Find two pairs where neither is the case ⟨x1,y1⟩⪯⟨x2,y2⟩, ⟨x1,y1⟩⪰⟨x2,y2⟩. [duplicate]

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This question already has an answer here:

Is $le$ defined by $langle 𝑥_1, 𝑦_1 rangle le langle 𝑥_2, 𝑦_2 rangle$, if $𝑥_1 le 𝑥_2 land𝑦_1 ge 𝑦_2$ a linear order? [on hold]

1 answer

can you help me find any two pairs where neither ⟨x1,y1⟩⪯⟨x2,y2⟩, ⟨x1,y1⟩⪰⟨x2,y2⟩. Neither should be the case.

edited 2 days ago

amWhy

191k27223437

asked 2 days ago

Jack

173

marked as duplicate by amWhy, jgon, Shailesh, Kelvin Lois, Chinnapparaj R 2 days ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1

(1,2) and (2,1)
– Jean Marie
2 days ago

1

How are you defining $preceq$ here?
– dbx
2 days ago

⪯ is defined by relationship ⟨𝑥1, 𝑦1⟩ ⪯ ⟨𝑥2, 𝑦2⟩, if 𝑥1 ≤ 𝑥2 ∧ 𝑦1 ≥ 𝑦2
– Jack
2 days ago

can the pairs commented by Jean Marie be used for my definition?
– Jack
2 days ago

3

Jack, this is a duplicate question of your earlier question. DO NOT REPOST questions, for whatever reason ("I didn't get an adequate answer" nor "My question got closed", nor for any other reason).
– amWhy
2 days ago

|
show 2 more comments

up vote
-2
down vote

favorite

This question already has an answer here:

Is $le$ defined by $langle 𝑥_1, 𝑦_1 rangle le langle 𝑥_2, 𝑦_2 rangle$, if $𝑥_1 le 𝑥_2 land𝑦_1 ge 𝑦_2$ a linear order? [on hold]

1 answer

can you help me find any two pairs where neither ⟨x1,y1⟩⪯⟨x2,y2⟩, ⟨x1,y1⟩⪰⟨x2,y2⟩. Neither should be the case.

edited 2 days ago

amWhy

191k27223437

asked 2 days ago

Jack

173

marked as duplicate by amWhy, jgon, Shailesh, Kelvin Lois, Chinnapparaj R 2 days ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1

(1,2) and (2,1)
– Jean Marie
2 days ago

1

How are you defining $preceq$ here?
– dbx
2 days ago

⪯ is defined by relationship ⟨𝑥1, 𝑦1⟩ ⪯ ⟨𝑥2, 𝑦2⟩, if 𝑥1 ≤ 𝑥2 ∧ 𝑦1 ≥ 𝑦2
– Jack
2 days ago

can the pairs commented by Jean Marie be used for my definition?
– Jack
2 days ago

3

Jack, this is a duplicate question of your earlier question. DO NOT REPOST questions, for whatever reason ("I didn't get an adequate answer" nor "My question got closed", nor for any other reason).
– amWhy
2 days ago

|
show 2 more comments

up vote
-2
down vote

favorite

This question already has an answer here:

Is $le$ defined by $langle 𝑥_1, 𝑦_1 rangle le langle 𝑥_2, 𝑦_2 rangle$, if $𝑥_1 le 𝑥_2 land𝑦_1 ge 𝑦_2$ a linear order? [on hold]

1 answer

can you help me find any two pairs where neither ⟨x1,y1⟩⪯⟨x2,y2⟩, ⟨x1,y1⟩⪰⟨x2,y2⟩. Neither should be the case.

edited 2 days ago

amWhy

191k27223437

asked 2 days ago

Jack

173

This question already has an answer here:

Is $le$ defined by $langle 𝑥_1, 𝑦_1 rangle le langle 𝑥_2, 𝑦_2 rangle$, if $𝑥_1 le 𝑥_2 land𝑦_1 ge 𝑦_2$ a linear order? [on hold]

1 answer

can you help me find any two pairs where neither ⟨x1,y1⟩⪯⟨x2,y2⟩, ⟨x1,y1⟩⪰⟨x2,y2⟩. Neither should be the case.

This question already has an answer here:

Is $le$ defined by $langle 𝑥_1, 𝑦_1 rangle le langle 𝑥_2, 𝑦_2 rangle$, if $𝑥_1 le 𝑥_2 land𝑦_1 ge 𝑦_2$ a linear order? [on hold]

1 answer

order-theory

edited 2 days ago

amWhy

191k27223437

asked 2 days ago

Jack

173

edited 2 days ago

amWhy

191k27223437

asked 2 days ago

Jack

173

edited 2 days ago

amWhy

191k27223437

edited 2 days ago

amWhy

191k27223437

edited 2 days ago

amWhy

191k27223437

asked 2 days ago

Jack

173

asked 2 days ago

Jack

173

asked 2 days ago

Jack

173

marked as duplicate by amWhy, jgon, Shailesh, Kelvin Lois, Chinnapparaj R 2 days ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by amWhy, jgon, Shailesh, Kelvin Lois, Chinnapparaj R 2 days ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1

(1,2) and (2,1)
– Jean Marie
2 days ago

1

How are you defining $preceq$ here?
– dbx
2 days ago

⪯ is defined by relationship ⟨𝑥1, 𝑦1⟩ ⪯ ⟨𝑥2, 𝑦2⟩, if 𝑥1 ≤ 𝑥2 ∧ 𝑦1 ≥ 𝑦2
– Jack
2 days ago

can the pairs commented by Jean Marie be used for my definition?
– Jack
2 days ago

3

Jack, this is a duplicate question of your earlier question. DO NOT REPOST questions, for whatever reason ("I didn't get an adequate answer" nor "My question got closed", nor for any other reason).
– amWhy
2 days ago

|
show 2 more comments

1

(1,2) and (2,1)
– Jean Marie
2 days ago

1

How are you defining $preceq$ here?
– dbx
2 days ago

⪯ is defined by relationship ⟨𝑥1, 𝑦1⟩ ⪯ ⟨𝑥2, 𝑦2⟩, if 𝑥1 ≤ 𝑥2 ∧ 𝑦1 ≥ 𝑦2
– Jack
2 days ago

can the pairs commented by Jean Marie be used for my definition?
– Jack
2 days ago

3

Jack, this is a duplicate question of your earlier question. DO NOT REPOST questions, for whatever reason ("I didn't get an adequate answer" nor "My question got closed", nor for any other reason).
– amWhy
2 days ago

(1,2) and (2,1)
– Jean Marie
2 days ago

How are you defining $preceq$ here?
– dbx
2 days ago

⪯ is defined by relationship ⟨𝑥1, 𝑦1⟩ ⪯ ⟨𝑥2, 𝑦2⟩, if 𝑥1 ≤ 𝑥2 ∧ 𝑦1 ≥ 𝑦2
– Jack
2 days ago

can the pairs commented by Jean Marie be used for my definition?
– Jack
2 days ago

Jack, this is a duplicate question of your earlier question. DO NOT REPOST questions, for whatever reason ("I didn't get an adequate answer" nor "My question got closed", nor for any other reason).
– amWhy
2 days ago

|
show 2 more comments

3 Answers
3

active

oldest

votes

up vote
0
down vote

$⟨𝑥1, 𝑦1⟩ ⪯ ⟨𝑥2, 𝑦2⟩, iff 𝑥1 ≤ 𝑥2 ∧ 𝑦1 ≥ 𝑦2$

Imagining the graph of this:

So we write $vec{A} ⪯ vec{B}$ when either $vec{A}$ appears to the left(or not to the right of) of $vec{B}$ AND when $vec{A}$ appears above (or not below) $vec{B}$.

So then consider $vec{A}=(0,0)$ and $vec{B}=(1,1)$.

$vec{A} ⪯ vec{B}$ is false because $0=y_1 <y_2=1$

$vec{B} ⪯ vec{A}$ is false because $1=x_1 > x_2=0$

$vec{A}$ is not above $vec{B}$ and $vec{B}$ is not to the left of $vec{A}$.

answered 2 days ago

Mason

1,6521325

so ⟨1, 0⟩ ⪯ ⟨0, 1⟩ and ⟨0, 1⟩ ⪯ ⟨1, 0⟩?
– Jack
2 days ago

Wait what? Those aren't the points I give. I may be reading this wrong but do $(0,0)$ and $(1,1)$ fit the bill?
– Mason
2 days ago

Then I'm not sure I understand. Can you give me any two pairs?
– Jack
2 days ago

@Jack+ anyone: Do $(0,0)$ and $(1,1)$ fit the criterion of the problem? Because you should be able to find more if it meets your needs but I am worried I am misreading this problem.
– Mason
2 days ago

Well, my answer would be ⟨1, 2⟩ ⪯ ⟨2, 1⟩ and ⟨2, 1⟩ ⪰ ⟨1, 2⟩ but afaik ⟨1, 2⟩ ⪯ ⟨2, 1⟩ is true, right? I need both to be false, if I understand this correctly.
– Jack
2 days ago

|
show 2 more comments

up vote
0
down vote

We need to negate

$$((x_1 leq x_2) wedge (y_1 geq y_2)) vee ((x_2 leq x_1) wedge (y_2 geq y_1))$$

which is just the combination of the two relations $(x_1,y_1) preceq (x_2,y_2)$ and vice-versa.

By DeMorgan, this is equivalent to:
$$ neg ((x_1 leq x_2) wedge (y_1 geq y_2)) wedge neg ((x_2 leq x_1) wedge (y_2 geq y_1)) $$

$$ equiv((x_1 > x_2) vee (y_1 < y_2)) wedge ((x_2 < x_1) vee (y_2 < y_1)) $$

$(2,1)$ and $(1,0)$ satisfy the above, since $x_1 > x_2$ and $y_2 < y_1$. You can confirm that they also satisfy the original statement.

answered 2 days ago

dbx

1,537311

add a comment |

up vote
-1
down vote

Choose
$x1 < y1$ and
$x2 > y2$.

(1,2) and (2,1)
as Jean Marie commented.

answered 2 days ago

marty cohen

71.3k546123

1

so ⟨1, 2⟩ ⪯ ⟨2, 1⟩ and ⟨2, 1⟩ ⪰ ⟨1, 2⟩? afaik the first one is right (therefore can't be used)
– Jack
2 days ago

One of us is reading this problem wrong... and it could be me.
– Mason
2 days ago

No : the point is that you have neither (a) (1,2)⪯(2,1) nor (b) (2,1)⪯(1,2). Let us explain (a) for example : one should have $1 leq 2$ AND $2 leq 1$ ; but the second inequality is false...
– Jean Marie
2 days ago

$⟨1, 2⟩ ⪯ ⟨2, 1⟩$ because $1=x_1 le x_2=2$ and $2=y_1 ge y_2=1$
– Mason
2 days ago

1

marty How can you even begin to answer when you don't event know the set in which $x_1, x_2, y_1, y_2$ reside? Pretty careless answer, seems to me.
– amWhy
2 days ago

|
show 4 more comments

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
0
down vote

$⟨𝑥1, 𝑦1⟩ ⪯ ⟨𝑥2, 𝑦2⟩, iff 𝑥1 ≤ 𝑥2 ∧ 𝑦1 ≥ 𝑦2$

Imagining the graph of this:

So we write $vec{A} ⪯ vec{B}$ when either $vec{A}$ appears to the left(or not to the right of) of $vec{B}$ AND when $vec{A}$ appears above (or not below) $vec{B}$.

So then consider $vec{A}=(0,0)$ and $vec{B}=(1,1)$.

$vec{A} ⪯ vec{B}$ is false because $0=y_1 <y_2=1$

$vec{B} ⪯ vec{A}$ is false because $1=x_1 > x_2=0$

$vec{A}$ is not above $vec{B}$ and $vec{B}$ is not to the left of $vec{A}$.

answered 2 days ago

Mason

1,6521325

so ⟨1, 0⟩ ⪯ ⟨0, 1⟩ and ⟨0, 1⟩ ⪯ ⟨1, 0⟩?
– Jack
2 days ago

Wait what? Those aren't the points I give. I may be reading this wrong but do $(0,0)$ and $(1,1)$ fit the bill?
– Mason
2 days ago

Then I'm not sure I understand. Can you give me any two pairs?
– Jack
2 days ago

@Jack+ anyone: Do $(0,0)$ and $(1,1)$ fit the criterion of the problem? Because you should be able to find more if it meets your needs but I am worried I am misreading this problem.
– Mason
2 days ago

Well, my answer would be ⟨1, 2⟩ ⪯ ⟨2, 1⟩ and ⟨2, 1⟩ ⪰ ⟨1, 2⟩ but afaik ⟨1, 2⟩ ⪯ ⟨2, 1⟩ is true, right? I need both to be false, if I understand this correctly.
– Jack
2 days ago

|
show 2 more comments

up vote
0
down vote

$⟨𝑥1, 𝑦1⟩ ⪯ ⟨𝑥2, 𝑦2⟩, iff 𝑥1 ≤ 𝑥2 ∧ 𝑦1 ≥ 𝑦2$

Imagining the graph of this:

So we write $vec{A} ⪯ vec{B}$ when either $vec{A}$ appears to the left(or not to the right of) of $vec{B}$ AND when $vec{A}$ appears above (or not below) $vec{B}$.

So then consider $vec{A}=(0,0)$ and $vec{B}=(1,1)$.

$vec{A} ⪯ vec{B}$ is false because $0=y_1 <y_2=1$

$vec{B} ⪯ vec{A}$ is false because $1=x_1 > x_2=0$

$vec{A}$ is not above $vec{B}$ and $vec{B}$ is not to the left of $vec{A}$.

answered 2 days ago

Mason

1,6521325

so ⟨1, 0⟩ ⪯ ⟨0, 1⟩ and ⟨0, 1⟩ ⪯ ⟨1, 0⟩?
– Jack
2 days ago

Wait what? Those aren't the points I give. I may be reading this wrong but do $(0,0)$ and $(1,1)$ fit the bill?
– Mason
2 days ago

Then I'm not sure I understand. Can you give me any two pairs?
– Jack
2 days ago

@Jack+ anyone: Do $(0,0)$ and $(1,1)$ fit the criterion of the problem? Because you should be able to find more if it meets your needs but I am worried I am misreading this problem.
– Mason
2 days ago

Well, my answer would be ⟨1, 2⟩ ⪯ ⟨2, 1⟩ and ⟨2, 1⟩ ⪰ ⟨1, 2⟩ but afaik ⟨1, 2⟩ ⪯ ⟨2, 1⟩ is true, right? I need both to be false, if I understand this correctly.
– Jack
2 days ago

|
show 2 more comments

up vote
0
down vote

$⟨𝑥1, 𝑦1⟩ ⪯ ⟨𝑥2, 𝑦2⟩, iff 𝑥1 ≤ 𝑥2 ∧ 𝑦1 ≥ 𝑦2$

Imagining the graph of this:

So we write $vec{A} ⪯ vec{B}$ when either $vec{A}$ appears to the left(or not to the right of) of $vec{B}$ AND when $vec{A}$ appears above (or not below) $vec{B}$.

So then consider $vec{A}=(0,0)$ and $vec{B}=(1,1)$.

$vec{A} ⪯ vec{B}$ is false because $0=y_1 <y_2=1$

$vec{B} ⪯ vec{A}$ is false because $1=x_1 > x_2=0$

$vec{A}$ is not above $vec{B}$ and $vec{B}$ is not to the left of $vec{A}$.

answered 2 days ago

Mason

1,6521325

$⟨𝑥1, 𝑦1⟩ ⪯ ⟨𝑥2, 𝑦2⟩, iff 𝑥1 ≤ 𝑥2 ∧ 𝑦1 ≥ 𝑦2$

Imagining the graph of this:

So we write $vec{A} ⪯ vec{B}$ when either $vec{A}$ appears to the left(or not to the right of) of $vec{B}$ AND when $vec{A}$ appears above (or not below) $vec{B}$.

So then consider $vec{A}=(0,0)$ and $vec{B}=(1,1)$.

$vec{A} ⪯ vec{B}$ is false because $0=y_1 <y_2=1$

$vec{B} ⪯ vec{A}$ is false because $1=x_1 > x_2=0$

$vec{A}$ is not above $vec{B}$ and $vec{B}$ is not to the left of $vec{A}$.

answered 2 days ago

Mason

1,6521325

answered 2 days ago

Mason

1,6521325

answered 2 days ago

Mason

1,6521325

answered 2 days ago

Mason

1,6521325

so ⟨1, 0⟩ ⪯ ⟨0, 1⟩ and ⟨0, 1⟩ ⪯ ⟨1, 0⟩?
– Jack
2 days ago

Wait what? Those aren't the points I give. I may be reading this wrong but do $(0,0)$ and $(1,1)$ fit the bill?
– Mason
2 days ago

Then I'm not sure I understand. Can you give me any two pairs?
– Jack
2 days ago

@Jack+ anyone: Do $(0,0)$ and $(1,1)$ fit the criterion of the problem? Because you should be able to find more if it meets your needs but I am worried I am misreading this problem.
– Mason
2 days ago

Well, my answer would be ⟨1, 2⟩ ⪯ ⟨2, 1⟩ and ⟨2, 1⟩ ⪰ ⟨1, 2⟩ but afaik ⟨1, 2⟩ ⪯ ⟨2, 1⟩ is true, right? I need both to be false, if I understand this correctly.
– Jack
2 days ago

|
show 2 more comments

so ⟨1, 0⟩ ⪯ ⟨0, 1⟩ and ⟨0, 1⟩ ⪯ ⟨1, 0⟩?
– Jack
2 days ago

Wait what? Those aren't the points I give. I may be reading this wrong but do $(0,0)$ and $(1,1)$ fit the bill?
– Mason
2 days ago

Then I'm not sure I understand. Can you give me any two pairs?
– Jack
2 days ago

@Jack+ anyone: Do $(0,0)$ and $(1,1)$ fit the criterion of the problem? Because you should be able to find more if it meets your needs but I am worried I am misreading this problem.
– Mason
2 days ago

Well, my answer would be ⟨1, 2⟩ ⪯ ⟨2, 1⟩ and ⟨2, 1⟩ ⪰ ⟨1, 2⟩ but afaik ⟨1, 2⟩ ⪯ ⟨2, 1⟩ is true, right? I need both to be false, if I understand this correctly.
– Jack
2 days ago

so ⟨1, 0⟩ ⪯ ⟨0, 1⟩ and ⟨0, 1⟩ ⪯ ⟨1, 0⟩?
– Jack
2 days ago

Wait what? Those aren't the points I give. I may be reading this wrong but do $(0,0)$ and $(1,1)$ fit the bill?
– Mason
2 days ago

Then I'm not sure I understand. Can you give me any two pairs?
– Jack
2 days ago

@Jack+ anyone: Do $(0,0)$ and $(1,1)$ fit the criterion of the problem? Because you should be able to find more if it meets your needs but I am worried I am misreading this problem.
– Mason
2 days ago

Well, my answer would be ⟨1, 2⟩ ⪯ ⟨2, 1⟩ and ⟨2, 1⟩ ⪰ ⟨1, 2⟩ but afaik ⟨1, 2⟩ ⪯ ⟨2, 1⟩ is true, right? I need both to be false, if I understand this correctly.
– Jack
2 days ago

|
show 2 more comments

up vote
0
down vote

We need to negate

$$((x_1 leq x_2) wedge (y_1 geq y_2)) vee ((x_2 leq x_1) wedge (y_2 geq y_1))$$

which is just the combination of the two relations $(x_1,y_1) preceq (x_2,y_2)$ and vice-versa.

By DeMorgan, this is equivalent to:
$$ neg ((x_1 leq x_2) wedge (y_1 geq y_2)) wedge neg ((x_2 leq x_1) wedge (y_2 geq y_1)) $$

$$ equiv((x_1 > x_2) vee (y_1 < y_2)) wedge ((x_2 < x_1) vee (y_2 < y_1)) $$

$(2,1)$ and $(1,0)$ satisfy the above, since $x_1 > x_2$ and $y_2 < y_1$. You can confirm that they also satisfy the original statement.

answered 2 days ago

dbx

1,537311

add a comment |

up vote
0
down vote

We need to negate

$$((x_1 leq x_2) wedge (y_1 geq y_2)) vee ((x_2 leq x_1) wedge (y_2 geq y_1))$$

which is just the combination of the two relations $(x_1,y_1) preceq (x_2,y_2)$ and vice-versa.

By DeMorgan, this is equivalent to:
$$ neg ((x_1 leq x_2) wedge (y_1 geq y_2)) wedge neg ((x_2 leq x_1) wedge (y_2 geq y_1)) $$

$$ equiv((x_1 > x_2) vee (y_1 < y_2)) wedge ((x_2 < x_1) vee (y_2 < y_1)) $$

$(2,1)$ and $(1,0)$ satisfy the above, since $x_1 > x_2$ and $y_2 < y_1$. You can confirm that they also satisfy the original statement.

answered 2 days ago

dbx

1,537311

add a comment |

up vote
0
down vote

We need to negate

$$((x_1 leq x_2) wedge (y_1 geq y_2)) vee ((x_2 leq x_1) wedge (y_2 geq y_1))$$

which is just the combination of the two relations $(x_1,y_1) preceq (x_2,y_2)$ and vice-versa.

By DeMorgan, this is equivalent to:
$$ neg ((x_1 leq x_2) wedge (y_1 geq y_2)) wedge neg ((x_2 leq x_1) wedge (y_2 geq y_1)) $$

$$ equiv((x_1 > x_2) vee (y_1 < y_2)) wedge ((x_2 < x_1) vee (y_2 < y_1)) $$

$(2,1)$ and $(1,0)$ satisfy the above, since $x_1 > x_2$ and $y_2 < y_1$. You can confirm that they also satisfy the original statement.

answered 2 days ago

dbx

1,537311

We need to negate

$$((x_1 leq x_2) wedge (y_1 geq y_2)) vee ((x_2 leq x_1) wedge (y_2 geq y_1))$$

which is just the combination of the two relations $(x_1,y_1) preceq (x_2,y_2)$ and vice-versa.

By DeMorgan, this is equivalent to:
$$ neg ((x_1 leq x_2) wedge (y_1 geq y_2)) wedge neg ((x_2 leq x_1) wedge (y_2 geq y_1)) $$

$$ equiv((x_1 > x_2) vee (y_1 < y_2)) wedge ((x_2 < x_1) vee (y_2 < y_1)) $$

$(2,1)$ and $(1,0)$ satisfy the above, since $x_1 > x_2$ and $y_2 < y_1$. You can confirm that they also satisfy the original statement.

answered 2 days ago

dbx

1,537311

answered 2 days ago

dbx

1,537311

answered 2 days ago

dbx

1,537311

answered 2 days ago

dbx

1,537311

add a comment |

up vote
-1
down vote

Choose
$x1 < y1$ and
$x2 > y2$.

(1,2) and (2,1)
as Jean Marie commented.

answered 2 days ago

marty cohen

71.3k546123

1

so ⟨1, 2⟩ ⪯ ⟨2, 1⟩ and ⟨2, 1⟩ ⪰ ⟨1, 2⟩? afaik the first one is right (therefore can't be used)
– Jack
2 days ago

One of us is reading this problem wrong... and it could be me.
– Mason
2 days ago

No : the point is that you have neither (a) (1,2)⪯(2,1) nor (b) (2,1)⪯(1,2). Let us explain (a) for example : one should have $1 leq 2$ AND $2 leq 1$ ; but the second inequality is false...
– Jean Marie
2 days ago

$⟨1, 2⟩ ⪯ ⟨2, 1⟩$ because $1=x_1 le x_2=2$ and $2=y_1 ge y_2=1$
– Mason
2 days ago

1

marty How can you even begin to answer when you don't event know the set in which $x_1, x_2, y_1, y_2$ reside? Pretty careless answer, seems to me.
– amWhy
2 days ago

|
show 4 more comments

up vote
-1
down vote

Choose
$x1 < y1$ and
$x2 > y2$.

(1,2) and (2,1)
as Jean Marie commented.

answered 2 days ago

marty cohen

71.3k546123

1

so ⟨1, 2⟩ ⪯ ⟨2, 1⟩ and ⟨2, 1⟩ ⪰ ⟨1, 2⟩? afaik the first one is right (therefore can't be used)
– Jack
2 days ago

One of us is reading this problem wrong... and it could be me.
– Mason
2 days ago

No : the point is that you have neither (a) (1,2)⪯(2,1) nor (b) (2,1)⪯(1,2). Let us explain (a) for example : one should have $1 leq 2$ AND $2 leq 1$ ; but the second inequality is false...
– Jean Marie
2 days ago

$⟨1, 2⟩ ⪯ ⟨2, 1⟩$ because $1=x_1 le x_2=2$ and $2=y_1 ge y_2=1$
– Mason
2 days ago

1

marty How can you even begin to answer when you don't event know the set in which $x_1, x_2, y_1, y_2$ reside? Pretty careless answer, seems to me.
– amWhy
2 days ago

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up vote
-1
down vote

Choose
$x1 < y1$ and
$x2 > y2$.

(1,2) and (2,1)
as Jean Marie commented.

answered 2 days ago

marty cohen

71.3k546123

Choose
$x1 < y1$ and
$x2 > y2$.

(1,2) and (2,1)
as Jean Marie commented.

answered 2 days ago

marty cohen

71.3k546123

answered 2 days ago

marty cohen

71.3k546123

answered 2 days ago

marty cohen

71.3k546123

answered 2 days ago

marty cohen

71.3k546123

1

so ⟨1, 2⟩ ⪯ ⟨2, 1⟩ and ⟨2, 1⟩ ⪰ ⟨1, 2⟩? afaik the first one is right (therefore can't be used)
– Jack
2 days ago

One of us is reading this problem wrong... and it could be me.
– Mason
2 days ago

No : the point is that you have neither (a) (1,2)⪯(2,1) nor (b) (2,1)⪯(1,2). Let us explain (a) for example : one should have $1 leq 2$ AND $2 leq 1$ ; but the second inequality is false...
– Jean Marie
2 days ago

$⟨1, 2⟩ ⪯ ⟨2, 1⟩$ because $1=x_1 le x_2=2$ and $2=y_1 ge y_2=1$
– Mason
2 days ago

1

marty How can you even begin to answer when you don't event know the set in which $x_1, x_2, y_1, y_2$ reside? Pretty careless answer, seems to me.
– amWhy
2 days ago

|
show 4 more comments

1

so ⟨1, 2⟩ ⪯ ⟨2, 1⟩ and ⟨2, 1⟩ ⪰ ⟨1, 2⟩? afaik the first one is right (therefore can't be used)
– Jack
2 days ago

One of us is reading this problem wrong... and it could be me.
– Mason
2 days ago

No : the point is that you have neither (a) (1,2)⪯(2,1) nor (b) (2,1)⪯(1,2). Let us explain (a) for example : one should have $1 leq 2$ AND $2 leq 1$ ; but the second inequality is false...
– Jean Marie
2 days ago

$⟨1, 2⟩ ⪯ ⟨2, 1⟩$ because $1=x_1 le x_2=2$ and $2=y_1 ge y_2=1$
– Mason
2 days ago

1

marty How can you even begin to answer when you don't event know the set in which $x_1, x_2, y_1, y_2$ reside? Pretty careless answer, seems to me.
– amWhy
2 days ago

so ⟨1, 2⟩ ⪯ ⟨2, 1⟩ and ⟨2, 1⟩ ⪰ ⟨1, 2⟩? afaik the first one is right (therefore can't be used)
– Jack
2 days ago

One of us is reading this problem wrong... and it could be me.
– Mason
2 days ago

No : the point is that you have neither (a) (1,2)⪯(2,1) nor (b) (2,1)⪯(1,2). Let us explain (a) for example : one should have $1 leq 2$ AND $2 leq 1$ ; but the second inequality is false...
– Jean Marie
2 days ago

$⟨1, 2⟩ ⪯ ⟨2, 1⟩$ because $1=x_1 le x_2=2$ and $2=y_1 ge y_2=1$
– Mason
2 days ago

marty How can you even begin to answer when you don't event know the set in which $x_1, x_2, y_1, y_2$ reside? Pretty careless answer, seems to me.
– amWhy
2 days ago

|
show 4 more comments

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