Mixing
Determine if $R subseteq A times A$ is reflexive, transitive, symmetric, antisymmetric
$begingroup$
Let A be the set of bit strings $a = a_1a_2 ldots a_9$ of length 9. Let $R subset A times A$ be the set of pairs $(a, b)$ such that $a_1 = b_1$ or $a_2 = b_2$.
Decide whether or not the relation $R$ is
- reflexive
- transitive
- symmetric
- antisymmetric
- an equivalence relation.
I do not understand this question. I know that $A times A = {(a_1a_2a_3a_4a_5a_6a_7a_8a_9, a_1a_2a_3a_4a_5a_6a_7a_8a_9)}$. Here's my attempt:
$R$ is reflexive since $aRa$ where $a = a_1a_2a_3a_4a_5a_6a_7a_8a_9$ is true since $a_1 = a_1$.- I'm not sure about this one
- If $R$ is symmetric, aRb is true then $bRa$ is true. Since $aRb$ is true where $a_1 = b_1$ or $a_2 = b_2$ then $bRa$ is true because $b_1 = a_1$ or $b_2 = a_2$.
- It cannot be antisymmetric if it's symmetric
elementary-set-theory
edited Jan 20 at 12:30
Henrik
6,03592030
asked Jan 20 at 11:52
llamaro25llamaro25
528
$endgroup$
add a comment |
$begingroup$
Let A be the set of bit strings $a = a_1a_2 ldots a_9$ of length 9. Let $R subset A times A$ be the set of pairs $(a, b)$ such that $a_1 = b_1$ or $a_2 = b_2$.
Decide whether or not the relation $R$ is
- reflexive
- transitive
- symmetric
- antisymmetric
- an equivalence relation.
I do not understand this question. I know that $A times A = {(a_1a_2a_3a_4a_5a_6a_7a_8a_9, a_1a_2a_3a_4a_5a_6a_7a_8a_9)}$. Here's my attempt:
$R$ is reflexive since $aRa$ where $a = a_1a_2a_3a_4a_5a_6a_7a_8a_9$ is true since $a_1 = a_1$.- I'm not sure about this one
- If $R$ is symmetric, aRb is true then $bRa$ is true. Since $aRb$ is true where $a_1 = b_1$ or $a_2 = b_2$ then $bRa$ is true because $b_1 = a_1$ or $b_2 = a_2$.
- It cannot be antisymmetric if it's symmetric
elementary-set-theory
edited Jan 20 at 12:30
Henrik
6,03592030
asked Jan 20 at 11:52
llamaro25llamaro25
528
$endgroup$
$begingroup$
Your description of $Atimes A$ is wrong. You should have a pair of $9$-element strings, not nine pairs. Also, you shouldn't have pairs of the same element, you can have two different elements of $A$ in a pair.
$endgroup$
– Michael Burr
Jan 20 at 12:00
$begingroup$
Two strings $a$ and $b$ are equivalent if $a_1=b_1$ or $a_2=b_2$.
$endgroup$
– Wuestenfux
Jan 20 at 12:02
add a comment |
$begingroup$
Let A be the set of bit strings $a = a_1a_2 ldots a_9$ of length 9. Let $R subset A times A$ be the set of pairs $(a, b)$ such that $a_1 = b_1$ or $a_2 = b_2$.
Decide whether or not the relation $R$ is
- reflexive
- transitive
- symmetric
- antisymmetric
- an equivalence relation.
I do not understand this question. I know that $A times A = {(a_1a_2a_3a_4a_5a_6a_7a_8a_9, a_1a_2a_3a_4a_5a_6a_7a_8a_9)}$. Here's my attempt:
$R$ is reflexive since $aRa$ where $a = a_1a_2a_3a_4a_5a_6a_7a_8a_9$ is true since $a_1 = a_1$.- I'm not sure about this one
- If $R$ is symmetric, aRb is true then $bRa$ is true. Since $aRb$ is true where $a_1 = b_1$ or $a_2 = b_2$ then $bRa$ is true because $b_1 = a_1$ or $b_2 = a_2$.
- It cannot be antisymmetric if it's symmetric
elementary-set-theory
edited Jan 20 at 12:30
Henrik
6,03592030
asked Jan 20 at 11:52
llamaro25llamaro25
528
$endgroup$
Let A be the set of bit strings $a = a_1a_2 ldots a_9$ of length 9. Let $R subset A times A$ be the set of pairs $(a, b)$ such that $a_1 = b_1$ or $a_2 = b_2$.
Decide whether or not the relation $R$ is
- reflexive
- transitive
- symmetric
- antisymmetric
- an equivalence relation.
I do not understand this question. I know that $A times A = {(a_1a_2a_3a_4a_5a_6a_7a_8a_9, a_1a_2a_3a_4a_5a_6a_7a_8a_9)}$. Here's my attempt:
$R$ is reflexive since $aRa$ where $a = a_1a_2a_3a_4a_5a_6a_7a_8a_9$ is true since $a_1 = a_1$.- I'm not sure about this one
- If $R$ is symmetric, aRb is true then $bRa$ is true. Since $aRb$ is true where $a_1 = b_1$ or $a_2 = b_2$ then $bRa$ is true because $b_1 = a_1$ or $b_2 = a_2$.
- It cannot be antisymmetric if it's symmetric
elementary-set-theory
elementary-set-theory
edited Jan 20 at 12:30
Henrik
6,03592030
asked Jan 20 at 11:52
llamaro25llamaro25
528
edited Jan 20 at 12:30
Henrik
6,03592030
asked Jan 20 at 11:52
llamaro25llamaro25
528
edited Jan 20 at 12:30
Henrik
6,03592030
edited Jan 20 at 12:30
Henrik
6,03592030
edited Jan 20 at 12:30
Henrik
6,03592030
6,03592030
asked Jan 20 at 11:52
llamaro25llamaro25
528
asked Jan 20 at 11:52
llamaro25llamaro25
528
asked Jan 20 at 11:52
llamaro25llamaro25
528
528
$begingroup$
Your description of $Atimes A$ is wrong. You should have a pair of $9$-element strings, not nine pairs. Also, you shouldn't have pairs of the same element, you can have two different elements of $A$ in a pair.
$endgroup$
– Michael Burr
Jan 20 at 12:00
$begingroup$
Two strings $a$ and $b$ are equivalent if $a_1=b_1$ or $a_2=b_2$.
$endgroup$
– Wuestenfux
Jan 20 at 12:02
add a comment |
$begingroup$
Your description of $Atimes A$ is wrong. You should have a pair of $9$-element strings, not nine pairs. Also, you shouldn't have pairs of the same element, you can have two different elements of $A$ in a pair.
$endgroup$
– Michael Burr
Jan 20 at 12:00
$begingroup$
Two strings $a$ and $b$ are equivalent if $a_1=b_1$ or $a_2=b_2$.
$endgroup$
– Wuestenfux
Jan 20 at 12:02
$begingroup$
Your description of $Atimes A$ is wrong. You should have a pair of $9$-element strings, not nine pairs. Also, you shouldn't have pairs of the same element, you can have two different elements of $A$ in a pair.
$endgroup$
– Michael Burr
Jan 20 at 12:00
$begingroup$
Your description of $Atimes A$ is wrong. You should have a pair of $9$-element strings, not nine pairs. Also, you shouldn't have pairs of the same element, you can have two different elements of $A$ in a pair.
$endgroup$
– Michael Burr
Jan 20 at 12:00
$begingroup$
Two strings $a$ and $b$ are equivalent if $a_1=b_1$ or $a_2=b_2$.
$endgroup$
– Wuestenfux
Jan 20 at 12:02
$begingroup$
Two strings $a$ and $b$ are equivalent if $a_1=b_1$ or $a_2=b_2$.
$endgroup$
– Wuestenfux
Jan 20 at 12:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$A mathrm{x} A={(a,b) | a=a_1...a_9 , b=b_1...b_9, a,b in A}$
i) Reflexive: is $(a,a) in R$, where $a=a_1...a_9$? Yes, because $a_1=a_1$.
ii) Transitive: if $(a,b) in R$ and $(b,c) in R$ do we have $(a,c) in R$? No, take for example $a_1=b_1$, $a_2 neq b_2$ and $b_1 neq c_1$, $b_2 = c_2$. We have $a_1 neq c_1$ and $a_2 neq c_2$, thus $(a,c) notin R$.
iii) Symmetric: if $(a,b) in R$ it means that $a_1=b_1$ or $a_2=b_2$. Therefore we have $b_1=a_1$ or $b_2=a_2$ i.e. $(b,a) in R$.
iv) Antisymmetric: if $(a,b) in R$ and $(b,a) in R$ then we can have $a neq b$: take for example $a_3 neq b_3$. Thus it's not antisymmetric
v) It can't be an equivalence relation since it's not transitive.
answered Jan 20 at 12:06
user289143user289143
1,002313
$endgroup$
$begingroup$
@Arthur Indeed, but not in this case (edited the answer)
$endgroup$
– user289143
Jan 20 at 12:18
$begingroup$
thank you for the answer :)
$endgroup$
– llamaro25
Jan 20 at 12:23
add a comment |
$begingroup$
Indeed, its not transitive: $a=011$, $b=001$ and $c=101$. Then $aequiv b$ ($a_1=0=b_1$) and $bequiv c$ $(b_2=1=c_2)$, but $anotequiv c$ ($a_1ne c_1, a_2ne c_2$).
answered Jan 20 at 12:13
WuestenfuxWuestenfux
4,7941513
$endgroup$
$begingroup$
the explanation is very concise. thank you!
$endgroup$
– llamaro25
Jan 20 at 12:23
add a comment |
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2 Answers
2
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2 Answers
2
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$begingroup$
$A mathrm{x} A={(a,b) | a=a_1...a_9 , b=b_1...b_9, a,b in A}$
i) Reflexive: is $(a,a) in R$, where $a=a_1...a_9$? Yes, because $a_1=a_1$.
ii) Transitive: if $(a,b) in R$ and $(b,c) in R$ do we have $(a,c) in R$? No, take for example $a_1=b_1$, $a_2 neq b_2$ and $b_1 neq c_1$, $b_2 = c_2$. We have $a_1 neq c_1$ and $a_2 neq c_2$, thus $(a,c) notin R$.
iii) Symmetric: if $(a,b) in R$ it means that $a_1=b_1$ or $a_2=b_2$. Therefore we have $b_1=a_1$ or $b_2=a_2$ i.e. $(b,a) in R$.
iv) Antisymmetric: if $(a,b) in R$ and $(b,a) in R$ then we can have $a neq b$: take for example $a_3 neq b_3$. Thus it's not antisymmetric
v) It can't be an equivalence relation since it's not transitive.
answered Jan 20 at 12:06
user289143user289143
1,002313
$endgroup$
$begingroup$
@Arthur Indeed, but not in this case (edited the answer)
$endgroup$
– user289143
Jan 20 at 12:18
$begingroup$
thank you for the answer :)
$endgroup$
– llamaro25
Jan 20 at 12:23
add a comment |
$begingroup$
$A mathrm{x} A={(a,b) | a=a_1...a_9 , b=b_1...b_9, a,b in A}$
i) Reflexive: is $(a,a) in R$, where $a=a_1...a_9$? Yes, because $a_1=a_1$.
ii) Transitive: if $(a,b) in R$ and $(b,c) in R$ do we have $(a,c) in R$? No, take for example $a_1=b_1$, $a_2 neq b_2$ and $b_1 neq c_1$, $b_2 = c_2$. We have $a_1 neq c_1$ and $a_2 neq c_2$, thus $(a,c) notin R$.
iii) Symmetric: if $(a,b) in R$ it means that $a_1=b_1$ or $a_2=b_2$. Therefore we have $b_1=a_1$ or $b_2=a_2$ i.e. $(b,a) in R$.
iv) Antisymmetric: if $(a,b) in R$ and $(b,a) in R$ then we can have $a neq b$: take for example $a_3 neq b_3$. Thus it's not antisymmetric
v) It can't be an equivalence relation since it's not transitive.
answered Jan 20 at 12:06
user289143user289143
1,002313
$endgroup$
$begingroup$
@Arthur Indeed, but not in this case (edited the answer)
$endgroup$
– user289143
Jan 20 at 12:18
$begingroup$
thank you for the answer :)
$endgroup$
– llamaro25
Jan 20 at 12:23
add a comment |
$begingroup$
$A mathrm{x} A={(a,b) | a=a_1...a_9 , b=b_1...b_9, a,b in A}$
i) Reflexive: is $(a,a) in R$, where $a=a_1...a_9$? Yes, because $a_1=a_1$.
ii) Transitive: if $(a,b) in R$ and $(b,c) in R$ do we have $(a,c) in R$? No, take for example $a_1=b_1$, $a_2 neq b_2$ and $b_1 neq c_1$, $b_2 = c_2$. We have $a_1 neq c_1$ and $a_2 neq c_2$, thus $(a,c) notin R$.
iii) Symmetric: if $(a,b) in R$ it means that $a_1=b_1$ or $a_2=b_2$. Therefore we have $b_1=a_1$ or $b_2=a_2$ i.e. $(b,a) in R$.
iv) Antisymmetric: if $(a,b) in R$ and $(b,a) in R$ then we can have $a neq b$: take for example $a_3 neq b_3$. Thus it's not antisymmetric
v) It can't be an equivalence relation since it's not transitive.
answered Jan 20 at 12:06
user289143user289143
1,002313
$endgroup$
$A mathrm{x} A={(a,b) | a=a_1...a_9 , b=b_1...b_9, a,b in A}$
i) Reflexive: is $(a,a) in R$, where $a=a_1...a_9$? Yes, because $a_1=a_1$.
ii) Transitive: if $(a,b) in R$ and $(b,c) in R$ do we have $(a,c) in R$? No, take for example $a_1=b_1$, $a_2 neq b_2$ and $b_1 neq c_1$, $b_2 = c_2$. We have $a_1 neq c_1$ and $a_2 neq c_2$, thus $(a,c) notin R$.
iii) Symmetric: if $(a,b) in R$ it means that $a_1=b_1$ or $a_2=b_2$. Therefore we have $b_1=a_1$ or $b_2=a_2$ i.e. $(b,a) in R$.
iv) Antisymmetric: if $(a,b) in R$ and $(b,a) in R$ then we can have $a neq b$: take for example $a_3 neq b_3$. Thus it's not antisymmetric
v) It can't be an equivalence relation since it's not transitive.
answered Jan 20 at 12:06
user289143user289143
1,002313
edited Jan 20 at 12:20
answered Jan 20 at 12:06
user289143user289143
1,002313
answered Jan 20 at 12:06
user289143user289143
1,002313
answered Jan 20 at 12:06
user289143user289143
1,002313
1,002313
$begingroup$
@Arthur Indeed, but not in this case (edited the answer)
$endgroup$
– user289143
Jan 20 at 12:18
$begingroup$
thank you for the answer :)
$endgroup$
– llamaro25
Jan 20 at 12:23
add a comment |
$begingroup$
@Arthur Indeed, but not in this case (edited the answer)
$endgroup$
– user289143
Jan 20 at 12:18
$begingroup$
thank you for the answer :)
$endgroup$
– llamaro25
Jan 20 at 12:23
$begingroup$
@Arthur Indeed, but not in this case (edited the answer)
$endgroup$
– user289143
Jan 20 at 12:18
$begingroup$
@Arthur Indeed, but not in this case (edited the answer)
$endgroup$
– user289143
Jan 20 at 12:18
$begingroup$
thank you for the answer :)
$endgroup$
– llamaro25
Jan 20 at 12:23
$begingroup$
thank you for the answer :)
$endgroup$
– llamaro25
Jan 20 at 12:23
add a comment |
$begingroup$
Indeed, its not transitive: $a=011$, $b=001$ and $c=101$. Then $aequiv b$ ($a_1=0=b_1$) and $bequiv c$ $(b_2=1=c_2)$, but $anotequiv c$ ($a_1ne c_1, a_2ne c_2$).
answered Jan 20 at 12:13
WuestenfuxWuestenfux
4,7941513
$endgroup$
$begingroup$
the explanation is very concise. thank you!
$endgroup$
– llamaro25
Jan 20 at 12:23
add a comment |
$begingroup$
Indeed, its not transitive: $a=011$, $b=001$ and $c=101$. Then $aequiv b$ ($a_1=0=b_1$) and $bequiv c$ $(b_2=1=c_2)$, but $anotequiv c$ ($a_1ne c_1, a_2ne c_2$).
answered Jan 20 at 12:13
WuestenfuxWuestenfux
4,7941513
$endgroup$
$begingroup$
the explanation is very concise. thank you!
$endgroup$
– llamaro25
Jan 20 at 12:23
add a comment |
$begingroup$
Indeed, its not transitive: $a=011$, $b=001$ and $c=101$. Then $aequiv b$ ($a_1=0=b_1$) and $bequiv c$ $(b_2=1=c_2)$, but $anotequiv c$ ($a_1ne c_1, a_2ne c_2$).
answered Jan 20 at 12:13
WuestenfuxWuestenfux
4,7941513
$endgroup$
Indeed, its not transitive: $a=011$, $b=001$ and $c=101$. Then $aequiv b$ ($a_1=0=b_1$) and $bequiv c$ $(b_2=1=c_2)$, but $anotequiv c$ ($a_1ne c_1, a_2ne c_2$).
answered Jan 20 at 12:13
WuestenfuxWuestenfux
4,7941513
answered Jan 20 at 12:13
WuestenfuxWuestenfux
4,7941513
answered Jan 20 at 12:13
WuestenfuxWuestenfux
4,7941513
answered Jan 20 at 12:13
WuestenfuxWuestenfux
4,7941513
4,7941513
$begingroup$
the explanation is very concise. thank you!
$endgroup$
– llamaro25
Jan 20 at 12:23
add a comment |
$begingroup$
the explanation is very concise. thank you!
$endgroup$
– llamaro25
Jan 20 at 12:23
$begingroup$
the explanation is very concise. thank you!
$endgroup$
– llamaro25
Jan 20 at 12:23
$begingroup$
the explanation is very concise. thank you!
$endgroup$
– llamaro25
Jan 20 at 12:23
add a comment |
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$begingroup$
Your description of $Atimes A$ is wrong. You should have a pair of $9$-element strings, not nine pairs. Also, you shouldn't have pairs of the same element, you can have two different elements of $A$ in a pair.
$endgroup$
– Michael Burr
Jan 20 at 12:00
$begingroup$
Two strings $a$ and $b$ are equivalent if $a_1=b_1$ or $a_2=b_2$.
$endgroup$
– Wuestenfux
Jan 20 at 12:02