Mixing
Proving symmetry and transitivity of a relation
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Let R be the relation {(1,1),(1,2),(2,2),(1,3),(3,3)} on the set {1,2,3}.
I am having difficulty proving that it is symmetrical and transitive. I know for symmetry we have to prove if xRy then yRx (if x related to y then y is related to x) and with transitivity we have to show xRy and yRz then xRz. I am having trouble with the set of points and showing that they are or are not symmetrical and transitive.
I already proved reflexive because (1,1)R(1,1).
I was thinking if (1,1)R(1,2) then (1,2)R(1,1) --showing symmetry--must be true because those points are in the set, but I think my logic is too simple.
analysis proof-verification relations equivalence-relations symmetry
asked Jan 31 at 21:36
AnneAnne
105
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add a comment |
$begingroup$
Let R be the relation {(1,1),(1,2),(2,2),(1,3),(3,3)} on the set {1,2,3}.
I am having difficulty proving that it is symmetrical and transitive. I know for symmetry we have to prove if xRy then yRx (if x related to y then y is related to x) and with transitivity we have to show xRy and yRz then xRz. I am having trouble with the set of points and showing that they are or are not symmetrical and transitive.
I already proved reflexive because (1,1)R(1,1).
I was thinking if (1,1)R(1,2) then (1,2)R(1,1) --showing symmetry--must be true because those points are in the set, but I think my logic is too simple.
analysis proof-verification relations equivalence-relations symmetry
asked Jan 31 at 21:36
AnneAnne
105
$endgroup$
add a comment |
$begingroup$
Let R be the relation {(1,1),(1,2),(2,2),(1,3),(3,3)} on the set {1,2,3}.
I am having difficulty proving that it is symmetrical and transitive. I know for symmetry we have to prove if xRy then yRx (if x related to y then y is related to x) and with transitivity we have to show xRy and yRz then xRz. I am having trouble with the set of points and showing that they are or are not symmetrical and transitive.
I already proved reflexive because (1,1)R(1,1).
I was thinking if (1,1)R(1,2) then (1,2)R(1,1) --showing symmetry--must be true because those points are in the set, but I think my logic is too simple.
analysis proof-verification relations equivalence-relations symmetry
asked Jan 31 at 21:36
AnneAnne
105
$endgroup$
Let R be the relation {(1,1),(1,2),(2,2),(1,3),(3,3)} on the set {1,2,3}.
I am having difficulty proving that it is symmetrical and transitive. I know for symmetry we have to prove if xRy then yRx (if x related to y then y is related to x) and with transitivity we have to show xRy and yRz then xRz. I am having trouble with the set of points and showing that they are or are not symmetrical and transitive.
I already proved reflexive because (1,1)R(1,1).
I was thinking if (1,1)R(1,2) then (1,2)R(1,1) --showing symmetry--must be true because those points are in the set, but I think my logic is too simple.
analysis proof-verification relations equivalence-relations symmetry
analysis proof-verification relations equivalence-relations symmetry
asked Jan 31 at 21:36
AnneAnne
105
asked Jan 31 at 21:36
AnneAnne
105
asked Jan 31 at 21:36
AnneAnne
105
asked Jan 31 at 21:36
AnneAnne
105
asked Jan 31 at 21:36
AnneAnne
105
105
add a comment |
add a comment |
1 Answer
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$begingroup$
I think you got a bit confused with the different ways to write these things. Here $xRy$ is the same thing as $(x,y)in R$. So you have $1R1,1R2,2R2,1R3,3R3$. This is the relation. So $(1,2)in R$ but $(2,1)notin R$ which already shows that the relation is not symmetric.
answered Jan 31 at 21:39
MarkMark
10.5k1622
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$begingroup$
that helped a lot, I solved it thank you!
$endgroup$
– Anne
Jan 31 at 21:42
add a comment |
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1 Answer
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$begingroup$
I think you got a bit confused with the different ways to write these things. Here $xRy$ is the same thing as $(x,y)in R$. So you have $1R1,1R2,2R2,1R3,3R3$. This is the relation. So $(1,2)in R$ but $(2,1)notin R$ which already shows that the relation is not symmetric.
answered Jan 31 at 21:39
MarkMark
10.5k1622
$endgroup$
$begingroup$
that helped a lot, I solved it thank you!
$endgroup$
– Anne
Jan 31 at 21:42
add a comment |
$begingroup$
I think you got a bit confused with the different ways to write these things. Here $xRy$ is the same thing as $(x,y)in R$. So you have $1R1,1R2,2R2,1R3,3R3$. This is the relation. So $(1,2)in R$ but $(2,1)notin R$ which already shows that the relation is not symmetric.
answered Jan 31 at 21:39
MarkMark
10.5k1622
$endgroup$
$begingroup$
that helped a lot, I solved it thank you!
$endgroup$
– Anne
Jan 31 at 21:42
add a comment |
$begingroup$
I think you got a bit confused with the different ways to write these things. Here $xRy$ is the same thing as $(x,y)in R$. So you have $1R1,1R2,2R2,1R3,3R3$. This is the relation. So $(1,2)in R$ but $(2,1)notin R$ which already shows that the relation is not symmetric.
answered Jan 31 at 21:39
MarkMark
10.5k1622
$endgroup$
I think you got a bit confused with the different ways to write these things. Here $xRy$ is the same thing as $(x,y)in R$. So you have $1R1,1R2,2R2,1R3,3R3$. This is the relation. So $(1,2)in R$ but $(2,1)notin R$ which already shows that the relation is not symmetric.
answered Jan 31 at 21:39
MarkMark
10.5k1622
answered Jan 31 at 21:39
MarkMark
10.5k1622
answered Jan 31 at 21:39
MarkMark
10.5k1622
answered Jan 31 at 21:39
MarkMark
10.5k1622
10.5k1622
$begingroup$
that helped a lot, I solved it thank you!
$endgroup$
– Anne
Jan 31 at 21:42
add a comment |
$begingroup$
that helped a lot, I solved it thank you!
$endgroup$
– Anne
Jan 31 at 21:42
$begingroup$
that helped a lot, I solved it thank you!
$endgroup$
– Anne
Jan 31 at 21:42
$begingroup$
that helped a lot, I solved it thank you!
$endgroup$
– Anne
Jan 31 at 21:42
add a comment |
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