determine total order am i correct The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Discrete Mathematics: Is relation $f$ a function?Finding the smallest relation that is reflexive, transitive, and symmetricWhy is the listing $3 , 2 , 2,3 , 1, 1,2 , 1,3 , 1,2,3$ considered a total order?Should I repeat the element of a composite of a relation?Confusion on in order problem with combinationsFinding pairs with respect to lexicographic order that meet a condition from a set?Characteristics of relations. Are these relations correct?Am I correct about the transitive closure of this relation?Find the set ∆XIs there any work on partition a partial order set into minimum number total order subsets?
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determine total order am i correct
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Discrete Mathematics: Is relation $f$ a function?Finding the smallest relation that is reflexive, transitive, and symmetricWhy is the listing $3 , 2 , 2,3 , 1, 1,2 , 1,3 , 1,2,3$ considered a total order?Should I repeat the element of a composite of a relation?Confusion on in order problem with combinationsFinding pairs with respect to lexicographic order that meet a condition from a set?Characteristics of relations. Are these relations correct?Am I correct about the transitive closure of this relation?Find the set ∆XIs there any work on partition a partial order set into minimum number total order subsets?
$begingroup$
T on N×N so that (a,b)T(c,d) iff ab less than cd
let's abstract as (a,b)=x,(c,d)=y, there is xTy or yTx .
((0,0)=0,(1,1)=1) , ((0,0)=0,(1,2)=2) , ((0,0)=0,(1,3)=3) , ((0,0)=0,(1,4)=4)....so on
((1,1)=1,(1,2)=2) , ((1,1)=1,(1,3)=3) , ((1,1)=1,(1,4)=4)....so on
((2,1)=2,(1,3)=3) , ((2,1)=2,(1,4)=4))....so on
am i correct?all are comparable.and it is a total order.
discrete-mathematics
asked Mar 25 at 7:31
tobytoby
84
$endgroup$
add a comment |
$begingroup$
T on N×N so that (a,b)T(c,d) iff ab less than cd
let's abstract as (a,b)=x,(c,d)=y, there is xTy or yTx .
((0,0)=0,(1,1)=1) , ((0,0)=0,(1,2)=2) , ((0,0)=0,(1,3)=3) , ((0,0)=0,(1,4)=4)....so on
((1,1)=1,(1,2)=2) , ((1,1)=1,(1,3)=3) , ((1,1)=1,(1,4)=4)....so on
((2,1)=2,(1,3)=3) , ((2,1)=2,(1,4)=4))....so on
am i correct?all are comparable.and it is a total order.
discrete-mathematics
asked Mar 25 at 7:31
tobytoby
84
$endgroup$
$begingroup$
Usespacesforreadibility.
$endgroup$
– William Elliot
Mar 25 at 8:49
$begingroup$
thx i corrected it
$endgroup$
– toby
Mar 25 at 12:21
add a comment |
$begingroup$
T on N×N so that (a,b)T(c,d) iff ab less than cd
let's abstract as (a,b)=x,(c,d)=y, there is xTy or yTx .
((0,0)=0,(1,1)=1) , ((0,0)=0,(1,2)=2) , ((0,0)=0,(1,3)=3) , ((0,0)=0,(1,4)=4)....so on
((1,1)=1,(1,2)=2) , ((1,1)=1,(1,3)=3) , ((1,1)=1,(1,4)=4)....so on
((2,1)=2,(1,3)=3) , ((2,1)=2,(1,4)=4))....so on
am i correct?all are comparable.and it is a total order.
discrete-mathematics
asked Mar 25 at 7:31
tobytoby
84
$endgroup$
T on N×N so that (a,b)T(c,d) iff ab less than cd
let's abstract as (a,b)=x,(c,d)=y, there is xTy or yTx .
((0,0)=0,(1,1)=1) , ((0,0)=0,(1,2)=2) , ((0,0)=0,(1,3)=3) , ((0,0)=0,(1,4)=4)....so on
((1,1)=1,(1,2)=2) , ((1,1)=1,(1,3)=3) , ((1,1)=1,(1,4)=4)....so on
((2,1)=2,(1,3)=3) , ((2,1)=2,(1,4)=4))....so on
am i correct?all are comparable.and it is a total order.
discrete-mathematics
discrete-mathematics
asked Mar 25 at 7:31
tobytoby
84
asked Mar 25 at 7:31
tobytoby
84
edited Mar 25 at 12:21
toby
asked Mar 25 at 7:31
tobytoby
84
asked Mar 25 at 7:31
tobytoby
84
asked Mar 25 at 7:31
tobytoby
84
84
$begingroup$
Usespacesforreadibility.
$endgroup$
– William Elliot
Mar 25 at 8:49
$begingroup$
thx i corrected it
$endgroup$
– toby
Mar 25 at 12:21
add a comment |
$begingroup$
Usespacesforreadibility.
$endgroup$
– William Elliot
Mar 25 at 8:49
$begingroup$
thx i corrected it
$endgroup$
– toby
Mar 25 at 12:21
$begingroup$
Usespacesforreadibility.
$endgroup$
– William Elliot
Mar 25 at 8:49
$begingroup$
Usespacesforreadibility.
$endgroup$
– William Elliot
Mar 25 at 8:49
$begingroup$
thx i corrected it
$endgroup$
– toby
Mar 25 at 12:21
$begingroup$
thx i corrected it
$endgroup$
– toby
Mar 25 at 12:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
T is not a linear (total) order because
neither (0,1)T(1,0) nor (1,0)T(0,1).
answered Mar 25 at 12:56
William ElliotWilliam Elliot
9,1812820
$endgroup$
$begingroup$
sorry,but if (0,1)T(1,0) appear,does it violate anti-symmetric which is property of total order?
$endgroup$
– toby
Mar 25 at 13:00
$begingroup$
It does not appear. @toby
$endgroup$
– William Elliot
Mar 26 at 0:48
add a comment |
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1 Answer
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1 Answer
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$begingroup$
T is not a linear (total) order because
neither (0,1)T(1,0) nor (1,0)T(0,1).
answered Mar 25 at 12:56
William ElliotWilliam Elliot
9,1812820
$endgroup$
$begingroup$
sorry,but if (0,1)T(1,0) appear,does it violate anti-symmetric which is property of total order?
$endgroup$
– toby
Mar 25 at 13:00
$begingroup$
It does not appear. @toby
$endgroup$
– William Elliot
Mar 26 at 0:48
add a comment |
$begingroup$
T is not a linear (total) order because
neither (0,1)T(1,0) nor (1,0)T(0,1).
answered Mar 25 at 12:56
William ElliotWilliam Elliot
9,1812820
$endgroup$
$begingroup$
sorry,but if (0,1)T(1,0) appear,does it violate anti-symmetric which is property of total order?
$endgroup$
– toby
Mar 25 at 13:00
$begingroup$
It does not appear. @toby
$endgroup$
– William Elliot
Mar 26 at 0:48
add a comment |
$begingroup$
T is not a linear (total) order because
neither (0,1)T(1,0) nor (1,0)T(0,1).
answered Mar 25 at 12:56
William ElliotWilliam Elliot
9,1812820
$endgroup$
T is not a linear (total) order because
neither (0,1)T(1,0) nor (1,0)T(0,1).
answered Mar 25 at 12:56
William ElliotWilliam Elliot
9,1812820
answered Mar 25 at 12:56
William ElliotWilliam Elliot
9,1812820
answered Mar 25 at 12:56
William ElliotWilliam Elliot
9,1812820
answered Mar 25 at 12:56
William ElliotWilliam Elliot
9,1812820
9,1812820
$begingroup$
sorry,but if (0,1)T(1,0) appear,does it violate anti-symmetric which is property of total order?
$endgroup$
– toby
Mar 25 at 13:00
$begingroup$
It does not appear. @toby
$endgroup$
– William Elliot
Mar 26 at 0:48
add a comment |
$begingroup$
sorry,but if (0,1)T(1,0) appear,does it violate anti-symmetric which is property of total order?
$endgroup$
– toby
Mar 25 at 13:00
$begingroup$
It does not appear. @toby
$endgroup$
– William Elliot
Mar 26 at 0:48
$begingroup$
sorry,but if (0,1)T(1,0) appear,does it violate anti-symmetric which is property of total order?
$endgroup$
– toby
Mar 25 at 13:00
$begingroup$
sorry,but if (0,1)T(1,0) appear,does it violate anti-symmetric which is property of total order?
$endgroup$
– toby
Mar 25 at 13:00
$begingroup$
It does not appear. @toby
$endgroup$
– William Elliot
Mar 26 at 0:48
$begingroup$
It does not appear. @toby
$endgroup$
– William Elliot
Mar 26 at 0:48
add a comment |
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$begingroup$
Usespacesforreadibility.
$endgroup$
– William Elliot
Mar 25 at 8:49
$begingroup$
thx i corrected it
$endgroup$
– toby
Mar 25 at 12:21