Implications of multiple ways to order four numbers
$begingroup$
Consider two sets $A,B$ composed of two real numbers each.
These four real numbers are in $[0,1]$.
Consider other two real numbers $cin [0,1]$, $din [0,1]$.
Assume there exists a way of ordering the two numbers in each set $A,B$ such that
$$
begin{cases}
w^A_1+w^B_1=c\
w^A_2+w^B_2=d
end{cases}
$$
where
$w^A_h$ denotes the $h$th element of set $A$ once we have ordered its two elements
$w^B_h$ denotes the $h$th element of set $B$ once we have ordered its two element
Claim: if such an ordering is not unique, then it should be that the two numbers in $A$ are equal and/or that the two numbers in $B$ are equal.
Is this claim correct? If yes, how can I prove it? If not, can you provide a counterexample?
linear-algebra combinatorics permutations combinations linear-programming
asked Dec 22 '18 at 17:40
STFSTF
791422
$endgroup$
add a comment |
$begingroup$
Consider two sets $A,B$ composed of two real numbers each.
These four real numbers are in $[0,1]$.
Consider other two real numbers $cin [0,1]$, $din [0,1]$.
Assume there exists a way of ordering the two numbers in each set $A,B$ such that
$$
begin{cases}
w^A_1+w^B_1=c\
w^A_2+w^B_2=d
end{cases}
$$
where
$w^A_h$ denotes the $h$th element of set $A$ once we have ordered its two elements
$w^B_h$ denotes the $h$th element of set $B$ once we have ordered its two element
Claim: if such an ordering is not unique, then it should be that the two numbers in $A$ are equal and/or that the two numbers in $B$ are equal.
Is this claim correct? If yes, how can I prove it? If not, can you provide a counterexample?
linear-algebra combinatorics permutations combinations linear-programming
asked Dec 22 '18 at 17:40
STFSTF
791422
$endgroup$
1
$begingroup$
If such an ordering exists, we have $w_2^A+w_2^B=d=1-c=1-w_1^A-w_1^Bimplies w_1^B+w_2^B=0implies B={0,0}$ which is a contradiction. So the ordering doesn't exist.
$endgroup$
– Shubham Johri
Dec 22 '18 at 19:14
$begingroup$
Yes, sorry, I have deleted the summing up to one. Thanks for the observation.
$endgroup$
– STF
Dec 22 '18 at 19:30
add a comment |
$begingroup$
Consider two sets $A,B$ composed of two real numbers each.
These four real numbers are in $[0,1]$.
Consider other two real numbers $cin [0,1]$, $din [0,1]$.
Assume there exists a way of ordering the two numbers in each set $A,B$ such that
$$
begin{cases}
w^A_1+w^B_1=c\
w^A_2+w^B_2=d
end{cases}
$$
where
$w^A_h$ denotes the $h$th element of set $A$ once we have ordered its two elements
$w^B_h$ denotes the $h$th element of set $B$ once we have ordered its two element
Claim: if such an ordering is not unique, then it should be that the two numbers in $A$ are equal and/or that the two numbers in $B$ are equal.
Is this claim correct? If yes, how can I prove it? If not, can you provide a counterexample?
linear-algebra combinatorics permutations combinations linear-programming
asked Dec 22 '18 at 17:40
STFSTF
791422
$endgroup$
Consider two sets $A,B$ composed of two real numbers each.
These four real numbers are in $[0,1]$.
Consider other two real numbers $cin [0,1]$, $din [0,1]$.
Assume there exists a way of ordering the two numbers in each set $A,B$ such that
$$
begin{cases}
w^A_1+w^B_1=c\
w^A_2+w^B_2=d
end{cases}
$$
where
$w^A_h$ denotes the $h$th element of set $A$ once we have ordered its two elements
$w^B_h$ denotes the $h$th element of set $B$ once we have ordered its two element
Claim: if such an ordering is not unique, then it should be that the two numbers in $A$ are equal and/or that the two numbers in $B$ are equal.
Is this claim correct? If yes, how can I prove it? If not, can you provide a counterexample?
linear-algebra combinatorics permutations combinations linear-programming
linear-algebra combinatorics permutations combinations linear-programming
asked Dec 22 '18 at 17:40
STFSTF
791422
asked Dec 22 '18 at 17:40
STFSTF
791422
edited Dec 22 '18 at 19:30
STF
asked Dec 22 '18 at 17:40
STFSTF
791422
asked Dec 22 '18 at 17:40
STFSTF
791422
asked Dec 22 '18 at 17:40
STFSTF
791422
791422
1
$begingroup$
If such an ordering exists, we have $w_2^A+w_2^B=d=1-c=1-w_1^A-w_1^Bimplies w_1^B+w_2^B=0implies B={0,0}$ which is a contradiction. So the ordering doesn't exist.
$endgroup$
– Shubham Johri
Dec 22 '18 at 19:14
$begingroup$
Yes, sorry, I have deleted the summing up to one. Thanks for the observation.
$endgroup$
– STF
Dec 22 '18 at 19:30
add a comment |
1
$begingroup$
If such an ordering exists, we have $w_2^A+w_2^B=d=1-c=1-w_1^A-w_1^Bimplies w_1^B+w_2^B=0implies B={0,0}$ which is a contradiction. So the ordering doesn't exist.
$endgroup$
– Shubham Johri
Dec 22 '18 at 19:14
$begingroup$
Yes, sorry, I have deleted the summing up to one. Thanks for the observation.
$endgroup$
– STF
Dec 22 '18 at 19:30
1
1
$begingroup$
If such an ordering exists, we have $w_2^A+w_2^B=d=1-c=1-w_1^A-w_1^Bimplies w_1^B+w_2^B=0implies B={0,0}$ which is a contradiction. So the ordering doesn't exist.
$endgroup$
– Shubham Johri
Dec 22 '18 at 19:14
$begingroup$
If such an ordering exists, we have $w_2^A+w_2^B=d=1-c=1-w_1^A-w_1^Bimplies w_1^B+w_2^B=0implies B={0,0}$ which is a contradiction. So the ordering doesn't exist.
$endgroup$
– Shubham Johri
Dec 22 '18 at 19:14
$begingroup$
Yes, sorry, I have deleted the summing up to one. Thanks for the observation.
$endgroup$
– STF
Dec 22 '18 at 19:30
$begingroup$
Yes, sorry, I have deleted the summing up to one. Thanks for the observation.
$endgroup$
– STF
Dec 22 '18 at 19:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Say the ordering is not unique. Then at-least one of $A,B$ can be ordered in two ways. Without loss of generality, let us assume that set is $A$. Keeping the order of $B$ intact, we have
$implies w_1^A+w_1^B=w_2^A+w_1^B=cimplies w_1^A=w_2^A$
Therefore, both the elements of $A$ are identical. We can assume that the ordering of $B$ is not unique and land at a similar conclusion for $B$.
Edit. As suggested by the OP, I considered the case when the ordering of either $A$ or $B$ was not unique, but forgot to consider the case when both their orderings changed. In that case, $c=d$ and the elements of $A,B$ need not be equal.
answered Dec 22 '18 at 19:16
Shubham JohriShubham Johri
5,477818
$endgroup$
$begingroup$
It could be that $w_1^A+w_1^B=c$ and $w_2^A+w_2^B=d$ and that $w_1^A+w_1^B=d$ and $w_2^A+w_2^B=c$ which implies $c=d$ (but the elements of $A$ can be different between each other and the elements of $B$ can be different between each other).
$endgroup$
– STF
Dec 22 '18 at 20:02
$begingroup$
Is it fair to say that if $cneq d$ then your proof is correct?
$endgroup$
– STF
Dec 22 '18 at 20:02
$begingroup$
@STF Yes, you are correct. I considered the case when the ordering of either $A$ or $B$ was not unique, but forgot to consider the case when both their orderings were not unique. In that case, $c=d$ and the elements of $A,B$ need not be equal.
$endgroup$
– Shubham Johri
Dec 22 '18 at 22:23
$begingroup$
Could you help with 4 elements per set if you have some time? math.stackexchange.com/questions/3049836/…
$endgroup$
– STF
Dec 23 '18 at 9:48
$begingroup$
Sure, I'll give it a try
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:53
add a comment |
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$begingroup$
Say the ordering is not unique. Then at-least one of $A,B$ can be ordered in two ways. Without loss of generality, let us assume that set is $A$. Keeping the order of $B$ intact, we have
$implies w_1^A+w_1^B=w_2^A+w_1^B=cimplies w_1^A=w_2^A$
Therefore, both the elements of $A$ are identical. We can assume that the ordering of $B$ is not unique and land at a similar conclusion for $B$.
Edit. As suggested by the OP, I considered the case when the ordering of either $A$ or $B$ was not unique, but forgot to consider the case when both their orderings changed. In that case, $c=d$ and the elements of $A,B$ need not be equal.
answered Dec 22 '18 at 19:16
Shubham JohriShubham Johri
5,477818
$endgroup$
$begingroup$
It could be that $w_1^A+w_1^B=c$ and $w_2^A+w_2^B=d$ and that $w_1^A+w_1^B=d$ and $w_2^A+w_2^B=c$ which implies $c=d$ (but the elements of $A$ can be different between each other and the elements of $B$ can be different between each other).
$endgroup$
– STF
Dec 22 '18 at 20:02
$begingroup$
Is it fair to say that if $cneq d$ then your proof is correct?
$endgroup$
– STF
Dec 22 '18 at 20:02
$begingroup$
@STF Yes, you are correct. I considered the case when the ordering of either $A$ or $B$ was not unique, but forgot to consider the case when both their orderings were not unique. In that case, $c=d$ and the elements of $A,B$ need not be equal.
$endgroup$
– Shubham Johri
Dec 22 '18 at 22:23
$begingroup$
Could you help with 4 elements per set if you have some time? math.stackexchange.com/questions/3049836/…
$endgroup$
– STF
Dec 23 '18 at 9:48
$begingroup$
Sure, I'll give it a try
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:53
add a comment |
$begingroup$
Say the ordering is not unique. Then at-least one of $A,B$ can be ordered in two ways. Without loss of generality, let us assume that set is $A$. Keeping the order of $B$ intact, we have
$implies w_1^A+w_1^B=w_2^A+w_1^B=cimplies w_1^A=w_2^A$
Therefore, both the elements of $A$ are identical. We can assume that the ordering of $B$ is not unique and land at a similar conclusion for $B$.
Edit. As suggested by the OP, I considered the case when the ordering of either $A$ or $B$ was not unique, but forgot to consider the case when both their orderings changed. In that case, $c=d$ and the elements of $A,B$ need not be equal.
answered Dec 22 '18 at 19:16
Shubham JohriShubham Johri
5,477818
$endgroup$
$begingroup$
It could be that $w_1^A+w_1^B=c$ and $w_2^A+w_2^B=d$ and that $w_1^A+w_1^B=d$ and $w_2^A+w_2^B=c$ which implies $c=d$ (but the elements of $A$ can be different between each other and the elements of $B$ can be different between each other).
$endgroup$
– STF
Dec 22 '18 at 20:02
$begingroup$
Is it fair to say that if $cneq d$ then your proof is correct?
$endgroup$
– STF
Dec 22 '18 at 20:02
$begingroup$
@STF Yes, you are correct. I considered the case when the ordering of either $A$ or $B$ was not unique, but forgot to consider the case when both their orderings were not unique. In that case, $c=d$ and the elements of $A,B$ need not be equal.
$endgroup$
– Shubham Johri
Dec 22 '18 at 22:23
$begingroup$
Could you help with 4 elements per set if you have some time? math.stackexchange.com/questions/3049836/…
$endgroup$
– STF
Dec 23 '18 at 9:48
$begingroup$
Sure, I'll give it a try
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:53
add a comment |
$begingroup$
Say the ordering is not unique. Then at-least one of $A,B$ can be ordered in two ways. Without loss of generality, let us assume that set is $A$. Keeping the order of $B$ intact, we have
$implies w_1^A+w_1^B=w_2^A+w_1^B=cimplies w_1^A=w_2^A$
Therefore, both the elements of $A$ are identical. We can assume that the ordering of $B$ is not unique and land at a similar conclusion for $B$.
Edit. As suggested by the OP, I considered the case when the ordering of either $A$ or $B$ was not unique, but forgot to consider the case when both their orderings changed. In that case, $c=d$ and the elements of $A,B$ need not be equal.
answered Dec 22 '18 at 19:16
Shubham JohriShubham Johri
5,477818
$endgroup$
Say the ordering is not unique. Then at-least one of $A,B$ can be ordered in two ways. Without loss of generality, let us assume that set is $A$. Keeping the order of $B$ intact, we have
$implies w_1^A+w_1^B=w_2^A+w_1^B=cimplies w_1^A=w_2^A$
Therefore, both the elements of $A$ are identical. We can assume that the ordering of $B$ is not unique and land at a similar conclusion for $B$.
Edit. As suggested by the OP, I considered the case when the ordering of either $A$ or $B$ was not unique, but forgot to consider the case when both their orderings changed. In that case, $c=d$ and the elements of $A,B$ need not be equal.
answered Dec 22 '18 at 19:16
Shubham JohriShubham Johri
5,477818
edited Dec 22 '18 at 22:25
answered Dec 22 '18 at 19:16
Shubham JohriShubham Johri
5,477818
answered Dec 22 '18 at 19:16
Shubham JohriShubham Johri
5,477818
answered Dec 22 '18 at 19:16
Shubham JohriShubham Johri
5,477818
5,477818
$begingroup$
It could be that $w_1^A+w_1^B=c$ and $w_2^A+w_2^B=d$ and that $w_1^A+w_1^B=d$ and $w_2^A+w_2^B=c$ which implies $c=d$ (but the elements of $A$ can be different between each other and the elements of $B$ can be different between each other).
$endgroup$
– STF
Dec 22 '18 at 20:02
$begingroup$
Is it fair to say that if $cneq d$ then your proof is correct?
$endgroup$
– STF
Dec 22 '18 at 20:02
$begingroup$
@STF Yes, you are correct. I considered the case when the ordering of either $A$ or $B$ was not unique, but forgot to consider the case when both their orderings were not unique. In that case, $c=d$ and the elements of $A,B$ need not be equal.
$endgroup$
– Shubham Johri
Dec 22 '18 at 22:23
$begingroup$
Could you help with 4 elements per set if you have some time? math.stackexchange.com/questions/3049836/…
$endgroup$
– STF
Dec 23 '18 at 9:48
$begingroup$
Sure, I'll give it a try
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:53
add a comment |
$begingroup$
It could be that $w_1^A+w_1^B=c$ and $w_2^A+w_2^B=d$ and that $w_1^A+w_1^B=d$ and $w_2^A+w_2^B=c$ which implies $c=d$ (but the elements of $A$ can be different between each other and the elements of $B$ can be different between each other).
$endgroup$
– STF
Dec 22 '18 at 20:02
$begingroup$
Is it fair to say that if $cneq d$ then your proof is correct?
$endgroup$
– STF
Dec 22 '18 at 20:02
$begingroup$
@STF Yes, you are correct. I considered the case when the ordering of either $A$ or $B$ was not unique, but forgot to consider the case when both their orderings were not unique. In that case, $c=d$ and the elements of $A,B$ need not be equal.
$endgroup$
– Shubham Johri
Dec 22 '18 at 22:23
$begingroup$
Could you help with 4 elements per set if you have some time? math.stackexchange.com/questions/3049836/…
$endgroup$
– STF
Dec 23 '18 at 9:48
$begingroup$
Sure, I'll give it a try
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:53
$begingroup$
It could be that $w_1^A+w_1^B=c$ and $w_2^A+w_2^B=d$ and that $w_1^A+w_1^B=d$ and $w_2^A+w_2^B=c$ which implies $c=d$ (but the elements of $A$ can be different between each other and the elements of $B$ can be different between each other).
$endgroup$
– STF
Dec 22 '18 at 20:02
$begingroup$
It could be that $w_1^A+w_1^B=c$ and $w_2^A+w_2^B=d$ and that $w_1^A+w_1^B=d$ and $w_2^A+w_2^B=c$ which implies $c=d$ (but the elements of $A$ can be different between each other and the elements of $B$ can be different between each other).
$endgroup$
– STF
Dec 22 '18 at 20:02
$begingroup$
Is it fair to say that if $cneq d$ then your proof is correct?
$endgroup$
– STF
Dec 22 '18 at 20:02
$begingroup$
Is it fair to say that if $cneq d$ then your proof is correct?
$endgroup$
– STF
Dec 22 '18 at 20:02
$begingroup$
@STF Yes, you are correct. I considered the case when the ordering of either $A$ or $B$ was not unique, but forgot to consider the case when both their orderings were not unique. In that case, $c=d$ and the elements of $A,B$ need not be equal.
$endgroup$
– Shubham Johri
Dec 22 '18 at 22:23
$begingroup$
@STF Yes, you are correct. I considered the case when the ordering of either $A$ or $B$ was not unique, but forgot to consider the case when both their orderings were not unique. In that case, $c=d$ and the elements of $A,B$ need not be equal.
$endgroup$
– Shubham Johri
Dec 22 '18 at 22:23
$begingroup$
Could you help with 4 elements per set if you have some time? math.stackexchange.com/questions/3049836/…
$endgroup$
– STF
Dec 23 '18 at 9:48
$begingroup$
Could you help with 4 elements per set if you have some time? math.stackexchange.com/questions/3049836/…
$endgroup$
– STF
Dec 23 '18 at 9:48
$begingroup$
Sure, I'll give it a try
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:53
$begingroup$
Sure, I'll give it a try
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:53
add a comment |
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$begingroup$
If such an ordering exists, we have $w_2^A+w_2^B=d=1-c=1-w_1^A-w_1^Bimplies w_1^B+w_2^B=0implies B={0,0}$ which is a contradiction. So the ordering doesn't exist.
$endgroup$
– Shubham Johri
Dec 22 '18 at 19:14
$begingroup$
Yes, sorry, I have deleted the summing up to one. Thanks for the observation.
$endgroup$
– STF
Dec 22 '18 at 19:30