Textbook answer: No.
The binary relation:
(X, P), where X = $\mathbb{R}\times \mathbb{R}$ and $(a,b)P(s,t)$ iff $a > s$ and $b > t$
Linear order: Relation P that is asymmetric, transitive, and complete.
Complete: For all $(a,b),(s,t) \in$ X, $(a,b)P(s,t)$ or $(s,t)P(a,b)$ or both and $(a,b) \neq (s,t)$
The reason this binary relation is not a strict linear order is because it is not complete, right?
However, I cannot seem to understand why it is not a strict linear order. It is definitely irreflexive and transitive, right? For completeness, isn't the Cartesian plane naturally ordered? Or am I missing something?