Is this relation symmetric, anti-symmetric or neither? [closed]
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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 Nov 19 at 20:21
happysaint
62
closed as unclear what you're asking by amWhy, Shailesh, Leucippus, max_zorn, KReiser Nov 20 at 6:57
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
0
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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 Nov 19 at 20:21
happysaint
62
closed as unclear what you're asking by amWhy, Shailesh, Leucippus, max_zorn, KReiser Nov 20 at 6:57
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
Nov 19 at 20:35
Also, in defining $R$ we should say $(x,y) in mathbb Z times mathbb Z$ or something equivalent.
– hardmath
Nov 20 at 3:30
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 Nov 19 at 20:21
happysaint
62
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 Nov 19 at 20:21
happysaint
62
asked Nov 19 at 20:21
happysaint
62
asked Nov 19 at 20:21
happysaint
62
asked Nov 19 at 20:21
happysaint
62
asked Nov 19 at 20:21
happysaint
62
62
closed as unclear what you're asking by amWhy, Shailesh, Leucippus, max_zorn, KReiser Nov 20 at 6:57
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.
closed as unclear what you're asking by amWhy, Shailesh, Leucippus, max_zorn, KReiser Nov 20 at 6:57
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
Nov 19 at 20:35
Also, in defining $R$ we should say $(x,y) in mathbb Z times mathbb Z$ or something equivalent.
– hardmath
Nov 20 at 3:30
add a comment |
I think you mean "(5,3) ∈ R, and (3,5)∈ R"
– Arthur
Nov 19 at 20:35
Also, in defining $R$ we should say $(x,y) in mathbb Z times mathbb Z$ or something equivalent.
– hardmath
Nov 20 at 3:30
I think you mean "(5,3) ∈ R, and (3,5)∈ R"
– Arthur
Nov 19 at 20:35
I think you mean "(5,3) ∈ R, and (3,5)∈ R"
– Arthur
Nov 19 at 20:35
Also, in defining $R$ we should say $(x,y) in mathbb Z times mathbb Z$ or something equivalent.
– hardmath
Nov 20 at 3:30
Also, in defining $R$ we should say $(x,y) in mathbb Z times mathbb Z$ or something equivalent.
– hardmath
Nov 20 at 3:30
add a comment |
2 Answers
2
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up vote
1
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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 Nov 19 at 21:40
amWhy
191k27223438
answered Nov 19 at 20:30
hamam_Abdallah
36.9k21533
Simply put. +1.
– amWhy
Nov 19 at 20:31
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 Nov 19 at 20:26
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
Nov 19 at 20:28
@amWhy $5R3$ and $3R5$ but $5neq3$ doesn't disprove antisymmetry? (He means $in R$, not $in Bbb Z$, presumably.) Have I missed something?
– Arthur
Nov 19 at 20:29
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
Nov 19 at 20:30
@amWhy He uses non-conventional and not entirely correct terminology, but to me it's clearly there.
– Arthur
Nov 19 at 20:33
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
Nov 19 at 20:36
|
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 Nov 19 at 21:40
amWhy
191k27223438
answered Nov 19 at 20:30
hamam_Abdallah
36.9k21533
Simply put. +1.
– amWhy
Nov 19 at 20:31
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 Nov 19 at 21:40
amWhy
191k27223438
answered Nov 19 at 20:30
hamam_Abdallah
36.9k21533
Simply put. +1.
– amWhy
Nov 19 at 20:31
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 Nov 19 at 21:40
amWhy
191k27223438
answered Nov 19 at 20:30
hamam_Abdallah
36.9k21533
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 Nov 19 at 21:40
amWhy
191k27223438
answered Nov 19 at 20:30
hamam_Abdallah
36.9k21533
edited Nov 19 at 21:40
amWhy
191k27223438
edited Nov 19 at 21:40
amWhy
191k27223438
edited Nov 19 at 21:40
amWhy
191k27223438
191k27223438
answered Nov 19 at 20:30
hamam_Abdallah
36.9k21533
answered Nov 19 at 20:30
hamam_Abdallah
36.9k21533
answered Nov 19 at 20:30
hamam_Abdallah
36.9k21533
36.9k21533
Simply put. +1.
– amWhy
Nov 19 at 20:31
add a comment |
Simply put. +1.
– amWhy
Nov 19 at 20:31
Simply put. +1.
– amWhy
Nov 19 at 20:31
Simply put. +1.
– amWhy
Nov 19 at 20:31
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 Nov 19 at 20:26
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
Nov 19 at 20:28
@amWhy $5R3$ and $3R5$ but $5neq3$ doesn't disprove antisymmetry? (He means $in R$, not $in Bbb Z$, presumably.) Have I missed something?
– Arthur
Nov 19 at 20:29
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
Nov 19 at 20:30
@amWhy He uses non-conventional and not entirely correct terminology, but to me it's clearly there.
– Arthur
Nov 19 at 20:33
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
Nov 19 at 20:36
|
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 Nov 19 at 20:26
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
Nov 19 at 20:28
@amWhy $5R3$ and $3R5$ but $5neq3$ doesn't disprove antisymmetry? (He means $in R$, not $in Bbb Z$, presumably.) Have I missed something?
– Arthur
Nov 19 at 20:29
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
Nov 19 at 20:30
@amWhy He uses non-conventional and not entirely correct terminology, but to me it's clearly there.
– Arthur
Nov 19 at 20:33
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
Nov 19 at 20:36
|
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 Nov 19 at 20:26
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 Nov 19 at 20:26
Arthur
108k7103186
edited Nov 19 at 20:36
answered Nov 19 at 20:26
Arthur
108k7103186
answered Nov 19 at 20:26
Arthur
108k7103186
answered Nov 19 at 20:26
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
Nov 19 at 20:28
@amWhy $5R3$ and $3R5$ but $5neq3$ doesn't disprove antisymmetry? (He means $in R$, not $in Bbb Z$, presumably.) Have I missed something?
– Arthur
Nov 19 at 20:29
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
Nov 19 at 20:30
@amWhy He uses non-conventional and not entirely correct terminology, but to me it's clearly there.
– Arthur
Nov 19 at 20:33
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
Nov 19 at 20:36
|
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
Nov 19 at 20:28
@amWhy $5R3$ and $3R5$ but $5neq3$ doesn't disprove antisymmetry? (He means $in R$, not $in Bbb Z$, presumably.) Have I missed something?
– Arthur
Nov 19 at 20:29
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
Nov 19 at 20:30
@amWhy He uses non-conventional and not entirely correct terminology, but to me it's clearly there.
– Arthur
Nov 19 at 20:33
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
Nov 19 at 20:36
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
Nov 19 at 20:28
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
Nov 19 at 20:28
@amWhy $5R3$ and $3R5$ but $5neq3$ doesn't disprove antisymmetry? (He means $in R$, not $in Bbb Z$, presumably.) Have I missed something?
– Arthur
Nov 19 at 20:29
@amWhy $5R3$ and $3R5$ but $5neq3$ doesn't disprove antisymmetry? (He means $in R$, not $in Bbb Z$, presumably.) Have I missed something?
– Arthur
Nov 19 at 20:29
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
Nov 19 at 20:30
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
Nov 19 at 20:30
@amWhy He uses non-conventional and not entirely correct terminology, but to me it's clearly there.
– Arthur
Nov 19 at 20:33
@amWhy He uses non-conventional and not entirely correct terminology, but to me it's clearly there.
– Arthur
Nov 19 at 20:33
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
Nov 19 at 20:36
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
Nov 19 at 20:36
|
show 4 more comments
I think you mean "(5,3) ∈ R, and (3,5)∈ R"
– Arthur
Nov 19 at 20:35
Also, in defining $R$ we should say $(x,y) in mathbb Z times mathbb Z$ or something equivalent.
– hardmath
Nov 20 at 3:30