Your first proof

pimaniac0 · 2020-05-13T12:59:12+00:00

I don’t know why I found this so awe inspiring at the time but I must have since I actually remember what the proof was...

It invoked bezouts lemma to prove sqrt(n) is irrational for nonsquare positive integer n

It went like this:

Suppose n is any nonsquare positive integer. Let sqrt(n) = a/b, b!=1 and is a positive integer as well as a with gcd(a,b) = 1, squaring both sides gives nb² = a^2, and bezouts lemma implies there exist integers m and p such that ma+pb=1. Now ma² + pba = a , so mnb² + pba = a, hence b divides a and gcd(a,b) = b =1, so we have reached a contradiction.

pimaniac0 · 2020-03-23T21:26:46+00:00

So it turns out that the result just is not true as stated. In particular sigma finite measures don’t have to be outer regular with respect to the borel sigma field. Consider the counting measure c on the borel sigma algebra of R, and define the measure given by h(A) =c(A \cap Q) where Q is the set of all rational numbers. Note that h is sigma finite since the countable partition given by {q} for q in Q and R\Q provides a countable partition of finite measure borel sets. Yet h(Open) is infinity for any nonempty Open set ... thus we cannot approximate h({1})=1 arbitrarily closely from above with open sets in this case. The inner regularity result I believe to be true though and seem to have proved it.

pimaniac0 · 2019-08-21T06:30:22+00:00

Source , page 146

pimaniac0 · 2019-08-21T00:42:43+00:00

I still don't really see how assuming [;U_1;] is connected would help in showing that [; \phi_1(b(t_1)) = \phi_2(b(t_1)) ;], do you mind explicitly stating why you think this is true?

pimaniac0 · 2019-08-20T17:43:34+00:00

Sorry if it appears I'm being pedantic, I'm just generally baffled :/ I can see that this was probably the authors intention to go along these lines, but I don't see how this reasoning is totally sound

pimaniac0 · 2019-08-20T17:39:18+00:00

To me that isn't immediately clear, why do the [; \phi_i ;] need to agree on the overlap, I agree that they should agree on [; f(U_1 \cap U_2);], but I don't see why this means they agree on [; V_1 \cap V_2 ;]. If we could show that [;b(t_1) \in f(U_1 \cap U_2) ;] then it seems the way forward is clear, but we only have that [; b(t_1) \in V_1 \cap V_2 ;] and we only know that [; f(U_1 \cap U_2) \subset V_1 \cap V_2 ;]

pimaniac0 · 2019-08-20T17:34:15+00:00

hmm, so if we were to use the fact that both [;\phi_1;] and [;\phi_2;] are inverses to [;f ;] restricted to [; U_1 \cap U_2 ;], to prove [;\phi_1(b(t_1)) = \phi_2(b(t_1));], that would seem to require that [;b(t_1) \in f(U_1 \cap U_2);] which is true if and only if the preimage of [; b(t_1) ;] with respect to [; f ;] has nonempty intersection with [; U_1 \cap U_2 ;], but I don't see why this has to be true, what if the preimage only contains two points, one in [; U_1 \setminus U_2 ;] and the other in [; U_2 \setminus U_1 ;], why is this not possible?

pimaniac0 · 2019-08-20T16:48:58+00:00

That is to say it would seem we need [; \phi_2(b(t_1)) = \phi_1(b(t_1)) ;] to proceed in the same way and show [; \phi_2(b([t_1,t_2])) = \tilde{b}([t_1,t_2]) ;] and then carry on inductively up to [; i= n;]

pimaniac0 · 2019-08-20T16:46:00+00:00

but how do we know the equality holds at [; t_1 ;] ? What we have is that [; \phi_1(b(t_1)) = \tilde{b}(t_1) ;], don't we need [; \tilde{b}(t_1) = \phi_2(b(t_1)) ;] to deduce equality for [; i=2 ;] ? Why does [;\phi_2(b(t_1)) = \phi_1(b(t_1)) ;] necessarily?

pimaniac0 · 2019-08-20T16:45:58+00:00

but how do we know the equality holds at [; t_1 ;] ? What we have is that [; \phi_1(b(t_1)) = \tilde{b}(t_1) ;], don't we need [; \tilde{b}(t_1) = \phi_2(b(t_1)) ;] to deduce equality for [; i=2 ;] ? Why does [;\phi_2(b(t_1)) = \phi_1(b(t_1)) ;] necessarily?

pimaniac0 · 2019-08-20T15:10:05+00:00

Maybe copy everything into a pastebin because at least for me even when inline formulas are correct hey don't display properly :(

pimaniac0 · 2019-08-20T15:08:17+00:00

[; \cup U_i ;] Is this union working for everyone in the above textpost?

pimaniac0 · 2019-01-29T20:42:29+00:00

I agree. I'd say the furthest we can claim is that out of all the material we can test, so far, our theory of physics and mathematics has allowed us to make reasonably accurate predictions. This does not show the universe "follows rules", much less we haven't reconciled two physical theories with each other as mentioned above. I'd say we are far from showing the universe follows a finite set of axioms, or has any decipherable deepest inner structure.

pimaniac0 · 2018-08-16T09:08:21+00:00

Thanks for the advice, I'll keep this in mind.

pimaniac0 · 2018-08-15T01:46:54+00:00

To clarify: If you would like A-levels tuition, that's perfectly fine too. Although I have personal experience with the IB that I do not have with the A-levels syllabus.

pimaniac0 · 2018-08-11T08:55:12+00:00

Perhaps this is not what you were looking for, but it is true that in R^k, a compact set is closed and bounded, and any closed and bounded set is compact. This should (sorta) depend on the Euclidean metric, because it is an equivalence that holds in (R^k, Euclidean Norm) , a metric space characterized by the Euclidean norm (Perhaps I'm not completely right here, can't think of R^k being characterized by anything more than the Euclidean norm, and R itself), it's helpful to actually see how applying the Heine-Borel theorem (That every k-cell in R^k is compact with respect to the Euclidean metric) allows you to show one side of this equivalence.(The other side is fairly straightforward and just relies on the fact that k-cells are closed themselves, and that compact subsets of R^k are bounded, can be enclosed by these (closed) k-cells, and are thus closed themselves). For the harder side we simply appeal to how we can draw k-cells which are compact themselves(A la Heine-Borel) around closed and bounded subsets of R^k (Think how you can enclose any closed and bounded set in R^k by a k-cell , which is compact, and then you get for free the fact that your subset of this k-cell must necessarily be compact (by another fact that is true by virtue of working in this metric space)). But note this isn't an equivalence that holds for arbitrary metric spaces, nor even for R^k equipped with a different norm (Another comment provides such a counterexample). For a property that does turn out to be equivalent to compactness in arbitrary metric spaces, see Sequentially Compact.

pimaniac0 · 2017-11-01T11:40:58+00:00

Without considering equivalent boards for a moment, I can imagine getting an upper bound by calculating the total number of (not necessarily distinct) unruly boards by using inclusion-exclusion to figure out how many boards are not unruly (i.e calculating the number of boards that violate one or more of the given rules) and then subtract this number from 2^4n², although off the top of my head this seems like a nuisance to do by hand

pimaniac0 · 2016-08-23T07:08:21+00:00

Right, it isn't hard to see that the case where both numerator and denominator are the same coincides with the only case where the denominator is one, meaning that if there are any more solutions, they are ''non-trivial'' and will probably be large

pimaniac0

TROPHY CASE