Mathematics - 2023-05-22

12:00 AM

Otherwise you have to use inclusion/exclusion carefully.

So I'm starting the putnam question. I'm only going to be able to start it, but I wanted to know if this is the right direction. I was going to compute the area of the quarter of the ellipse in the positive quadrant by taking an integral in polar coordinates. Then once I get this area I can split this area in half. To do this will depend on finding the $m$ which splits my area in half

Ted Shifrin

12:21 AM

Good luck with that.

D.C. the III

Thought I'd run it by you first to make sure I wasn't spinning my wheels

monoidaltransform

12:57 AM

If $X$ is a (infinite dimensional) CW complex, are all integer cohomology groups finitely generated?

In case of $S^{\infty}$ and $R^{\infty}$ this is clear, because they are contractible

one potato two potato

How about infinite wedge sum of spheres?

I remember f.g. holds for manifolds at least

monoidaltransform

Thats just direct sum of cohomology of spheres, right? @onepotatotwopotato

one potato two potato

yes

monoidaltransform

ah right, direct sum of f.g. generated modules need not be f.g.

robjohn

1:49 AM

@D.C.theIII are the axes of the ellipse parallel to the coordinate axes?

D.C. the III

yes. $\frac{x^2}{4} + y^2 \leq 1$

robjohn

The area in the first quadrant is $\frac\pi4ab=\frac\pi2$

It's just a mapped disc and you can use the Jacobian to scale the area

D.C. the III

Is there any reason in particular you know that so quickly or is this something I should have known?

robjohn

The area of an ellipse is $\pi ab$ where $a$ and $b$ are the semi-major and semi-minor axes.

D.C. the III

So you would use a change of coordinates between an ellipse and a unit circle say?

robjohn

1:55 AM

this gives $\pi r^2$ when $a=b=r$

@D.C.theIII yes

D.C. the III

It just so happens that you've done it so many times that you know the area of an ellipse by heart now...

@robjohn interesting because where it falls in Ted's book is before the chapter on change of variables I did so didn't think it would apply.

The whole question is asking for me to find the value of $m$ such that the line $y = mx$ bisects the region of the ellipse in the first quadrant

robjohn

ah, so half the area.

D.C. the III

yes. That's why I was thinking "how can I represent bisecting this region"?

robjohn

what slope does that when you have a circle?

D.C. the III

in that case is $m = 1$

robjohn

1:59 AM

so, take that picture and expand out by a factor of $2$ in the $x$ direction.

what would the slope of the scaled line be?

D.C. the III

hmmmmm.......$1/2$

robjohn

and that would be the answer

D.C. the III

well this is rather anti-climatic.....this was in a chapter about determinants

robjohn

You could mention the Jacobian of the matrix $\begin{bmatrix}2&0\\0&1\end{bmatrix}$

which is the transform from the unit circle to that ellipse

D.C. the III

and up to this point I had learned to solve for the coefficients of a curve: circle, parabola, etc when having a set of given points and setting up a linear problem to solve for the coeffcients of the given curve

Well actually I did do the change of variables as well, but didn't think it would apply. But I will look at the "formal" way of doing it w.r..t mentioning the Jacobian

you did do something I should be getting better at: simplifying the question into something easier to solve and then translating those ideas to a more complex scenario.

robjohn

2:09 AM

@D.C.theIII a lot of progress in math is started that way.

D.C. the III

I keep on forgetting to apply the simple idea.

Rome wasn't built in a day

Ted Shifrin

Just linear map and area/volume. Pay attention to where it is in the book — a hint people taking the Putnam exam don’t get.

If you go back to Chapter 2, the ellipse is presented as the image of a circle under a linear map.

copper.hat

part of Rome was built in a day

D.C. the III

ah yes $(a \cos t, b \sin t)$

I may not end up doing the Putnam, but this will help when I do take the GRE math

I'll be more robust with my discussion on it tomorrow, but for now it is back to Beyond Multiple Linear Regression

leslie townes

the basic scaling stuff is very much at the heart of what area is. don't think jacobian as much as basic area stuff. anything is a sum of little squares or rectangles, and think of how areas of those scale.

as a bonus problem, compute the arc length of the ellipse with semiaxes a and b :)

robjohn

2:23 AM

@leslietownes nasty

Ted Shifrin

@robjohn What do we expect? He is, after all, munchkin’s progeny.

robjohn

2:44 AM

@TedShifrin true

goedelite

3:26 AM

I am reading Graham and Green's Topology published by Dover, I like its conciseness in contrast to a more encyclopedic text such as that by Munkres. I find many misprint or what seem to me to be misprings in G&G. I wonder if it is worth trying to overcome them, or is there another text as concise but with more careful printing?

I seem also to create misprints!

copper.hat

3:42 AM

Misprints throw me completely.

leslie townes

4:30 AM

that's kind of weird, usually dover reprints things that are, if not a 'classic,' at least classic enough to have gone through more than one printing, and at least one round of fixes (and not on dover's dime but like a real publisher's budget).

i did not know that graham greene wrote a topology book, but he was sure was prolific, why not

Koro

8:03 AM

12 hours ago, by Koro

(i.e., if $(b_n)$ is a sequence of non negative integers and $b: Z^+\to [0,\infty]: n\mapsto b_n$, then $\int b \,dc=\sum_n b_n$, where c is the counting measure.)

Here is my solution for this: For any $k\in \mathbb N$, $\{1\}, \{2\},..., \{k\}$, $Z^+- \{1,2,...,k\}$ is a partition $P$ of $Z^+$. So we have $L(b, P)=\sum_{i=1}^k b_i$. It follows that $\int f dc\ge \sum_{i=1}^k b_i$ for all $k$. Letting $k\to \infty$, we get $\int f dc\ge \sum_i b_i$.

For the reverse inequality, let $P=\{A_1,...,A_k\}$ be any partition of $Z^+$. We have $L(b, P)= \sum_{i=1}^k (\inf_{A_i} b) c(A_i)\le \sum_{i=1}^\infty b_i$, where the last inequality is by non negativity of $b_i$'s. It follows that $\int f\, dc\le \sum_{i=1}^\infty b_i.$

This is the definition that I'm using for integration:

12 hours ago, by Koro

Let $(X, S, \mu)$ be a measure space. Let $A_1,A_2,...,A_k$ be an S- partition P of X: this means that $Ai$'s are elements of the sigma algebra S and that $\bigcup{1\le i\le k} Ai=X$. Let $f:X\to [0,\infty]$ be S-measurable. Lower Lebesgue sum is defined as $L(f, P)=\sum_{j=1}^k \inf_{x\in A_j} f(x)\mu (A_j)$. Then, $\int f , d\mu := \sup \{L(f,P): P$ is an $S-$ partition of $X$}

One good thing about this is that the proof for 'integration w.r.t. to the counting measure is summation' does not require monotone convergence theorem.

Koro

9:40 AM

0

Q: What is the use of taking $t\in (0,1)$ in proving monotone convergence theorem?

Let $(X, S, \mu)$ be a measure space.Let $0\le f_1\le...$ be an increasing sequence of $S$ measurable functions. Define $f:X\to [0,\infty]$ by $f=\lim_n f_n$. Then $\lim_{k\to \infty} \int f_k = \int f.$ Proof: $f$ is measurable as it is limit of measurable functions. $\forall x\in X, f_k(x)\le f...

Sourav Ghosh

9:57 AM

let X be a finite set with n elements and 2< m <2^n. Is the characterization of m for which a topology of m elements exists possible?

@AlessandroCodenotti

For n=2 , m=2, 3,4 all are possible.

one potato two potato

In the next semester, functional analysis and algebra 2 courses will be held at the same time. I need to choose one of them. Quite a disappointing decision of the department.

Sourav Ghosh

Indiscrete, two particular point topology (homeomorphic), discrete.

For n=3 , only m=7 is not possible.

@leslietownes

Is there any existential argument that m=7 is possible for n=4 ?

@AlessandroCodenotti please :)

Jakobian

10:56 AM

@onepotatotwopotato I think the obvious choice here is functional analysis

:-)

@SouravGhosh what?

one potato two potato

@Jakobian Agree

Algebra 2 prof is different from the current algebra 1 prof (why?). Syllabus says it deals with some commutative algebra, homological algebra, etc. not very tempting.

Sourav Ghosh

11:12 AM

@Jakobian See my MO post mathoverflow.net/q/447280/483536

porridgemathematics

can someone help me fill in the blanks here (this should follow readily but im not currently seeing why) : let $(u_{v})_{v \in \mathcal{V}}$ be a collection of subharmonic functions (in particular, upper semin-continuous), defined on an open subset $U \subset \mathbb{C}$, and let $(D_j)_{j=1}^{\infty}$ be a countable base of relatively compact open subsets of $U$.

Jakobian

oh, I understand now

porridgemathematics

Then for every $k \in \mathbb{N}$, there exists $v_{jk} $ in the collection, such that $\sup_{D_{j}} v_{jk} > \sup_{D_j} u^{\ast} - \frac{1}{k}$, where $u = \sup_{v \in \mathcal{V}} u_{v}$ and $u^{\ast}(z) = \inf (\sup_{y \in N} u(y))$, where the infimum is taken over all neighbourhoods of $z$

in other words, $u^{\ast}(z) = \limsup_{y \rightarrow z} u(y)$

(the upper semicontinuous regularization of $u$)

it should just follow from definitions

kauray

if $f(a, b)$ represents the frequency of features a and b occurring together and $f(a)$ and $f(b)$ represent the frequency of individual occurrences of features a and b, what does the term $f(a,b) - (f(a)* f(b)$ interpreted as or physically signify?

porridgemathematics

maybe this works: (the result is local, so it is sufficient to replace $D_j$ with arbitrarily small relatively compact disks), fix a point $z \in U$, then by upper semicontinuity of $u^{\ast}$, there exists some $r > 0$ for which $\overline{\Delta(z , r)} \subset U$, and $u^{\ast}(w) < u^{\ast}(z) + \frac{1}{3k}$ on $\Delta(z,r)$, i.e. $\sup_{\Delta(z,r)} u^{\ast} \leq u^{\ast}(z) + \frac{1}{3k}$

now choose $v_{jk}$ in the collection satisfying $v_{jk}(z)$ is within $\frac{1}{3k}$ of $u^{\ast}(z)$, and shrink the disk using upper semicontinuity of $v_{jk}$ so that $\sup_{D_{jk}} v_{jk} < v_{jk}(z) + \frac{1}{3k}$ (where $D_{jk}$ is the smaller disk centered at $z$)

kinda ugly though

Koro

12:04 PM

Is there a Borel measurable function $f:[0,1]\to (0,\infty) $, whose lower Riemann integral is $0$?

noballpointpen

12:17 PM

Could anyone give me any hint what is anything I can do to prove $\vdash (\exists x) \mathscr C^*$ when I have $\vdash (\exists x) \mathscr C$ and $\vdash \mathscr C$ if and only if $\vdash \mathscr C^*$.
I am bashing my head... this one seems to be impossible. If we assume not $\vdash (\exists x) \mathscr C^*$ then we can deduce that not $\vdash \mathscr C^*$, otherwise we could introduce existential quantifier. But it doesn't seem to point to anything interesting...

Koro

12:27 PM

nvm, I got such a function.

noballpointpen

1:09 PM

@Koro hey. How do I post a question from MSE in the chat so that it is formatted nicely? Wrapping it in a usual link formatting doesn't seem to work. Do I just leave a link raw?

Ah... he's gone. Ok. If anyone is interested:
https://math.stackexchange.com/questions/4704284/dual-formula-in-first-order-a-common-syntactical-proof
:/

Prithu Biswas

1:21 PM

My avatar just changed =O

Jakobian

1:54 PM

@PrithuBiswas look at starred messages, the one by PM 2Ring

Akiva Weinberger

2:40 PM

Challenge problem: For $n=1$, $2$, $3$, and $4$, exhibit polynomials $p(x)$ of degree $n$ with leading coefficient $1$ for which $\max\limits_{|x|\le1}|p(x)|$ is as small as possible.

noballpointpen

2:50 PM

Am I right that for the first degree it is the trivial $y=x$?

Soumik Mukherjee

3:24 PM

@AkivaWeinberger $x, x^2-\frac{1}{2}, x^3-x, x^4-x^2+\frac{1}{8}$?

Ted Shifrin

@goedelite I know only of Gamelin and Greene. Check your authors?

Mad

@TedShifrin i hope your well prof!

Sourav Ghosh

4:00 PM

No book in point set topology can substitute Munkres's book.

Sourav Ghosh

4:22 PM

Set of all limit points of a set in a topological space need not be a closed set.

A limit need not a sequential limit

Metric spaces are well-behaved :)

Done! Counter for 1)X Indiscrete space and A=\{a\}

2) X co-countable space and A proper uncountable set.

Hi @Koro kemon আছো?

Ted Shifrin

4:40 PM

@Jakobian Wow. Fascinating. At least mine is safe :)

Soumik Mukherjee

5:00 PM

@SouravGhosh Is Koro bengali?

robjohn

@TedShifrin No custom icons are affected. Just the default ones created based on PI.

Ted Shifrin

Yes, so I deduced from the link PM2 posted.

Sourav Ghosh

5:28 PM

f is uniformly continuous iff f preserve quasi-cauchy sequences.

Ted Shifrin

Never heard of such a thing. What's that?

Sourav Ghosh

(x_n) is quasi-cauchy iff $ d(x_n, x_{n+1}) $ converges to 0.

Ted Shifrin

Oh, quite non-Cauchy.

Sourav Ghosh

For example, $x_n=\sqrt{n} $ in euclidean metric.

Ted Shifrin

Or $x_n = 1+1/2+\dots+1/n$.

Sourav Ghosh

5:32 PM

Yes.

x_n=\log(n)

Ted Shifrin

So why is this a useful notion?

Sourav Ghosh

uniformly continuous map respect cauchy sequence. But cauchy sequence doesn't respect uniformly continuous map.

Ted Shifrin

I don't get your English.

Sourav Ghosh

f is uniformly continuous then f(x_n) cauchy whenever (x_n) cauchy.

Ted Shifrin

Yes.

Sourav Ghosh

5:37 PM

Converse not true. f(x) =x^2 on \Bbb{R}

Ted Shifrin

Oh, because Cauchy sequences must live in a bounded subset.

I see.

I've never heard this notion before. And I survived :P

Sourav Ghosh

You are very clever.

Ted Shifrin

Well, off for several hours. Bubye.

user223626865

Cya.

noballpointpen

I also had an argument that involved Cauchy sequences and uniform continuity recently. If $E$ is bounded and $f$ is uniformly continuous then $f(E)$ is bounded. Am I right that we can prove it by defining a continuous extension for the limit points not in $E$ by looking at this property?

Sourav Ghosh

5:41 PM

Even more is true. E is totally bounded and f is u.c then f(E)is totally bounded.

In case of \Bbb{R} ( or any finite dimensional NLS) totally bounded is equivalent to boundedness.

noballpointpen

Yes, I forgot to mention that the exercise was about real-valued functions defined on $E \subseteq R^1$ :)

Sourav Ghosh

@noballpointpen Then closure of a bounded set is compact.

Uniformly continuous functions are continuous.

Continuous image of a compact set is bounded.

noballpointpen

I actually already proved it. Yes, used these facts. Just asked if the notion with Cauchys was correct.

I started with the mention that if $E$ is closed, then the conclusion is ready. Then we assume that $E$ is not closed. All by Heine-Borel.

Going off, too. Bye.

Sourav Ghosh

@noballpointpen Yes. Uniformly continuous function has unique uniform extension over the closure of the domain.

In case of metric space, f:A\subset X \to Y 1) f is uniformly continuous and 2) Y complete then f has unique uniform extension over closure (A).

geocalc33

6:37 PM

how do you write the n-th root of something in mathjax?

without doing something like $a^{1/n}$

$\sqrt[n]{a}$

robjohn

Either of those work

Xander Henderson

7:03 PM

Or "the $n$-th root of $a$".

Jakobian

@SouravGhosh I think many books can substitute it

user10478

Is this notation correct? I know the analogous would be fine for a summation but the intersection and big intersection next to each other look strange.

$W\ \cap\ \bigcap\limits_{i = 1}^{n - 2} C_i$

Sourav Ghosh

@Jakobian May be. But I have found Munkres more interesting than any other book.

Jakobian

as far as I know this is a pretty standard treatment of topology

and depending on what is interesting to you, I think Engelking might be even more interesting

not to mention books such as Encyclopedia of General Topology where information is tightly packed and restricted to only necessary information, where you can quickly learn a lot of interesting topology concepts

if someone is interested in eagle-eye view on things there's also History of General Topology

btw, I think all three of those books are great when it comes to general topology

Xander Henderson

7:27 PM

@user10478 This is fine. Nothing wrong with the notation.

Jakobian

and after exploring Munkres you can jump into Engelking if you have the time to spare

user10478

@XanderHenderson Thanks

Xander Henderson

If you don't like it, $$\bigcap_{i=1}^{n-2} C_i \cap W$$ means the same thing (whether you interpret the $\cap W$ to be part of the argument of the big intersection, or not).

Or if it really bothers you, $$W \cap \left( \bigcap_{i=1}^{n-2} C_i \right). $$

Sourav Ghosh

When a zero dimensional space is discrete?

Rational topology, sorgenfrey line, Furstenberg topology (though homeomorphic to rational topology) are zero dimensional spaces but not discrete.

user10478

Would probably opt for the parentheses in most cases but I already have a couple nested so I'll probably just use the initial notation to be economical since it's fine.

Sourav Ghosh

7:34 PM

All the above spaces are not locally connected.

@Jakobian First I thought Engelking wrote a history book.I will try :) But Munkres is able to motivate me.

Jakobian

a Hausdorff locally connected zero-dimensional space $X$ has open components, for $x\in X$ taking such component $U$ we see $x\in U$, taking a clopen set $V\subseteq U$ containing $x$ we see $V = U$. Thus $U = \{x\}$ is open for all $x\in X$.

if we don't assume Hausdorff, something like indiscrete two-point space is an obstruction

shintuku

7:56 PM

all prime ideals in integers modulo n are maximal right?

leslie townes

shin: yes

shintuku

cool cool thanks

leslie townes

conceptually this ties into a cluster of other results you may be aware of. an ideal in a ring R is prime if the quotient of R by it is an integral domain, and maximal if the quotient of R by it is a field. the proper nontrivial ideals of Z are the sets nZ. that Z/nZ is only an integral domain or a field when n is prime is just a restatement of some basic arithmetic about factors of n.

alternatively, of the quotients of Z by these nZ things are finite, and all finite integral domains are fields.

shintuku

yeah i'm working through this stuff, i'm trying to get intuition around the fact that I maximal iff R/I field

leslie townes

i should put scare quotes above around my "just" in "is just a restatement of" above. the definition-checking here is pretty close to the surface, but also pretty important, both in this case and in general. its part of the ABCs of the isomorphism theorems, which again i do not mean in any derogatory way, just, this stuff is fundamental and comes before a lot of other things. like ABCs.

when beginning algebra books talk about the 'fundamental' isomorphism theorems or whatever i think that is all they mean by 'fundamental.' at least, they aren't huge results of independent interest (the way the 'fundamental' theorem of algebra arguably is, or even the 'fundamental' theorem of calculus) as much as they are starting points for getting situated with the notions they're about.

Akiva Weinberger

8:21 PM

5 hours ago, by Soumik Mukherjee

@AkivaWeinberger $x, x^2-\frac{1}{2}, x^3-x, x^4-x^2+\frac{1}{8}$?

@SoumikMukherjee Yes I think

What maximums do those attain

(@noballpointpen)

shintuku

at leslie: importantly i had not noticed until you mentioned it that there's a distinction between prime and maximal ideals only in nonfinite rings

Soumik Mukherjee

9:12 PM

@AkivaWeinberger $1, \frac{1}{2}, \frac{2}{3\sqrt{3}}, \frac{1}{8}$

Shaun

Hi :) I have a false proof I vaguely remember about recurring decimals. A contradiction is obtained in this proof by mirroring a proof $1=0.\overline{9}$; you know: $S=0.99 . . .$ gives $10S=9.99...$, so $10S=9+S$, meaning $S=1$. I think the error in the false proof is in assuming something exists. Anyway, I can't seem to find it. Does anyone know which false proof I have in mind?

leslie townes

9:34 PM

it's not really a "false" proof, it's just maybe a bit weird to implicitly assume stuff about the arithmetic of infinite decimals (e.g. that allows you to compute the expansion of 10S in terms of that of S, for example, and lets you deduce that 9.999... = 9 + S) in a context where maybe the meaning of infinite decimals hasn't been established.

every step in that argument is a true statement, it's just a question of what it shows you, if you don't understand what infinite decimals "mean" in the first instance. in that case, it's not so much a proof that 0.999.... is 1, but maybe only a proof that if you assume that arithmetic of infinite decimals behaves the way it's used in this argument, then you have to accept that 0.999... is 1 as a consequence of those assumptions.

which if you don't phrase it that way, but present it as an "explanation" of "why" 0.999... is 1, maybe is misleading. but not exactly false. everything in that argument is consistent with how the arithmetic of infinite decimals does in fact work. :)

semi related: the "explanation" why "a negative times a negative is a positive" as follows: 0 = (-a) * (-b + b) = (-a)*(-b) + (-a)*b = (-a)*(-b) - ab and add ab to both sides. this implicitly assumes various "facts" about arithmetic of the same approximate "status of being" as the thing being proved (e.g. the distributive law). but is a fine relative proof - that if you want the "laws" about arithmetic used in the argument to be true, then the rule for multiplying signs is forced from that.

Shaun

9:51 PM

I didn't say that proof was false, @leslietownes. I'm trying to find a "proof" that looks like it!

I think the false proof I have in mind is to do with $4/3$. I'm not sure.

Akiva Weinberger

@SoumikMukherjee Oh, sorry - your third polynomial is incorrect

shintuku

10:09 PM

what are rings other than Z[x] that have prime ideals that aren't maximal? hopefully, nice and simple rings

Z has (0) but just that one

Shaun

I wonder how many natural numbers are of the form $24(n^2-1)$. I've been thinking about the theorem that says, for any prime $p\ge 5$, we have $24\mid p^2-1$. A friend half-joked that this could assist in looking for primes, since we could narrow things down to natural numbers of the form $24(n^2-1)$.

leslie townes

@Shaun is it about arithmetic, or something else?

Shaun

I mean, the proportion of natural numbers . . .

@leslietownes Arithmetic, and decimal expansion in particular.

leslie townes

oh, hrm. i can think of other things about fiddling with "infinite sums" generally, but the bugs tend to rely on the things involved not actually being numbers, and + not being actual +.

Shaun

Maybe my memory is off, though; what do you have in mind, @leslietownes?

leslie townes

10:19 PM

well, a lot of stuff involving series of matrices or operators, which is basically just actual math that only seems mysterious if you haven't thought of it before. or "mazur's swindle" in topology. en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle

oh, there's also all of those "proofs" or false proofs of copper's favorite sum, 1 + 2 + 3 + ... = -1/12. could it be one of those?

there's a genre of, "here's X way of making sense of 1 + 2 + ... = -1/12, but here's why this is shaky and not as simple as normal convergent series: applying the same logic of X way to something else, we get [something goofy]."

Shaun

@leslietownes That's interesting . . .

Now that you mention convergence, it could be a proof that assumes a sum is convergent, but hidden somehow in a decimal expansion.

leslie townes

mm, there are lots of bogus proofs in the realm of 'silently rearrange a conditionally convergent series and assume that it won't change the result'. maybe one of those?

Shaun

@leslietownes Okay. Yeah, maybe. What would be an example of that, then?

geocalc33

Find all functions $f:\Bbb R^+\to\Bbb R^+$ s.t. for all $x,y \in \Bbb R^+$ the following is valid:

$$ f(x^y)=\sqrt[y]{f(x)}$$

Any hints on this?

Shaun

@leslietownes You mean the Riemann Series Theorem, right? en.m.wikipedia.org/wiki/…

leslie townes

10:32 PM

not specifically. i mean, the kinds of "paradoxes" you get when you ignore the fact that rearrangement can change sums.

look at e.g. slides 4 and 5 of math.iupui.edu/~ccowen/ButlerAHslides.pdf

the (-1 + 1) + (-1 + 1) + ... = rearrange and get goofy stuff, familiar from the mazur swindle, is maybe also a less subtle example of this

Soumik Mukherjee

@AkivaWeinberger yes, I think it would be $x^3-\frac{3}{4}x$

leslie townes

oh, this might be the same example that is on that wikipedia page :) but, i don't mean the rearrangement theorem or any other generality at all. i mean, examples of specific rearrangements of sums that converge to different things

seems like a handy way of hiding a "paradox," although maybe not one involving decimal expansions (which are absolutely convergent :~( )

Shaun

Thank you for the help, @leslietownes. I'll look things up a little more but I'm winding down for the day, so of you have any more ideas, just ping me :)

leslie townes

will do. let me know if you remember it :)

Ted Shifrin

11:10 PM

What wreckage has Munchkin wreaked this Monday?

leslie townes

we shall see. i'm guessing she snuck one or more toys to school, that's her latest thing. she even has other kids doing it.

Ted Shifrin

11:38 PM

Musical toys? See who ends up with what when the music stops?

leslie townes

11:50 PM

we think it started as a way of showing her friends that she was a badass who didn't care about the rules. just "look what i brought from home and am not allowed to have, and we're all doing this while the teachers are none the wiser." then some kids did begin bringing similarly small toys, i think for purposes of trade in some kind of toy economy.

she tends to arrive home with more than she left with. i don't think she actually trades her toys away.

for "toys" here don't think anything of value or that you could buy in a store, think a component of something like that, that could fit in a pocket or lunch bag or shoe. e.g. one lego man from a lego set.

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$Mathematics$ Mathematics