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  • 8
    $\begingroup$ 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
  • 3
    $\begingroup$ 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 Maybe someday we can get rid of this outdated crutch.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


This question is free for all to use


$$ b= {ra + s \over 2^A} \\ c= {rb + s \over 2^B} = a\\ $$ $$ b\cdot a = {ra + s \over 2^A}\cdot {rb + s \over 2^B} \\ 2^S = (r+{s \over a})\cdot (r+{ s \over b} ) \\ {2^S\over r^N} = (1+{s \over ra})\cdot (1+{ s \over rb} ) $$ $$ S \log2-N \log r = \log(1+{s \over ra})+\log(1+{ s \over rb}) \\ S \log2-N \log r \lt {s \over ra}+{ s \over rb} = {s\over r}({1 \over a}+{1 \over b}) $$ let $a_h$ be the harmonic mean of $a$ and $b$, then one of the values $a$ and $b$ must be smaller than $a_h$ and one larger. Let $a \lt a_h$ then $a_h$ is an upper bound for $a$ and we have $$ (a \lt ) \qquad a_h \lt {s\over r}{1 \over S \log2-N \log r} \\ $$ Because the denominator on the rhs is not "small" the value for $a_h$ and $a$ on the lhs cannot be "large", and indeed it is possible to find an upper bound for $a_h$ depending on $N$, $s$, and $r$ using the known results of Baker and the improvements of the lower bounds for $S \log 2 - N \log r$ found so far.


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: $$ 2^{S_2} 3^{S_3} = \left(5+ \frac1{a_1} \right) \left(5+ \frac1{a_2} \right) \left(5+ \frac1{a_3} \right) \\ \text{ rhs in } (125,216) \\ \text{ lhs } \in \{128,144,162,192,216 \} \\ $$

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


Series $3$: until December $2$

  1. Let $n$ be an integer, $n > 2$. Given that for real numbers $a_{ij}$, where $1 ≤ i,j ≤ n$, for any $x_1, \cdots, x_n \in \ ${$-1,1$}, the following condition holds:

$$\sum_{1 ≤ i,j ≤ n} a_{ij}x_i x_j \in \text{{1,1}}$$

Evaluate in terms of $n$ the greatest number of such pairs $i,j$ that $a_{ij} \ne 0$ and $i < j$.

  1. A sphere inscribed in the tetrahedron $ABCD$ is tangent to the faces $BCD, CDA, DAB$ and $ABC$ respectively at the points $A', B', C', D'$. Prove that if the line segments (simplexes) $AA'$ and $BB'$ intersect, $CC'$ and $DD'$ also intersect.

  2. A prime number is given with $p > 2$. Let us call a positive integer $n$ pretty, if and only if the sum of the remainders when $n, n^2, n^3, \cdots, n^{p-1}$ is divided by $p$ is equal to $\frac{1}{2}p(p-1)$. Prove that in the set {$1, \cdots, p - 1$}, there are an odd number of pretty numbers.

Attention: We call $r$ the remainder from dividing an integer $a$ by a positive integer $b$, where $r \in$ {$0, 1, 2, \cdots, b-1$}, and that the number $a-r$ is divisible by $b$.

  1. Let $A_{m,n}$ denote the set of vectors $(k, l)$, where $0 \le k \le m-1$ and $0 \le l \le n-1$, and $m,n$ are integers. We call the function $f:A_{m,n} \to A_{m,n}$ good if and only if both conditions are met:

(1) $f$ is a multivalued function.

(2) if $v,w \in A_{m,n}$ and $v +f(v)-w-f(w) =(am,bn)$ for certain integers $a,b$, that $v = w$.

Find all pairs of positive integers $m, n$ for which there exist good functions $f:A_{m,n} \to A_{m,n}$.


Since $P$ is a quadratic polynomial and $P(0) = 0$, one can write $$ P(q) = aq+ bq^2$$ for some real numbers $a, b$. The conditions $$P'(\bar{q}_P)+\bar{q}_P=1, \frac{P(\hat{q}_P)}{\hat{q}_P}+\hat{q}_P=1$$ translates to \begin{align*}\tag{1} a + (2b+1) \bar q_P &= 1, \\ a+ (b+1) \hat q_P &= 1 \end{align*} or $$ \bar q_P = \frac{1-a}{2b+1}, \ \ \hat q_P = \frac{1-a}{b+1}.$$ We will assume that our conditions are that $\bar q_P ,\hat q_P>0$ instead of $\ge 0$ (when one of them is zero, $J(P) = 0$. Thus it suffices to check that the maximum of $J$ is positive, which we will do later).

The above conditions together with (1) imply that $b\neq -1$ and $b\neq -1/2$.

Next, note that $\bar q_P = \hat q_P$ only when $a=1$. The lines $$ a=1, \ \ b=-1, \ \ b = -1/2$$ split the $a-b$ plane into six region. Only two of them $$R_1 = \{a<1, b>-1/2\}, \ \ R_2 = \{ a>1, b<-1\}$$ gives BOTH positive $\bar q_P, \hat q_P$.

In each region either $q^*_P = \bar q_P$ or $\hat q_P$. For example, in $R_1$ we have $\hat q_P > \bar q_P$, giving $q^*_P = \bar q_P$.

Then one can check in each region: For example in $R_1$,

\begin{align*} J(aq+ bq^2) &= \int_0^{\frac{1-a}{2b+1}} (a+2bq-c)(1+q-\lambda (a+bq)-(1-\lambda)(a+2bq))dq \\ &=\int_0^{\frac{1-a}{2b+1}} (a+2bq-c)(1+q-(a+bq)+\lambda bq)dq \\ &=\int_0^{\frac{1-a}{2b+1}} (a-c+2bq)(1-a +(1 +(\lambda -1)b)q) dq \\ &=\int_0^{\frac{1-a}{2b+1}} \bigg(a(1-a) + \big(2(1-a)b +a(1 +(\lambda -1)b) \big)q + 2b(1 +(\lambda -1)b)q^2\bigg) dq\\ &= \frac{a(1-a)^2}{2b+1} + \frac{2(1-a)b +a(1 +(\lambda -1)b)}{2(2b+1)^2} \end{align*}


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.


First I will define the metric on Iso(M)

Let $M$ be a finite dimensional Riemannian manifold and $\operatorname{Iso}(M)$ be its set of isometries. It can be shown that $\operatorname{Iso}(M)$ is a finite dimensional manifold with a metric as defined below:

Consider $(n + 1)$ points on $M$ so close together that $n$ of them lie in an normal neighborhood of the other, and if the points are linearly independent (i.e. not in the same $(n-1)$-dimensional geodesic hypersurface). Then the distance $d(f, \tilde f)$ between two isometries $f$ and $\tilde f$ will be defined as the maximum of the distance $d_i[f(x), \tilde f(x)]$ as $x$ ranges over the given set of $n+1$ points. This distance can be shown to satisfy the usual metric axioms. Here $d_i$ is of course the induced metric on $M$ (Riemannian distance fucntion)

Given $\operatorname{Iso}(M)$ is now a metric space with metric $d$ as defined, we thus get a natural metric topology for $\operatorname{Iso}(M)$. That is open sets are all subsets that can be realized as the unions of open balls of form $B(f_0, r) = \{f \in \operatorname{Iso}(M): d(f_0,f)< r\}$ where $f_0 \in \operatorname{Iso}(M)$ and $r>0$.

$\ $

Q) I am trying to prove that $$\mathscr M: \operatorname{Iso}(M) \times \operatorname{Iso}(M) \rightarrow \operatorname{Iso}(M), \, (f,g) \mapsto f \circ g$$ is continuous in the metric topology of $\operatorname{Iso}(M)$.

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Ans) It is easy to show that $\operatorname{Iso}(M)$ is a group (which I have shown before) and I am assuming the continuity of the inversion $$\iota: \operatorname{Iso}(M) \to \operatorname{Iso}(M), \, f \mapsto f^{-1},$$ so I will use these two facts in the following answer. Now let $S$ denote the set of the $n+1$ points defining the metric on $\operatorname{Iso}(M)$. For $g, f, \tilde{f} \in \operatorname{Iso}(M)$ we have $$d(g \circ f, g \circ \tilde{f}) = \max_{x \in S} d_i(g(f(x)), g(\tilde{f}(x))) = \max_{x \in S} d_i(f(x), \tilde{f}(x)) = d(f, \tilde{f})$$ since $g$ is an isometry, so the map $$\mathscr{M}(g, -): \operatorname{Iso}(M) \to \operatorname{Iso}(M), \, f \mapsto g \circ f$$ is a $d$-isometry and in particular continuous for any $g \in \operatorname{Iso}(M).$ Furthermore $$f \circ g = \left((f \circ g)^{-1}\right)^{-1} = (g^{-1} \circ f^{-1})^{-1}, \text{ i.e. } \mathscr{M}(f,g) = \iota(\mathscr{M}(g^{-1}, \iota(f)),$$ so the continuity of the inversion and the above shows that $$\mathscr{M}(-,g): \operatorname{Iso}(M) \to \operatorname{Iso}(M), \, f \mapsto f \circ g$$ is continuous for $g \in \operatorname{Iso}(M)$ since $\mathscr{M}(-,g) = \iota \circ \mathscr{M}(g^{-1}, -) \circ \iota$.

Now let $(f_n)_{n \in \mathbb{N}}, (g_n)_{n \in \mathbb{N}}$ be sequences in $\operatorname{Iso}(M)$ and $f,g \in \operatorname{Iso}(M)$ such that $$\lim_{n \to \infty} d(f_n, f) = \lim_{n \to \infty} d(g_n, g) = 0.$$ Then the continuity of $\mathscr{M}(-,g)$ implies $\lim_{n \to \infty} d(f_n \circ g, f \circ g) = 0$. Furthermore we can use the triangle inequality of $d$ and the fact that $\mathscr{M}(f_n, -)$ is a $d$-isometry for any $n \in \mathbb{N}$ to obtain $$d(f_n \circ g_n, f \circ g) \leq d(f_n \circ g_n, f_n \circ g) + d(f_n \circ g, f \circ g) = d(g_n, g) + d(f_n \circ g, f \circ g),$$ so $\lim_{n \to \infty} d(f_n \circ g_n, f \circ g) = 0$. Hence $$\mathscr{M}: \operatorname{Iso}(M) \times \operatorname{Iso}(M) \to \operatorname{Iso}(M), \, (f,g) \mapsto f \circ g$$ is sequentially continuous and therefore continuous on the metric space $\operatorname{Iso}(M) \times \operatorname{Iso}(M)$.

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Q) So now the only thing left to do would be to prove the assumption. That is I need to show that the inversion map $$\iota: \operatorname{Iso}(M) \to \operatorname{Iso}(M), \, f \mapsto f^{-1},$$ is indeed continuous in the metric topology of Iso(M)

That is I need a proof that if {$f_k$} $\rightarrow$ $f$ then {$f_k^{-1}$} $\rightarrow$ $f^{-1}$

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$\ $


This answer is available for anyone to use


This slot is available for anyone to use


$$ \begin{array} {rll}a&= 2^{a-1}+5 \\ a-6 &= 2^{a-1}-1 \\ a-6 &= 32 \cdot 2^{a-6}-1 \\ b &= 32 \cdot 2^b-1 \\ \end{array}$$

$$ \begin{array} {rll}a= 2^{a-1}+5 \\ a-1 = 2^{a-1}+4 \\ b = 2^b+4 \\ b-4 = 16 \cdot 2^{b-4} \\ c = 16 \cdot 2^c \\ c \log2 = 16 \log2 \cdot \exp(c \log2) \\ d = 16 \log2 \cdot \exp( d) \\ -d \exp(-d) = 16 \log2 \\ -d = LW(16 \log2) & = -1.44956906065 + 2.16153171200*I \end{array}$$

$$ \begin{array} {rll} a_1 &= 1+2 \\ a_2 &= 1+(2^2+4)&=1+4(2^0+1)\\ a_3 &= 1+(2^{2^2+4}+4)&=1+4(2^{2^2+2}+1)&=1+4(4 \cdot 2^{2^2}+1)\\ a_4 &= 1+(2^{2^{2^2+4}+4}+4)&=1+4(2^{2^{2^2+4}+2}+1)\\ a_{k+1} &= 1+ (2^{a_k-1} + 4) \\ a_{k+2} &= 1+ (2^{a_{k+1}-1} + 4) &= 1+ 2^{2^{a_k}+4} + 4 \\ a_{k+3} &= 1+ (2^{a_{k+1}-1} + 4) &= 1+ 2^{2^{a_k}+4} + 4 \\ \end{array} $$

This slot is available for use.


$\mathcal{Y}\sim \mathcal{N}(\mu=\frac{X}{X+K_m}, \sigma^2) \Rightarrow $

In this answer I assume $V_{max}$ is known to be (without loss of generality) 1. As confirmed in the comments you are using the following model:

$\mathcal{Y}\sim \mathcal{N}(\mu=\frac{X}{X+K_m}, \sigma^2)$

The corresponding likelihood function is

$L(K_m, \sigma) = p_{\mathcal{Y^N}}(Y^N|K_m, \sigma, X^N) = \prod_{i=1}^Np_{\mathcal{N}}(Y^N|\mu=\frac{X^N_i}{X^N_i+K_m}, \sigma^2)$,

where $p_\mathcal{N}$ is the density of the normal distribution.

Now, you would like to know the distribution of a random variable $\mathcal{\hat{K_m}}$ that is the maximum likelihood estimate,

$ML_{\hat{K_m}}(X^N,Y^N) = \operatorname*{argmax}\limits_{K_m} \operatorname*{max}\limits_{\sigma} p_{\mathcal{Y^N}}(Y^N|K_m, \sigma, X^N)$ $ = \operatorname*{argmax}\limits_{K_m} \operatorname*{max}\limits_{\sigma} \prod_{i=1}^Np_{\mathcal{N}}(Y^N_i|\mu=\frac{X^N_i}{X^N_i+K_m}, \sigma^2)$

$ = \operatorname*{argmax}\limits_{K_m} \operatorname*{max}\limits_{\sigma} \sum_{i=1}^N\log(p_{\mathcal{N}}(Y^N_i|\mu=\frac{X^N_i}{X^N_i+K_m}, \sigma^2))$

$ = \operatorname*{argmax}\limits_{K_m} \operatorname*{max}\limits_{\sigma} \sum_{i=1}^N\log(\frac{1}{\sqrt{2\pi\sigma^2}}) - \frac{\left(Y^N_i-\frac{X^N_i}{X^N_i+K_m}\right)^2}{2\sigma^2}$

$ = \operatorname*{argmin}\limits_{K_m} \sum_{i=1}^N \left(Y^N_i-\frac{X^N_i}{X^N_i+K_m}\right)^2$,

obtained for draws of draws of size $N$ from $\mathcal{Y}$, $\mathcal{Y^N}$, for any $N$, $X^N$, $\sigma$.

You then sampled $K_m$ for some fixed $K$, $X^N$, $K_m$ and $\sigma$ by first sampling $\mathcal{Y^N}$ accordingly and then applying above ML estimator. Based on this, you think that $\mathcal{K_m}$ follows a log normal distribution.

It is known that, for any differentiable function $f: \mathbb{R}^N \to \mathbb{R}$ and $\mathcal{Y} = f(\mathcal{X})$,
$p_\mathcal{Y}(y) = \int_x \delta(f(x)-y) p_\mathcal{X}(x)\mathrm{d}x$ , where $\delta$ is the Dirac delta.

And that for any monotonic function $g: \mathbb{R} \to \mathbb{R}$ and $\mathcal{Y} = f(\mathcal{X})$,
$p_\mathcal{Y}(y) = p_\mathcal{X}(g^{-1}(y)) \left|\frac{\mathrm{d}}{\mathrm{d}y} g^{-1}(y) \right|$

We can use this to try to derive a closed form for the density of the distribution of $\mathcal{\hat{K_m}}$:

$p_{\mathcal{\hat{K_m}}}(\hat{K_m})=\int \delta (\hat{K_m}-ML_{\hat{K_m}}(X^N,Y^N)) p_{\mathcal{Y^N}}(Y^N) \mathrm{d} Y^N$

$\overset{\tiny{\text{if i'm lucky}}}{=}\int \delta(\frac{\mathrm{d}}{\mathrm{d} \hat{K_m}} \sum_{i=1}^N \left(Y^N_i-\frac{X^N_i}{X^N_i+\hat{K_m}}\right)^2) p_{\mathcal{Y^N}}(Y^N) \mathrm{d} Y^N$

$=\int \delta(\sum_{i=1}^N \frac{X^N_i(\hat{K_m}Y^N_i+X^N_i(Y^N_i-1))}{(\hat{K_m}+X^N_i)^3}) p_{\mathcal{Y^N}}(Y^N) \mathrm{d} Y^N$

But I don't how to find a simpler form for that.

For $N=1$ this is a bit simpler:

$p_{\mathcal{\hat{K_m}}}(\hat{K_m})=p_\mathcal{Y}(g^{-1}(\hat{K_m})) \left|\frac{\mathrm{d}}{\mathrm{d}\hat{K_m}} g^{-1}(\hat{K_m}) \right| = p_\mathcal{Y}(\frac{X}{X+\hat{K_m}}) \left|\frac{\mathrm{d}}{\mathrm{d}\hat{K_m}} \frac{X}{X+\hat{K_m}} \right|= p_\mathcal{Y}(\frac{X}{X+\hat{K_m}}) \left|- \frac{X}{(X+\hat{K_m})^2} \right|= p_{\mathcal{N}}(\frac{X}{X+\hat{K_m}}|\mu=\frac{X^N_i}{X^N_i+K_m}, \sigma^2) \frac{X}{(X+\hat{K_m})^2} $

This contains $ML_{\hat{K_m}}(X^N,Y^N)$ for which we might be able to find and algebraic expression for: $ML_{\hat{K_m}}(X^N,Y^N)

$ 0 = \left.\frac{\mathrm{d}}{\mathrm{d} K_m} \sum_{i=1}^N \left(Y^N_i-\frac{X^N_i}{X^N_i+K_m}\right)^2\right|_\hat{K_m}$ $ = \sum_{i=1}^N \left.\frac{\mathrm{d}}{\mathrm{d} K_m} \left(Y^N_i-\frac{X^N_i}{X^N_i+K_m}\right)^2\right|_\hat{K_m}$ $ = \sum_{i=1}^N \frac{X^N_i(\hat{K_m}Y^N_i+X^N_i(Y^N_i-1))}{(\hat{K_m}+X^N_i)^3}$

For $N=1$, $0=\frac{x(\hat{K_m}y+x(y-1))}{(\hat{K_m}+x)^3}$ solves $\hat{K_m}=x(\frac{1}{y}-1)$ whereas the solution for $N=2$ has a few more terms From where I don't know how to continue.


Let's prove, without using any set theory, that the natural numbers are the natural numbers!

Specifically, we're going to show that the natural numbers, defined as the inductive data type generated by one constant and one unary operation, are the natural numbers, defined as the free monoid with one generator. We'll do this by sheer symbol manipulation, without recourse to any underlying foundational theory.

Starting definitions

Define a general presentation as an algebraic theory. When we talk about general presentations, we are interested in the initial algebra of the underlying theory, and we call it the algebra presented by that presentation. Given two general presentations $P$ and $Q$, we say that $Q$ is a conservative extension of $P$ if every axiom (that is, every term (generator) and every equation) of $P$ is also an axiom of $Q$, and the algebras generated by $P$ and $Q$ are isomorphic (when considered as algebras for the underlying theory of $P$).

$\newcommand{Nat}{\mathbb{N}}\newcommand{Nind}{\Nat_\mathrm{ind}}\newcommand{Nmon}{\Nat_\mathrm{mon}}$In order to show that "the natural numbers are the natural numbers", we will consider two general presentations $\Nind$, representing the first definition of the natural numbers, and $\Nmon$, representing the second, and we will show that there is a third general presentation, $\Nat_\mathrm{ind+mon}$, which is a conservative extension of both.

First, here is $\Nind$:

  • $\newcommand{Zero}{\mathrm{Zero}}\Zero : \Nat$
  • $\newcommand{PlusOne}{\mathrm{PlusOne}}\PlusOne : \Nat \to \Nat$

In other words, $\Nind$ consists of a generator (or term) called $\Zero$ of arity zero; a generator called $\PlusOne$ of arity one; and no equations.

The general presentation $\Nmon$ is more complex. Here it is:

  • $\Zero : \Nat$
  • $\newcommand{Plus}{\mathrm{Plus}}\Plus : \Nat \times \Nat \to \Nat$
  • $\newcommand{One}{\mathrm{One}}\One : \Nat$
  • $\Plus(\Zero, x) = x$
  • $\Plus(x, \Zero) = x$
  • $\Plus(x, \Plus(y, z)) = \Plus(\Plus(x, y), z)$

The general presentation $\Nmon$ consists of three generators (two of arity 0, one of arity 2) and three equations. Notice that the axioms of $\Nmon$, aside from $\One$, are simply the axioms of a monoid. Adding $\Nmon$ to the general presentation turns it into a presentation of a monoid with one generator.

How can we prove things?

Now, before we continue, here's a question: how can we identify conservative extensions of $\mathbb{N}_\mathrm{ind}$ without recourse to an underlying foundational theory?

Well, the Tietze transformations are a collection of rules for transforming a group presentation into another equivalent group presentation. The new group presentation is equivalent in the sense that the group generated by the new presentation is isomorphic to the group generated by the old presentation, and, furthermore, the isomorphism preserves any generators which were not affected by the transformation.

There are four Tietze transformations:

  • Adding a generator: You may add a generator, along with an equation asserting that the new generator equals some term.
  • Removing a generator: If an equation asserts that some generator equals some term, and that generator does not appear in the right-hand side of that equation, or anywhere in any other equation, then you may remove that generator and that equation.
  • Adding an equation: You may add an equation, if you can prove that equation from the other equations.
  • Removing an equation: You may remove an equation, if you can prove that equation from the other equations.

For the purposes of adding and removing an equation, a proof is not allowed to refer to outside axioms, or even to use first-order logic; the proof must be performed entirely using the substitution and reflexive properties of equality.

The Tietze transformations work just fine for general presentations, too. Specifically, if one applies the "add a generator" or "add an equation" transformation to a general presentation, the new presentation will be a conservative extension of the old one. If one applies the "remove a generator" or "remove an equation" transformation, the converse happens: the new presentation will be conservatively extended by the old one.

Given an underlying theory such as ZFC, hopefully the assertions I made in the above paragraph would not be too difficult to prove. For now, we simply take the four transformations as axioms.

The Tietze transformations will get us part of the way where we want to go, but in order to go the entire way, we will need to use two additional transformations:

  • Adding a function: You may add a function, along with a collection of equations which constitute a primitive recursive definition of that function.
  • Removing a function: You may perform the reverse of adding a function.

There will be an example of just what I mean by this below.

In addition, inductive proofs of equality will be permitted. I apologize for not giving a definition of what constitutes an inductive proof of equality.

The proof

We now have all we need to show that there is a general presentation $\Nat_{\mathrm{ind+mon}}$ which is a conservative extension of both $\Nind$ and $\Nmon$.

We start with $\Nind$, which consists of only these two axioms:

  1. $\Zero : \Nat$
  2. $\PlusOne : \Nat \to \Nat$

We will now apply the "adding a function" transformation. We add a function symbol and two equations:

  1. $\Plus : \Nat \times \Nat \to \Nat$
  2. $\Plus(\Zero, y) = y$
  3. $\Plus(\PlusOne(x), y) = \PlusOne(\Plus(x, y))$

Notice that the definition of $\Plus$ is by cases, and the cases are perfectly exhaustive: every possible pair of terms of $\Nind$ fits exactly one of the cases. Furthermore, although the definition of $\Plus$ is recursive, the recursion is primitive recursion.

We now desire to apply the "adding an equation" transformation:

  1. $\Plus(x, \Zero) = x$

This can be proven inductively by noting that $\Plus(\Zero, \Zero) = \Zero$ and $\Plus(\PlusOne(x), \Zero = \PlusOne(\Plus(x, \Zero)) = \PlusOne(x)$.

We apply the "adding an equation" transformation again:

  1. $\Plus(x, \Plus(y, z)) = \Plus(\Plus(x, y), z)$

Once again, we can do a proof by induction, noting that $\Plus(\Zero, \Plus(y, z)) = \Plus(y, z) = \Plus(\Plus(\Zero, y), z)$ and that

$$\Plus(\PlusOne(x), \Plus(y, z)) = \PlusOne(\Plus(x, \Plus(y, z))) = \PlusOne(\Plus(\Plus(x, y), z)) = \Plus(\PlusOne(\Plus(x, y)), z) = \Plus(\Plus(\PlusOne(x), y), z).$$

Next, we apply the "adding a generator" transformation:

  1. $\One : \Nat$
  2. $\One = \PlusOne(\Zero)$

Finally, we apply the "adding an equation" transformation again:

  1. $\PlusOne(x) = \Plus(\One, x)$

We have created a general presentation with 10 axioms which is a conservative extension of $\Nind$. This general presentation is called $\Nat_\mathrm{ind + mon}$.

Next, it only remains to show that $\Nat_\mathrm{ind + mon}$ is a conservative extension of $\Nmon$ as well. In order to do this, we will start by listing the axioms of $\Nat_\mathrm{ind + mon}$ again, but in a different order:

  1. $\Zero : \Nat$
  2. $\Plus : \Nat \times \Nat \to \Nat$
  3. $\One : \Nat$
  4. $\Plus(\Zero, y) = y$
  5. $\Plus(x, \Zero) = x$
  6. $\Plus(x, \Plus(y, z)) = \Plus(\Plus(x, y), z)$
  7. $\PlusOne : \Nat \to \Nat$
  8. $\PlusOne(x) = \Plus(\One, x)$
  9. $\Plus(\PlusOne(x), y) = \PlusOne(\Plus(x, y))$
  10. $\One = \PlusOne(\Zero)$

Equation 10 can be proven from the other equations: $\One = \Plus(\One, \Zero) = \PlusOne(\Zero)$. So we may remove it.

Next, equation 9 can also be proven from the other equations: $\Plus(\PlusOne(x), y) = \Plus(\Plus(\One, x), y) = \Plus(\One, \Plus(x, y)) = \PlusOne(\Plus(x, y))$. So we may remove it as well.

At this point, only axioms 1 through 8 remain. We can use the "removing a function" rule to remove axioms 7 and 8, leaving only axioms 1 through 6. These axioms are $\Nmon$.

This completes the proof that both $\Nind$ and $\Nmon$ are both conservatively extended by a single presentation $\Nat_\mathrm{ind + mon}$.

To restate, we have shown that the natural numbers, defined as the inductive data type generated by one constant and one unary operation, are the natural numbers, defined as the free monoid with one generator.

The question

Surely I'm not the first person to think of all this.

Has anyone studied these "generalized Tietze transformations" before? How powerful are they? Are they sufficiently powerful to prove, say, the fundamental theorem of arithmetic? (We would do that by showing that the positive integers, as usually defined, are isomorphic to the free monoid on countably infinitely many generators, with the monoid operation being multiplication.)


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After the answer of @JohnOmielan a further answer is not really needed. But I'd like to comment on a simpler view, just a bit empirical, which would show, at least: indicate, that an argument for the non-existence of cycles with $m+1$ where $2^m \gt 3^n \gt 2^{m-1}$ is not so easy...

If we reformulate the lhs and the rhs a bit, saying $$ lhs=\frac{2^m}{3^n} \\ rhs = \prod_{k=1}^n \left(1+\frac1{3x_k} \right) $$ and we check the possibility of existence of a nontrivial cycle with $x_k>1$ simply by getting an upper bound for the minimal $x_1$ due to the (rough) mean value $$x_m = { 1 \over 2^{m/n} -3 } \qquad \text{ with } x_1 \lt x_m $$ then we'll find, that for many $n$ and their related value for $m$ we find, that $x_m$ has a value high enough that after $5 \le x_1 \lt x_m$ the minimal element $x_1$ could have a reasonable value in the positive integers $6j \pm 1$ (which of course we must check individually whether that value allows actually a cycle).
So for some $n$ we find $x_m$ too small to allow a cycle with an allowed value in $x_1$ at all, for instance:


Let $M \ge 2$. Inspired by the Collatz iteration / algorithm ($M=2$), I tried the following function:

$$C_s(a) = \cases { a \over s & if $a \equiv 0 \mod s$ \\ (s+1)a+\left((s-a) \pmod s \right) & if $a \not \equiv 0 \mod s$\\ } $$

$$C_s(a) = \cases { a \over s & if $a \equiv 0 \mod s$ \\ as+ (a - (a \mod s)) & if $a \not \equiv 0 \mod s$\\ } $$

$$\begin{array} {} a&=ks+b \\ x & = (s+1)(ks+b)+\left((s-(ks+b)) \pmod s \right) \\ & = (s+1)ks +sb+b -b \\ & =s( (s+1)k +b) \\ \end{array} $$

$$C_M(n) = \cases {n \over M & if $n \equiv 0 \mod M$, \\ (M+1)n+\left((M-n) \pmod M \right) & \text{ otherwise }\\ } $$

We might view, unproven, the Collatz iteration, as an algorithm to solve the following diophantine equation:

$$\forall n \in \mathbb{N} \exists x,y,s,l \in \mathbb{N}_0: xM^s = (M+1)^l n + y$$


$$s+l = \text{ stopping time until first duplicate value in a cycle is detected }$$

If the Collatz iteration stops at the number $c$ because the next value would be a duplicate in the iteration, then:

$$ c = \frac{a}{\hat{a}}n + \frac{b}{\hat{b}}$$

This answer is free for anyone to use.


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$$2^S-3^N =3^{N-1}(1/a+1/b+...+1/h) +.... \geq 2^S/ 2^{S/10/l2} \\ 1- 3^N/2^S \geq e^{-S/10} \\ 1-e^{-S/10} \geq 3^N/2^S \\ \log(1-e^{-S/10}) \geq N \log(3) - S \log(2) $$

$$2^S-3^N \geq 2^S e^{-S/10} \\ 1- 3^N/2^S \geq e^{-S/10} \\ 1-e^{-S/10} \geq 3^N/2^S \\ \log(1-e^{-S/10}) \geq N \log(3) - S \log(2) $$ $$|2^S-3^N| \geq 2^S e^{-S/10}= 1.808^{S}$$

You can use Ellison’s estimate, $$|2^S-3^N| \geq 2^S e^{-S/10}= 1.808^{S}$$, which holds for $S \geq 12$ with $ S \neq 13, 14, 16, 19, 27$ and all $N$.
This is based on results by Pillai and Baker. Reference: “WJ Ellison, On a theorem of S. Sivasankaranarayana Pillai, S´eminaire de th´eorie des nombres de Bordeaux, 1970 (1971), pp. 1–10.”


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