A logic assignment requires me to create a model in which most X's are smarter than most Y's, but most Y's are such that it is not the case that most X's are smarter than it.

It's easy to do this when we grant that $Sxy$ = "x smarter than y" is a non-commutative predicate eg. $Sxy => -Syx$.

However it seems impossible to do this if we assume that $Sxy$ is transitive ie. $(Sxy + Syz) => Sxz$

My question is regarding whether "smarter than" is, in reality, transitive.

My logic professor argued that it wasn't because we can imagine different types of smartness such that A > B, B > C and C > A. I disagree because:

  • Different types of smartness demand different predicates eg. Pxy = x is smarter than y at visual problems, Cxy = x is smarter than y at chess problems, etc

  • To try and rescue this idea, if we declared "smarter than" to mean "smarter than [in any capacity]" (ie. Sxy = Pxy or Cxy or ...), then we lose its non-commutativity and it becomes pretty meaningless.

He also suggested thinking about it in terms of a competition in which A usually beats B and B usually beats C and C usually beats A. There is no contradiction here, but again, "x beats y" is a different predicate with different semantics.

Therefore I continue to believe that "smarter than", reasonably, is a transitive relationship.

Who is correct??

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    $\begingroup$ Depends on how you define "smarter than," but it can be much like a competition. Given that it is not defined, and might not be transitive, you certainly can't conclude that it is. $\endgroup$ – Thomas Andrews Jun 3 '15 at 3:20
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    $\begingroup$ I think this is a linguistic question--not a logic question. You could certainly think up some measure of smartness such that it's not transitive--the same goes for anything that is subjective in nature. If you go by something like a competition then you have the very real problem that Joe is smarter than Alice doesn't always hold (i.e. it depends on when we measure who is deemed smarter--although such a scalar score would give an ordering to the elements). $\endgroup$ – Jared Jun 3 '15 at 3:22
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    $\begingroup$ A mathematical answer requires you to furnish a mathematical definition of the relation. In the absence of a precise definition, how can you expect an answer to be correct or incorrect? $\endgroup$ – MPW Jun 3 '15 at 3:29
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    $\begingroup$ How can we assert that x is smarter than y, yet NOT be asserting that x has smartness in a greater quantity than y? If we are stating that x has a greater quantity of smartness than y, then smartness is a linear scale and hence transitive. $\endgroup$ – Brendan Hill Jun 3 '15 at 3:49
  • $\begingroup$ Even if you assume that "smarter than" is transitive (and I believe, contrary to what Thomas Andrews suggests, that you are supposed to assume that) there is still a simple model that satisfies the requirements. $\endgroup$ – MJD Jun 3 '15 at 6:44

Your argument is irrelevant. As you said, your professor argued that it wasn't because we can imagine non-transitive smartness. That is a red herring. There is a simple and transitive smartness relation that has the properties that were asked for. So it doesn't matter whether or not we are willing to consider nontransitive relations, because there is a transitive solution.

Let the smartness $S(X_n) = S(Y_n) = \frac 1n$ for $n\in\{1,2,\ldots\}$. Say that $A$ is smarter than $B$ if $S(A)>S(B)$.
1. This is evidently transitive. (You prove this.) 2. But it also satisfies the requirements. (You prove this.) 3. Something in your reasoning was wrong. Once you understand the example, you should try to understand your error.

  • $\begingroup$ Isn't it easier to assume X = Y = |N, and "smarter than" is just "<" relation? :-) $\endgroup$ – Veky Jul 19 '15 at 13:55
  • $\begingroup$ Yes, but I wanted smarter people to have larger numbers than less smart people, instead of smaller numbers. $\endgroup$ – MJD Aug 10 '15 at 4:21
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    $\begingroup$ May I ask why? Smartness is not a number, it can be modelled as a number. And those numbers can be compared in any way we want. It seems to me you're confusing the map with the territory. ;-) $\endgroup$ – Veky Aug 11 '15 at 5:09

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