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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 Maybe someday we can get rid of this outdated crutch.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

17 Answers 17


$\def\AA{\mathbb{A}}$ $\newcommand\BB{\mathbb{B}}$

$\AA$ and $\BB$ both work.


Comment My initial thoughts go in the following direction, but by second read I see that this concerns likely only a subset of impossible solutions. But perhaps the ansatz itself helps as initial idea...

Error detected - needs update

(taken from my two earlier comments)
By subtracting $8$ on each side one gets $7⋅(7^X−1)=3⋅(3^Y−1)+5⋅(5^Z−1)$ and $(X,Y,Z)=(x−1,y−1,z−1)$. Then divide all by $7⋅3⋅5$.

For instance we get on the lhs $(7^X−1)/15$ and on the rhs equivalent terms.

This gives restrictions for the modular classes for the exponents, and possibly one can proceed with further modular considerations (for instance on primefactorizations in lhs and rhs).

For instance the lhs has then always $2^5⋅5$ as factor, the terms in the rhs always $2^4$ and $2^3$ as factor. Since the constant occurence of $2^5$ as minimum means a strong selection for the $5^Z−1$-term: only $Z=24j$ is possible and by this the selected have all the primefactors $3⋅13⋅31⋅31$. This can similarly be checked for the other terms and often it is possible to derive from this contradictions.

By the corresponding primefactors of 2 and 13 in the lhs and rhs I find that in the lhs we can only have valid solutions in steps of 12 in the exponent. The primefactor decomposition of the first two exponents give

 X,    value,                primefactors
[12, 922752480,            "2^5.3.13 . "]
[24, 12772082092037760960, "2^6.3.13 .  xxx.73.193.409.1201"]

This means, with all following admissible exponents we shall have the set of primefactors of the first row.

Similarly we look at the $5^z$ term, in already required $24$ steps:

 Z    value                primefactors
[24, 2838316417875744,     "2^5.3.13   .31.313.601.390001"]
[48, 16917684184764290..., "2^6.3.13   .31.313.601.390001.17.11489.152587500001"]

Again, in all following admissible exponents we shall have the set of primefactors in the first row. But interesting: $7^{12k}-1 $ as well as $5^{24j}-1$ have the primefactor $3$ - but of course this cannot happen in the third term:

 Y    value                primefactors
[24, 8069415328,           "2^5  .13      .41.73.6481"]
[48, 227904123076...     , "2^6  .13      .41.73.6481   ."]

So, if I've not made a silly mistake, this is a proof of nonexistence of solutions aside $(X,Y,Z)=(0,0,0)$ resp $(x,y,z)=(1,1,1)$. This answer space is free to use :)


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I have a simple inversion routine for lower triangular matrices (assumed to be invertible, no errorchecks) in Pari/GP

I get, dependent on counting, for multiplications either 1/6*x^3 + 1/2*x^2 + 1/3*x (where x means dimension $n$) or 1/6*x^3 - 1/3*x ; see below.

 \\ invert a lower triangular matrix; no errorchecks! 
  su_m=su_d=0; \\ inserted for documenting the multiplications & divisions
       tmp[c,c]=1/m[c,c]; su_d++;
                   / m[r,r];
               su_m+= (r-c);
  \\ return(m);  \\ commented out, we need the counters of mult. & divs!


\\ document the multiplications, divisions in a list for m=1..20   
for(m=1,20,print( triinv(PPow(1,m))));   \\ PPow gives a Pascalmatrix of size m x m


\\ protocol
\\ dim mu di  "mu"=multiplications "di"=divisions
\\ [1  0  1]
\\ [2  1  3]
\\ [3  4  6]
\\ [4 10 10]
\\ [5 20 15]
\\ [6 35 21]

Find formula:

\\ for multiplications we get the sequence [0,1,4,10,20,...]
\\ for divisions [1,3,6,10,...]
polinterpolate([1,4,10,20]) \\ check formula for multiplications
\\ %7 = 1/6*x^3 + 1/2*x^2 + 1/3*x   \\ result by Pari/GP


\\ %8 = 1/6*x^3 - 1/6*x      \\ result by Pari/GP

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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?


From this you have bounds for your $r(a_1)$-parameter: $$ r(a_1) = { E \over N+E } = 1-{ 1 \over E/N+1 } \implies $$

$$E_{min}=1-\{N \gamma_1 \} + N \gamma_1 \\ E_{max}=N $$

$ \displaystyle \lim_{N \to \infty} r(a_1)_{E_{min}} = 1-{ 1 \over (1-\{N \gamma_1 \} + N \gamma_1 )/N+1 } = 1-\frac 1{\gamma_1+1} = 1- \log_3(2) = \log_3(1.5) $

$ \displaystyle \lim_{N \to \infty} r(a_1)_{E_{max}} = { E_{max} \over N + E_{max} } = { N \over 2N } = \frac 12 $


{\lceil N \cdot \gamma_1 \rceil \over N+E} \le \frac E{N+E} = r(a_1) \le \frac N{N+E} $$

Let us assume the "Syracuse"-style notation of the Collatz-iteration $$ a_{k+1}= {3a_k+1\over 2^{A_1}} \qquad a_k \small \text{ from the odd integers} $$ and for a $N$-fold iterated transformation the short, vectorial, notation: $$ a_{N+1}=T(a_1;[A_1,A_2,...,A_N]) $$ So let $N$ denote the (N)umber of steps $3x+1$ and $S$ denote the (S)um of the exponents $A_k$, which is also the number of $x/2$-steps.

Then, to convert this into the version of $(3x+1)/2$ and $x/2$ -stepping, we introduce $E$ the number of even steps without the $(3x+1)/2$ steps, so $E=S-N$.

With that, I understand your $r(a_1)$ as $r(a_1)=E/(N+E) = (S-N)/S = 1- N/S$.

We can observe,

  • that the trival cycle $1 = T(1;[2,2,2,...2])$ to any length $N$, has the values $S=2N$ and $E=N$ and $r(a_1)= 1-N/S = 1-N/(2N)=1/2 $
  • that the first cycle in the negative numbers $-1= T(-1;[1,1,1,1,...,1])$ to any length $N$, has $S=N$, $E=0$, and $r(a_1)= 0$
  • that the second cycle in the negative numbers $-5=T(-5;[1,2,1,2,1,2,...,1,2])$ to any even length $N=2n$, has $S=3n$, $E=n$ and $r(a_1) = E/(N+E)= n/(3n) = 1/3 $

Now to have a cycle of any length, and other than $T(a_1;[2,2,2,...,2])$ we can use the well known multiplication-formula for the $N$ members of an expected cycle $a_k$ ($k=1..N$) $$ a_2 \cdot a_3 \cdot ... \cdot a_N \cdot a_1 = \left({3a_1+1\over 2^{A_1}}\right) \left({3a_2+1\over 2^{A_2}}\right) \cdots \left({3a_N+1\over 2^{A_N}}\right)$$ This can be rearranged to $$ 2^S = 2^{A_1+A_2+...A_N} =\left(3+{1\over a_1}\right) \left(3+{1\over a_2}\right) \cdots \left(3+{1\over a_N}\right)$$ We see, that the rhs must be at least as large as the smallest perfect power of $2$ larger than $3^N$, but at most as $4^N = 2^{2N}$ so we get for the lhs (writing $\gamma=\log_2(3)$, and further below $\gamma_1=\log_2(3)-1$): $$ 2^{\lceil N \cdot \gamma \rceil} \le 2^S \le 2^{2N} $$ which in terms of $S$ means $$ \lceil N \cdot \gamma \rceil \le S \le 2N \qquad \text{where } S \in \mathbb N^+ $$ and in terms of $E$ instead $$ \lceil N \cdot \gamma \rceil -N =\lceil N \cdot \gamma_1 \rceil \le E \le N $$ From this you have bounds for your $r(a_1)$-parameter: $$ r(a_1) = { E \over N+E } \implies \\ {\lceil N \cdot \gamma_1 \rceil \over N+E} \le \frac E{N+E} = r(a_1) \le \frac N{N+E} $$ Well, this formula, in which $E$ must be evaluated after $N$ is given and might be checked for calculation of $r(a_1)$ between $\lceil N \gamma_1 \rceil$ and $N$, looks not very nice to me, so I'd reconsider the choice for the ratio-parameter $r()$.

FREE REAL ESTATE $$ {}{}{}{}{} $$


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Test newly available syntax using data from this question.

$n$ approximation solution
$1$ $2.84071 +1.69496 \,i$ $2.84550 +1.68429 \,i$
$2$ $3.34357 +1.44741 \,i$ $3.35044 +1.43963 \,i$
$3$ $3.64315 +1.33808 \,i$ $3.65025 +1.33195 \,i$
$4$ $3.85317 +1.27397 \,i$ $3.86019 +1.26882 \,i$
$5$ $4.01366 +1.23065 \,i$ $4.02051 +1.22614 \,i$
$6$ $4.14297 +1.19882 \,i$ $4.14966 +1.19475 \,i$
$7$ $4.25095 +1.17409 \,i$ $4.25749 +1.17037 \,i$
$8$ $4.34346 +1.15414 \,i$ $4.34987 +1.15068 \,i$
$9$ $4.42428 +1.13756 \,i$ $4.43056 +1.13432 \,i$
$10$ $4.49595 +1.12348 \,i$ $4.50212 +1.12041 \,i$
$20$ $4.95332 +1.04533 \,i$ $4.95880 +1.04314 \,i$
$30$ $5.21027 +1.00868 \,i$ $5.21539 +1.00685 \,i$
$40$ $5.38829 +0.98573 \,i$ $5.39318 +0.98411 \,i$
$50$ $5.52410 +0.96940 \,i$ $5.52881 +0.96791 \,i$
$60$ $5.63366 +0.95690 \,i$ $5.63824 +0.95551 \,i$
$70$ $5.72535 +0.94687 \,i$ $5.72982 +0.94556 \,i$
$80$ $5.80411 +0.93855 \,i$ $5.80849 +0.93730 \,i$
$90$ $5.87307 +0.93147 \,i$ $5.87738 +0.93027 \,i$
$100$ $5.93438 +0.92534 \,i$ $5.93862 +0.92418 \,i$
$200$ $6.32917 +0.88910 \,i$ $6.33302 +0.88818 \,i$
$300$ $6.55383 +0.87067 \,i$ $6.55748 +0.86985 \,i$
$400$ $6.71067 +0.85862 \,i$ $6.71420 +0.85787 \,i$
$500$ $6.83097 +0.84980 \,i$ $6.83439 +0.84908 \,i$
$600$ $6.92840 +0.84289 \,i$ $6.93176 +0.84222 \,i$
$700$ $7.01021 +0.83726 \,i$ $7.01351 +0.83661 \,i$
$800$ $7.08066 +0.83252 \,i$ $7.08391 +0.83189 \,i$
$900$ $7.14249 +0.82845 \,i$ $7.14570 +0.827838 \,i$
$1000$ $7.19756 +0.82489 \,i$ $7.20073 +0.824289 \,i$

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