Mixing
Is this relation symmetric, anti-symmetric or neither? [on hold]
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The relation R on the set ℤ is defined by the rule R = {(x,y) ∈ ℤ : xy+y is even}.
For example: (5,3) ∈ ℤ, and (3,5)∈ ℤ, both would be even so this would be symmetric.
For a counterexample: (3,4) ∈ ℤ but (4,3) would not be an element of the set since 4(3)+3= 15, so this example would be anti-symmetric. Would that mean neither is the answer since it has symmetric and anti-symmetric examples?
discrete-mathematics
asked 2 days ago
happysaint
11
put on hold as unclear what you're asking by amWhy, Shailesh, Leucippus, max_zorn, KReiser 2 days ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
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The relation R on the set ℤ is defined by the rule R = {(x,y) ∈ ℤ : xy+y is even}.
For example: (5,3) ∈ ℤ, and (3,5)∈ ℤ, both would be even so this would be symmetric.
For a counterexample: (3,4) ∈ ℤ but (4,3) would not be an element of the set since 4(3)+3= 15, so this example would be anti-symmetric. Would that mean neither is the answer since it has symmetric and anti-symmetric examples?
discrete-mathematics
asked 2 days ago
happysaint
11
put on hold as unclear what you're asking by amWhy, Shailesh, Leucippus, max_zorn, KReiser 2 days ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
I think you mean "(5,3) ∈ R, and (3,5)∈ R"
– Arthur
2 days ago
Also, in defining $R$ we should say $(x,y) in mathbb Z times mathbb Z$ or something equivalent.
– hardmath
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The relation R on the set ℤ is defined by the rule R = {(x,y) ∈ ℤ : xy+y is even}.
For example: (5,3) ∈ ℤ, and (3,5)∈ ℤ, both would be even so this would be symmetric.
For a counterexample: (3,4) ∈ ℤ but (4,3) would not be an element of the set since 4(3)+3= 15, so this example would be anti-symmetric. Would that mean neither is the answer since it has symmetric and anti-symmetric examples?
discrete-mathematics
asked 2 days ago
happysaint
11
The relation R on the set ℤ is defined by the rule R = {(x,y) ∈ ℤ : xy+y is even}.
For example: (5,3) ∈ ℤ, and (3,5)∈ ℤ, both would be even so this would be symmetric.
For a counterexample: (3,4) ∈ ℤ but (4,3) would not be an element of the set since 4(3)+3= 15, so this example would be anti-symmetric. Would that mean neither is the answer since it has symmetric and anti-symmetric examples?
discrete-mathematics
discrete-mathematics
asked 2 days ago
happysaint
11
asked 2 days ago
happysaint
11
asked 2 days ago
happysaint
11
asked 2 days ago
happysaint
11
asked 2 days ago
happysaint
11
11
put on hold as unclear what you're asking by amWhy, Shailesh, Leucippus, max_zorn, KReiser 2 days ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by amWhy, Shailesh, Leucippus, max_zorn, KReiser 2 days ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
I think you mean "(5,3) ∈ R, and (3,5)∈ R"
– Arthur
2 days ago
Also, in defining $R$ we should say $(x,y) in mathbb Z times mathbb Z$ or something equivalent.
– hardmath
2 days ago
add a comment |
I think you mean "(5,3) ∈ R, and (3,5)∈ R"
– Arthur
2 days ago
Also, in defining $R$ we should say $(x,y) in mathbb Z times mathbb Z$ or something equivalent.
– hardmath
2 days ago
I think you mean "(5,3) ∈ R, and (3,5)∈ R"
– Arthur
2 days ago
I think you mean "(5,3) ∈ R, and (3,5)∈ R"
– Arthur
2 days ago
Also, in defining $R$ we should say $(x,y) in mathbb Z times mathbb Z$ or something equivalent.
– hardmath
2 days ago
Also, in defining $R$ we should say $(x,y) in mathbb Z times mathbb Z$ or something equivalent.
– hardmath
2 days ago
add a comment |
2 Answers
2
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up vote
1
down vote
It is not symmetric because
$$(3,4)in R text{ but } (4,3)notin R$$
it is not antisymmetric because
$$(2,4)in R text{ and } (4,2)in R text{ and } 2ne 4$$
edited 2 days ago
amWhy
191k27223437
answered 2 days ago
hamam_Abdallah
36.5k21533
Simply put. +1.
– amWhy
2 days ago
add a comment |
up vote
0
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Yes, it would mean that it is neither. You've found one counterexample pair $(5,3)$ which shows that the relation isn't antisymmetric and one pair $(4,3)$ which shows it isn't symmetric.
Don't think of a single pair like $(5,3)$ as symmetric and $(4,3)$ as antisymmetric. A whole relation can have such a property, but not a single pair (or a pair of pairs). A single pair can, however, disprove that a relation has such a property, like in this case.
answered 2 days ago
Arthur
108k7103186
The OP hasn't proved that the relation isn't antisymmetric. It can be implied by the fact that $(3, 5) in R$ and $(5, 3) in R$, but $3neq 5$, but neither the OP, nor you, showed explicitly that hence the relation is antisymmetric.
– amWhy
2 days ago
@amWhy $5R3$ and $3R5$ but $5neq3$ doesn't disprove antisymmetry? (He means $in R$, not $in Bbb Z$, presumably.) Have I missed something?
– Arthur
2 days ago
I don't see the OP nor you explicitly stating that the presence of $(3, 5), (5, 3) in R$ means the relation is antisymmetric. Neither of you states the definition of antisymmetry, nor that this pair shows a violation because $3neq 5$. You did so in a comment. If you put your comment in your answer, I'll think more favorably on your answer.
– amWhy
2 days ago
@amWhy He uses non-conventional and not entirely correct terminology, but to me it's clearly there.
– Arthur
2 days ago
The OP suggests that because $(3, 4) in R$ but not $(4, 3)$, that makes it antisymmetric. You have not addressed this misunderstanding about what anit-symmetric means.
– amWhy
2 days ago
|
show 4 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
It is not symmetric because
$$(3,4)in R text{ but } (4,3)notin R$$
it is not antisymmetric because
$$(2,4)in R text{ and } (4,2)in R text{ and } 2ne 4$$
edited 2 days ago
amWhy
191k27223437
answered 2 days ago
hamam_Abdallah
36.5k21533
Simply put. +1.
– amWhy
2 days ago
add a comment |
up vote
1
down vote
It is not symmetric because
$$(3,4)in R text{ but } (4,3)notin R$$
it is not antisymmetric because
$$(2,4)in R text{ and } (4,2)in R text{ and } 2ne 4$$
edited 2 days ago
amWhy
191k27223437
answered 2 days ago
hamam_Abdallah
36.5k21533
Simply put. +1.
– amWhy
2 days ago
add a comment |
up vote
1
down vote
up vote
1
down vote
It is not symmetric because
$$(3,4)in R text{ but } (4,3)notin R$$
it is not antisymmetric because
$$(2,4)in R text{ and } (4,2)in R text{ and } 2ne 4$$
edited 2 days ago
amWhy
191k27223437
answered 2 days ago
hamam_Abdallah
36.5k21533
It is not symmetric because
$$(3,4)in R text{ but } (4,3)notin R$$
it is not antisymmetric because
$$(2,4)in R text{ and } (4,2)in R text{ and } 2ne 4$$
edited 2 days ago
amWhy
191k27223437
answered 2 days ago
hamam_Abdallah
36.5k21533
edited 2 days ago
amWhy
191k27223437
edited 2 days ago
amWhy
191k27223437
edited 2 days ago
amWhy
191k27223437
191k27223437
answered 2 days ago
hamam_Abdallah
36.5k21533
answered 2 days ago
hamam_Abdallah
36.5k21533
answered 2 days ago
hamam_Abdallah
36.5k21533
36.5k21533
Simply put. +1.
– amWhy
2 days ago
add a comment |
Simply put. +1.
– amWhy
2 days ago
Simply put. +1.
– amWhy
2 days ago
Simply put. +1.
– amWhy
2 days ago
add a comment |
up vote
0
down vote
Yes, it would mean that it is neither. You've found one counterexample pair $(5,3)$ which shows that the relation isn't antisymmetric and one pair $(4,3)$ which shows it isn't symmetric.
Don't think of a single pair like $(5,3)$ as symmetric and $(4,3)$ as antisymmetric. A whole relation can have such a property, but not a single pair (or a pair of pairs). A single pair can, however, disprove that a relation has such a property, like in this case.
answered 2 days ago
Arthur
108k7103186
The OP hasn't proved that the relation isn't antisymmetric. It can be implied by the fact that $(3, 5) in R$ and $(5, 3) in R$, but $3neq 5$, but neither the OP, nor you, showed explicitly that hence the relation is antisymmetric.
– amWhy
2 days ago
@amWhy $5R3$ and $3R5$ but $5neq3$ doesn't disprove antisymmetry? (He means $in R$, not $in Bbb Z$, presumably.) Have I missed something?
– Arthur
2 days ago
I don't see the OP nor you explicitly stating that the presence of $(3, 5), (5, 3) in R$ means the relation is antisymmetric. Neither of you states the definition of antisymmetry, nor that this pair shows a violation because $3neq 5$. You did so in a comment. If you put your comment in your answer, I'll think more favorably on your answer.
– amWhy
2 days ago
@amWhy He uses non-conventional and not entirely correct terminology, but to me it's clearly there.
– Arthur
2 days ago
The OP suggests that because $(3, 4) in R$ but not $(4, 3)$, that makes it antisymmetric. You have not addressed this misunderstanding about what anit-symmetric means.
– amWhy
2 days ago
|
show 4 more comments
up vote
0
down vote
Yes, it would mean that it is neither. You've found one counterexample pair $(5,3)$ which shows that the relation isn't antisymmetric and one pair $(4,3)$ which shows it isn't symmetric.
Don't think of a single pair like $(5,3)$ as symmetric and $(4,3)$ as antisymmetric. A whole relation can have such a property, but not a single pair (or a pair of pairs). A single pair can, however, disprove that a relation has such a property, like in this case.
answered 2 days ago
Arthur
108k7103186
The OP hasn't proved that the relation isn't antisymmetric. It can be implied by the fact that $(3, 5) in R$ and $(5, 3) in R$, but $3neq 5$, but neither the OP, nor you, showed explicitly that hence the relation is antisymmetric.
– amWhy
2 days ago
@amWhy $5R3$ and $3R5$ but $5neq3$ doesn't disprove antisymmetry? (He means $in R$, not $in Bbb Z$, presumably.) Have I missed something?
– Arthur
2 days ago
I don't see the OP nor you explicitly stating that the presence of $(3, 5), (5, 3) in R$ means the relation is antisymmetric. Neither of you states the definition of antisymmetry, nor that this pair shows a violation because $3neq 5$. You did so in a comment. If you put your comment in your answer, I'll think more favorably on your answer.
– amWhy
2 days ago
@amWhy He uses non-conventional and not entirely correct terminology, but to me it's clearly there.
– Arthur
2 days ago
The OP suggests that because $(3, 4) in R$ but not $(4, 3)$, that makes it antisymmetric. You have not addressed this misunderstanding about what anit-symmetric means.
– amWhy
2 days ago
|
show 4 more comments
up vote
0
down vote
up vote
0
down vote
Yes, it would mean that it is neither. You've found one counterexample pair $(5,3)$ which shows that the relation isn't antisymmetric and one pair $(4,3)$ which shows it isn't symmetric.
Don't think of a single pair like $(5,3)$ as symmetric and $(4,3)$ as antisymmetric. A whole relation can have such a property, but not a single pair (or a pair of pairs). A single pair can, however, disprove that a relation has such a property, like in this case.
answered 2 days ago
Arthur
108k7103186
Yes, it would mean that it is neither. You've found one counterexample pair $(5,3)$ which shows that the relation isn't antisymmetric and one pair $(4,3)$ which shows it isn't symmetric.
Don't think of a single pair like $(5,3)$ as symmetric and $(4,3)$ as antisymmetric. A whole relation can have such a property, but not a single pair (or a pair of pairs). A single pair can, however, disprove that a relation has such a property, like in this case.
answered 2 days ago
Arthur
108k7103186
edited 2 days ago
answered 2 days ago
Arthur
108k7103186
answered 2 days ago
Arthur
108k7103186
answered 2 days ago
Arthur
108k7103186
108k7103186
The OP hasn't proved that the relation isn't antisymmetric. It can be implied by the fact that $(3, 5) in R$ and $(5, 3) in R$, but $3neq 5$, but neither the OP, nor you, showed explicitly that hence the relation is antisymmetric.
– amWhy
2 days ago
@amWhy $5R3$ and $3R5$ but $5neq3$ doesn't disprove antisymmetry? (He means $in R$, not $in Bbb Z$, presumably.) Have I missed something?
– Arthur
2 days ago
I don't see the OP nor you explicitly stating that the presence of $(3, 5), (5, 3) in R$ means the relation is antisymmetric. Neither of you states the definition of antisymmetry, nor that this pair shows a violation because $3neq 5$. You did so in a comment. If you put your comment in your answer, I'll think more favorably on your answer.
– amWhy
2 days ago
@amWhy He uses non-conventional and not entirely correct terminology, but to me it's clearly there.
– Arthur
2 days ago
The OP suggests that because $(3, 4) in R$ but not $(4, 3)$, that makes it antisymmetric. You have not addressed this misunderstanding about what anit-symmetric means.
– amWhy
2 days ago
|
show 4 more comments
The OP hasn't proved that the relation isn't antisymmetric. It can be implied by the fact that $(3, 5) in R$ and $(5, 3) in R$, but $3neq 5$, but neither the OP, nor you, showed explicitly that hence the relation is antisymmetric.
– amWhy
2 days ago
@amWhy $5R3$ and $3R5$ but $5neq3$ doesn't disprove antisymmetry? (He means $in R$, not $in Bbb Z$, presumably.) Have I missed something?
– Arthur
2 days ago
I don't see the OP nor you explicitly stating that the presence of $(3, 5), (5, 3) in R$ means the relation is antisymmetric. Neither of you states the definition of antisymmetry, nor that this pair shows a violation because $3neq 5$. You did so in a comment. If you put your comment in your answer, I'll think more favorably on your answer.
– amWhy
2 days ago
@amWhy He uses non-conventional and not entirely correct terminology, but to me it's clearly there.
– Arthur
2 days ago
The OP suggests that because $(3, 4) in R$ but not $(4, 3)$, that makes it antisymmetric. You have not addressed this misunderstanding about what anit-symmetric means.
– amWhy
2 days ago
The OP hasn't proved that the relation isn't antisymmetric. It can be implied by the fact that $(3, 5) in R$ and $(5, 3) in R$, but $3neq 5$, but neither the OP, nor you, showed explicitly that hence the relation is antisymmetric.
– amWhy
2 days ago
The OP hasn't proved that the relation isn't antisymmetric. It can be implied by the fact that $(3, 5) in R$ and $(5, 3) in R$, but $3neq 5$, but neither the OP, nor you, showed explicitly that hence the relation is antisymmetric.
– amWhy
2 days ago
@amWhy $5R3$ and $3R5$ but $5neq3$ doesn't disprove antisymmetry? (He means $in R$, not $in Bbb Z$, presumably.) Have I missed something?
– Arthur
2 days ago
@amWhy $5R3$ and $3R5$ but $5neq3$ doesn't disprove antisymmetry? (He means $in R$, not $in Bbb Z$, presumably.) Have I missed something?
– Arthur
2 days ago
I don't see the OP nor you explicitly stating that the presence of $(3, 5), (5, 3) in R$ means the relation is antisymmetric. Neither of you states the definition of antisymmetry, nor that this pair shows a violation because $3neq 5$. You did so in a comment. If you put your comment in your answer, I'll think more favorably on your answer.
– amWhy
2 days ago
I don't see the OP nor you explicitly stating that the presence of $(3, 5), (5, 3) in R$ means the relation is antisymmetric. Neither of you states the definition of antisymmetry, nor that this pair shows a violation because $3neq 5$. You did so in a comment. If you put your comment in your answer, I'll think more favorably on your answer.
– amWhy
2 days ago
@amWhy He uses non-conventional and not entirely correct terminology, but to me it's clearly there.
– Arthur
2 days ago
@amWhy He uses non-conventional and not entirely correct terminology, but to me it's clearly there.
– Arthur
2 days ago
The OP suggests that because $(3, 4) in R$ but not $(4, 3)$, that makes it antisymmetric. You have not addressed this misunderstanding about what anit-symmetric means.
– amWhy
2 days ago
The OP suggests that because $(3, 4) in R$ but not $(4, 3)$, that makes it antisymmetric. You have not addressed this misunderstanding about what anit-symmetric means.
– amWhy
2 days ago
|
show 4 more comments
I think you mean "(5,3) ∈ R, and (3,5)∈ R"
– Arthur
2 days ago
Also, in defining $R$ we should say $(x,y) in mathbb Z times mathbb Z$ or something equivalent.
– hardmath
2 days ago