Recent content by JSuarez

In Logic, you may apply quantifiers over symbols that refer to other entities: first-order variables refer to individuals within a class; second-order variables (also known as predicate or class variables) refer to classes of individuals, and you may go on from here. Now, it is indeed known...

In the non-cooperative scenario, this game has no pure Nash equilibrium. The mixed one is attained when all players pick one of the N numbers randomly, using a uniform distribution.

By the way, have you considered multiplying the latter by x, and then differentiating again?

Well, I'm not going to offer any positive suggestions or even try to cheer you up so, if you want, you may stop reading right here. I taught for almost 20 years; taught Linear Algebra and Calculus, pretty much like you are doing now and also more advanced courses, in Logic, Set Theory and...

No, Russell's paradox asks if A is a member of itself (A \in A), not a subset of itself (A \subseteq A). Your entire question rests solely in the confusion between element and subset; in fact, all sets are, trivially, subsets of themselves.

Ups! Got a bit carried away, haven't I? :redface: You're right: the number of elements in the preimage is given by: \frac{k!}{j!(k-j)!}j^j(k-j)^{k-j} But, as the preimage is a subset of A(k), who has k^k elements, we have: k^k \geq \frac{k!}{j!(k-j)!}j^j(k-j)^{k-j} Which is...

It seems to me that this thread veered a little off-course, regarding the OP's question. To him/her, let me say that it would help if you provide more details about what you discussed in class; something related to "Truth in Mathematics" doesn't necessarily have to be a Logic-related topic...

A purely combinatorial proof: The cases k = 0 or k = 1 are immediate, so let k \ge 2, 2 \leq j \leq k, A a k-element set, C \subseteq A a j-subset of A, A\left(k\right) the set of all length k sequences over A and B\left(k\right) the set of length k binary sequences. Take any elements a...

Suppose that \alpha is a sentence of FOL with equality and a binary function and "feed" it to the decision algorithm of Presburger Arithmetic (PrA) (which is called quantifier elimination); if the answer is "yes", then you may be sure that \alpha is a theorem of PrA, but not a logically valid...

And what looks like homework, smells like homework and tastes like homework is...

The purpose is that do Carmo's is interested in differential forms defined on a manifold (or in an open set of it), not just on a single point. Therefore, given a manifold M and p \in M and R3 is its tangent space at p, then the dual basis to the canonical one is indeed dx_i(p) and is exactly...

Why is it so obvious that that statement denotes a proposition? Note that if it doesn't, then the problem disappears.

The book's assertion is correct; I don't know the authors's explanation, so I cannot comment on it, but it seems that your reasoning ends with a contradiction, which is precisely what happens when we try to ascribe a truth-value to that particular statement, so I don't understand what is exactly...

You are confusing equations with functions. The equation: x4 + 24x2 + 1 = 0 Asks if there is some point (in this case, a real one) such that the above equality is true, not that it's true for all points. But when you write: d2/dx2( x4 + 24x2 + 1) = d2/dx2(0) You are stating that...

No, this is true only for the trivial (full) topology. For an extreme couterexample, consider the other trivial topology \left\{\emptyset,X\right\}.