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12:20 AM
Hi chat.
Does the Weierstrass theorem (boundness of a continuous function over of a closed set) depend on AC?
I could provide a proof, but it depends on AC.
 
that's not true
so could you state the whole theorem
 
$f:[a,b]\to \Bbb R$ continuous $\implies f[a,b] = [c,d]$
In calculus, the extreme value theorem states that if a real-valued function f is continuous on the closed interval [a,b], then f must attain a maximum and a minimum, each at least once. That is, there exist numbers c and d in [a,b] such that: f ( c ) ≥ f ( x ) ≥ f ( d ) for all x ∈ [ a , b ] {\displaystyle f(c)\geq f(x)\geq f(d)\quad {\text{for all }}x\in [a,b...
 
that's a closed interval
 
12:45 AM
oops
yeah, over a closed interval
I proved that $f[a,b] \supset [f(a), c[$ where $c = \operatorname{sup} f(x)$
So I used choice to choose a sequence of x's such that f at that point gets arbitrarily small
and by continuity...
I want to slice $[a,b]$ in a partition of intervals such that the restriction of the function is monotone on all of them. But... I don't think I can do this without AC
The problem is to find how to approach "c["
Any tip is appreciated.
 
1:02 AM
@Ultradark discrete topology makes it into a 0-dimensional manifold, which is next to useless.
 
5
Q: What is the status of the extreme value theorem in forms of constructive mathematics, such as Smooth Infinitesimal Analysis?

Mikhail KatzIn certain intuitionistic frameworks the extreme value theorem cannot be proved. Depending on the exact framework, counterexamples can be constructed as well; see for example pp. 294-295 in Troelstra, A. S.; van Dalen, D. Constructivism in mathematics. Vol. I. An introduction. Studies in Logic a...

 
@anakhro okay fine
I guess that's not a weak vector field
I'm having trouble finding the definition for a weak vector field
ah, it's called Leray-Hopf Weak solutions
 
1:37 AM
Never heard of a "weak vector field"
 
 
2 hours later…
3:17 AM
well they exist
 
3:36 AM
@anakhro right, but what I am interested in is to find a way to define an actual infinity x using finite data without x popping up somewhere in the description. As far I am aware from discussions with user21820 and Leaky, there isn't a known way to do that (thus is an open question) which is one reason why things like axiom of infinity cannot be derived as any known axiomatic system that can turn that into a theorem, will itself have to assume some infinite object to exist in the first place
 
4:09 AM
Thus I suspect the relationship between circles and a sequence of polygons arranged in ascending order of lengths (as all of these shapes can be defined using finite data) may be able to define an actual infinity in some way without have to first assume the actual infinity exist
in particular, the sequence of polygons can be encoded using something analogous to {0,s} in the Hilbert axioms, based on your discussion about for potential infinity, and the circle (which can also be encoded with finite data using Hilbert axioms) can then be a representation of $\omega$ for example
 
4:29 AM
But I guess the trouble here is still to justify in this framework, is that a program is needed to prove that there exists some linear ordering < such that polygons < circle, and there is no known computable way to do so (since such program will necessarily need to compare every n-gon with the circle and returns either 0 or 1, and this program obviously cannot terminate due to the potential infinity of the sequence of polygons)
So unless there exists a notion of < defined using finite data where n-gon < circle and n-gon < n+1-gon, there seemed to be no way to define an actual infinity using finite data
Area based on radius may seemed an attractive way to establish a <, and the required values of the sine function in this formula can be constructed easily by geometry
Still, we need a finite and natural way to establish:
$$\forall n \in \Bbb{N}: \frac{1}{2} n \sin (\frac{2\pi}{n}) > \pi$$
If we can do that, we are done since we can then axiomatically define $\omega$ with finite data as follows:
1. Use an axiomatic geometry (e.g. Hilbert axioms) to define a circle and its area $\pi r^2$
2. $\pi$ is then defined to be the ratio of its circumference to diameter
3. Use the same axioms of geometry to define n-gons for each n, triangles and their areas $\frac{1}{2} ab \sin \theta$
4. Use Peano axioms and the signature {0,s} to define n and induction
5. Sine can be defined in terms of triangles, and some identities it needs to satisfy
6. Establish the proof that $\frac{1}{2} (n+1) \sin (\frac{2\pi}{n+1}) < \frac{1}{2} n \sin (\frac{2\pi}{n}) $ (to be figured out)
Compute the relevant values of $\sin (\frac{2\pi}{n})$ for each $n$ by constructing the corresponding triangles with the geometric axioms, or otherwise
7. Finally complete the well ordering by showing that $\forall n \in \Bbb{N}: \frac{1}{2} n \sin (\frac{2\pi}{n}) > \pi$ thus establishing the concept of the ordinal $\omega$ with no reference to the axiom of infinity
(actually I may have the inequalities in the wrong way around)
Under this framework, it should then be possible to define $\omega$ using only $\Bbb{N}$ (which is a potential infinity and hence not assumed to be a set, and this potential infinity is often treated as predicative in predicative mathematics), finite data and geometry axioms
Actually typo: Defining $\omega$ using only geometry axioms, Peano axioms and finite data
2
Q: Area in axiomatic geometry

MarcLet's say we have axiomatic geometry as defined by Hilbert's axioms. For line segments, angles, triangles, squares, etc. we have the notion of congruency to determine whether two of them are "the same". But this doesn't seem sufficient to determine whether two figures of different shape have the...

 
5:29 AM
hmm...
$\pi = \frac{C}{D} = \lim_{n\to \infty} \frac{2 r n\sin (\frac{2\pi}{n})}{2r} = \lim_{n \to \infty} n\sin (\frac{2\pi}{n})$
11
A: How do we know the ratio between circumference and diameter is the same for all circles?

John DoumaTo show that $\pi$ is constant we must show that given two circles of diameters $d_1$ and $d_2$ and circumferences $c_1$ and $c_2$, respectively, that $\frac{c_1}{d_1}=\frac{c_2}{d_2}$. If $d_1=d_2$ then the two circles are congruent because one can be placed upon the other and they will line up...

right, so proving the constancy of $\pi$ only needs a potential infinity
thus area of circle can be established within this framework
 
for all you complex gurus out there: what conditions must $f(z)$ satisfy such that $\int_{\infty}^{\infty}f(z)dz = \int_{-\infty + i\alpha}^{\infty + i\alpha}f(z)dz$, for all real $\alpha$?
 
5:47 AM
Yesterday I was telling my friend that my chiropodist improved my posture. He pointed out that it was in fact my chiropractor. I stand corrected. — user67049 Feb 25 '14 at 17:58
@dm__ it must... vanish around the two real infinities I guess
you're essentially using the contour integral trick
 
but the entire path is being shifted up, seems like it would have to be more restrictive
or down
 
do you know the contour integral trick?
you make a finite rectangle and let the width go to infinity
 
ohhh, right, it's been a while. and so there would have to be no poles between the horizontal lines $Im[z]=0$ and $Im[z]=\alpha$, yes?
or perhaps 'odd' in such a way that any residues cancel?
 
yeah...
 
great, that makes sense to me. probably very basic, but it's been a while. thanks!
@LeakyNun what is meant by "relative complement is a contravariant functor"?
 
6:08 AM
it's a fancy way of saying $\forall S,T \in P(X): S \subseteq T \implies T^c \subseteq S^c$
where $S^c = X\setminus S$
 
haha nice XD
 
6:32 AM
Last night dream, read about a strange space called a Cluster Space:
It appears those spaces are in general so complicated to study that the chapter of a book that introduces them, have the title "I Hate Cluster Spaces"
According to the dream, a Cluster space is described as follows:
> A space where Carbons (represented by hexagons) contains carbons and so on ad infinum in a progressively random manner. The boundaries of the hexagons fuses to give a trechet tile like appearance such that the closer to the centre of the space, the more distorted from hexagonal it is
More descriptively, that particular cluster space that is shown in the dream (illustrated roughly in the above picture, as the actual structure is dense and fractal like in some way) is constructed as follows:
1. The underlying structure is some off centered concentric hexagons of alternating color
2. Next, a few slim hexagons were inserted into this structure
3. Then for every intersection between these rays, it is yellow, otherwise it is orange
4. The same procedure is repeated in the deeper layers, but with more rays, resulting in more fragmented intersections
5. Random yellow and orange hexagons were then added which became more random and dense the closer it is to the center
 
the perfect representation of a contravariant functor
 
6:48 AM
6. Eventually, the added hexagons became so dense (they will look something like 2D version of ford circles except more disordered) that the intersections started to behave chaotically, and then the hexagonal edge became more distored from a regular hexagon shape, and something like this results (with a few isolated hexagons visible due to coincidental intersections)
Mar 13 at 10:15, by Akiva Weinberger
user image
So in a way, Cluster Space seemed to be what happens when fractals, dense sets and chaotic dynamics all meshed together
Also correction: The density of the hexagons in step 5 only need to obey the following rules:
1. For every hexagon, the probability of it being surrounded by any hexagons of any size is the same
2. The hexagons together form a dense set made of unions of hexagons
The dream also describe that the experience of travelling within a Cluster Space is very similar to travelling inside a hyperbolic honeycomb, except that the closer the person is to the center of the honeycomb, the more distorted and erractic the hyperbolic space is. It has some kind of self similarity at a rough scale, but that breaks down once the chaotic region is reached
 
@MatheinBoulomenos this is gold
 
 
1 hour later…
8:08 AM
Hello Math. I'd like to advertise a rather math-oriented question recently posted on the physics site:
0
Q: What is the link between the rotating wave approximation and the algebraic representation of a dynamical system?

DanielSankIn analyzing a system of two coupled oscillators, I noticed a rather interesting correspondence between the so-called "rotating wave approximation" (RWA) for solving differential equations and the structure of the algebraic representation of the matrix form of those equations. Background The sy...

 
8:35 AM
Now that I am only checking arXiv for stuff that I might be interested in knowing (rather than stuff that might be useful to know for my research), it is much faster to browse through everything since I have been able to give almost everything else a negative score on arXivist.
 
 
2 hours later…
10:45 AM
@Semiclassical Aha!! Thank you very much!!
 
11:26 AM
Is this answer wrong, when it claims that degree 1 polynomial irreducible over integral domain, but indeed $2x-2=2(x-2)$ is reducible polynomial in $Z[x]$?
 
@Silent Yes, that is an error in the answer.
 
@Silent no, the answer is correct
due to the weird definition of irreducible in the question
 
@MatheinBoulomenos Ahh, I missed that the OP had defined irreducible in a way that is not compatible with the usual definition
 
11:46 AM
ok! :) Thank u both
Of course it is not quite the situation here, but I recalled my Economics days, where two giants used to speak totally contradicting stuff, but they both seemed to be so true that I used to agree with them both!
 
12:19 PM
@Secret that doesn't really make much sense, tbh
"define an actual infinity x using finite data without x popping up somewhere in the description"
 
ok let me elaborate:
What I want is to find a way to define a notion of infinity that is predicative. By predicative, I mean let $x$ be the infinite object in question, then $x$ has to be able to be derived in terms of some computable function of the natural numbers, without having to presuppose $x$ to exist anywhere in the process
That is, I want to construct $x$ using only natural numbers and a process that will terminate in a finite string of instructions (this is what I meant by "finite data")
The simplest possible version of this goal will be to able to first construct $x$ in terms of some finite collection of natural numbers and strings, and then to show that for all natural numbers $n$, $n<x$
and I felt that the relationship between a circle and n-gons sharing the same radii will provide a way to achieve this (since we can define both a circle and the n-gons each using a sentence of finite length, of which no words in that sentence is an infinite concept)
sorry typo: first construct some representation of x in terms of some finite collection of natural numbers and strings
and I am suspecting a circle will be such representation, of which $x$ can then be extracted from it afterwards
The current sketch of this proof and construction is then:
1. Pick some axiomatic system of geometry
2. Construct n-gons for each n and a circle of radius 1
3. Pick some predicate P which takes in n-gons or circles, and return some truth values (i.e. P can be the area of these objects)
4. Define a irreflexive, transitive and antisymmetric relation < that relates P(n-gons) and P(circle)
 
12:34 PM
I think you should add a step 0: "define rigorously every word that I posit as a mathematical concept". Otherwise you are going to run into the problem which you apparently suffer from, which is that you are trying to make mathematical judgements of very vague english concepts.
 
I think the definition will came by as the construction proceeds. Natural numbers can be defined using peano axioms and the induction schema, and the geometric shapes will be constructed from say hilbert axioms
 
You will already run into an issue where you state that you want a process that constructs an "infinity" that will terminate in finite time. Unless you redefine "infinity", in which case, I highly recommend using a different word.
 
In that case, I think "infinity" is indeed a bad word. I have not thought of a good name yet for that object
But the point is, I want my system to be able to prove the following:
1. For all n in N: P(n-gon)<P((n+1)-gon)
2. For all n in N: P(n-gon)<P(circle)
and I want that proof to be computable in that if I wrote a program for that, the proof will terminate with the above two results
That's the plan for the proof
 
1:23 PM
I don't see how that really relates to your concept of faux-finity.
 
The idea is that if these two proofs can be established, then what we basically have is:
1. For all natural numbers n, n < n+1
2. We have constructed some x and proved that for all n, n < x
Compared this with $n < \omega$ for all natural numbers $n$
Thus the faux-finite x can be said to have some properties of $\omega$ including it being a supremum of all objects indexed in a well ordered fashion by natural numbers
P(circle) thus represents x in some way (hopefully...)
 
But proving 1 or 2 includes making infinitely many computations.
Because you are just using induction.
Which presupposes the existence of N.
 
But induction schema is only a potential infinity, it is compatible with the requirement of predicative mathematics
Predicative mathematics is ok with N except that we cannot show that N is a set
 
This doesn't show "x" is a set either.
It effectively is just an enumeration of polygons.
And then you vaguely claiming that a circle is the limit of the n-gons as n-->infty
 
well, I am less worried about whether we can show that x is a set. Showing that x > n is sufficient to have a mathematical object that behaves like an actual infinity. I don't really need that to be a set, I only need that object to be predicative
As for "vaguely claiming that a circle is the limit of the n-gons as n-->infty", I think that can be rigorously dealt with by picking a suitable P, say P is "area of the shape" and then the limiting process will be taken cared of by < and the induction schema (and the continuity axioms in Hilbert axioms), I think...
 
1:34 PM
Your end goal of this all being?
 
A mathematical object $x$ that behaves like an actual infinity, and can be proved to be larger than any element in the collection N
and a way to encode such x using geometric axioms
thus producing a foundation system with something that behave like an actual infinity, bypassing the need of an axiom of infinity
hence getting as close as possible to the notion of a predicative actual infinity
 
But you presuppose the axiom of infinity...........................................................
 
But how? axiom of infinity want inductive sets to be infinite, but x will not even be a set (it will not be provable in this proposed system that x is a set (or x is not a set))
 
@Secret "For all natural numbers n, n < n+1".
Are there the natural numbers or not?
 
Well, they will be defined using the Peano axioms, thus they are natural numbers. But Peano axioms don't have an axiom of infinity, they only have induction schema
without axiom of infinity, N cannot be proved to be a set if I recalled correctly
 
1:44 PM
Someone, correct me if I am wrong: to make a statement "$\forall x \in X\,.\,P(x)$" in predicate logic, $X$ needs to be a set.
 
@LeakyNun
 
Hi. What does $Th(\mathbb N)$ mean?
 
@LeylaAlkan is this mathematical logic?
 
The theory of $\Bbb N$, the set of all first order statements in the language you're using which are true in $\Bbb N$
 
@anakhro yes
 
1:50 PM
Then what Alessandro said.
 
Okay thanks
 
Just to be clear $\Bbb N$ is a structure of the language $\{+,\cdot,<,0,1\}$ here when they write $\mathrm{Th}(\Bbb N)$, not of $L_C$
 
Question
 
Say we have a compact subset of $\Bbb R^n$
Consider its convex hull
Is that necessarily equal to the intersection of all half-spaces containing the set?
Intuitively I wanna say yes
Clearly that intersection is convex (it's the intersection of convex things) and thus contains the convex hull
so the hard part is the other direction
Is there a notation for the convex hull of a set $X$?
${\rm Conv}(X)$ I guess
 
2:00 PM
I'd say "Hull(X)" would be more unambiguous.
 
So I guess we need to show that, if $Y$ is convex, and $X\subseteq Y\subseteq\text{intersection of half-spaces containing $X$}$ and $Y$ is convex then $Y=\text{intersection of half-spaces containing $X$}$
Or alternatively
If a point in $\Bbb R^n$ is in every half-space containing $X$, then it's in every convex set containing $X$
Also this is clearly true if $X$ is just a collection of $n+1$ points (where the convex hull of $X$ is a simplex)
Hm. We can make another conjecture: a point is in the convex hull of $X$ iff there's a collection of $n+1$ points in $X$ whose simplex contains our point
Dunno if we need that though
The contrapositive of the thing I wrote earlier is: if there exists a convex set containing $X$ that misses $p$, then there exists a half-space containing $X$ that misses $p$
Oh you know what
 
By half space, you mean that one of the factors of $\mathbb R^n$ is divided by a line (and you take only one half of that space).
 
I think this is equivalent to a statement about $S^{n-1}$
@anakhro Cut $\Bbb R^n$ by a hyperplane (not necessarily one aligned with the axes)
But the statement about $S^{n-1}$ I was gonna say is, if $X\subseteq S^{n-1}$, then either the center of the sphere is in the convex hull of $X$, or there's a hemisphere of the sphere containing $X$
If there's no such hemisphere then the center is in the convex hull
This feels really obvious
I must be missing something simple 'cause I can't make this a rigorous proof
Something about dotting everything in $X$ with a vector?
OK here's an easier question
Say we're given $n+1$ points in $\Bbb R^n$, OK? We're given their coordinates
Is there an easy test to see if the origin is in their convex hull?
I have a vague sense that this is what linear programming is about
Unfortunately, I know nothing about linear programming
 
2:22 PM
So one way is to show for any convex set $C\supseteq X$, there exists a hyperplane $H\subset\mathbb R^n$ such that $H\cap C = \{\}$.
The other way seems harder.
You'd have to show every hyperplane $H$ missing $X$ also misses a convex set containing $X$.
 
2:36 PM
well... in type theory, natural numbers can be defined as an inductive type using 0 and the successor function. Thus each time S is appplied to 0, S0, SS0, SSS0 etc. will be new objects of type nat
thus N need not be a set
 
You are predicating on N implicitly though
"for all n in N"
 
ah right, that dreaded "quantifying over the universe"
hmm...
 
I would highly recommend just being more formal with how you define your faux-finity, and then move on from there.
 
I see, sounds like more reading to do, ok then
 
Well that part you don't really need reading for.
That's your part. The "I want to see if this thing exists in math!"
Only after that do you do reading to figure out if it does.
 
2:53 PM
I am not sure if showing S0 < x, SS0 < x, SSSO < x, ... by induction will predicating implicitly over N (since there will obviously be some nonstandard elements in some model of N that is obviously not included and hence it will not be quantifying all of N).

The set theory analogue of what I said here is basically trying to find a proof where we can construct a computable increasing sequence in the naturals and also computably showed that some element x exists and x > a for the members of this sequence. This is not known to be possible in set theory which is why the axiom of infinity is no
Thus fax-finity, if it must be phrased in set theoretic language, is to find a set theory such that there is a theorem that behaves like the axiom of infinity
 
I thought you didn't want to show your faux-finity was a set?
 
(see edited)
And yes, I don't want faux finity to be a set, because I don't think set theory is expressive enough to define faux finity
I am suspecting it might be possible in type theory because the language of types is more expressive (I think?) since not all objects have to have membership relations
whereas in set theory, everything is a set
Hmm...
Let x be faux-finity. Then x satisfy the following:
1. There is a computable sequence K consists of objects of type nat such that the resulting sequence is linearly ordered under some transitive, irreflexive and antisymmetric relation <
2. Pick the smallest object k in K, we can show that k < x, and hence by induction, objects y in K satisfy y < x
3. We can construct x using only objects of type nat and the language of first order logic and geometric axioms
sorry 1 should read: We can construct a computable sequence K consists of (...)
4. (and we cannot use any object that has not been constructed or proved to exist)
 
K is a set?
 
3:10 PM
no it is an inductive type. Hmm... maybe I should rewrite the above requirements into a program and it will be clear on what I want to do:
Declare x as Type FauxFin
x := geometry axiom 1 ( geometry axiom 2 ( ... (geometry axiom m (peano axiom ( 0))))
i = 1
while i > 0:
    Declare f(i) as Type K
    Prove f(i) < x
    i++
Hence by induction, x is greater than objects of Type K
where the while loop does the induction and f is computable
So put it in another way, we compute x, and then we compute f(i) inductively, and then show that f(i) < x by induction
x will be computable with the help of the geometry axioms
Thus even though the induction itself is an infinite loop, we can terminate it at any step and will always get f(i) < x thus x behaves like an actual infinity
 
Test
$\varpi$
 
$\varbeta$
 

Sandbox

Where you can play with chat features (except flagging) and ch...
 
does that make any sense?
 
Not to me.
 
3:25 PM
Hi @MatheinBoulomenos.
 
I think the only other ones are $\vartheta$, $\varepsilon$, and $\varphi$
(the latter two of which look much better than $\epsilon$ and $\phi$, at least to me)
 
$\varkappa$
 
Fun fact, LaTeX has no $\Alpha$ or $\Beta$ because they'd look and typeset identically to $A$ and $B$
 
$\varsigma$
 
3:26 PM
I like varkappa
varsigma is not really different
 
I think I knew about varsigma and forgot about it. I didn't know about varkappa
 
It's the same letter, just used differently.
 
Varsigma is how Greeks write sigmas at the end of a word
 
Yeah.
 
Σσς
=Sss
 
3:27 PM
It's interesting hearing how people pronounce the greek letters.
 
It's different in the US and the UK, an they're both very different to how they're pronounced in Modern Greek
(and Ancient Greek as well I think)
 
Let $(P,\leq)$ be a poset. If every chain and antichain of $P$ are of finite then is it true that $P$ is also finite?
 
There is no version of Greek that pronounced Euclid like we do
or Zeus, Ares, Hades, etc
 
How do you pronounce "xi", @AkivaWeinberger?
 
3:31 PM
Is that a long i or a short i
 
Like cheese
 
So long.
 
I might do /ksai/ ("fly") sometimes, I dunno
Similarly, phi is usually /fi/ but occasionally /fai/
Pi is always /pai/ if only because English already has a letter called /pi/
 
So more akin to the modern greek than the ancient.
 
Any idea @MatheinBoulomenos?
 
3:35 PM
US beta is /ˈbeɪtə/, UK beta is /ˈbiːtə/. (The two dots indicate vowel length, which doesn't really exist in American English). Wikipedia says Ancient Greek pronounced it /bɛ̂ːta/ whereas Modern Greek pronounces it /ˈvita/.
And it makes the /v/ sound in Modern Greek
So US "face", UK "feet", Ancient Greek "fess" but longer and also a falling tone, Modern Greek "feet"
for the vowels
Wait, Ancient Greek was tonal?
 
All my french professors do it modern greek style, and all the english professors do ancient.
 
@user170039 just to clarify, does antichain for you mean every two elements are incomparable?
 
Oh, it's a pitch-accent system like Japanese
(as opposed to a tone system like Chinese)
 
@B.Mehta Yes.
 
So I'm doing a Jewish version of "Secret Santa" (since Purim is coming up)
(Usually this is done for Hanukkah but whatever)
and I dunno what to get
 
3:39 PM
Are you jewish?
 
so I'm walking around an outdoor mall to see if anything catches my eye
@anakhro Yes
Last time I did something like this was last year and I got something kinda crappy
 
It's the thought that counts.
 
a dry-erase board (about the same size as printer paper) and some dry-erase markers
 
ya ur right that's total crap go beat that person up
 
No I got it for the other person
I bought it
 
3:41 PM
Oh
O
 
The thing I actually got was a book (that I had read)
 
Don't beat yourself up then. :(
 
so that was kinda crappy (but I didn't tell him I had read the book)
Proofs and Refutations by Lakatos
 
@user170039 then yes, by infinite Ramsey
 
Do you like Lakatos's ideas?
I was talking to someone about him recently.
I was only told roughly what his ideas were, but it seemed like he took issue with "turtles all the way down"?
 
3:43 PM
I lied, I didn't actually read it, it's just that I already owned a copy
That's what I meant to say
I got partway through it and couldn't finish
 
Huh.
 
It felt like the whole thing was gonna be about $V-E+F=2$ and I don't know how you can write a whole book about that
 
@B.Mehta I actually don't know about it. But I think I have a proof of the same using AC. Would you mind to listen to it?
 
Like, you know, refining what counts as a "polyhedron" since if you aren't precise about that then there exist counterexamples
 
Go ahead
 
3:45 PM
Random:
$\pi$ is irrational proof: $p\neq 0, \pi = \frac{p}{q} \implies q \pi = p \implies \sin (q \pi) = \sin (p) \implies 0 = \sin (p)$ contradiction. Therefore $\pi$ is irrational
 
You just reduced it to a harder problem @Secret
 
ooops
 
Why can't the sine of an integer be zero?
 
@user170039 hello
 
There might actually be a route in that direction
maybe looking at sine's Taylor series
The only things they sell in this mall are shoes, clothes, and jewelry (or at least that's what it looks like)
and food
and you shouldn't trust me to buy anything like that for anyone
 
3:48 PM
@AkivaWeinberger really really ancient greek was tonal, by the time of Plato, it was long lost
 
@AkivaWeinberger what is your price range?
 
Roughly $15
 
What interests does the person you are gifting to have?
 
I barely know them
 
Maybe you should learn a bit more about them.
The more you learn about them, the better of a chance you have of making them smile.
 
3:54 PM
Suppose I have $k\ge n+1$ points in $\Bbb R^n$, and that the origin is in their convex hull. Does there exist a subset of $n+1$ of those points whose convex hull contains the origin?
Well I can assume they span space
WLOG, the first $n$ of them are the standard basis vectors
Hm what if I take $n$-tuple-wise determinants
(or whatever you call "pairwise" for $n$-tuples)
 

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