## New evidence makes it look like Arbery went on the site to get a drink of water from a faucet. What did cons think he was going to steal…?

You don’t walk onto a construction site unless you belong there. Theft of materials and tools is huge, happens all the time on sites, it sucks, mostly for the ones that don’t secure their crap. A drink of water is a nice excuse to be scoping out what to come back for later.

Oh look at you joke when in the middle of the night people steal entire palates of material. They of course have to know what is where before, minimize lights and time on site.

## Number theory – is this valid evidence for the Sylvester-Schur theorem for \$ n ge 9 \$ and \$ x ge 2n \$?

Below is my effort to provide an alternative argument for the Sylvester-Schur theorem that is simpler than that of Paul Erdős.

Please let me know if I made mistakes or if a point is unclear.

The argument uses the following lemma:

lemma: For integers $$n ge 9$$ and $$x ge 2n$$::

$${x select n}> dfrac {x!} {(x – pi (n))!}$$

(1) For $$n ge 8$$It's easy to show that:

$$n> ( ln n) ^ 2$$

Note: You can find details here $$n> ( ln n) ^ 2$$

(2) For $$n ge 9$$, it follows $$(2 ln n)> 4.25506$$ and since $$n> ln n$$::

$$2n ln (n) – left ( dfrac {n + 2} {n + 1} right) ln n – 1.25506n> 4.25506n – left ( dfrac {n + 2} {n + 1} right) ln n – 1.25506n> n$$

(3) $$2n ln n – left ( dfrac {n + 2} {n + 1} right) ln n – 1.25506n> ( ln n) ^ 2$$

(4) $$(2n-1) – dfrac {1.25506n} { ln n}> ln n + dfrac {1} {n + 1}$$

(5) Multiply $$n + 1$$ on both sides
$$(n + 1) left ((2n-1) – dfrac {1.25506n} { ln n} right)> (n + 1) ln n + 1$$

(6) Multiply $$ln n$$ on both sides:

$$(n + 1) ((n-1) ln n – 1.25506n)> ( ln n) ((n + 1) ln n -n + 1)$$

(7) $$n + 1> left ( dfrac { ln n} {(n-1) ln n – 1.25506n} right) left ((n + 1) ln n – (n-1) ln e right)$$

(8th) $$sqrt (n – 1 – frac {1.25506n} { ln n}) { left (2n – dfrac {1.25506n} { ln n} right) left (2n-1 – dfrac { 1.25506n} { ln n} right) points (n + 1)}> (n + 1)> sqrt (n – 1 – frac {1.25506n} { ln n}) { dfrac {n ^ {n + 1}} {e ^ {n-1}}}$$

(9) Since $$n! le dfrac {n ^ {n + 1}} {e ^ {n-1}}$$, it follows:

$$sqrt (n – 1 – frac {1.25506n} { ln n}) { left (2n – dfrac {1.25506n} { ln n} right) left (2n-1 – dfrac {1.25506n} { ln n} right) points (n + 1)}> (n + 1)> sqrt (n – 1 – frac {1.25506n} { ln n}) {n!}$$

Note: See sentence here for the argument that $$n! le dfrac {n ^ {n + 1}} {e ^ {n-1}}$$.

(10) Since $$x ge 2n$$::

$$left (x – dfrac {1.25506n} { ln n} right) left (x-1 – dfrac {1.25506n} { ln n} right) dots (xn + 1) ge left (2n – dfrac {1.25506n} { ln n} right) left (2n-1 – dfrac {1.25506n} { ln n} right) dots (n + 1)> n !$$

(11) As for $$n> 1$$, $$pi (n) < dfrac {1.25506n} { ln n}:$$

$$(x- pi (n)) (x-1- pi (n)) points (n + 1)> n!$$

Note: See 3.6, Sentence 2, Corollary 1, by Rosser & Schoenfeld, 1962

(12) $$dfrac {(x) (x-1) points (x-n + 1)} {n!}> (x) (x-1) points (x- pi (n) +1)$$

(13) It follows:

$${x choose n}> frac {x!} {(x – pi (n))!}$$

sentence: For integers $$n ge 9, x ge 2n$$there is a prime number $$p> n$$ so that $$p | dfrac {x!} {(x-n)!}$$

(1) Leave $$x, n$$ be integers so that $$n ge 9$$ and $$x ge 2n$$.

(2) Suppose there is no prime number greater than $$n$$ that divides $$dfrac {x!} {(x-n)!}$$

(3) Leave $$text {lcm} (x-n + 1, x-n + 2, dots, x)$$ be their least common multiple.

(4) It can be shown that:

$$text {lcm} (x-n + 1, x-n + 2, dots, x) ge {x choose n}$$

Note: The argument for this can be found here

(5) From the assumption in (2) it follows:
$$text {lcm} (x-n + 1, x-n + 2, dots, x) < dfrac {x!} {(x – pi (n))!}$$

Note: where $$pi (n)$$ is the prime count function.

(6) But then we have a contradiction since e.g. $$n ge 9, x ge 2n$$follows from the lemma:

$${x select n}> dfrac {x!} {(x – pi (n))!}$$

Feel free to let me know if a point is unclear or if there is a better way to make the same argument.

## real analysis – evidence verification for \$ alpha, beta \$ Holder spaces

To let $$X subseteq mathbb {R} ^ n$$ be a compact set.

We also have $$f: X rightarrow mathbb {R}$$ so that $$f$$ is $$alpha$$ Holder. which means that it exists $$0 < alpha leq 1 ;$$ and $$K> 0 ;$$ so that $$left | f (x) -f (y) right | leq K left Vert x-y right Vert ^ { alpha} ;$$ for all $$x, y in X$$

I want to show that for $$0 < beta < alpha leq 1 ;$$, $$f$$ is too $$beta$$ Holder.

I tried to prove it by dividing myself into cases:

If $$left Vert x-y right Vert leq1, ;$$ $$forall$$ $$x, y in X$$ we understand that $$f$$ is too $$beta$$ Owner right because $$left | f (x) -f (y) right | leq K left Vert x-y right Vert ^ { alpha}

Otherwise, $$X$$ is compact and therefore limited so that we can designate:

$$M = sup_ {x, y in X} left Vert x-y right Vert ^ { alpha}> 0$$

$$L = sup_ {x, y in X} left Vert x-y right Vert ^ { beta}> 0$$

and choose $$K _ { beta} = frac {M cdot K} {L} +1> 0$$

We understand that now $$left | f (x) -f (y) right | leq K _ { beta} left Vert x-y right Vert ^ { beta}$$ $$forall$$ $$x, y in X$$.

## Solution check – The square root of the prime number is irrational. Is this valid evidence?

I know there are similar questions, but I want to know if this is valid evidence.

A prime number has 1 and itself as a divisor. So the subset of a prime number is $$D (p) = {1, p }$$

Now I want to prove it $$sqrt {p} = frac {a} {b}$$ to get the opposite.

Likewise $$gcd (a, b) = 1$$.

Let’s align it.

$$p = frac {a ^ 2} {b ^ 2}$$

And divide by $$b ^ 2$$

$$frac {p} {b ^ 2} = frac {a ^ 2} {b ^ 4}$$

In this case $${b ^ 2}$$ is also a divisor of $$p$$, which does not fit the definition of prime numbers. The square root of a prime number must be irrational.

## Evidence Assistants – Coding Theory Optimal code with a given maximum length

Why do we need at least 2 code words of maximum length in optimal codes?

Why do they only differ in their prefix?

Could someone give me more insight?

You have to prove that for a certain optimal code C with maximum length L there are at least 2 code words with maximum length that only differ in their prefix.

I have nothing to start with, so a good start would be good.

I didn't provide a definition of the optimal code in my script, and I don't know if it needs to be a block code that I see in most definitions.

## Incompleteness – If PA is consistent, does PA prove for \$ n \$ that "\$ n \$ does not code for evidence of inconsistency"?

I'm still struggling to differentiate between What is proven Where. I think I have a good understanding of the theory and the meta-theory, but then I'm stumped every now and then, so I'm afraid that something fundamental is missing in my understanding.

By CON (PA) I mean the statement "PA is consistent" (meta theory), while con (PA) is the sentence in the language of arithmetic that expresses "there is no evidence of a contradiction" (theory).

To get to the point, I specifically ask myself the following: We know that con (PA) is really a number-theoretic statement that asks for the existence of a witness who solves a diophantine equation. According to Goedel II, PA con (PA) cannot prove.

Let us now assume that CON (PA) applies. Is it true that for everyone $$n$$, PA proves $$(*)$$ ""$$n$$ is not a contradiction code "? It is clear that PA cannot prove the universal degree (as it would then prove con (PA)).

This seems to be a theorem scheme in which each $$n$$is an actual natural number (as defined in the meta theory).

My reasoning is: if there were any $$n$$ so PA hasn't proven $$(*)$$ then we would have a model $$M$$ to satisfy that $$n ^ M$$ is the code to prove inconsistency. In meta theory we could decode $$n$$ in a proof that is now supposed to show that CON (PA) is wrong, a contradiction.

Surely it is not possible that this witness does not meet the standard? In other words, it is true that $$n ^ M = n$$ where the right hand $$n$$ is the actual natural as defined in the meta theory? If not, that would be a problem for my reasoning, because in such a case the witness would not decode one indeed Proof in meta theory.

## Liberals even have ONE credible piece of evidence for Trump's "accusers". Is there no, unlike Biden?

Anyone who does even a fleeting background check of his story and of * her * – personally – will quickly find that she is mentally disturbed.

She has changed her story countless times, often told contradictory versions, put all sorts of garbage on the Internet under different names, quickly deleted these blogs and accounts …… and supported BERNIE during the race … … as she did invented the claims against Biden. — who are 27 years old at best.

Since then there have been internet recordings, tweets and likes in which she has praised Biden countless times, even after she has stopped working for him.

Her newly discovered aversion to him only seemed to develop when he kicked Bernie's As in the elections. ….. undoubtedly funded by a Super Pac who repays your mortgage.

The Times examined its history and found no evidence to support it, and a whole host of evidence to question its credibility.

Her story went from an uncomfortable hug to a feeling to knees trapped between her legs, and in the latest version, he is said to have fingered her in the halls of the congress.

Really … did she go around with no panties?

She claims to have submitted numerous reports to one and the other, but has not kept a single copy of these mythical reports, and each authority has now denied and confirmed that they have such reports and have never heard of such reports.

She also claims to have spoken to this person and that person and other Biden employees at the time and each of them has DENIED their stories … and Neither of them even remembers mentioning anything like that, even in passing.

and finally she did a lie detector test ……. and failed.

How many damn times and how many ways does she have to be proven a liar for this story to die?

But the Republicans will of course continue to whip this dead horse and at the same time TRUMP will fully and completely pass on the 27 allegations against HIM.

## Complexity Theory – Dissolution of Exponential Lower Limit … Alternative Evidence?

I read the exponential lower bound of the resolution proof system using Haken's bottleneck method for the pigeonhole principle, as presented in Arora and Baraks Computer complexity: a modern approachChapter 15. However, I do not like how the evidence is presented in the book, and I have some difficulty in following it.

Does anyone know of any alternative sources that provide the same evidence? I know there are different techniques to display exponential lower bounds for the resolution, but I want something that is based on the pigeonhole principle. It's just that the wording in this book is really confusing.

## ag.algebraic geometry – results that are believed to be true without partial / detailed evidence

The background to this question is the lecture by Kevin Buzzard.

The slides of the lecture can be found here.

One of the points in the lecture is that people believe that some results are true, but the evidence is not mentioned anywhere. He says this leads to wrong conclusions, but I'm currently not interested in wrong conclusions. All I care about are results that are believed to be true, but without partial / detailed evidence.

What are results that are believed to be true without partial / detailed evidence?

## real analysis – evidence in differential geometry

I have some "doubts" about the differential geometry, not the concepts, but how to prove a few sentences (I find it really difficult), such as:

1. How can I prove / What is the best way to prove / What is the canonical way to prove that $$(U, Phi)$$ is a local coordinate system?
2. How can I prove it? $$S neq Emptyset Subseteq R ^ 3$$ is a surface?
3. How can I prove it? $$S$$ is a parameterized surface?
4. To let $$S$$ be a regular surface and $$f: S subseteq R ^ 3 longmapsto R ^ n$$. How can I prove that it can be differentiated?
5. How can I prove that a particular application is a dipheomorphism?
6. How can I prove that a certain surface can be oriented?

I know there are a lot of questions. The thing is, I just don't know how to show it. So it would be nice to have a method to solve these kinds of questions.