question on showing transitivity of a relation
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Define a relation on Z as xRy if |x−y|<1.
I have shown this relation is symetric and reflexive and i am pretty sure its
transitive because this is the equality relation isnt it? thats my first question and my second is how to show it is transitive.
I attempted a direct proof but i dont know how to link the two inequalities together to get that |x−z|<1 (im trying to show if xRy and yRz then xRz).
Any help would be appreciated, thanks! I am looking for the proof of this last property (transitivity).
relations
asked Dec 3 at 3:31
Carlos Bacca
178116
add a comment |
up vote
0
down vote
favorite
Define a relation on Z as xRy if |x−y|<1.
I have shown this relation is symetric and reflexive and i am pretty sure its
transitive because this is the equality relation isnt it? thats my first question and my second is how to show it is transitive.
I attempted a direct proof but i dont know how to link the two inequalities together to get that |x−z|<1 (im trying to show if xRy and yRz then xRz).
Any help would be appreciated, thanks! I am looking for the proof of this last property (transitivity).
relations
asked Dec 3 at 3:31
Carlos Bacca
178116
Thanks, but its a relation on z so only for integer values
– Carlos Bacca
Dec 3 at 4:14
There are no integers satisfying this relation.
– user58697
Dec 3 at 4:41
what about x=4 y=4?
– Carlos Bacca
Dec 3 at 4:43
Sorry I forgot to say distinct. Modulo is non-negative; the only non-negative less than 1 is 0.
– user58697
Dec 3 at 4:49
So is it transitive?
– Carlos Bacca
Dec 3 at 4:53
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Define a relation on Z as xRy if |x−y|<1.
I have shown this relation is symetric and reflexive and i am pretty sure its
transitive because this is the equality relation isnt it? thats my first question and my second is how to show it is transitive.
I attempted a direct proof but i dont know how to link the two inequalities together to get that |x−z|<1 (im trying to show if xRy and yRz then xRz).
Any help would be appreciated, thanks! I am looking for the proof of this last property (transitivity).
relations
asked Dec 3 at 3:31
Carlos Bacca
178116
Define a relation on Z as xRy if |x−y|<1.
I have shown this relation is symetric and reflexive and i am pretty sure its
transitive because this is the equality relation isnt it? thats my first question and my second is how to show it is transitive.
I attempted a direct proof but i dont know how to link the two inequalities together to get that |x−z|<1 (im trying to show if xRy and yRz then xRz).
Any help would be appreciated, thanks! I am looking for the proof of this last property (transitivity).
relations
relations
asked Dec 3 at 3:31
Carlos Bacca
178116
asked Dec 3 at 3:31
Carlos Bacca
178116
edited Dec 3 at 18:18
asked Dec 3 at 3:31
Carlos Bacca
178116
asked Dec 3 at 3:31
Carlos Bacca
178116
asked Dec 3 at 3:31
Carlos Bacca
178116
178116
Thanks, but its a relation on z so only for integer values
– Carlos Bacca
Dec 3 at 4:14
There are no integers satisfying this relation.
– user58697
Dec 3 at 4:41
what about x=4 y=4?
– Carlos Bacca
Dec 3 at 4:43
Sorry I forgot to say distinct. Modulo is non-negative; the only non-negative less than 1 is 0.
– user58697
Dec 3 at 4:49
So is it transitive?
– Carlos Bacca
Dec 3 at 4:53
add a comment |
Thanks, but its a relation on z so only for integer values
– Carlos Bacca
Dec 3 at 4:14
There are no integers satisfying this relation.
– user58697
Dec 3 at 4:41
what about x=4 y=4?
– Carlos Bacca
Dec 3 at 4:43
Sorry I forgot to say distinct. Modulo is non-negative; the only non-negative less than 1 is 0.
– user58697
Dec 3 at 4:49
So is it transitive?
– Carlos Bacca
Dec 3 at 4:53
Thanks, but its a relation on z so only for integer values
– Carlos Bacca
Dec 3 at 4:14
Thanks, but its a relation on z so only for integer values
– Carlos Bacca
Dec 3 at 4:14
There are no integers satisfying this relation.
– user58697
Dec 3 at 4:41
There are no integers satisfying this relation.
– user58697
Dec 3 at 4:41
what about x=4 y=4?
– Carlos Bacca
Dec 3 at 4:43
what about x=4 y=4?
– Carlos Bacca
Dec 3 at 4:43
Sorry I forgot to say distinct. Modulo is non-negative; the only non-negative less than 1 is 0.
– user58697
Dec 3 at 4:49
Sorry I forgot to say distinct. Modulo is non-negative; the only non-negative less than 1 is 0.
– user58697
Dec 3 at 4:49
So is it transitive?
– Carlos Bacca
Dec 3 at 4:53
So is it transitive?
– Carlos Bacca
Dec 3 at 4:53
add a comment |
1 Answer
1
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oldest
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up vote
0
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If your relation is in fact on $Bbb{Z}$, then $xRy$ is the same as writing $x=y$
because if $xne y$ then the distance between x and y is obviously larger than 1.
"=" is transistive thus R is also transistive because the two relations are equivalent
Proof:
If $xne y$ then $y = x+k$, with $kin Bbb{Z}^*$
And $|x-y| = |k| ge 1$, thus no two distinct integers verify this relation
If $xRy$ and $yRz$ that means that $x=y=z$ thus $xRz$ also.
So yes the relation is transistive
answered Dec 3 at 9:59
TheD0ubleT
35618
Thanks, i understand that it is transitive but how woud you show this
– Carlos Bacca
Dec 3 at 18:19
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If your relation is in fact on $Bbb{Z}$, then $xRy$ is the same as writing $x=y$
because if $xne y$ then the distance between x and y is obviously larger than 1.
"=" is transistive thus R is also transistive because the two relations are equivalent
Proof:
If $xne y$ then $y = x+k$, with $kin Bbb{Z}^*$
And $|x-y| = |k| ge 1$, thus no two distinct integers verify this relation
If $xRy$ and $yRz$ that means that $x=y=z$ thus $xRz$ also.
So yes the relation is transistive
answered Dec 3 at 9:59
TheD0ubleT
35618
Thanks, i understand that it is transitive but how woud you show this
– Carlos Bacca
Dec 3 at 18:19
add a comment |
up vote
0
down vote
If your relation is in fact on $Bbb{Z}$, then $xRy$ is the same as writing $x=y$
because if $xne y$ then the distance between x and y is obviously larger than 1.
"=" is transistive thus R is also transistive because the two relations are equivalent
Proof:
If $xne y$ then $y = x+k$, with $kin Bbb{Z}^*$
And $|x-y| = |k| ge 1$, thus no two distinct integers verify this relation
If $xRy$ and $yRz$ that means that $x=y=z$ thus $xRz$ also.
So yes the relation is transistive
answered Dec 3 at 9:59
TheD0ubleT
35618
Thanks, i understand that it is transitive but how woud you show this
– Carlos Bacca
Dec 3 at 18:19
add a comment |
up vote
0
down vote
up vote
0
down vote
If your relation is in fact on $Bbb{Z}$, then $xRy$ is the same as writing $x=y$
because if $xne y$ then the distance between x and y is obviously larger than 1.
"=" is transistive thus R is also transistive because the two relations are equivalent
Proof:
If $xne y$ then $y = x+k$, with $kin Bbb{Z}^*$
And $|x-y| = |k| ge 1$, thus no two distinct integers verify this relation
If $xRy$ and $yRz$ that means that $x=y=z$ thus $xRz$ also.
So yes the relation is transistive
answered Dec 3 at 9:59
TheD0ubleT
35618
If your relation is in fact on $Bbb{Z}$, then $xRy$ is the same as writing $x=y$
because if $xne y$ then the distance between x and y is obviously larger than 1.
"=" is transistive thus R is also transistive because the two relations are equivalent
Proof:
If $xne y$ then $y = x+k$, with $kin Bbb{Z}^*$
And $|x-y| = |k| ge 1$, thus no two distinct integers verify this relation
If $xRy$ and $yRz$ that means that $x=y=z$ thus $xRz$ also.
So yes the relation is transistive
answered Dec 3 at 9:59
TheD0ubleT
35618
edited Dec 4 at 8:50
answered Dec 3 at 9:59
TheD0ubleT
35618
answered Dec 3 at 9:59
TheD0ubleT
35618
answered Dec 3 at 9:59
TheD0ubleT
35618
35618
Thanks, i understand that it is transitive but how woud you show this
– Carlos Bacca
Dec 3 at 18:19
add a comment |
Thanks, i understand that it is transitive but how woud you show this
– Carlos Bacca
Dec 3 at 18:19
Thanks, i understand that it is transitive but how woud you show this
– Carlos Bacca
Dec 3 at 18:19
Thanks, i understand that it is transitive but how woud you show this
– Carlos Bacca
Dec 3 at 18:19
add a comment |
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Thanks, but its a relation on z so only for integer values
– Carlos Bacca
Dec 3 at 4:14
There are no integers satisfying this relation.
– user58697
Dec 3 at 4:41
what about x=4 y=4?
– Carlos Bacca
Dec 3 at 4:43
Sorry I forgot to say distinct. Modulo is non-negative; the only non-negative less than 1 is 0.
– user58697
Dec 3 at 4:49
So is it transitive?
– Carlos Bacca
Dec 3 at 4:53