# If a weak preference relation is complete and transitive, why is the strict preference relation negatively transitive?

My textbook says that "if a weak preference relation is complete and transitive, the strict preference relation MUST be asymmetric and negatively transitive". Now, I think I understand why it must be asymmetric - all strict preference relations are asymmetric, correct? However, I don't understand why it is negatively transitive. If the weak preference is transitive, couldn't the strict preference also be transitive?

• Please don't make more work for other people by vandalizing your posts. By posting on the Stack Exchange (SE) network, you've granted a non-revocable right, under the CC BY-SA 3.0 license for SE to distribute that content. By SE policy, any vandalism will be reverted. If you want to know more about deleting a post, please take a look at: How does deleting work? Feb 9 '19 at 6:50

Definition 1. A binary relation $$\mathcal{R}$$ is transitive if: $$x \mathcal{R} y \text{ and } y \mathcal{R} z \implies x \mathcal{R} z.$$
Definition 2. A binary relation $$\mathcal{R}$$ is negatively transitive if: $$x \not\mathcal{R} y \text{ and } y \not\mathcal{R} z \implies x \not\mathcal{R} z.$$
(Those are forward slashes over the $$\mathcal{R}$$'s meaning "not". The LaTeX "not" here isn't quite working as well as I'd like.)
The strict preference relation $$\succ$$ is both transitive and negatively transitive.
• It is transitive because $$(x \succ y \text{ and } y \succ z)$$ implies $$x \succ z.$$
• It is negatively transitive because $$(x \not\succ y \text{ and } y \not\succ z)$$ is equivalent to $$(y \succsim x \text{ and } z \succsim y)$$, which, by the transitivity of $$\succsim$$, implies $$z \succsim x$$ or $$x \not\succ z$$.