# 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? [closed]

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!

• In what set are $x1, y1, x2, y2$ elements? One can only define a linear order on a specified set. – Namaste Nov 19 '18 at 22:49

A linear order needs two elements to be comparable i.e. for $$\left,\left$$ we have either $$\left\preceq\left$$ or $$\left\preceq\left.$$ If you can find two pairs where neither is the case, you have proven that the relation can't be a linear order.
• try $<1,1>$ and $<2,2>$ – weee Nov 19 '18 at 21:52