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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
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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 Maybe someday we can get rid of this outdated crutch.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


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$$ b= {ra + s \over 2^A} \\ c= {rb + s \over 2^B} = a\\ $$ $$ b\cdot a = {ra + s \over 2^A}\cdot {rb + s \over 2^B} \\ 2^S = (r+{s \over a})\cdot (r+{ s \over b} ) \\ {2^S\over r^N} = (1+{s \over ra})\cdot (1+{ s \over rb} ) $$ $$ S \log2-N \log r = \log(1+{s \over ra})+\log(1+{ s \over rb}) \\ S \log2-N \log r \lt {s \over ra}+{ s \over rb} = {s\over r}({1 \over a}+{1 \over b}) $$ 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.


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Question title: Can we name every object in every model of a FO theory?

Self answer:

Consider that we have a name for every object, in every model of a language.

Since we have a name for every object, consider an infinite set of formulae comprising of a property for every name in the language $(1)$:

\begin{equation}\tag{1}\{\lnot P_1,\lnot P_2,\lnot P_3\dots\}\end{equation}

Then consider those same formula in a theory $T$, which also contains the sentence $(2)$.

\begin{equation}\tag{2}\exists x \lnot P_x\end{equation}

Now each propositional atom cannot be satisfied unless that atom is false, so it cannot be the case that $\exists x \lnot P_x$ is true.

However, by compactness, since each finite subset is satisfiable, the whole thing is.

This is a contradiction, so it must not be the case that for a particular language, that we can have a name for every object, in every model of it.

That is to say, for each language, we can not have a symbol for every object, in every model of it.


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*}


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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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Question: Epitrochoids and adjacent loop touching

I'll replace OP's $\alpha$ with $n$ (to avoid visual confusion with $a$). Also, I'll use a version of the curve such that each "half-leaf" is traced by $t$ in the range of consecutive integer multiples of $\pi/(n-1)$.

$$P = (\;\cos t + a \cos n t \;,\; \sin t + a \sin n t\;) \tag{1}$$

enter image description here

With this version of the curve, we observe that when adjacent loops are tangent, the point of tangency lies on the $x$-axis, and the tangent vector at that point is horizontal. Thus, a necessary condition for tangency is that the $y$-components of $P$ and $\frac{d}{dt}P$ simultaneously vanish: $$\sin t + a \sin n t = 0 = \cos t + na \cos n t \tag{2}$$ for some $\dfrac{\pi}{n-1}\leq t < \dfrac{2\pi}{n-1}$. Also, we observe that $|a|>1$ causes loops to surround the origin, precluding tangency; so we may assume $|a|\leq 1$.

Our goal is to eliminate $t$ from the system in $(2)$. One route ventures through the complex realm. Recall the expressions for cosine and sine in terms of the complex exponential: $$\cos\theta = \frac12\left(e^{i\theta}+e^{-i\theta}\right)\qquad \sin\theta = \frac1{2i}\left(e^{i\theta}-e^{-i\theta}\right) \tag{3}$$ Defining $\omega = \exp(it)$ and $\omega_n = \exp(int)$, our system becomes $$\begin{align} \omega_n(\omega^2 - 1) + \phantom{n}a \omega(\omega_n^2-1) &= 0 \\ \omega_n(\omega^2+1)+ na\omega(\omega_n^2+1) &= 0 \end{align} \tag{4}$$ which has four solutions of the form $$\omega = \frac{p+q}{\sqrt{n^2-1}} \qquad \omega_n = -\frac{p+nq}{na\sqrt{n^2-1}} \qquad\qquad \begin{array}{l} p^2 := n^2 (1-a^2) \geq 0 \\ q^2 := 1-n^2a^2\;\; \text{(possibly $<0$)} \end{array} \tag{5}$$ (taking all combinations of signs on $p$ and $q$). Since $\omega^n=\omega_n$, we can write

$$na\left(p+q\right)^n = -(p+nq)(n^2-1)^{(n-1)/2} \tag{$\star$}$$

So, for a given $n$, candidate values of $a$ that cause tangent loops satisfy the four sign variants of $(\star)$.

Interestingly, $p^2-q^2=n^2-1$ and $p^2-n^2q^2 = n^2a^2(n^2-1)$, so that can square the equation (potentially introducing extraneous $a$ candidates) and rewrite it a bit more symmetrically as

$$\left(\frac{p+q}{p-q}\right)^{n} = \frac{nq+p}{nq-p} \tag{6}$$

In this form, we see that the simultaneous substitutions $p\to-p$ and $q\to-q$ preserve the relation, so there are really only two sign combinations to worry about and we can fix, say, the sign of $p$. Moreover, the substitution $q\to-q$ effectively reciprocates each side of the equation, while also changing the sign on the right-hand side.

Consequently, we can fix the signs of both $p$ and $q$ and express the sole sign ambiguity thusly:

$$\left(\frac{p+q}{p-q}\right)^{n} = \pm \frac{nq+p}{nq-p} \tag{$\star\star$}$$

We see immediately that $a=\pm1$ and $a=\pm 1/n$ satisfy $(\star\star)$ by making either $p$ or $q$ vanish.


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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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