In http://web.mit.edu/neboat/www/6.046-fa09/rec8.pdf, I see that they pad a 2SAT clause $$(x\vee y)$$ to make it a 3SAT clause by writing $$(x\vee y\vee p) \wedge (x\vee y\vee \neg p)$$. Why doesn't $$(x\vee y\vee y)$$ work instead? Isn't the truth table identical?
Sometimes we insist that the three literals in a 3SAT clause belong to different variables. This ensures, for example, that a random assignment satisfies a clause with probability exactly $$7/8$$. The translation $$x \lor y \lor y$$ doesn't satisfy this condition, but the other one does.