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I need to tell if the relation $\le$ defined by relationship $\langle 𝑥_1, 𝑦_1 \rangle \le \langle 𝑥_2, 𝑦_2 \rangle$, if $𝑥_1 \le 𝑥_2$ $\land$ $𝑦_1 \ge 𝑦_2$ linear order?

I have already proven that the relation is an order, but I need to decide if it is a linear order and why. Thanks for any help!

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  • $\begingroup$ In what set are $x1, y1, x2, y2$ elements? One can only define a linear order on a specified set. $\endgroup$ – Namaste Nov 19 '18 at 22:49
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A linear order needs two elements to be comparable i.e. for $\left<x_1,y_1\right>,\left<x_2,y_2\right>$ we have either $$\left<x_1,y_1\right>\preceq\left<x_2,y_2\right>$$ or $$\left<x_2,y_2\right>\preceq\left<x_1,y_1\right>.$$ If you can find two pairs where neither is the case, you have proven that the relation can't be a linear order.

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    $\begingroup$ so ⟨1, 2⟩ ⪯ ⟨2, 1⟩ and ⟨2, 1⟩ ⪰ ⟨1, 2⟩? $\endgroup$ – Jack Nov 19 '18 at 20:16
  • $\begingroup$ that's the same thing. $\endgroup$ – weee Nov 19 '18 at 20:20
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    $\begingroup$ can you help me find any pairs? $\endgroup$ – Jack Nov 19 '18 at 20:21
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    $\begingroup$ try $<1,1>$ and $<2,2>$ $\endgroup$ – weee Nov 19 '18 at 21:52
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    $\begingroup$ but ⟨1, 2⟩ ⪯ ⟨2, 1⟩ is true, right? $\endgroup$ – Jack Nov 19 '18 at 21:55

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