When talking about the theorem Gowers here wrote that the factorization of 36 is 2 x 2 x 3 x 3. But the composite number 36 by that notation looks like a product of a number and a composite number (2 x 18), a product of a composite number and a composite number (4 x 9), or a product of a composite number and a prime number (12 x 3). A notation like x(2, 2, 3, 3) or (2, 2, 3, 3)x doesn’t get used all that often when talking about the Fundamental Theorem of Arithmetic. And thus, it’s not so obvious, because authors don’t use appropriate notation to theorem all that often when talking about it or make it clear that the theorem implies that for any composite number, there exists of a k-ary product of prime numbers only, where k is some natural number.

]]>$ A=\left( \begin{array}{ccc}

Id & 0 \\

X & Id

\end{array} \right)$

where the rows of $X$ are all the rows with five $1$ and two $0$

like $(1111100)^t,(1111010)^t,(1110110)^t…(1101011)^t…..$

The line separator of $A$ are the line of

$ A=\left( \begin{array}{ccc}

X & Id

\end{array} \right)$

and they have 15 times $1$ and 12 times $0$

So maybe we have change it a litlle bit, I’ve got few idees… but I’m waiting for reactions, it would seem a bit ridiculous to go on, while the comments are not even put in latex^^

]]>That’s a tragedy, really. I think it would very much be worth reading.

Thank you for revealing the common structure (i.e. the prisoner’s dilemma) of such agreements to those of us that didn’t immediately see it (in hindsight, good math often seems so obvious).

The case has occasionally been made that Germany is currently a sweatshop of Europe at least in comparison to some of its immediate neighbours like France: With an ever-growing sector of temporary/lent work, late retirement and very low wages in jobs that do not require a qualification (e.g. working the fields), Germany is making it very hard for a country like France to maintain its desirably worker-friendly status. If France eventually gives in, both German and French workers will end up losing.

(To those who can read German — I’m afraid I could not find a translation) I can also recommend this slightly dated piece: http://www.monde-diplomatique.de/pm/2013/09/13.mondeText.artikel,a0004.idx,0

]]>1) a is prime means there exists no number between 2 and a-1 that divides it

2) there exists a prime factorization of every integer (algo exists for calculating it)

3) let a prime factorization be pi0 * pi1 * … pin

let a different prime factorization be pj0 * pj1 … pjm

pi0 * … pin = pj0 * … pjm

pi0 must divide pj0 * … pjm

thus pi0 must divide some pj, since it itself cannot be written as a product of 2 or more primes

this is a contradiction, so there cannot be a different pj0 … pjm

Let $M\in \mathcal M_n(R)$, with $n>1$, be a lower triangular invertible matrix with all coefficient $0$ or $1$, and suppose that for each integer $0<i\leq j\leq n$

$M_{i,j}=1\Leftrightarrow M_i-M_j\in \left\{0,1\right\}^n$(*)

where $M_i$ and $M_j$ are the line number $i$ and $j$ of $M$

we will say that $M$ is triangular connectable.

If a line in a matrix is such that by removing it we get two equal rows, or a full-0 row : we will say that it is a line-separator:

Sep-FUNC:

In a triangular connectable $n\times n$ matrix without the full-1 line ,there is a line separator $s$ such that $||s||_1\leq n/2$

This does not need the $\wedge$ product which give to matrices of caracteristical fonction of intersection closed Familly a very NOT-VECTORIAL shape, if I can speak this way…but the Sep-FUNC get reed of this Boole ring product and the condition (*) is the vestige of the "inclusion", no more intersection or union…^^

to give a summary of the reason why this is stronger than FUNC, we will say that a version of FUNC still holds in a union closed familly $\mathcal A$ with the same**** size than the ground set $X$, we than see that $\mathcal X=\left\{\mathcal A_x\space|\space x\in X\right\}$ is also closed for the union. and FUNC is equivalent to the fact that there exist a $x\in X$ such that $x$ is aboundant and such that $\mathcal A_x$ is a generator of $\mathcal X$…

****same size up to trivial "lines" and "row" but this is in the upper comment

If we than indexate correctly $X=\left\{x_1…,x_n\right\}$, and $\mathcal A=\left\{A_1,…A_n\right\}$ such that the $Z_2$-matrix, $A$ such that $A_{i,j}=0\Leftrightarrow x_i\in A_j$ we get a anti-caracteristic matrix, and this matrix is a connectable matrix, and the rows of this matrix correspunding to a generators are exactly the separated rows

(of course we have the same statement for lines, please note that I pretend that $M$ is connectable if and only if $M^t$ is connectable, everything is exposed in the upper comment)

I think this conjecture is quite intersting

]]>0) preliminaries and main triangular property

let’s index matrices in $\mathbb N$ with sets $A$ and $B$ with NO ORDER. A classical matrix will then be obtain by giving bijections from $a:[1,|A]\cap \mathbb N|\to A$ and $b:[1,|B|]\cap\mathbb N\to B$

IN BOTH CASES we will note for any matrix $M$, $lin(M)$(resp.$row(M)$) the set of line of $M$ (resp. rows of $M$)

If $M$ is NO-ORDER-indexed we will note that $lin(M)$ (resp.$row(M)$) define a unique NO-ORDERED-indexed matrix, and we will say that $M$ is $line-separated$ ,(resp. $row-separated$)

we will say $separated$ when the matrix is both $line-separated$ and $row-separated$).

we note that $row(lin(M))=lin(row(M)):=M^*$ and if we note $A^*$ and $B^*$ the (non-ordered) sets of indexation of $M^*$, each line $l\in lin(M^*)$ is a element of $\mathbb Z_2^{A^*}$, which is a Boole ring, and we can canonically associate the product $\wedge$ in this ring with the intersection in $\mathcal P(A^*)$

The same can be said with rows, and we can considere what we will call the line-cloture (resp. row-cloture) of a NO-ORDERED-matrix $M$, it will be the $separated$ matrix $L(M)$ (resp. $R(M)$ such that $lin(L(M^*))$ (resp. $row(R(M))$) is closed for $\wedge$ and is the smalest set containing $lin(M^*)$ (resp.$row(M^*)$)

We can easyly see that $R$ and $C$ commute, and I will give a demonstration that if $M$ has the line full-$0$ and not the line full $1$, than $R(L(M))=L(R(M))$ is a NON-ORDERED square matrix, and that you can order the indexed set such that it is a triangular matrix (I will give lter a demonstration and more very interesting properties)

lets finish with terminology!

if we call $l(M)\subset lin(M)\subset L(M)$ the smalest separated NO-ORDRED matrix to have the property $L(l(M))=L(M)$ we will call $l(M)$ the set of Line Generators, the obvious definition handle for rows and we will call $r(M)$ the set of Row-Générators.

$a\in l(M)$ if and only if for any $x,y\in L(M)$, such that $x\ne a\ne y$ we have $x\wedge y\ne a$

another way to see $l(M)$ is that they are the elements $l$ of $L(M)$ such that $L(M)\setminus \left\{l\right\}$ is closed for the product of lines, we will say that a element of $l(M)$ is minimal element if no-one in $lin(M)$ is a divisor (Then for the inclusion it correspond to maximal element, since we chosed the $\wedge$-cloture, corresponding to intersection, if we did the same job with the union (corresponding to “$\vee$”) the definition of our minimal would be the minimal corresponding subset of inclusion)

for example the set if $B$= {$a$,$b$,$c$,$d$,$e$} and $A$= {{$a$},{$b$},{$a,c$},{$a,c,d$},$\emptyset$} $\subset \mathcal P(B)$ can be AFTER having ordered $A$ and $B$… give the ordered matrix (= “usual” matrix) :

$N= \left( \begin{array}{ccc}

1 & 0 & 0 & 0 & 0\\

1 & 1 & 0 & 0 & 0\\

1 & 0 & 1 & 0 & 0\\

1 & 0& 1& 1 & 0 \\

0 & 0 & 0 & 0 & 0

\end{array} \right)$

if $P$ and $Q$ are permutation matrices $PNQ$ will define the same NO-ORDERED-matrix, $M$, in our case $M$ is indexed not by ordered sets but by $A$ and $B$ themselfs!

$M$ is here, line-closed and row-closed ,

$L(M)=lin(M)$ and $R(M)=row(M)$

we can give an ordered representation of $l(M)$ as

$ \left( \begin{array}{ccc}

1 & 1 & 0 & 0 & 0\\

1 & 0 & 1 & 0 & 0\\

1 & 0& 1& 1 & 0 \\

0 & 0 & 0 & 0 & 0

\end{array} \right)$

and we can see that $r(M)=M$ : every row is a generator in this example!

the minimal lines are

$(1,0,1,1,0)$ and $(1,1,0,0,0)$ in $N$

and we can see that they are the one that intersect the unique $1$ coefficient of a row that have only one $1$

this is a caractarisation of minimal lines in a row-closed separated matrix.

to see this suppose that there is a minimal line in a row-closed separated matrix that intersect not such a row, tht meen has $1$ as soon as the minimal line has a $1$ in the corresponding row, and that mean that this line is a divisor of our minimal line, so by minimality , the two lines are equal, but we said the matrix was separated so it is absurd.

We can use this statement to proove by induction on the number of rows (by “removing” a minimal line) that a row-closed separated matrix with row $0$, and without row $1$, is line-closed if and only if it is a NO-ORDERED square matrix (then it has a triangular $n\times n$ representation, of rank $n-1$, because all the diagonal coefficient are $1$ except the commun coefficient of the $0$-row and the $0$-line.

We will call BISTABLE any separated NO-ORDERED matrix which is line-closed and row-closed, and which contain line full $0$ line and row

We can call $\mathbb B$ the set of BISTABLE matrix and $\mathbb B_n$ Bistable matrix of size $n\times n$

If we have a NO-ORDERED matrix $M$, and $l\in lin(M)$ (resp. $r\in row(M)$) we a unique NON-ORDERED matrix $M-l$ (resp. $M-r$ such that $lin(M-l)=lin(M)\setminus \left\{l\right\}$ (resp $row(M-r)=lin(M)\setminus \left\{r\right\}$)

We will also call $\mathbb T$ the (ordered) lower-triangular representation of BISTABLE (NO-ORDERED)-matrix

FUNC is equivalent to the fact that there exist in such a matrix a line-generator with more $0$ than $1$

1)quick definitions and immediate properties

$T$ will be a member of $\mathbb T$ and we will note $T^*$ the corresponding NO-ORDERED matrix in $\mathbb B$ obtained by “forgeting” the indexation (the notation $T^*$ has already been used it for the separated NO-ORDERED matrix of a NO-ORDERED matrix, but as soon as member of $\mathbb B$ are separated, it is not confusing)

we get these property easily by induction and they can themeself be usefull for induction

definition a)

We say that a $x\in lin(T)\cup row(T)$ is a $separator$ when $T^*-x$ is not separated

property a)

separators are generators and generators are separators

definition b)

if $a$ is a line of $T^*$ and $b$ a row of $T^*$, we note $a\wedge b$ the common coeficient of $a$ and $b$. and we will aslo assume that $a\wedge 1=1\wedge a=a\wedge b=b\wedge 1$ and that $0=0\wedge a=a\wedge 0=b\wedge 0=0\wedge b$

we call $T_*=(lin(T^*)\cup row(T^*)\cup \left\{0,1\right\})/\mathcal Z$ the square monoïd

were $\mathcal Z$ is the identificaton $0$=full-0 line=full-0 row

property b)

$(T_*,\wedge)$ is a monoïd

definition c)

if $i\in \mathbb Z_2$ and $a$ is a line (resp. row) we way that $b$ is $i$-perpendicular to $a$ if and only if $b$ is a row (resp. line) and $a\wedge b=i$

We call $a^{\perp}:=\bigwedge_{b\wedge a=1}b$ the product of all rows (resp.line) that are $1$-perpendicular to $a$.

properties c1)

$a\wedge b=a\Leftrightarrow b^{\perp}\wedge a^{\perp}=b^{\perp}$

property c2)

$(a^{\perp})^{\perp}=a$

note1

in terms of intersection-closed separated subset $\mathcal A$ of size $n$ with a groud set $X$ of size $n$ such that $\emptyset \in \mathcal A$ and such that $\mathcal A_{\emptyset}=\emptyset$, the ortogonal of any $x\in X$ is $\mathcal A_x=\left\{a\in \mathcal A\space |\space x\in a\right\}$

and property c1) become $a\subset b\Leftrightarrow \mathcal A_b \subset \mathcal A_a$

note2

in if $a$ is the line nuber $j$ in $T$ than $a^{\perp}$ is the row number $j$

2_very intersting properties of $\mathbb T$ !!

we can give a impressively better result than the main (triangular) property

This result will hold in a bigger set than $\mathbb T$, and we will caracterize this set few interseting equivalent ways!

lets $D$ be a diagonal matrix in $\mathcal M(\mathbb Z_2)$ than

I will say that $P$ a principal matrix of $M$ if $P=(D.M.D)^*$

lets note by $\mathbb P$ the set of all principal matrices of matrices in $\mathbb T$

Very obvioussly members of $\mathbb P$ are lower-triangular matrices.

Quite obvioussly to if $P\in \mathbb P$ and $a\in lin(P)$, we have $(P-a)-a^{\perp}=(P-a^{\perp})-a\in mathbb P$

wait… it is obvious if we call $a^{\perp}$ the row that intesect $a$ on the diagonal… but it mayght not be so clear that the definition and property in c) still work!

in fact it does and $\mathbb P$ is the bigest set of lower-triangular matrix that has those properties , we can say this WITHOUT USING $\wedge$ anymore (!!)

For any lower trangular matrix in $\mathbb R$, $A$ of size $n$, for any lets call $A_i$ the line numer $i$ in $A$, $A^j$ the row number $j$ and $A_i^j$ the $i,j$ coefficient of $A$.

lets note by $\mathbb P^0\subset GL_n(\mathbb R)$ the set of lower-triangular matrices $P$ such that

for any $0<i\leq j\leq n$ integers,

$P_i^j\in \left\{0,1\right\}$ and $ P_i^j=1 \Leftrightarrow P_j-P_j\in \left\{0,1\right\}^n$

if you see $P\in \mathbb P^0$ like matrix with coeficient in $\mathbb Z_2$ you get all the invertible matrix in $\mathbb P$ (we removed the full-0 line and row of $\mathbb P_0$

This is a way to say that $\mathbb P$ is the set of rinagular $\mathbb Z_2$ matrices closed under the $a\mapsto a^{\perp}=\bigwedge_{b\wedge a=1}b$ application where $\wedge$ and $P_*$ can be defined like in definition c) but where $P_*$ is no longer a monoid, because it might not be closed for the $\wedge$

3_connectable lines/rows

let $M$ be the NO-ORDERED matrix indexed by $A$ and $B$ and let $x\in lin(M)$ with $|x|$ the number of $1$ in $x$.

If there exist $a$ bijection from $[1,|A|]\cap \mathbb N\to A$ and $b$ bijection from $[1,|B|]\cap \mathbb N\to B$

such that

1)the matrix $M'=(M_{a(i),b(j)})_{i,j\leq n}$ is trigonal

2) $a^{-1}(x)=|x|$

we say that $x$ is connectable in $M$

(in other terms you have a orderded matrix that represent $M$, which is lower triangular and where $x$ is the line where there is only $1$ under the diagonal, in this diagonal ordered matrix we say that this line is "connected")

a connectable matrix is a matrix where all line are connectable

property e)

$P$ is connectable separated matrix $P^*\in \mathbb P$

(easy by induction)

corollary :

if $P$ is connectable $P^t$ is also connectable

if $T$ is Bistable , then $T$ is connectable

important property f):

if $P\in \mathbb P_n$ and $a\in lin(P)$ and if $mult(a):=\left\{x\in lin(P),\space \exists y\in 2^n, x=a\wedge y\right\}$ then

$|a|=|mult(a)|-1$

(if $P$ is bistable we can write $mult(a)=a\wedge lin(P)$, we note that $MULT(P)=\left\{mult(a),\space a\in lin(P)\right\}$ is a monoide for intersection, and that $a\mapsto mult(a)$ is an isomorphism of monoïd.

note for any monoïd $P,\wedge$, we can also define

MULT'(P) is always a monoïde when the basic monoïd $P$ is idempotent

if we call $divi(a):=\left\{x\in P,\space \exists y\in P, a=x\wedge y\right\}$ we see that $a\mapsto divi(a)$ is an ismorphism of monoïd between $lin(P)$ and $DIVI'(P)=\left\{divi(a),\space a\in P\right\}$ but for the Union (only because of the idempotence $a\wedge a=a$)

$DIVI(P)=(DIVI'(lin(P)), \cup)$ is isomorphic to $(MULT(row(P)),\cap)$

this is given by property c)

4) a week FUNC!

in a bistable if $a$ and $b$ are 0-perpendicular than $|a|+|b|<n$

(it is funny to see that it still holds if $a$ and $b$ are not "perpendicular", but verify $a\wedge b=0$ in $T_*$, (see property and definition b) in paragraph 1))

this a direct corollary of the connectability of a bistable.

we sayd that FUNC is equivalent to the fact that there is a line-generator $a$ in a bistable $B$ such that $|a|<n/2$(**)

this is also equivalent to the fact that there is a line-generator $a$ and a row-generator $b$ such that $|a|<n/2$ and $|b|<n/2$(***)

what would then hapend if we have not (**): we would have

for all genertor line $a\in l(M)$, $ |a|\geq n/2$ , and then each row $r\in row(M)$ that would intersect any generator-line in a $0$ would verify $|r|<n/2$, but all rows are 0-perpendicular to at list one generator line, because if it was not the case, this row would be the full-1 row and as soon as we have the full-0 line this is not possible!

So a contrexemple to FUNC (**) would give us matrix with all rows verifying FUNC,

if we call the equivalent (***) version of FUNC the "and-FUNC", we than have, the $or-FUNC$ :

there is a line-generator $a$ or a row-generator $b$ such that $|a|<n/2$ or $|b|<n/2$

the weeker-FUNC that we HAVE is better than this it say that if the line-generator are not Frankl the row-generators are "uniformally-Frankl", but I think it is not a new result.

Note also that the duality of rows and line, as well as the duality for MULT and DIVI, give, with the property f) the DUAL-FUNC

if $(P,.)$ is a finite commutative idempotent monoïd, and if $g$ is a generator (that mean that $P-\left\{g\right\}$ is still a monoïd, then

$card(g.P)\leq card(P)/2$

(this is exactly the lattice version)

So a intersection closed (or union closed) familly that would not verify the DUAL-lattice-FUNC=FUNC) would verify FUNC uniformilly.

again that is not new but we have it with the tool that I built.

And with this tool I think that we "see" whats happening… therefore it is now time to give a stronger conjecture that the tools ask naturally : we note that bistable matrix are particular connectable matrix, and that the generator are exactly the separators then….

5) Separator-FUNC

If $P\in \mathbb P_n$ there exists $a\in P$ such that $a$ is separator and such that $|a|<n/2$

]]>For other rings however this makes perfect sense. Consider for example the ring of polynomials. What can you divide x + 3 by? ]]>

So why use imaginary numbers in your argument at all?

]]>I looked at https://en.wikipedia.org/wiki/Modulus_(algebraic_number_theory) but like many other wikipedia articles on math it was not very helpful. ]]>

The UK party manifestos describe the laws and policies which the parties promise to implement if they win a majority of seats and form a government.

It’s simply not possible for a UK party to win a majority of seats in the EU parliament or to form an EU government, so there’s no point in including EU laws or policies in their manifestos.

]]>It seems to me that they are held more accountable in their role as PM by the fact that they can be removed by the MPs that that the general population elected than by the fact that they are an MP. Having both levels of accountability doesn’t seem bad, of course.

I suppose something similar applies to the cabinet, except that members of the house of lords can serve there too. These ought to be quite important in setting the political agenda.

applicative — it sounds like the sensible thing there would be for the parties to include EU policies in their manifestos.

]]>The 28 members of the European Commission are directly sent by their respective governments. These governments are democratically legitimated – therefore the 28 members are democratically elected as well. The president of the European Commission is elected by the European Parliament after being proposed by the European Council.

On one hand the members of the European Parliament are directly elected by the people of Europe. On the other hand the European Council consists of the head of state of each member state (democratically elected), the President of the European Commission (elected by the European Council) and the President of the European Commission (elected by the European Parliament). Therefore the European Parliament and the European Council are democratically legitimated.

Lastly, the Council of the European Union directly consists of 28 ministers of the member states (one per state). As each meeting has a different topic, different ministers are sent by their national government each time. These ministers are democratically legitimated in their state, therefore the Council of the European Union is democratically legitimated.

As one can see, the four major EU institutions are either directly or indirectly democratically legitimated.

The three minor institutions of the European Union, the Court of Justice of the European Union, European Central Bank and the European Court of Auditors are all equally staffed and/or elected by the governments of the 28 member states and therefore democratically legitimated.

So, please explain how the EU and its power could be described as tyranny.

]]>I concede that Hitler’s rise to power is something of a chicken-and-egg problem: he rose to power because of the failings of his predecessors. But how did those attain power?

Your point about the democracies of Western Europe avoiding causing misery to their own populations illustrates I think that Amartya Sen’s argument best applies to comparatively small countries. I think it is revealing that it is usually the Nordic countries, with populations of about 5 to 10 million people, that top happiness surveys. (I guess smaller democracies are intrinsically dependent on their larger neighbours.) Finally, the electorate need not express their opinion through the ballot paper: to quote Blackadder, “These are volatile times, Your Highness. [T]here are tremendous rumblings in Prussia – although that might have something to do with the sausages.”

Further, I do not think that the uninformed majority vote randomly, certainly not in countries where the tabloid press is similarly strong to the tabloid press in the UK – a quick check on circulation numbers of tabloid newspapers (data from wikipedia) shows that the right-wing papers have a much larger circulation. There are also recent analyses like “Misperceiving Bullshit as Profound Is Associated with Favorable Views of Cruz, Rubio, Trump and Conservatism” (http://journals.plos.org/plosone/article?id=10.1371%2Fjournal.pone.0153419). And depending on how large you think this minority is, a small bias of that ilk will decide the outcome of elections.

]]>