R$ and $(y,z) in R Rightarrow (x,z) in R$.
Begin by defining a subset $R subseteq A times A$ and translate the necessary properties into propositional logic and it should, within reason, all come together pretty smoothly, at least insofar as being an equivalence relation.
To be not an equivalence relation, one or more of the above properties must be false.
answered Dec 16 '18 at 12:58
Eevee TrainerEevee Trainer
6,53811237
$endgroup$
$begingroup$
Hi, I know the definition and I understand it, but even though I tried I don't know how to make such formula just using only given things that I can use
$endgroup$
– Karol
Dec 16 '18 at 13:09
add a comment |
$begingroup$
Hint:
Recall the definition of an equivalence relation $R$ on a set $A$. $R$ will be a subset of $A times A$. It also satisfies three properties:
- Reflexivity: Each element is related to itself, i.e. $(x,x) in R$.
- Symmetry: If two elements are related if and only if they're related going "both ways." (Not the best words for it but the formula clarifies it.) Explicitly, $(x,y) in R Leftrightarrow (y,x) in R$.
- Transitivity: Self-explanatory. $(x,y) in R$ and $(y,z) in R Rightarrow (x,z) in R$.
Begin by defining a subset $R subseteq A times A$ and translate the necessary properties into propositional logic and it should, within reason, all come together pretty smoothly, at least insofar as being an equivalence relation.
To be not an equivalence relation, one or more of the above properties must be false.
answered Dec 16 '18 at 12:58
Eevee TrainerEevee Trainer
6,53811237
$endgroup$
$begingroup$
Hi, I know the definition and I understand it, but even though I tried I don't know how to make such formula just using only given things that I can use
$endgroup$
– Karol
Dec 16 '18 at 13:09
add a comment |
$begingroup$
Hint:
Recall the definition of an equivalence relation $R$ on a set $A$. $R$ will be a subset of $A times A$. It also satisfies three properties:
- Reflexivity: Each element is related to itself, i.e. $(x,x) in R$.
- Symmetry: If two elements are related if and only if they're related going "both ways." (Not the best words for it but the formula clarifies it.) Explicitly, $(x,y) in R Leftrightarrow (y,x) in R$.
- Transitivity: Self-explanatory. $(x,y) in R$ and $(y,z) in R Rightarrow (x,z) in R$.
Begin by defining a subset $R subseteq A times A$ and translate the necessary properties into propositional logic and it should, within reason, all come together pretty smoothly, at least insofar as being an equivalence relation.
To be not an equivalence relation, one or more of the above properties must be false.
answered Dec 16 '18 at 12:58
Eevee TrainerEevee Trainer
6,53811237
$endgroup$
Hint:
Recall the definition of an equivalence relation $R$ on a set $A$. $R$ will be a subset of $A times A$. It also satisfies three properties:
- Reflexivity: Each element is related to itself, i.e. $(x,x) in R$.
- Symmetry: If two elements are related if and only if they're related going "both ways." (Not the best words for it but the formula clarifies it.) Explicitly, $(x,y) in R Leftrightarrow (y,x) in R$.
- Transitivity: Self-explanatory. $(x,y) in R$ and $(y,z) in R Rightarrow (x,z) in R$.
Begin by defining a subset $R subseteq A times A$ and translate the necessary properties into propositional logic and it should, within reason, all come together pretty smoothly, at least insofar as being an equivalence relation.
To be not an equivalence relation, one or more of the above properties must be false.
answered Dec 16 '18 at 12:58
Eevee TrainerEevee Trainer
6,53811237
answered Dec 16 '18 at 12:58
Eevee TrainerEevee Trainer
6,53811237
answered Dec 16 '18 at 12:58
Eevee TrainerEevee Trainer
6,53811237
answered Dec 16 '18 at 12:58
Eevee TrainerEevee Trainer
6,53811237
6,53811237
$begingroup$
Hi, I know the definition and I understand it, but even though I tried I don't know how to make such formula just using only given things that I can use
$endgroup$
– Karol
Dec 16 '18 at 13:09
add a comment |
$begingroup$
Hi, I know the definition and I understand it, but even though I tried I don't know how to make such formula just using only given things that I can use
$endgroup$
– Karol
Dec 16 '18 at 13:09
$begingroup$
Hi, I know the definition and I understand it, but even though I tried I don't know how to make such formula just using only given things that I can use
$endgroup$
– Karol
Dec 16 '18 at 13:09
$begingroup$
Hi, I know the definition and I understand it, but even though I tried I don't know how to make such formula just using only given things that I can use
$endgroup$
– Karol
Dec 16 '18 at 13:09
add a comment |
1
$begingroup$
Do you know what is an equivalence relation ?
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 12:56
2
$begingroup$
A binary relation $R$ on a set $A$ is an equivalence relation iff three properties of $R$ do hold. Thus, two steps : (i) express the three properties with three formulas with quantifiers and connectives; (ii) express the fact that the conjunction of the three fromula is false.
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 12:59
$begingroup$
@MauroALLEGRANZA, hi, I know that I should show that the conjunction is false but with given things for me to use I don't know how.
$endgroup$
– Karol
Dec 16 '18 at 13:22