Showing a relation is transitive
$begingroup$
Problem:
Let $A = {(a,b)}$, determine whether the relation $R = {(b,a)}$ is transitive.
Claim:
No, $R$ is not transitive.
Proof:
Since $a,bin A$ and $ain R$ but $b,anotin R$.
I am not sure if this is a sufficient proof. Any suggestions would help.
proof-verification equivalence-relations
asked Jan 16 at 6:37
SnorrlaxxxSnorrlaxxx
1706
$endgroup$
add a comment |
$begingroup$
Problem:
Let $A = {(a,b)}$, determine whether the relation $R = {(b,a)}$ is transitive.
Claim:
No, $R$ is not transitive.
Proof:
Since $a,bin A$ and $ain R$ but $b,anotin R$.
I am not sure if this is a sufficient proof. Any suggestions would help.
proof-verification equivalence-relations
asked Jan 16 at 6:37
SnorrlaxxxSnorrlaxxx
1706
$endgroup$
3
$begingroup$
Do you mean $A = {a, b}$, a two element set containing $a$ and $b$, instead o the one element set ${(a, b)}$ containing the single ordered pair $(a, b)$?
$endgroup$
– Theo Bendit
Jan 16 at 6:48
add a comment |
$begingroup$
Problem:
Let $A = {(a,b)}$, determine whether the relation $R = {(b,a)}$ is transitive.
Claim:
No, $R$ is not transitive.
Proof:
Since $a,bin A$ and $ain R$ but $b,anotin R$.
I am not sure if this is a sufficient proof. Any suggestions would help.
proof-verification equivalence-relations
asked Jan 16 at 6:37
SnorrlaxxxSnorrlaxxx
1706
$endgroup$
Problem:
Let $A = {(a,b)}$, determine whether the relation $R = {(b,a)}$ is transitive.
Claim:
No, $R$ is not transitive.
Proof:
Since $a,bin A$ and $ain R$ but $b,anotin R$.
I am not sure if this is a sufficient proof. Any suggestions would help.
proof-verification equivalence-relations
proof-verification equivalence-relations
asked Jan 16 at 6:37
SnorrlaxxxSnorrlaxxx
1706
asked Jan 16 at 6:37
SnorrlaxxxSnorrlaxxx
1706
asked Jan 16 at 6:37
SnorrlaxxxSnorrlaxxx
1706
asked Jan 16 at 6:37
SnorrlaxxxSnorrlaxxx
1706
asked Jan 16 at 6:37
SnorrlaxxxSnorrlaxxx
1706
1706
3
$begingroup$
Do you mean $A = {a, b}$, a two element set containing $a$ and $b$, instead o the one element set ${(a, b)}$ containing the single ordered pair $(a, b)$?
$endgroup$
– Theo Bendit
Jan 16 at 6:48
add a comment |
3
$begingroup$
Do you mean $A = {a, b}$, a two element set containing $a$ and $b$, instead o the one element set ${(a, b)}$ containing the single ordered pair $(a, b)$?
$endgroup$
– Theo Bendit
Jan 16 at 6:48
3
3
$begingroup$
Do you mean $A = {a, b}$, a two element set containing $a$ and $b$, instead o the one element set ${(a, b)}$ containing the single ordered pair $(a, b)$?
$endgroup$
– Theo Bendit
Jan 16 at 6:48
$begingroup$
Do you mean $A = {a, b}$, a two element set containing $a$ and $b$, instead o the one element set ${(a, b)}$ containing the single ordered pair $(a, b)$?
$endgroup$
– Theo Bendit
Jan 16 at 6:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Well, it is transitive (with $A={a,b}$). The definition is
$$forall a,b,cin A: aRb wedge bRcRightarrow aRc$$
where $Rsubseteq Atimes A$.
The point is that if the premise is not fulfilled, the relation is transitive (since then the implication becomes true).
For instance, $R={(a,b)}$ (where $A={a,b}$) and $R={(a,b),(a,c)}$ are all transitive, but $R={(a,b),(b,c)}$ is not since $(a,c)$ is missing.
answered Jan 16 at 7:56
WuestenfuxWuestenfux
5,3631513
$endgroup$
add a comment |
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$begingroup$
Well, it is transitive (with $A={a,b}$). The definition is
$$forall a,b,cin A: aRb wedge bRcRightarrow aRc$$
where $Rsubseteq Atimes A$.
The point is that if the premise is not fulfilled, the relation is transitive (since then the implication becomes true).
For instance, $R={(a,b)}$ (where $A={a,b}$) and $R={(a,b),(a,c)}$ are all transitive, but $R={(a,b),(b,c)}$ is not since $(a,c)$ is missing.
answered Jan 16 at 7:56
WuestenfuxWuestenfux
5,3631513
$endgroup$
add a comment |
$begingroup$
Well, it is transitive (with $A={a,b}$). The definition is
$$forall a,b,cin A: aRb wedge bRcRightarrow aRc$$
where $Rsubseteq Atimes A$.
The point is that if the premise is not fulfilled, the relation is transitive (since then the implication becomes true).
For instance, $R={(a,b)}$ (where $A={a,b}$) and $R={(a,b),(a,c)}$ are all transitive, but $R={(a,b),(b,c)}$ is not since $(a,c)$ is missing.
answered Jan 16 at 7:56
WuestenfuxWuestenfux
5,3631513
$endgroup$
add a comment |
$begingroup$
Well, it is transitive (with $A={a,b}$). The definition is
$$forall a,b,cin A: aRb wedge bRcRightarrow aRc$$
where $Rsubseteq Atimes A$.
The point is that if the premise is not fulfilled, the relation is transitive (since then the implication becomes true).
For instance, $R={(a,b)}$ (where $A={a,b}$) and $R={(a,b),(a,c)}$ are all transitive, but $R={(a,b),(b,c)}$ is not since $(a,c)$ is missing.
answered Jan 16 at 7:56
WuestenfuxWuestenfux
5,3631513
$endgroup$
Well, it is transitive (with $A={a,b}$). The definition is
$$forall a,b,cin A: aRb wedge bRcRightarrow aRc$$
where $Rsubseteq Atimes A$.
The point is that if the premise is not fulfilled, the relation is transitive (since then the implication becomes true).
For instance, $R={(a,b)}$ (where $A={a,b}$) and $R={(a,b),(a,c)}$ are all transitive, but $R={(a,b),(b,c)}$ is not since $(a,c)$ is missing.
answered Jan 16 at 7:56
WuestenfuxWuestenfux
5,3631513
answered Jan 16 at 7:56
WuestenfuxWuestenfux
5,3631513
answered Jan 16 at 7:56
WuestenfuxWuestenfux
5,3631513
answered Jan 16 at 7:56
WuestenfuxWuestenfux
5,3631513
5,3631513
add a comment |
add a comment |
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$begingroup$
Do you mean $A = {a, b}$, a two element set containing $a$ and $b$, instead o the one element set ${(a, b)}$ containing the single ordered pair $(a, b)$?
$endgroup$
– Theo Bendit
Jan 16 at 6:48