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This sandbox is intended for saving drafts of long, complex posts, especially posts whose composition takes a long time. It serves to localize to one thread the front-page "bumps" caused by edits to drafts of such posts, so that they may be easily ignored. Also, it helps to guard against losing longly-composed posts due to system crashes.

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Proper Use of the Sandbox

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  7. Do not create new such sandboxes. The point of having a unique such sandbox is that it minimizes the noise on the front page when the sandbox is edited. If there were multiple sandboxes they will frequently occupy numerous front page slots, pushing other topics off the front page, and increasing noise.

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    $\begingroup$ I have added a [sandbox] tag to allow people ignore it more easily (via software support of ignoring tags), and since it seems that we have two sandboxes now, a tag may seem a bit more in place here. $\endgroup$ – Asaf Karagila Jul 18 '12 at 8:35
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    $\begingroup$ (+1) For thinking outside the (sand)box. $\endgroup$ – cardinal Jul 18 '12 at 19:40
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    $\begingroup$ At the suggestion of the moderators, I have gone and changed the associated owners of all the answers here to the Community user. This way, the original owners will not receive excess pings for each time another user uses the draft space for their work. Enjoy! $\endgroup$ – Grace Note Oct 5 '12 at 14:45
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    $\begingroup$ To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. $\endgroup$ – leo Dec 17 '12 at 18:03
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    $\begingroup$ PSA: Between the creation of this sandbox (in July 2012) and today (December 2015), technology has advanced. Something like StackEdit (or others, it's simply the only one I know) essentially solves all the limitations of this sandbox. You can have multiple concurrent drafts, you don't have to worry about polluting meta's front page, you can leave your draft untouched for days and expect it to still be there, you don't have to explicitly clear up your draft when you're done... Maybe someday we can get rid of this outdated crutch. $\endgroup$ – Najib Idrissi Dec 2 '15 at 14:07

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Existence of a one-step-cycle: $$ 2^{S_2} 3^{S_3} = \left(5+ \frac1{a_1} \right) \text{ in } [5,6] \\ \implies (S_2,S_3)=(1,1) \implies a_1=1 $$ Existence of a two-step-cycle: $$ 2^{S_2} 3^{S_3} = \left(5+ \frac1{a_1} \right) \left(5+ \frac1{a_2} \right)\\ \implies \text{ lhs in } [25,36] \implies \text { solutions for lhs } \in \{27,32,36 \} \\ \to (a) \qquad 2 = 5(\frac1{a_1}+\frac1{a_2})+ \frac1 {a_1a_2} \\ 4a_m^2 - 20a_m +25 = + 27\\ a_m = (\sqrt{27}+5)/2 $$ Existence of a three-step-cycle: $$ 2^{S_2} 3^{S_3} = \left(5+ \frac1{a_1} \right) \left(5+ \frac1{a_2} \right) \left(5+ \frac1{a_3} \right) \\ \text{ rhs in } (125,216) \\ \text{ lhs } \in \{128,144,162,192,216 \} \\ $$

$$ 2^{S_2} 3^{S_3} = \left(5+ \frac1{a_1} \right) \left(5+ \frac1{a_1} \right) \cdots \left(5+ \frac1{a_N} \right) $$

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Question title: Can first order logic be extended to include infinite conjunctions?

Can first order logic (FOL) be extended in some way where infinite conjunctions are permissible? Specifically, can it be extended and still refute statements in a finite number of steps?

I would like this question to be answered using Tarskian semantics, where names refer to objects external to the logic.



Self answer:

Suppose that an infinite conjunction is legal syntax in a first order theory. In first order logic, we are free to negate any formula we can reason about, so a theory which can reason about infinite disjunctions will necessarily reason about their negation.

\begin{equation}\tag{1} \{P_a,P_b,\dots\} \end{equation}

\begin{equation}\tag{2} \lnot(P_a\lor P_b\lor\dots) \end{equation}

Consider a first order theory which contains the infinite set of formulae $(1)$ and the negation of an infinite disjunction $(2)$. Any interpretation which satisfies $(2)$ will necessarily make the set of formulae $(1)$ unsatisfiable. However, every finite subset of formulae will be satisfiable, which by compactness means that the whole thing is.

This is a contradiction, so it must not be the case that an infinite conjunction can be made legal syntax in a first order theory.



Question title: What did Hilary Putnam mean by this following quote of his?

In Putnam's paper "The logic of quantum mechanics", he states:

There is nothing really answering to the Copenhagen idea that two kinds of description (classical and quantum mechanical) must always be used for the description of physical reality (one kind for the ways to be used for the 'observer' and the other for the 'system'), nor to the idea that measurement changes what is measured in a indescribable way (or even brings into existence), nor to the 'quantum potential', 'pilot waves', ect. of the hidden variable theorists. These no more than Reichenbach's 'universal forces'.

Precisely what did he mean by this? In particular, what did he mean by the "idea that measurement changes what is measured in a indescribable way (or even brings into existence)"? How does one clearly define the the issue raised in the first point, and has it been resolved today?


Question title: Has it become too hard to write self answer questions?

Recently I posted a self answer question which I had been working on in the Sandbox for drafts of long, complex posts for $2$ months (August $10^{th}$ to October $10^{th}$), which underwent $44$ revisions before it was posted. I also vetted the answer in the logic room to make sure it was correct before posting.

Despite the effort involved the question still got closed, and received $4$ downvotes within a short timeframe.

I was able to get the question reopened by addressing the critical feedback in the comments. The process of learning new content to edit the question, asking for it to be reopened, and now this meta post has taken a lot of time. I also went to constructive feedback to ask for possible explanations regarding the downvotes, since no comments were left.

Should self answer questions really require this much involvement? I learnt a lot from writing this question and from the feedback I got, but the closure and downvotes were rather unwelcome and time consuming.

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(4) Now let us look for which (odd!) $n$ the expression $f(n)$ contains small primefactors.

  • Assume $p=3, \lambda=2$, find $n$ :
    Remember: $$\begin{align} \{f(n),p\} &= (\{2n : \lambda\} &-\{n : \lambda\})&\cdot (1 + \{n,p\})&-\{n,p\} \\ \end{align}$$ For every odd $n$ the leading parenthese must evaluate to $1$, so $2n$ must be divisible by $\lambda=2$ but not $n$, so simply $n$ must be odd: $$\begin{align} n \text{ odd:}\\ \{f(n),3\}&= (\{2 \cdot n : 2\} &-\{n : 2\})&(1 + \{n,3\})&-\{n,3\} \\ &= (1&-0)&(1 + \{n,3\})&-\{n,3\} \\ &= &&1 + \{n,3\}&-\{n,3\} \\ &= 1 \end{align}$$ and the smallest $n$ is simply $n=1$. The next $n$ are from $(1,3,5,7,9,...)$ and always $\{f(n),3\}=1$ which is the same as $f(n) = 3 \cdot x$ where $x$ does not contain the primefactor $3$.

  • Assume $p=5$, $\lambda=4$ The first parenthese $(\{2 \cdot n : 4\} -\{n : 4\}) $evaluates to $1$ when $\{n,2\}=1$, so for $n \in \{2,6,10,... \}$ we have $ \{ f(n),5\} = 1 $ and for odd $n$ we have $\{f(n),5\}=-\{n,5\}$. We already know, that $n$ must be odd, so we will never have $5$ as primefactor in the result.

  • Assume $p=7$, $\lambda=3$ The first parenthese $(\{2 \cdot n : 3\} -\{n : 3\}) $ evaluates always to $0$ so we have always $\{f(n),7\}=-\{n,7\}$ and we will never have $7$ as primefactor in the result.

  • Assume $p=11, \lambda=10=2\cdot 5$
    For the leading parenthese to equal $1$, $2n$ must be divisible by $\lambda=10$ but not $n$, so simply $n$ must be divisible by $5$ and be odd. $$ \begin{align} \text{for } n = 5+ k \cdot 10 : \\ \{f(n),3\}&= (\{2 \cdot n : 2\} &-\{n : 2\})&(1 + \{n,3\})&-\{n,3\} \\ &= (1&-0)&(1 + \{n,3\})&-\{n,3\} \\ &= &&1 + \{n,3\}&-\{n,3\} \\ &= 1 \end{align}$$ and the smallest $n$ is simply $n=5$.

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(3) The more involved problem of $f(n) = {2^n+1 \over n}$ being integer gives even more restrictions. We have

  • $n$ must be odd, because otherwise the denominator would not divide the numerator
  • since $n$ is odd, $\lambda(p)$ must be even with exactly one primefactor $2$ : $\lambda(p) = 2 x$ where $x$ is odd. This defines a restriction on allowable primes: $\small (p,\lambda(p))=(3,2),(11,10),(13,6),(19,18),\cdots$ are allowed but not $\small (5,4),(17,4),\cdots$

We have moreover that -if the expression $f(n)$ is integer- it is also squarefree because: $$ \left\{{2^n+1 \over n},p\right\}= (1+ \{n,p\})-\{n,p\}=1$$

$ \qquad \qquad$ as far as is does not contain Wieferich primes.


(2) Now for the question on $2^n+1$ we refer to that $2^n+1=(2^{2n}-1)/(2^n-1)$ and since we look only at odd and non-wieferich primes $p$ we can write $\alpha=1$: $$\small \begin{align} \{2^n+1,p\} &=\{2^{2n}-1,p\} - \{2^{n}-1,p\} \\ &= \{2n : \lambda\}\left(1+ \{2n,p\}\right) - \{n : \lambda\}\left(1+ \{n,p\}\right) \qquad \text{ can be simplified } \\ &= \left(\{2n : \lambda\}-\{n : \lambda\}\right) \left(1+ \{n,p\}\right)\\ \end{align}$$ This gives three little observations:

  • primes p with odd $\lambda$ cannot occur in $2^n+1$ : we have then $\left(\{2n : \lambda\}-\{n : \lambda\}\right)=0$ because if odd $\lambda$ divides $2n$ then it also divides $n$ and the parenthese evaluates to zero.The same result of course if odd $\lambda$ does not divide $2n$ .

  • if $n$ is odd, then $\lambda=2 c$ for some odd $c$ and having $p=1+k \lambda$

  • if $n$ is even, say $n=2^a c$ then $\lambda=2^{a+1} d$ for some odd $c,d$

For a prime $p$ which actually occurs in $2^n+1$ its exponent in the primefactorization is exactly $1+ \{n,p\}$.


(1) First notice that $$ 2^n+1 = {2^{2n}-1\over 2^n-1 }$$ Then for divisibility of $2^n-1 $ by a primefactor $p$ there are nice little rules due to Fermat's little theorem, Euler's totient rules, and the so-called LTE ("Lifting the exponent")-rules. In an essay where I looked at this I preferred the following notations for better writing: $$\begin{array} {}\text{divisibility: } \qquad &\{a : p \} = \begin{cases} 1 & \text{ if } p \mid a \\ 0 & \text{ if } p \not \mid a \end{cases} \\ \text{valuation: } & \{a , p \} = \nu_p(a) \\ \text{order: } & \lambda_2(p) \underset{def}= \text{minimal } m>0 \text{ such that } \{2^m-1 : p \}=1 \\ \text{offset (of expon.):} & \alpha_2(p) \underset{def}= \{2^{\lambda_2(p)}-1,p\} \ge 1 \end{array} $$ For better readability, if the index and the argument of $\lambda_2(p)$ and $\alpha_2(p)$ are clear by the formula where they occur, I'll omit their reference for instance in $ \alpha = \{2^\lambda-1,p\}$

Then, for the exponent of a prime $p$ in $ 2^n-1 $ we have the general rule $$ \{2^n-1, p\} = \{n : \lambda\}\left(\alpha+ \{n,p\}\right) \qquad \text {with } \alpha \ge 1 \tag 1 $$ Note, that for base $2$ in $2^n-1$ all $\alpha_2(p)$ are $1$ with up to today two known exceptions $p=1093$ and $p=3511$ ("Wieferich primes") where $\alpha_2(p)=2$. It is an open problem whether there exist more Wieferich-primes and what $\alpha$ they have.
Examples:

$$ \begin{align} \{2^n-1,3\} &= \{n:2\}\left( 1+ \{n,3\} \right)&(\lambda,\alpha)=(2,1)\\ \{2^n-1,5\} &= \{n:4\}\left( 1+ \{n,5\} \right) &(\lambda,\alpha)=(4,1)\\ \{2^n-1,7\} &= \{n:3\}\left( 1+ \{n,7\} \right) & (\lambda,\alpha)=(3,1)\\ \{2^n-1,19\} &= \{n:18\}\left( 1+ \{n,19\} \right) & (\lambda,\alpha)=(18,1)\\ \vdots & \text{ the two known exceptional} & \text{ Wiefer.primes} \\ \{2^n-1,1093\} &= \{n:364\}\left( 2+ \{n,1093\} \right) & (\lambda,\alpha)=(364,2)\\ \{2^n-1,3511\} &= \{n:1755\}\left( 2+ \{n,3511\} \right) & (\lambda,\alpha)=(1755,2)\\ \vdots \end{align}$$


$$ \{2^n+1,p\} = \{2^{2n}-1,p\}-\{2^n-1,p\} $$ $$\begin{array}{} \{2^n-1,3\} &= \{ n :2 \}(1+\{n,3\}) \\ \{2^n+1,3\} &= \{ 2n :2 \}(1+\{2n,3\})-\{ n :2 \}(1+\{n,3\})\\ &= (1-\{n :2 \}) (1+\{n,3\}) \end{array}$$ $$\begin{array}{} \{2^n-1,19\} &= \{ n :18 \}(1+\{n,19\}) \\ \{2^n+1,19\} &= \{ n :9 \}(1+\{n,19\})-\{ n :18 \}(1+\{n,19\})\\ &= \{n:9\}(1-\{n :2 \}) (1+\{n,19\}) \end{array}$$

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