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

Question

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

Answer 1

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

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

Answer 2

$⟨1, 1⟩ ⪯ ⟨2, 2⟩, \iff 1 ≤ 2 ∧ 1 ≥ 2$

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 $\vec{A}$.

Answer 3

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.