What is the notion of truth used in Godels incompleteness theorem? First-order logic is complete & sound. The notion of truth used here is model-theoretic.

Informally Godels incompleteness theorem says that for a sufficiently strong formal language there are truths that cannot be proved. How is truth encoded in the language? Is this actually correct, should it be that there are statements which cannot be either proved or disproved - that is they are undecidable. Does this mean that they actually have no truth status?

"True" here means "true in the standard model." For example, the standard model of PA is the "actual" natural numbers. I agree that this is confusing. A more model-agnostic way to state the incompleteness theorem is that there are statements that are true in some models but false in others (hence that are neither provable nor disprovable).

In this thread I was wondering about what I believe are similar issues of truth in 'the standard model' and provability.

I will use the language of Gensler's little book Godel's Theorem Simplified.

For what Gensler calls System C, he gives an outline for how to construct a wff in system C. The system C the symbols are /, (, ), - , n (for a variable), nn (another variable),...etc.).

The reference numbers for system C formulas (like Godel numbers) are defined too.

For example the ref # for / is 1, the reference # for the variable n is 8, etc.

Then Gensler gives instructions for how to create a formula called F which states

(F) "There is no system C proof for the son of the system C formula with reference # n ".

Here n is a variable. So the system C formula with reference # n exists, call that system C formula XX, and F is the statement that the system C son of XX, SonXX, is not a theorem in system C).

F is a string of symbols of system C.The son of F is obtained by taking the reference number (an  actual whole number like 112278 etc) for F, writing that whole number as the system C string //.../ (with reference # of F many /'s appearing) and replacing each occurrence the variable n in (F) with that many ///.../.

The result is a new system C formula, the son of F, called G.

The 'standard model' interpretation of G is then

(G) "There is no system C proof for the son of the system C formula with reference # equal to the reference # of F."

In other words, there is no system C proof of G itself.

Then Gensler and other sources indicate this standard interpretation of G is true, so it is a true 'ordinary arithmetic' statement, for which the corresponding system C formula has no proof.

It seems to me that what you can really say is this:

If you assume that the standard interpretation of G is a true statement in the ordinary standard model, then the corresponding system C formula is an unprovable system C formula. So you have a 'true' ordinary arith statement which cannot be proved in system C, BECAUSE you have ASSUMED that the standard model version of G is true.

If you assume that the standard interpretation of G is a false statement in the ordinary standard model, then the corresponding system C formula is an provable system C formula.So you have a 'false' ordinary arith statement which can be proved in system C, BECAUSE you have ASSUMED that the standard model version of G is false.

But how do you know that one of the two statements " G is a true statement in the ordinary standard model" or " G is a false statement in the ordinary standard model" must be valid?

Isn't the best you can say that IF you could determine that
 G were a true statement in the ordinary standard model" or " G is a false statement in the ordinary standard model" THEN you can draw conclusions about non provability or provability of G as a system C formula?


Evaluate $\lim_{x\rightarrow 0}(1-1/x)^x$ using l'Hospital's rule. The problem is $$\lim_{x\rightarrow 0}(1-1/x)^x$$

I took the natural log and got:

$x\ln (1-(1/x))  $

What do i do now? My brain is clogged it seems I forgot how to simplify the expression inside $\ln$, help?

Write:
$$ lny = \frac{ln [1-(1/x)]}{\frac{1}{x}} $$

Now, if you are looking for x goes to infinity, you have type $0/0$, and you have  and if x goes to zero you have type $\infty/\infty $. In both cases, L'Hopital applies: take derivatives and evaluate.


Hyperoctahedral group and odd permutations? I apologize in advance for possibly using faulty terminology as I am a group theory novice. I am interested in looking at Hyperoctahedral groups when viewed as permutations. According to Wikipedia, when $n=2$, one can obtain any of the permutations of the square (https://en.wikipedia.org/wiki/Hyperoctahedral_group), because one of it's group elements is an odd permutation (i.e the cycle on 4 elements). However, there is no odd permutation in the group when $n=3$. Does this hold for larger values of $n$ as well?

I'm not sure what exactly you are stating when you say that there is no odd permutation in the case $ n = 3 $, mostly because a group is not just a collection of permutations, it can have a lot of different structure.

Anywho, the wikipedia page also says the hyper octahedral group with $ n = 3 $ is isomorphic to $ S_4 \times S_2 $, and thus contains the element $ ( (1, 2, 3, 4) , e ) $, which you might consider in some sense to be an odd permutation.

It might be easier to visualize as just rotating a hypercube in one dimension, since they are defined as the group of symmetries of those objects. They all are composed of squares, so just rotate one of the squares four times, and there will be an element of order four (that rotation).


Quotient group $SL(2,\mathbb{R})/SO(2,\mathbb{R})$ I am confused by expressions like $\mathbb{R}/\mathbb{Z}$, especially if I see something of the sort $SL(2,\mathbb{R})/SO(2,\mathbb{R})$. I don't know what to make of it. I understand what a quotient of a group for an equivalence looks like, but I can't get my head around expressions like above.

Is a quotient like this always a group and what do elements in the set $SL(2,\mathbb{R})/SO(2,\mathbb{R})$ look like?

They are a lot of questions in one post. I will do my best to answer them.
First, let me recall the notion of group action:

Definition $1$. Let $X$ be a set and let $G$ be a group, a left action of $G$ on $X$ is a map $\cdot\colon G\times X\rightarrow X$ which satisfies the two following axioms:

$\forall x\in X,e\cdot x=x.$

$\forall(g,g')\in G\times G,\forall x\in X,g\cdot(g'\cdot x)=gg'\cdot x.$

This definition is equivalent to the following:

Definition $2$. Let $X$ be a set and let $G$ be a group, a left action is group morphism from $G$ to $\mathfrak{S}(X)$.

If $\cdot\colon G\times X\rightarrow X$ is a group action in the sense of definition $1$, then $g\mapsto \{x\mapsto g\cdot x\}$ is a group action in the sense of definition $2$. Conversely, if $\varphi\colon G\rightarrow\mathfrak{S}(X)$ is a group homorphism, then $g\cdot x=\varphi(g)(x)$ is a group action in the sense of definition $1$.
Example 1. If $H$ is a subgroup of $G$, then $H$ acts on $G$ by left translation i.e. $(h,g)\mapsto hg$ is a group action.
Now, let us define the notion of quotient set in this general setup:

Definition $3$. Let $G$ be a group acting on a set $X$, then $G/X$ is the set of all orbits, namely:
$$G/X:=\{G\cdot x;x\in X\}$$
where $G\cdot x:=\{g\cdot x;g\in G\}$ is the orbit of $x$.

The set $G/X$ is a collection of subsets of $X$. In all generality, this is not a group, for example, $X$ may not be a group itself.
Example $2$. If $H$ is a subgroup of $G$, then $G/H:=\{gH;g\in G\}$.
Let us investigate the structure of $G/H$ when $G$ is a group and $H$ is a subgroup acting by left translation.

Proposition $1$. Assume that $H$ is normal in $G$ i.e. for all $g\in G$, $gHg^{-1}=H$, then $G/H$ is endowed with a unique group structure such that the canonical projection $\pi\colon G\twoheadrightarrow G/H$ is a group morphism.

Proof. Let define the operation on $G/H$ to be $gH\cdot g'H=gg'H$ or equivalently : $\pi(g)\pi(g')=\pi(gg')$. The key point is that this definition does not depend on the choice of $g$ and $g'$, namely:
$$gH=xH,g'H=x'H\Rightarrow gg'H=xx'H.$$
Indeed, there exists $h,h'\in H$ such that $x=gh$ and $x'=g'h'$, therefore, one has:
$$xx'H=ghg'\underbrace{h'}_{\in H}H=ghg'H=gg'\underbrace{g'^{-1}hg'}_{\in H}H=gg'H.$$
Now that the operation on $G/H$ is well-defined, I let you check the remaining properties (associativity, existence of identity element, existence of inverse). Whence the result. $\Box$
Conversely, if $G/H$ is a group such that $\pi$ is a group morphism, then $H$ is normal as the kernel of $\pi$.

What I like to emphasize is that when working when quotient sets it is crucial to specify the group action!

Maybe I'll close this answer mentioning a useful result:

Theorem. Let $G$ be a group acting on $X$, then for all $x\in X$, there is a bijective correspondence:
$$G/G_x\cong G\cdot x$$
where $G_x:=\{g\in G\textrm{ s.t. }g\cdot x=x\}$.

Proof. Consider $gG_x\mapsto g\cdot x$, in particular, check the well-definedness of this map. $\Box$
Remark 1. The set $G_x$ is a subgroup of $G$, but it may not be normal.
Example 3. $\mathbb{S}^n\cong SO(n+1)/SO(n)$ as sets, but even as smooth manifolds with a much-refined theorem.
Here, the action of $SO(n+1)$ on $\mathbb{S}^n$ is given by $(M,x)\mapsto Mx$ which is transitive, which means that each orbit under $SO(n+1)$ is the whole $\mathbb{S}^n$. If $x=(1,0,\ldots,0)$, then the stabilizer is given by:
$$SO(n+1)_x=\left\{\begin{pmatrix}1 & 0\\0& A\end{pmatrix};A\in SO(n)\right\}\underset{\textrm{abusive}}{=}SO(n).$$
Indeed, the first column must be $x$ and by orthogonality the first entry of each other column must be $0$. Whence the result by the theorem.

Coming back to your main question (and guessing what the action you are working with is):
Let $\mathbb{H}$ be the upper-half plane of $\mathbb{C}$, namely the set of points in $\mathbb{C}$ having a strictly positive imaginary part. Let $SL(2,\mathbb{R})$ acts on $\mathbb{H}$ by homography:
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\cdot z=\frac{az+b}{cz+d}.$$
Then this action is transitive and the stabilizer of $i$ is given by:
$$SL(2,\mathbb{R})_i=\left\{\begin{pmatrix}a & b\\-b&a\end{pmatrix};a,b\in\mathbb{R}\right\}=SO(2).$$
Whence using the theorem, $SL(2,\mathbb{R})/SO(2)\cong\mathbb{H}$ as sets and again even as smooth manifolds.

This quotient group consists of $2\times 2$ matrices with determinant 1 (because both $SL_2(\mathbb{R})$ and $SO_2(\mathbb{R})$ consist of matrices with determinant 1).

The quotient group is partitioned into equivalence classes of matrices that are not orthogonal, where elements of $SO_2(\mathbb{R})$ are treated as 0, and other elements are conceived of as their remainders mod an element of $SO_2(\mathbb{R})$.

A quotient of two groups $G/H$ is always a group as long as $H$ is a normal subgroup of $G$.


X discrete random variable, then 2X cannot have a binomial distribution I saw the following question and answer:

Q: True or false? Let X,Y be random discrete variables.
If X has a binomial distribution with parameters n,p and Y binomial distribution with n_1,p, then X+Y has a binomial distribution with parameters n_1+n,p.

A: False, take Y=X, then X+Y=2X and 2X cant have a binomial distribution since its even.

I was wondering does the argument of the answer holds (so the even -> no bin distr), do you guys have an idea? Thanks in advance

The pdf of $Y$ is given as follows $ P\left( Y=y \right)=P(2X=y)=P\left(X=\frac{y}{2} \right) = \begin{pmatrix}
 n \\ \frac{y}{2}
 \end{pmatrix} \left( p \right)^{\frac{y}{2}} \left( 1-p \right)^{n-\frac{y}{2}}, \ y= 0, \ 2, \ 4, \cdots, 2n $
(Robert V. Hogg, and Allen T. Craig, THE UNIVERSITY OF IOWA, Introduction to Mathematical
Statistics, Fourth Edition, page 128). As such, I have a feeling that this is still a binomial distribution.
Secondly, if $X \sim Bin(n, p)$, then $M_X (t)=(pe^t+1-p)^n$. Hence, $M_{2X}(t)=M_X(2t)=(pe^{2t}+1-p)^n$. If we choose $u=2t$, we have moment generating function of binomial distribution. I ask you colleagues to reconsider your answers. Good Luck,

In general, you are right indeed. If we take $X$ and $Y$ to be independent however, the statement is true.

To see this, remember that a binomial random variable with parameters $n,p$ is just saying how many of $n$ independent experiments with probability of succeeding $p$ succeeded.

Hence looking at $X+Y$ is like doing $n$ of those experiments, then doing $n_1$ of them, and adding up how many succeeded. This is exactly the same as doing $n+n_1$ experiments, and then counting how many worked.


Exercises EG.3 and EG.4 in Probability with martingales, by David Williams Let $G$ be the free group with two generators $a$, $b$. Start at time 0 with unit element 1, the empty word. At each second multiple the current word on the right by one of the four elements $a$, $a^{-1}$, $b$, $b^{-1}$ choosing each with probability 1/4 independently of previous choices. For instance the choices
  $a,a,b,a^{-1},a,b^{-1},a^{-1},a,b$ yield the reduced word of length 3, $aab$ at time 9.

  EG.3 Prove that the probability that the reduced word $1$ ever occurs at a positive time is $1/3$.

  EG.4 Now suppose elements $a$, $a^{-1}$, $b$, $b^{-1}$ are chosen with probability $\alpha$, $\alpha$, $\beta$, $\beta$ respectively, where $2\alpha+2\beta=1$. Prove that the conditional probability that the reduced word $1$ ever occurs at a positive time given that $a$ is chosen at time $1$ is the unique root in $(0,1)$  of the polynomial $3x^3 + (3-4\alpha^{-1})x^2+x+1$.

I know little about Markov Chains, and about limiting distributions but the chain is not of the form that a limiting distribution exists, so I am not able to make use of that facts. Chain I have in mind is of length of reduced word, states are $\{0,1,2,\dots,\}$. Now $p_{k\to k+1}=3/4,\ p_{k\to k-1}=1/4$ for $k\geq1$ and for $k=0,\ p_{0\to1}=1$.

I have only been able to solve this through guidance of Did.
$\def\P{\mathbb{P}}$
$\def\Ex{\mathbb{E}}$
$\def\Z{\text{\ensuremath{\mathbb{Z}}}}$
$\def\I{\textrm{I}}$
$\def\as{\textrm{a.s.}}$

EG.3 Use the Markov Chain theory developed in Markov chains, by Norris,
to calculate the hitting time to reach sate $0$ from state of length
$1$. Specifically, we model the length of reduced word as Markov Chain with length as it states.
With $p:=\P\left(\left\{ \text{hitting }0\text{ given length is }1\right\} \right)$
we have
\begin{align*}
p= & \frac{1}{4}+\frac{3}{4}\P\left(\left\{ \text{hitting }0\text{ given length is }2\right\} \right)\\
= & \frac{1}{4}+\frac{3}{4}\P\left(\left\{ \text{hitting }0\text{ given length is }1,\text{hitting 1 given length is 2}\right\} \right)\\
\overset{\left(1\right)}{=} & \frac{1}{4}+\frac{3}{4}\P\left(\left\{ \text{hitting }0\text{ given length is }1\right\} \right)\P\left(\left\{ \text{hitting }1\text{given length is }2\right\} \right)\\
\overset{\left(2\right)}{=} & \frac{1}{4}+\frac{3}{4}p^{2}.
\end{align*}
where in $\left(1\right):$ Strong Markov Property; Markov chains, by Norris
is used and in $\left(2\right):$ we exploit symmetry of setup that
\begin{align*}
\P\left(\left\{ \text{hitting }0\text{ given length is 1}\right\} \right)=\P\left(\left\{ \text{hitting }1\text{ given length is }2\right\} \right).
\end{align*}
Solving $p=\frac{1}{4}+\frac{3}{4}p^{2}$ we get $p=\frac{1}{3},1$
where the latter is discarded by observing the Markov Chain (higher
probability to visit states down the chain).

EG.4

Similar to previous exercise let $p_{a}$ be the probability of interest;
i.e.
\begin{align*}
p_{a}=\P\left(\left\{ \text{hitting $0$ given state is }a\right\} \right).
\end{align*}
Likewise generalize the definitions above and the argument of Strong
Markov Property above to get
\begin{align*}
p_{aa}= & \P\left(\left\{ \text{hitting $0$ given state is }aa\right\} \right)=p_{a}^{2}\\
p_{ab}= & p_{a}p_{b}\\
p_{ab^{-1}}= & p_{a}p_{b}^{-1}=p_{a}p_{b}
\end{align*}
where the last equality is true by the symmetry of setup. Now note
that hitting time
\begin{align*}
p_{a}= & \alpha+\alpha p_{aa}+2\beta p_{ab}\\
= & \alpha+p_{a}^{2}\alpha+2\beta p_{a}p_{b}.
\end{align*}
Similarly, we find $p_{b}=\beta+\beta p_{b}^{2}+2\alpha p_{a}p_{b}$
but from equation above we have
\begin{align*}
p_{b}=\frac{p_{a}-\alpha\left(1+p_{a}^{2}\right)}{2\beta p_{a}}.
\end{align*}
Substiting this into $p_{b}=\beta+\beta p_{b}^{2}+2\alpha p_{a}p_{b}$
we get after simplification
\begin{equation}
p_{a}^{2}\left(2a^{2}-4b^{2}+1\right)+3a^{2}p_{a}^{4}-a^{2}-4ap_{a}^{3}=0.\label{eq:EG4}
\end{equation}
Now setting $p_{a}=1$ above we get
\begin{align*}
\left(2a^{2}-4b^{2}+1\right)+3a^{2}-a^{2}-4a=4\left(a^{2}-b^{2}-a\right)+1=4\left(\frac{1}{2}\left(a-b\right)-a\right)+1=0.
\end{align*}
Thus dividing original polynomial by $\alpha^{2}\left(p_{a}-1\right)$ (i.e. rejecting
this solution choice) we get $3p_{a}^{3}+\left(3-4\alpha^{-1}\right)p_{a}^{2}+p_{a}+1=0$.
This indeed has a unique solution in $\left(0,1\right)$ as can be
checked easily.

Limiting distributions won't help you here, anyway. The long term probability of hitting a given state doesn't necessarily tell you anything about the probability of hitting it ever.

This is basically a combinatorics exercise. What is the probability of the $n$-th state being $1$? This is easier than it might seem at first. What strings of length $n$ would reduce to $1$? Let $a_n$ be the number of strings of length $n$ that reduce to $1$. I'm sure you can compute $a_n$ via some sort of recurrence relation.


Which is bigger $\tan^{-1}(\frac{1}{2}) $ or $\frac{1}{\sqrt{5}}$? A number theory textbook asked us to compare $\tan^{-1}(\frac{1}{2})$ and $\sqrt{5}$.  In fact, these are rather close:

\begin{eqnarray*}
\tan^{-1} \frac{1}{2} &=& 0.46364 \\ \\
\frac{1}{\sqrt{5}} &=& 0.44721
\end{eqnarray*}

So at least numerically I think we have the answer that the first one is bigger.  Momentarily, I thought we had an exact answer: $\tan^{-1}\frac{1}{\sqrt{2}} = \frac{\pi}{4} $, but that's totally different.  So we are left with:

$$ \tan^{-1} \frac{1}{2}  > \frac{1}{\sqrt{5}} > 0 $$

It's impressive that we could have so many decimal places, and I wonder if I should take the computer on faith for that.  And I noticed these two answers are close, so I also wonder if we estimate the difference... I don't have any conjecture in either case yet.

For now I just want proof of the inequality as stated above.

Suppose $\tan(x)=\frac{1}{2}$. Then in particular you know $\frac{\sin(x)}{\cos(x)}=\frac{1}{2}$. You also know $\sin^2(x)+\cos^2(x)=1$. Using the last two equalities you find that $\sin(x)=\frac{1}{\sqrt{5}}$. Since $x\geq\sin(x)$, you're done.

This seems like more of a real analysis question than number theory. The first method I thought of was taking the Taylor series for tangent($\sqrt{5}$) and comparing that to $\frac{1}{2}$ (and noting that tangent is an increasing function), but arctan has a nicer Taylor series than tangent does. Either way, what you can do is take the first n Taylor series terms, argue that the remaining terms add up to no more than some error term, and if you've taken a large enough n, then the error term will be small enough that you can say which is larger, even with the error term. That's the general real analysis strategy for comparing number obtained from analytic functions. There might be some clever number theory method I'm missing.


Triangle dissection, no shared edges It's possible to divide a triangle into smaller triangles such that no edge lengths are shared.  Alternately, no two internal triangles share two vertices. The top three are the known simplest solutions. The bottom three were found via doodling.

The triangles in the first solution could be subdivided the same way, so another criteria is added: no subdivided internal triangles are allowed.

What are some other solutions with 7 to 18 internal triangles?

EDIT. Of the prime 7 dissections, and the prime-8, no new solutions are found.

New solutions might be found by analyzing Schlegel diagrams of planar graphs, and picking three vertices of each face to be a triangle. If no two triangles share two faces, and if excised facial vertices can be moved between triangular vertices, then it's a potential solution. For example, this graph leads to the upper right solution, with triangles 234, 478, 456, 138, 125, 167.

One more solution, with nine of the triangles having area 1.

If you have an equilateral triangle in the solution, you may just copy into it any of the other solutions thus having number of solutions as you desire.


Finite additivity, atomlessness and countable additivity So, I'm trying to get my head around when you can have finitely but not countably additive probabilities.

The standard example of a finitely additive but not countably additive space is the following strange distribution over the natural numbers.  All finite sets get measure 0, but the whole space gets measure 1. This is finitely additive but not countably so, since a finite union of finite sets is finite, but a countable union needn't be so.

So this got me thinking that if you had an atomless space, examples of this form would be harder to come by. Does atomlessness plus finite additivity guarantee countable additivity? If not, what is missing?

I know that Villegas (1964) shows that for a comparative probability structure to be countably additive, the important properties of the structure are atomlessness and a certain kind of continuity. But I don't know how relevant that point is to the current question.

Here’s an example.

Define an equivalence relation $\sim$ on $\wp(\Bbb N)$ by $A\sim B$ iff $A\,\triangle\, B$ is finite, where $\triangle$ is symmetric difference, and let $\mathscr{B}=\wp(\Bbb N)/\sim$. For $A\subseteq\Bbb N$ denote by $[A]$ the $\sim$-equivalence class of $A$. Let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$. Note that for any $A\subseteq\Bbb N$, $A\in\mathscr{U}$ iff $[A]\subseteq\mathscr{U}$. Now define a $\{0,1\}$-valued measure $\mu$ on $\mathscr{B}$ by $\mu\big([A]\big)=1$ iff $A\in\mathscr{U}$. Then $\mathscr{B}$ is atomless, and $\mu$ is finitely additive. However, $\mu$ is not countably additive, since it is possible to partition $\Bbb N$ into countably infinitely many infinite sets, none of which is in $\mathscr{U}$.

Added: Michael Greinecker has pointed out that I’m using a notion of atomless that may be different from the one intended by Seamus. Here’s another example that may be preferable.

Let $d:\wp(\Bbb N)\to[0,1]$ be asymptotic density, and let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$. For $A\subseteq\Bbb N$ let $$\mu(A)=\mathscr{U}\text{-}\lim_n\frac{|A\cap\{1,\dots,n\}|}n\;.$$ (For basic information on $\mathscr{U}$-limits see this answer by Martin Sleziak.) Then $\mu$ is a finitely additive non-atomic probability measure on $\wp(\Bbb N)$ such that $\mu(A)=d(A)$ whenever $A$ has an asymptotic density.

The missing ingredient is Monotone Continuity, not atom-less-nes (the villegas 'continuity' condition mentioned by the OP perhaos), as described below, not atom-lessness

Monotone continuity this is the qualitative analog of Kolmogov's continuity condition described above by some of the answers.
The CP relation  need not be Atom-less, to be countably additive, nor is it a sufficient condition, where atomless, is defined, in the sense of Villegas, (1964, 1967):

$$\forall (A\in S;A>O):\exists (B\in S):(B\subsetneq  A)\land A>B>O.$$.
where $S$ is the algebra of events, that are under a usually total ordered CP relation $\geq|\leq|$ and $O$ is the impossible event.

To be countable additive, the relation does need to be it needs to be 'monotone continuous' in the sense Villegas (1964):
See theorem 2 page 1794 Villegas (1964).

$$(MS)[[A_{n}\uparrow A]\land\forall (n;n\in N^{+}):[B\geq A_{n}]]\iff B\geq A.$$.

Where $B,A\in S$, the algebra of events and $A_n\uparrow A$ is a monotonely increasing sequence of events $A_i\in S$ that convergent onto A

$$A_n\uparrow A\equiv \{A_n\}:[\forall(n):A_{n}\subset A_{n+1}]\,; \cup_{n=1}^{n=\infty}A_{n}=A.$$

This is presumably the 'kind of continuity the OP speaks of.

It is roughly also, entails continuity from above at 0, and thus represent-ability by a countable additive measure in the context of an already  numerically representable finitely additive CP relation over a sigma algebra.

Although, I have seen strengthened version in less conventional axiomat-izations which needed to explicitly state the axiom both for continuity from above at $O$ and below and at $\Omega$ where $\Omega$ denotes the certain event.

Otherwise, without monotone continuity, the relation must satisfy something of its ilk, or will not be countably additive full stop.

Over an infinite algebra, the  Comparative Probability relation may or may not, be representable by an  even finitely additive, depending on what other conditions are satisfied.Dispensing with monotone continuity or something of its kind,unfortunately, is not an option for a CP relation, if it is to be compatible with a countably additive representation.

And even if one desires that the relation to be neither not-atomic, nor defined over a sigma algebra, it requires $(MS)$ but it requires the addition of multiple other axioms $(1)$ and $(2)$in addition to quite possibly losing uniqueness.

-If one dispenses with atomless-ness, then to have a countable additive representation of the CP order, you will need to add $(MS)$,,$(1)$ and $(2)$ to the list of usual axioms below.

-total comparative CP order.
$$A\geq B \vee B\geq C$$

strict weak order,(transitivity etc).

-
-$$\text{non-negativity and non-triviality}\forall(\Omega_i\in S):\Omega \geq \Omega_i\geq\emptyset$$.
-$$\text{non-negativity and non-triviality) }\Omega> \emptyset$$.
   **-$$\text{monotone continuity}(MS)$$.*
-$$(1)\text{perfect separability}$$
-$$(2)\text{Scott's axiom plus archimedean condition}$$*

One will also need , not only $(MS)$, as above, but also
$$(1) \text{sigma algebra or perfect separability} \land (2)$$.
where,

$(2)$= $(2.2)$ Scott's finite addivity condition with an arch-imedean condition supplement$(2.1)$.

below, or use Scott's infinite condition  in addition. Which is much more complex, so I would not take that route.

This is,  as non-atomicity allows one to dispense with Scotts, finite condition,(or infinite, or $(2)$ which is the finite Scotts axiom with an archimedean condition.
 Moreover, one need not have perfect separability ,I think, which non-atomicy entails, in the case of a non-sigma algebra, countably addive but non , non-atomic representation.
 I am not going the type this (Scott's infinite condition )axiom here, it is rather complicated (see Z. Domotor's 1969 PHD thesis (Probalistic relational structures; see https://www.google.com.au/search?q=domotor+relational+systems&ie=utf-8&oe=utf-8&client=firefox-b&gfe_rd=cr&dcr=0&ei=dAfnWbDAOeLc8weYo5XQBQ

or use the conditions mentioned by Jaffray and Chauttenaff (1984) which are those of $(1)$ and $(2)$ below and discussed in "Archimedean Qualitative Probabilities", listed below in the references.

Roughly,for a countable additive  (non, non-atomic representation) that $(1)$ and $(2)$ must be added to the usual axioms. As  well as monotone continuity $(MS)$.

$$(1)$$

$(1)$ keeping the sigma algebra condition of Villegas or:
$(1a)$replace the the $\sigma$ algebra,with a 'perfect separability ' axiom, and possibly a strengthened archi-medean condition.
 $(2)$ replacing De-finetti additivity: $(F)$.

$(2)$
$$(F):A\cap C=B\cap C=\emptyset: A\geq B \iff A \cup C \geq B \cup C$$.

with a supplemented version  of Dana Scott's axiom/strong additivity, finite cancellation (strong additivity) . That is supplemented with an addition archimedean constraint.

.

where condition (2)  mentioned below is  called (A) in the picture from Qualitative Archimdedean PRobabilites (1984).

(2) or (A) in the cited picture, is the $(2.1)$ Archimedean axiom and $(2.1)$ Scotts axiom  when combined into one condition See  (chautenaff 1984) referenced below for more on this.

It is, the supplemented Scott's(1964) condition $(A)$ in Chauttenauff (1984).

I have posted a link and picture below.
$$(3)$\text{add monotone continuity }MS$$.

Or you use Scott's infinite axiom (I am not sure if you want to do that)!!!

Moreover,   if you dispense with  both the sigma algebra requirement, and atomless-ness, then you will need in addition to the above, a perfect separability requirement atomless-ness is not sufficient for a countably additive representation. It almost always is necessary for a unique representation.

-The sigma algebra relation can be dispensed with as well, but unless one wishes to add perfect  separability.

If you want to know the conditions that are necessary and sufficient for a non-necessarily unique countably additive representation, read Jaffray and chautenaff (1984) who give simpler set of axioms for a comparative probability structure that is necessary and sufficient for a countably additive representation, in particular a representation over an infinite sample space.

If one dispenses with atom-less-ness (in the traditional comparative probability sense; that is in the sense of Villegas (1964), then
$(1)$ one does require additional axioms to those of Villegas,

$(2)$  In addition, the representation will likely no longer be unique.

For example, if one  dispenses, with both the sigma algebra condition and atom-less-ness:

-(1)(then one must replace) s de-finetti's axiom in Villegas set of axioms:
$$A\cap C=B\cap C=\emptyset: A\geq B \iff A U C \geq B UC$$.
 with $(A),\land (B),

-If one keeps the sigma algebra condition

Then perfect separability is entailed, by one needs only (1), and (3)
See, Jaffray 'On the Extension of Additive Utilities to Infinite Sets'(1974) which are necessary and sufficient
space satisfy. Is another on not necessarily non-atomic or countably additive probability measures that may not even be on sigma algebras.

One good source is Fisburne (1986), the axioms of subjective probability

Fishburn, Peter C., [The axioms of subjective probability]; see
(http://dx.doi.org/10.1214/ss/1177013611), Stat. Sci. 1, 335-358 (1986). ZBL0604.60004. Theories of probability), Luce, Narens; maybe even fisburne 1986).

Villegas, C., see

On qualitative probability $\sigma$-algebras, Ann. Math. Stat. 35, 1787-1796 (1964). ZBL0127.34807.  but it  must be monotone continuous (or something like it).
Villegas, C., On qualitative probability, Am. Math. Mon. 74, 661-669 (1967). ZBL0154.42301.
3.Chateauneuf, Alain; Jaffray, Jean-Yves, [Archimedean qualitative probabilities].

(http://dx.doi.org/10.1016/0022-2496(84)90026-9), J. Math. Psychol. 28, 191-204 (1984). ZBL0558.60003. if you retain the sigma algebra requirement,  but forsake Atom-less-ness then in addition to possibility losing 'unique-ness':


Formal algebraic notation for RGB color scale? BACKGROUND

I'm trying to write down an algebraically rigorous (read: pedantic) expression for colors in the RGB "space". All I know is that I have a set
\begin{equation}
\mathcal{S} = \{R, G, B\}
\label{eq:set}
\end{equation}
of three qualitative "entities" (for lack of a better term), each of which can take on a value in the set
\begin{equation}
\mathcal{K} = [0, \cdots, 255]
\end{equation}
of possible coefficients. Note that I'm not making any assumptions as to the orthogonality of the elements of $\mathcal{S}$.

QUESTION

How can I write down the vector space $\mathcal{C}$ in which colors "live" in both a rigorous and compact way? I initially thought of writing
\begin{equation}
\mathcal{C} = \mbox{span}\{\mathcal{S}\},
\end{equation}
but that doesn't look right since $\mathcal{S}$ is a set of "labels" more than a set of vectors per se. Do I then have to go as far as writing down
\begin{equation}
\mathcal{C} = \mbox{span}\{ \mathcal{V} \} = \left\{\sum\limits_{c} \lambda_{c} v_{c} \mid c \in \mathcal{S}, \lambda_{c} \in \mathcal{K}, v_{c} \in \mathcal{V} \right\},
\end{equation}
where
\begin{equation}
\mathcal{V} = \left\{ [1,0,0], [0,1,0], [0,0,1] \right\} ?
\end{equation}

The problem is that this latest expression for $\mathcal{V}$

looks clumsy,
doesn't scale to large dimensions since it's a brute-force enumeration, and
does not say which dimension corresponds to which color since $\mathcal{S}$ is by default an unordered set---so we're left with yet another aspect of the problem to transcribe.

EDIT

Calling $\mathcal{C}$ a vector space has raised quite a few eyebrows. I shall therefore refer to it as a bounded, discretized, "space of vectors". Please let me know if you have a better term for it. The fact remains that vectors for colors live in that "space".

One immediate problem is that we don't even have a good compact notation for the set $\{0,1,2,\ldots,255\}$ in the first place (unless you've had too much axiomatic set theory the night before and claim that this set equals the number $256$ ... but that won't fly with a general readership).

It is fairly common in the context of color spaces to view $\{0,1,2,\ldots,255\}$ simply as an (imperfect) representation of the unit interval $[0,1]$, implicitly letting $n$ stand for $\frac{n}{255}$.

If this idealization works for you, you can describe the color space as simply
$$ [0,1]^3 $$
and leave it to convention which of the three channels denote red, green, and blue, respectively.

It won't give you a vector space, but that is to be expected, because the space fo 24-bit RGB colors doesn't actually support arbitrary sums and differences of colors, like a vector space ought to. It can be seen as a subset of the vector space $\mathbb R^3$, though -- so you have some justification for using vector space operations inside formulas as long as you remember to explain what happens if the final result falls outside the gamut.

(In particular $[0,1]^3$ is a convex subset of $\mathbb R^3$, so you never risk falling outside it by interpolating or averaging between already known colors).

A natural(?) way is to view a colour as a measure over the (visible) range of wave lengths (or frequencies), so that $\int_a^b\mathrm d\mu$ answers the question: How much energy is transmitted in the wavelength range from $a$ to $b$ per time unit? The characteristics of receptors in the human eye, or the characteristics of the construction of RGB displays reduce this to a measure on a finite space. In other words, you consider the set of functions $\{R,G,B\}\to [0,\infty)$ (perhaps even including $\infty$). These measures do not form a vector space, but you can add or scale or convex combine them, etc.

Additional constraints on real world monitors (and image file formats) impose an upper bound on the power per channel as well as quantization effects (but not linearly - see $\gamma$ correction!), but it depends whether you want to adhere to these before a final step in your processing.


Calculating Odds: $1-10$, $15$ times If a program selects a random number between 1 and 10, and does so 15 times, what are the odds of it picking the same number twice or more, and how many times on average would it pick the same number?

From running this program I can see that it chooses the same number twice or more 100% of the time, and averages ~3 of the same number.

But I need something more mathematical to show the powers that be.

Can anyone help me out? Is there a good formula for this kind of thing I can use in the future?

Assuming that you mean integers between $1$ and $10$, the same number must come up at least once since you’re making more than ten choices: if no duplicates have come up after ten, the next pick has to be a duplicate since you’ve run out of distinct numbers. For the second question, you’ll need to define more clearly what you mean, per Soke’s comment.

Note that it’s easy to compute the expected number of times that any specific number will get chosen. That’s just the expectation of the number of successes in repeated Bernoulli trials, which is in this case $\frac1{10}\cdot15=1.5$, unsurprisingly.

Not a sufficing answer, but a long comment + comments don't support formatting well.

I made a program that ran this problem 1,000,000 times and this was the result:

Results:

Number chosen 0 times in a row: 231934
Number chosen 1 times in a row: 357143
Number chosen 2 times in a row: 255615
Number chosen 3 times in a row: 112538
Number chosen 4 times in a row: 34018
Number chosen 5 times in a row: 7335
Number chosen 6 times in a row: 1220
Number chosen 7 times in a row: 178
Number chosen 8 times in a row: 19
Number chosen 9 times in a row: 0
Number chosen 10 times in a row: 0

Another run gave this result:

Results:

Number chosen 0 times in a row: 231651
Number chosen 1 times in a row: 356585
Number chosen 2 times in a row: 256375
Number chosen 3 times in a row: 112532
Number chosen 4 times in a row: 33983
Number chosen 5 times in a row: 7509
Number chosen 6 times in a row: 1196
Number chosen 7 times in a row: 154
Number chosen 8 times in a row: 14
Number chosen 9 times in a row: 0
Number chosen 10 times in a row: 1

Running this program multiple times just resulted in minor changes in the result, no surprises at all.

This does however disprove that "it chooses the same number twice or more 100% of the time"

Code can be found here. (Qt C++)

Edit:
Just realized just how sloppy that code-snippet is...


Find $x$ that minimizes $\max_{1 \leq i \leq 3} f_i(x)$ My question is:

  For $0 < a_1 \leq a_2 \leq a_3$ fixed, find $x>0$ that minimizes
  \begin{align*}
g(x) = \max_{1\leq i \leq 3} f_i(x).
\end{align*}
  where
  \begin{align*}
f_i(x) =  \left(\frac{a_i - x}{a_i + x}\right)^2
\end{align*}

I do not know how to start. Any idea, reference is welcome. Thank you!

I think I have a solution to my question.

  Answer: the minimizer is $x = \sqrt{a_1 a_3}$.

Let $0 < \underline{a} < \overline{a}$ be fixed. We study the relative position of $f_{\underline{a}}$ and $f_{\overline{a}}$ defined by
\begin{align*}
f_{\underline{a}}(x) = \left(\frac{\underline{a} - x}{\underline{a} + x}\right)^2, \qquad
f_{\overline{a}}(x) = \left(\frac{\overline{a} - x}{\overline{a} + x}\right)^2
\end{align*}
for $x>0$. Let $h$ be defined by $h = f_{\underline{a}} - f_{\overline{a}}$. Then,
\begin{align*}
h(x) &= \frac{(\underline{a}-x)^2(\overline{a} + x)^2 - (\overline{a} - x)^2(\underline{a}+x)^2}{(\underline{a}+x)^2(\overline{a}+x)^2}\\
&= \frac{\big[(\underline{a}-x)(\overline{a} + x) - (\overline{a} - x)(\underline{a}+x)\big]\big[(\underline{a}-x)(\overline{a} + x) + (\overline{a} - x)(\underline{a}+x)\big]}{(\underline{a}+x)^2(\overline{a}+x)^2}\\
&= \frac{4 x (\overline{a} - \underline{a})(x^2 - \underline{a}\overline{a})}{(\underline{a}+x)^2(\overline{a}+x)^2}.
\end{align*}
Hence, $f_{\underline{a}} < f_{\overline{a}}$ for $x \in (0, \sqrt{\underline{a}\overline{a}})$, $f_{\underline{a}} = f_{\overline{a}}$ for $x = \sqrt{\underline{a}\overline{a}}$, and $f_{\underline{a}} > f_{\overline{a}}$ for $x \in (\sqrt{\underline{a}\overline{a}}, +\infty)$. Since $0 < a_1 < a_2 < a_3$, we deduce that
\begin{align*}
&f_1 \leq f_2 \quad \text{if} \quad x \leq \sqrt{a_1 a_2}, \quad\text{otherwise} \quad f_1>f_2,\\
&f_1 \leq f_3 \quad \text{if} \quad x \leq \sqrt{a_1 a_3}, \quad\text{otherwise} \quad f_1>f_3, \\
&f_2 \leq f_3 \quad \text{if} \quad x \leq \sqrt{a_2 a_3}, \quad\text{otherwise} \quad f_2 > f_3.
\end{align*}
Consequently, the maximum is given by
\begin{align*}
g(x) = \begin{cases}
f_3(x) & \text{if }x \leq \sqrt{a_1 a_3}\\
f_1(x) & \text{if }x > \sqrt{a_1 a_3}.
\end{cases}
\end{align*}
On the other hand, for $i \in \{1, 2, 3\}$,
\begin{align*}
f_i'(x) = \frac{4 a_i (x- a_i)}{(a_i+x)^3}.
\end{align*}
Hence, $f_i$ is decreasing on $(0, a_i]$, and increasing on $(a_i, + \infty)$. This implies that $f_3$ is decreasing on $(0, \sqrt{a_1 a_3}]$, and $f_1$ is increasing on $(\sqrt{a_1a_3}, + \infty)$. Hence, $g(x)$ is the smallest for $x = \sqrt{a_1 a_3}$.

You can introduce a new variable $z$ and minimize $z$ subject to $$z \ge \left(\frac{a_i-x}{a_i+x}\right)^2$$ for all $i$.  This reformulation avoids the nondifferentiable $\max$ function.  Maybe you can try the method of Lagrange multipliers.


Ways to partition a sphere? first of all, sorry for the lack of terminology/ignorance on the subject, I just joined this website.

I need a sphere or sphere-like 3D shape, whose surface is partitioned into another geometric primitive, in some kind of grid. I would prefer these partitions to be hexagons, tiled into each other. The bigger the sphere surface is, the more hexagons are supposed to be present.

After some research, I found the truncated icosahedron, which looks quite similar to what I want, except it has some pentagons in there, which kills the premise I need to satisfy:

If i have an object in any given partition, I need to be able to travel that object all around the sphere, always passing through the center of the next partition, and it has to arrive the initial location in a straight line. The traveling direction is arbitrary but always is the middle of one of the edges of the geometric shape of the partition.

I need to be able, in a visualization sense, to have a whole line of the elementary geometric shapes be moved at once, as if it was a huge circular rubik puzzle.

EDIT:
http://en.wikipedia.org/wiki/Truncated_order-7_triangular_tiling
This might be what I am looking for to some extend.. ?

I know I am probably not explaining myself perfectly, but if anyone could help out it would be appreciated.

There is also the option to use other shapapes like
triangles with 72° angles, five meeting at a vertex
triangles with 90° angles, four meeting at a vertex
triangles with 120° angles, three meeting at a vertex
quadrilaterals with 120° angles, three meeting at a vertex
pentagons with 120° angles, three meeting at a vertex

It is impossible to partite a sphere only with hexagons but, for the purpose of a game, you may partite only its visible part. This kind of partition is done here. And pay attention: this reference is a joke.


Opposite of Fermat's Last Theorem? So Wiles' proof showed that no three positive integers $a$, $b$, and $c$ can solve the equation $a^n+b^n=c^n$ for any integer value of $n$ greater than $2$. Now what about the opposite?

What does this mean for any $a$ greater than $2$, and $x$, $y$ and $z$ are positive integers in the equation $a^x+a^y=a^z$. Is there any relation? Is it solvable?

Here's a less algebraic way to interpret your question: Look at things in base $a$. Then the equation $a^x+a^y=a^z$ looks like $1 \ldots 0 + 1 \ldots 0 = 1\ldots0$. (where the number of zeros is $x$, $y$, or $z$) This can only work if we're in base $2$ and the addition carries to produce a $1$ in the next digit.

It is clear that $1+a^{y-x}=a^{z-x}$ where $a=1$,$x=y=z$ is an obvious solution. On the other hand the minimun difference of powers of integers is of the forme $2k+1$ which corresponds to squares.So, (making abstraction of the previous division by zero!) this would gives $a=0$.


If $\lambda$ is an eigen value of an orthogonal matrix $A$, then show that $\frac{1}{\lambda}$ is also an eigen value of $A$ If $\lambda$ is an eigen value of an orthogonal matrix $A$, then show that $\frac{1}{\lambda}$ is also an eigen value of $A$, with the same set of eigen vectors.

I have proceeded like this:
If $A$ is orthogonal, $A^{-1}=A^{T}$
So, the characteristics equation : $0=|A-\lambda I_n|=|A^T - \lambda I_n|=|A^{-1} - \lambda I_n|=|A^{-1}(I_n-\lambda A)|=|A^{-1}|(-\lambda)^n|A-\frac{1}{\lambda}I_n| \implies |A-\frac{1}{\lambda} I_n|=0$
So, $\frac{1}{\lambda}$ is also an eigen value.
How can I show that it has the same set of eigen vectors?

$A$ is orthogonal if and only if $A^{-1}=A^{\top}$.

Suppose $\mathbf{v}\ne\mathbf{0}$ is an eigenvector of some orthogonal marix $A$ associated with its eigenvalue $\lambda$. Since $A$ is invertible, $\lambda\ne 0$. We have
$$
A\mathbf{v}=\lambda\mathbf{v}.
$$
Multiply $A^{\top}$ on both sides, and
$$
\mathbf{v}=A^{\top}A\mathbf{v}=\lambda A^{\top}\mathbf{v}.
$$
Since $\lambda\ne 0$, this implies that
$$
A^{\top}\mathbf{v}=\frac{1}{\lambda}\mathbf{v}.
$$
Since $\mathbf{v}\ne\mathbf{0}$, the above equation indicates that $\mathbf{v}$ is also an eigenvector of $A^{\top}$ associated with an eigenvalue $1/\lambda$.

Using this trick, in general, any eigenvector of an orthogonal matrix $A$ is also an eigenvector of $A^{\top}$, and vice versa. Therefore, an orthogonal matrix shares all of their eigenvectors with its transpose/inverse.

Suppose $~λ~$ is an eigen value of $~A~$.

If $~A~$ is orthogonal, $~A^{-1}= A'~\tag1$

Now, eigen values of $~A~$ and $~A'~$ are always same.

Also, eigen values of $~A^{-1}=\frac{1}{λ}~$

From $(1)$ eigen values of $~A' =~$ eigen values of $~A^{-1}~$

Thus, $~\frac{1}{λ}~$ is an eigen values of $~A'~$ and also of $~A~$.


Difference between units and dimensions Though this question may seem related to Physics, I think that at the very root this is a mathematical question and so I have posted this on math.stackexchange.

Background: Initially I thought that the terms-unit and dimension, refer to the same thing.
Physical quantities are categorised into fundamental/basic physical quantities and derived physical quantities. A fundamental physical quantity cannot be broken down into simpler physical quantitities, cannot be obtained from other fundamental quantities and all the known physical quantities can be obtained using fundamental quantities. There was a line in my book which stated

  Mass, length, time, thermodynamic temperature, electric current, amount of substance, luminous intensity are the seven fundamental quantities and are often called the seven dimensions of the world. Thus the dimension of mass is [M], that of length is [L] and so on. The dimensions of a derived physical quantity are the powers to which the units of the fundamental physical quantities have to be raised in order to represent that derived physical quantity completely.

This is very confusing and I am finding it difficult to understand the difference between the two terms-dimension and unit.

Question: What exactly is the difference between the meaning of the terms unit and dimension?

The difference is quite subtle and of little practical importance if done accordingly.

The difference is that a unit incorporates a property of scale while dimension doesn't. For example in the case of length it could be measured in meters, decimeters or kilometers, but nevertheless the dimension is length.

If you use unit-analysis instead of dimensional analysis you would have to take into account that different units of length only differs by a  (dimensionless) constant.

Using numbers in physics

Units are artificial constructs that allow people to express a dimension as a number. This allows physicians to use arithmetic calculations to predict an experimental result. Very powerful invention. And a dimension is an arbitrarily chosen physical property.

In English, a number can be also called a "quantity"; hence, a "physical quantity" is a number of physical units.

For example: "this ball will fall for a quite short span of time" can be more precisely expressed as "this ball will fall 3 seconds". Note that in both cases dimension is time, but only in the latter case it could be computed.

The "dimensions are powers" confusion

Alas your book (maybe Elements of Physics judging from a cursory Google Books search) uses different definitions. It assumes a base of fundamental quantities. Say [mass, length, time]. In such base a unit of an area can be described as [0, 2, 0] and unit of acceleration as [0, 1, -2]. They don't call the  fundamental quantities "dimensions". They don't call these vectors of exponents "dimensions". Dimension is defined as a single exponent that comprises that vector. Just one small number. Confusing, isn't it? As you see, with such definition, "dimensions are powers". To obtain a unit of area, you need to assign a dimension of 2 to a unit of length. So one of dimensions is literally 2 and other dimensions are zero. Personally, I think such definition of "dimension" is utter crap, especially when talking to mathematicians.


what is the definition of $=$? what is the definition of $=$?

Above is the question that I would like to be answered, below are some of my thoughts.

I've been thinking about what it means to say

$A = B$

I came to this from thinking about why it makes no sense to say something like

$5 = \begin{pmatrix}1, 2, 3\end{pmatrix}$

My answer to myself was that $=$ is context dependent, it is a relation on a certain set and so only makes sense when the elements on either side of the $=$ are in the said set.

Now, say we are in $\mathbb{N}$, then how do we define $=$?
My initial thought was to say that $A = B$  if, given any proposition regarding A, $P(A)$, with truth value $\alpha$, the truth value for $P(B)$ will always be $\alpha$.

However, what happens now if we take $P(\star) $ as $ \star = B$? Do we then need a definition for $=$ in order for $P(\star)$ to make sense? Is this an issue?

In logic, equality is not something we define, it's a primitive notion.  If $A=B$ then $A$ has property $P$ if and only if $B$ has property $P$—this is probably the most important thing to know about equality and although it doesn't define equality, it is assumptions like this that shape our understanding of what equality "is", just like the axioms of set theory shape our understanding of what it means to say $A \in B$ even though membership is a primitive notion that is not defined in terms of other things.

The converse statement "if for every property $P$, $A$ has property $P$ if and only if $B$ has property $P$, then $A=B$" is not true in general. (Unless you allow the property $P$ to have a parameter—then as you point out it's trivially true because $P(x)$ can say $x =B$.)  It's true in structures like the natural numbers, where every object is definable, but other structures may have indiscernible elements that cannot be distinguished by any logical property.

Just in terms of math I would say the equality sign of A=B means that A and B have the same displacement from a fixed point of reference (and of course displacement here is a vector quantity).


Definitions of the Chinese Remainder Theorem The Chinese Remainder Theorem can be stated in a few ways:

(i) If $N = N_1N_2\cdots N_k$ and the $N_i$ are pairwise coprime we have a canonical isomorphism $$\mathbb{Z}/N\mathbb{Z} \cong \mathbb{Z}/N_1\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/N_k\mathbb{Z}$$
(ii) For any $a_i \bmod N_i, \quad  i = 1,\ldots, k$ there exists an $a$ such that $a \equiv a_i \bmod N_i$ for all $i$.

My question is how are these definitions equivalent and how is the isomorphism in the first definition true?

Also, the value of $a$ can be found via the Euclidean Algorithm. That is, put $M_i = N/N_i$. By assumption $M_i$ and $N_i$ are relatively prime. Then find $X_i$ with $X_iM_i \equiv 1 \bmod N_i$ and put $$a = \sum_{i=1}^k a_iX_iM_i$$

I am also confused how this uses the euclidean algorithm and finds the value of $a$.

To get a decent proof, we need to be a bit more formal.

Suppose $N_0 = \prod_{i=1}^k N_i$ where the $N_i$ are pairwise coprime.

Notationally, we will define $a..b$ to be the set $\{a, a+1, a+2, \dots, b\}$

Define $[n]_i = \{x \in \mathbb Z : x \equiv n \pmod{N_i} \}$.
Then  $\mathbb{Z}_{N_i} = \{ [n]_i : n \in \mathbb Z\}$.

Theorem. Define $f:\mathbb{Z}_{N_0} \to \bigoplus_{i=1}^k \mathbb{Z}_{N_i}$ by $f([x]_0) = ([x]_1, [x]_2, \dots, [x]_k)$. Then $f$ is an isomorphism.

Proof.

The proof that $f$ is a well-defined homomorphism can be found in just about any good group theory text. We will skip that because I am too lazy to type the whole thing out.

For $i = 1..k$, define $e_k \in \prod_{i=1}^k N_i$ by
$e_1 = ([1]_1, [0]_2, \dots, [0]_k)$
$e_2 = ([0]_1, [1]_2, \dots, [0]_k)$
$\phantom{e_8}\;\;\vdots$
$e_k = ([0]_1, [0]_2, \dots, [1]_k)$

We will show that, for $i \in 1..k$, there exists an $m_i \in \mathbb Z$ such that $f([m_i]_0) = e_i$.

We start by defining $M_1 = \dfrac{N_0}{N_1}$
Note that, for $i \in 2..k$, $M_i \equiv 0 \pmod{N_i}$.
This means that $f([M_1]_0) = ([M_1]_1, [0]_2, \dots, [0]_k)$.

Also, since the $N_i$ are relatively prime, $\gcd(M_1,N_1) = 1$. So there exists an $X_1 \in \mathbb Z$ such that $X_1 M_1 \equiv 1 \pmod{N_1}$. Let $m_1 = X_1 M_1$. It follows that $f([m_1]_0) = ([1]_1, [0]_2, \dots, [0]_k)$.

Using the same reasoning but changing the things that need to be changed, it follows that for $i \in 1..k$, there exists an $m_i \in \mathbb Z$ such that $f([m_i]_0) = e_i$.

Now it's pretty easy to show that $f$ is onto. Let
$([a_1]_1, [a_2]_2, \dots, [a_k]_k) \in \bigoplus_{i=1}^k \mathbb{Z}_{N_i}$.
Define $a = a_1 m_1 + a_2 m_2 + \cdots + a_k m_k$. Because $f$ is a homomorphism,

\begin{align}
    f([a]_0)
    &= f\left( \left[\sum_{i=1}^k a_i m_i \right]_0  \right) \\
    &= \sum_{i=1}^k f\left( [a_i m_i]_0  \right) \\
    &= \sum_{i=1}^k a_i f\left([m_i]_0  \right) \\
    &= \sum_{i=1}^k a_i e_i \\
    &= ([a_1]_1, [a_2]_2, \dots, [a_k]_k)
\end{align}

So $f$ is onto. Since $|\mathbb Z_{N_0}| = |\bigoplus_{i=1}^k \mathbb{Z}_{N_i}| = N_0$. It follows that $f$ must also be one-to-one.

If, for a prime $p$, $p^{\alpha_i}|m_i$, with $\alpha_i$ being the largest exponent with that property, then $p^{\alpha}|\operatorname{lcm}(m_1,m_2,\dots,m_k)$, where $\alpha=\max\{\alpha_i\}$. Similarly, the greatest common divisor of several numbers is the product of the largest powers of the primes that divide all the given numbers.

Associativity allows one to proceed a step at a time with an inductive argument without putting all eggs into a basket at once. Jumping at the opportunity I'll prove the most basic case of $k=2$.

  Theorem 1 : Two simultaneous congruences
  $$\left\{\begin{array}{cccc}n & \equiv & n_1 & (\mod m_1) \\ n & \equiv & n_2 &  (\mod m_2)\end{array}\right.$$
  are only solvable when $n_1\equiv n_2 \ (\mod gcd(m_1,m_2))$. The solution is unique modulo $lcm(m_1,m_2)$.
  (When $m_1$ and $m_2$ are coprime their gcd is $1$. By convention, $a\equiv b\ (\mod 1)$ holds for any integers $a$ and $b$.)

Proof:
By a generalization of Euclid's algorithm, there are integers $s$ and $t$ such that
$$tm_1+sm_2=gcd(m_1,m_2).$$
Since $n_2−n_1\equiv 0 \ (\mod gcd(m_1,m_2))$, for some $t'$ and $s'$,
$$t'm_1+s'm_2=n_2−n_1.$$
Then $n=t'm_1+n_1=−s'm_2+n_2$ satisfies both congruences in the theorem. This proves the existence of a solution.

To prove the uniqueness part, assume $n$ and $N$ satisfy the two congruences. Taking the differences we see that
$N−n\equiv 0 \ (\mod m1)$ and $N−n\equiv 0\ (\mod m_2)$,
which implies $N−n\equiv 0 \ (\mod lcm(m_1,m_2))$.

As was previously stated, a more general theorem can now be proved by induction.

  Theorem 2 : The simultaneous congruences
  $$\left\{\begin{array}{cccc}n & \equiv & n_1 & (\mod m_1) \\ n & \equiv & n_2 &  (\mod m_2) \\ & & \vdots & \\ n & \equiv & n_k & (\mod m_k)\end{array}\right.$$
  are only solvable when $n_i=n_j \ (\mod gcd(m_i,m_j))$, for all $i$ and $j$, $i\neq j$. The solution is unique $\mod lcm(m_1,m_2,\dots,m_k)$.

Proof :
Theorem 1 serves the initial step verification. Assume the theorem holds for $k$ congruences and consider $k+1$ of them.

$$\left\{\begin{array}{cccc}n & \equiv & n_1 & (\mod m_1) \\ n & \equiv & n_2 &  (\mod m_2) \\ & & \vdots & \\ n & \equiv & n_k & (\mod m_k) \\ n & \equiv & n_{k+1} & (\mod m_{k+1})\end{array}\right.$$

Let $s$ be a solution of the first $k$ equations. Then the congruence $n\equiv s \ (\ mod lcm(m_1,m_2,\dots,m_k))$ has a solution. (Observe that every solution of the latter also satisfies the first $k$ congruences.) To be able to apply the already proven Theorem 1, we need to show that
$$s\equiv n_k+1 (mod gcd(lcm(m_1,\dots,m_k),m_{k+1})).$$
Let's write $g_u=gcd(m_,m_{k+1})$, $u=1,\dots,k$. Then we know that $n_u \equiv n_{k+1}\ (\mod g_u)$ for these values of u. But $n_u=s+t_um_u$, for some $t_u$, implying $n_{k+1}=s+t_um_u\ (\mod g_u)$ so that

$$s=n_{k+1} \ (\mod g_u), u=1,2,\dots ,k.$$

If so,

\begin{align} s & \equiv n_{k+1} \ (\mod lcm(g_1,\dots,g_k))\\ & \equiv n_{k+1}\ (\mod lcm(gcd(m_1,m_{k+1}),\dots ,gcd(m_k,m_{k+1}))) \\ & \equiv n_{k+1}\ (\mod gcd(lcm(m1,…,mk),mk+1)),\end{align}

because $lcm(gcd(m_1,m_{k+1}),
\dots,gcd(m_k,m_{k+1})) =gcd(lcm(m_1,\dots,m_k),m_{k+1})$.
Thus the system

$$\left\{\begin{array}{cccl}n & \equiv & s & (\mod lcm(m_1,m_2,\dots,m_k)) \\ n & \equiv & n_{k+1} &  (\mod m_{k+1})\end{array}\right.$$

has a solution which is unique modulo $lcm(lcm(m_1,m_2,\dots,m_k),m_{k+1})=lcm(m_1,m_2,\dots,m_k,m_{k+1})$.

It also satisfies the whole set of $k+1$ congruences.

  Corollary : The simultaneous congruences
  $$\left\{\begin{array}{cccl}n & \equiv & n_1 & (\mod m_1) \\ n & \equiv & n_2 &  (\mod m_2) \\ & & \vdots & \\ n & \equiv & n_k & (\mod m_k)\end{array}\right.$$
  where all $m_i$'s are pairwise coprime has a unique solution modulo $m_1\cdot m2\cdots m_k$.

If some $m_i$'s are not mutually prime, a solution may not exist unless the corresponding congruences agree. For example, the system

$$\left\{\begin{array}{cccc}n & \equiv & 1 & (\mod 2) \\ n & \equiv & 2 &  (\mod 4) \end{array}\right.$$
has no solutions, while the system

$$\left\{\begin{array}{cccc}n & \equiv & 1 & (\mod 2) \\ n & \equiv & \pm 1 &  (\mod 4)\end{array}\right.$$
does.


Let $G$ be a finite simple group. Suppose that $A, B < G$, $G = AB$ and $A$ is an Abelian group. Is it true that $A \cap B=1 $? Let $G$ be a finite simple group. Suppose that $A$ and $B$ are proper subgroups of $G$, $ G = AB$ and $A$ is an Abelian group. Is it true that $A \cap B=1 $ ?

I checked it with some examples and it seems to be true; but I have no proof for it.

Set $C:= A \cap B$. On the contray suppose that $C \neq 1$. So $C^B \neq 1$.
We have $C^G = C^{AB}$. Since $A$ is Abelian, $C^G = C^B$.
But $ C^G \trianglelefteq G$. Thus $C^B=C^G=G$.
For every $c \in C$ and $b \in B$, we have $c^b = b^{-1}cb \in B$, so $G = C^B \le B$, a contradiction.

Counter-example: Let $G= Z_p$ with $p$ prime number. G is simple and abelian, $G=G\cdot G$ and $G \cap G=G\neq 1$.


Does continuity on rays imply measurability? Let us define a function on $\mathbb{R}^n$ by defining it on all lines crossing the origin $y=tx$,  $t\in\mathbb{R}$. It is defined to be continuous on these lines. Is it known to be measurable?

Let A be non-mesurable subset of $\left\lbrace x\in \mathbb{R}^n : \|x\| = 1 \right\rbrace$,

Then $B = \left\lbrace tx\in \mathbb{R}^n : x\in A ,  t\in\mathbb{R} \right\rbrace$

And now define the function f as :

$f(x) = 0$ if $x \not\in B$ and $f(x) = {\|x\|}$ if $x\in B$

then $f^{-1}(\left\lbrace 1 \right\rbrace) = A$, that is non measurable


Generating functions to solve number of integer solution problem If I have $x_1 + x_2 + x_3 =10$ with $1\leq x_1 \leq 5, \; 2 \leq x_2 \leq 6, \;3 \leq x_3 \leq 9$

I know that I compute $(t^1+\dots + t^5)(t^2 +\dots + t^6)(t^3+\dots +t^9)$ and look at the coefficient of $t^{10}$ to find number of integer solutions.

But below I have $3a_2,5a_3$

Say I have $a_1+3a_2+5a_3=33$ with $1\leq a_1 \leq 11,\;3\leq a_2 \leq 18,\; 5 \leq a_3 \leq 13$.

Do I simply change it to $9 \leq 3a_2 \leq 54$ etc

and compute $(t^1+t^2+\dots +t^{10}+ t^{11})(t^9+t^{12}+\dots +t^{51}+t^{54})(t^{25}+t^{30}+\dots +t^{60}+t^{65})$ and look at coefficient $33$?

  Meaning do I count the exponents up in the increased rate?


Correct answer, wrong method in improper integral problem. Note: this is a problem from a class, but my solution has already been submitted and the assignment is locked for all students so this shouldn't break any sort of academic honesty guidelines.
$$\int_{0}^{\ln{3}}\frac{e^x}{(e^x-1)^\frac{2}{3}}dx$$
I understand that this is an improper integral given $\frac{e^0}{(e^0 - 1)^\frac{2}{3}} = \frac{1}{0}$, but I seem to have come to the right answer without using any improper integration techniques. Through two layers of u-substitution $u = e^x$ and $v = u - 1$, my answer comes to $3 \cdot 2^\frac{1}{3}$. Every online calculator I've checked with confirms that this is correct, but I don't understand why it seems to be correct without any use of improper integration techniques. This hints at some disconnect I've got with improper integrals or u-substitution, and I'm really at a loss for what it might be.
Thanks for any insight!

Since the integral is improper at the lower bound, it is defined as $\lim_{\epsilon\to0} \int_\epsilon^{\ln 3}$. Let us look at what happens:
$$
\int_{\epsilon}^{\ln{3}}\frac{e^x}{(e^x-1)^\frac{2}{3}}dx
= \{ u=e^x \}
= \int_{e^\epsilon}^{3} \frac{u}{(u-1)^\frac{2}{3}} \frac{du}{u}
= \int_{e^\epsilon}^{3} \frac{du}{(u-1)^\frac{2}{3}} \\
= \{v=u-1\}
= \int_{e^\epsilon-1}^{2} \frac{dv}{v^\frac{2}{3}}
= \int_{e^\epsilon-1}^{2} v^{-\frac{2}{3}} \, dv
= \left[ \frac{1}{3}v^{\frac{1}{3}} + C \right]_{e^\epsilon-1}^{2} \\
= \frac{1}{3} \cdot 2^{\frac{1}{3}} - \frac{1}{3} (e^\epsilon-1)^{\frac{1}{3}}
.
$$
Taking limits we find that
$\frac{1}{3} (e^\epsilon-1)^{\frac{1}{3}} \to 0 = \frac{1}{3} (e^0-1)^{\frac{1}{3}}$
as $\epsilon \to 0,$ i.e. the primitive function is defined and continuous at $0$. And that is the reason it works.
Another example is the improper integral $\int_0^1 \ln x \, dx$. For this we get
$$
\int_\epsilon^1 \ln x \, dx = \left[ x\ln x - x \right]_\epsilon^1
= (1\ln 1-1) - (\epsilon\ln\epsilon-\epsilon).
$$
Here the lower bound term is not defined for $\epsilon=0$ and we can not just insert $\epsilon=0$ but really need to take limits (which result in $0$ so the full integral has value $-1$).

Let $u = e^x -1$.  Then $du = e^x\, dx$.  Your integral becomes
$$\int_0^2 {du\over u^{2/3}}.$$
This integral integrates at zero because the power is less than 1.


Expected number of rolls to get all colors on 6-sided die colored in with 3 colors If I have a die that has 3 red sides, 2 blue sides, and 1 green side, how many rolls do I expect until every color has appeared at least once?

I have run some tests and I’m getting numbers around 7.31, but clearly I’m looking for a mathematical solution.

Thanks in advance.

Label the possible states of the game according to which colors have previously been seen.  Thus we consider $8$ possible states $\{r,b,g\}$ where each symbol can be either $0$ or $1$ according to whether the associated color has been seen.  Similarly, we denote by $E[r,b,g]$ the expected number of throws it will take to finish from the state $\{r,b,g\}$.  The answer you want is $E=E[0,0,0]$.  We'll proceed by backwards induction, starting with the observation that $E[1,1,1]=0$.

I.  Missing one color.  Say we are in state $\{1,1,0\}$, so we are only missing Green.  We throw the die.  We finish if we get a Green, probability $\frac 16$.  With probability $\frac 56$ we stay in the state $\{1,1,0\}$.  Thus $$E[1,1,0]=\frac 16\times 1+\frac 56\times (E[1,1,0]+1)\implies E[1,1,0]=6$$

Similarly,$$E[1,0,1]=\frac 26\times 1+\frac 46\times (E[1,0,1]+1)\implies E[1,0,1]=3$$  And $$E[0,1,1]=\frac 12\times 1+\frac 12\times (E[0,1,1]+1)\implies E[0,1,1]=2$$

II.  Missing two colors.  Say we are in state $\{1,0,0\}$.  As before we roll the die and see that $$E[1,0,0]=\frac 26\times (E[1,1,0]+1)+\frac 16\times (E[1,0,1]+1)+\frac 36\times(E[1,0,0]+1)$$ $$=\frac {14}6+\frac {4}6+\frac 36\times(E[1,0,0]+1)\implies E[1,0,0]=7$$

Similarly, we get:  $$E[0,1,0]=\frac {26}4\;\;\&\;\;E[0,0,1]=\frac {19}5$$

Finally, $$E=E[0,0,0]=\frac 36\times (E[1,0,0]+1)+\frac 26\times (E[0,1,0]+1)+\frac 16\times (E[0,0,1]+1)=7.3$$

Okay, the idea is simple.

Let the first color you hit be $x$ with probability $pr(x)$. On your next throw you will get the a different color with probability $(1-pr(x))$. The probability that you will hit the next color will be geometrically distributed, so that your mean number of throws will be $ \frac{1}{1-pr(x)} $.

Then, let the second color be $y$ with probability $pr(y)$. The probability that will hit the third color will be $(1-pr(x) -pr(y))$. Again the number of throws you will need is geometrically distributed. Such that the mean number of times that you will have to throw will be $ \frac{1}{1-pr(x)-pr(y)} $.

Now you just need to use some combination theory to solve the problem, as the colors you hit the first, second, and third time can have a different order.

Br,
Carl


Show that a metric space has uncountably many disjoint dense subsets if every ball is uncountable. I was wondering how to show that a metric space $X$ has uncountably many dense subsets $Y_{\alpha}$ such that $Y_{\alpha_1} \cap Y_{\alpha_2} = \emptyset$ if $\alpha_1 \neq \alpha_2$, under the assumption that every ball in $X$ is uncountable. When $X = (a,b)$ is a non-empty interval, this is fairly trivial because we can just choose an  (uncountable) Hamel basis $H \subset \mathbb{R}$ when ${\mathbb R}$ is a vector space over $\mathbb{Q}$, and then let $Y_\alpha = \alpha {\mathbb Q} \cap X$ for $\alpha \in H$. But what about an arbitrary metric space $X$ where every ball is uncountable?

Claim 1: Let $\kappa$ be an uncountable cardinal and $X$ a metric space in which each open ball has size $\kappa$. Let us call such a space $\kappa$-homogeneous. Then $X$ can be divided into $\kappa$ many dense subsets.

Proof: Clearly $|X| = \kappa$ and hence $X$ has a basis of size $\kappa$ (use rational radii balls), say $\{B_{\alpha} : \alpha < \kappa\}$. Inductively construct $\{D_{\alpha} : \alpha < \kappa\}$ by putting, at stage $\alpha$, one point from each $B_{\beta}$, $\beta < \alpha$ into every $D_{\beta}$, $\beta < \alpha$.

Claim 2: Let $X$ be a metric space in which every open ball is uncountable. Then there is a family of pairwise open balls $U$ of $X$, such that the union of the family is dense in $X$ and each $U$ is a $\kappa$-homogeneous metric space for some uncountable cardinal $\kappa$.

Proof: Every open ball in $X$ contains a homogeneous ball as there is no infinite decreasing sequence of cardinals. So take a maximal family of pairwise disjoint homogeneous open balls in $X$.

Claims 1 + 2 give your result.

The fact that $X$ is a metric space was important here. There are examples of Hausdorff spaces in which every ball is uncountable yet there is no dense codense subset. Such spaces are called irresolvable.

We can define our sets by transfinite recursion.  Let's take the set of countable ordinals as our indexing set, and proceed as follows:

Let us take the set of all open balls in $X$, and using the axiom of choice, take one point from each of the open balls, and define that set of points to be $Y_1$.

For any countable ordinal $\alpha$, let us take the set of all open balls in $X$, and using the axiom of choice, define $Y_\alpha$ by taking one point from each of the open balls, but not choosing a point that's already been chosen for any of the previous sets $Y_\beta$ for $\beta < \alpha$ (we can do this because each open ball is uncountable, and we've only taken out countably many points from each open ball for our previous sets, since there are only countably many ordinals less than $\alpha$).

Since there are uncountably many countable ordinals, we're done.

EDIT:  As @user2566092 points out, you could accidentally pick all the points in some open ball for $Y_1$, so subsequent sets $Y_\alpha$ wouldn't be able to choose any points for that open ball.  This can be remedied as follows: use the axiom of choice to well-order the set of all open balls in $X$, and let the order type of that well-ordered set be $\gamma$.  Let us define $Y_1$ be transfinite recursion.  Choose an arbitrary point $y_1$ in $B_1$.  For any $\alpha < \gamma$, first try choosing a point $y_\alpha$ in $B_\alpha$ that has already been chosen as one of the previous points $y_\beta$ for $\beta < \alpha$.  If no such point exists, then choose any point $y_\alpha$ in $B_\alpha$.  And then define $Y_1$ to be the set of all $y_\alpha$ for $\alpha < \gamma$.  (The fact that $Y_1$ exists relies on the axiom of choice, of course.)

We can similarly choose points for any $Y_\alpha$ (where $\alpha$ is a countable ordinal) in the same manner, maintaining the condition from before that the points you choose for $Y_\alpha$ can't be the same as any of the points that have been chosen for the sets $Y_\beta$ for $\beta < \alpha$.

EDIT 2:  Here is a proof that my construction works, under the assumption that $X$ is a separable metric space and the continuum hypothesis holds.  (I'm not sure whether one or both assumptions can be dropped.)

Assuming the continuum hypothesis, a separable metric space must have cardinality less than or equal to aleph_1, so $X$ has cardinality at most aleph_1, and thus so does the set $B$ of open balls on $X$.  Using the axiom of choice, well-order the set $B$, and let $\gamma$ be the order type of the well-ordering.  Note that $\gamma$ is less than or $\omega_1$.

Since $X$ is separable, it's second countable, so there's a countable subset $C$ of $B$ which is a basis for the topology on $X$,  i.e. if $U$ is an open set and $x$ is a point in $U$, there's an element of $C$ which contains $x$ and is a subset of $U$.  Let $\delta$ be the smallest ordinal $\alpha$ such that the set of all $B_\beta$ for $\beta < \alpha$ contains C.  Then $\delta < \gamma$, so $\delta$ is a countable ordinal.

Let $y_1$ be any element of $B_1$.  For any ordinal $\alpha < \gamma$, if any of the previous elements $y_\beta$ for $\beta < \alpha$, then let $y_\alpha$ be the $y_\beta$ in $B_\alpha$ with the least index.  If no such point exists, let $y_\alpha$ be any element of $B_\alpha$.  Then the set of all $y_\alpha$ for $\alpha < \delta$ is the same as the set of all $y_\alpha$ for $\alpha < \gamma$, because $C$ is a basis for the topology on $X$.  Thus this set $Y_1$ is a countable dense subset of $X$.  (It's countable because $\delta$ is countable.)  Since it's countable, it doesn't contain any open balls, since every open ball is uncountable.

We can do a similar construction for each $Y_\alpha$ (where $\alpha$ is a countable ordinal), maintaining the condition from before that the points you choose for $Y_\alpha$ can't be the same as any of the points that have been chosen for the sets $Y_\beta$ for $\beta < \alpha$.  Then each $Y_\alpha$ will be countable, and there are only countably many ordinals less than a given countable ordinal, so we'll never cover all the points in an open ball in the stages before we get to any $Y_\alpha$.


Can finite induction be independent of mathematical induction? Suppose we have a property $P(n)$ pertaining to natural numbers and $P(n) \to P(n+1)$.
We know by mathematical induction that $P(N)$ implies $P$ being true for all natural numbers greater than or equal to $N$.
It is an obvious consequence of this that for a finite number of natural numbers greater than or equal to $N$, $P$ is also true. I am wondering, however, that whether this "finite induction" can be applied even if we didn't know mathematical induction principle. This might seem trivial, just like $P(1) \to P(2)$, $P(2) \to P(3)$ and so on, but I don't know any logical rules approving this...

We can prove the meta-theorem that in any logical system which lets you derive $Q$ from $P$ and $P \implies Q$, there is a proof of any given $P(n)$ from a base case less than $n$ and the induction step $P(n) \implies P(n+1)$. It is not necessary that the logical system have induction as an axiom.
In that logical system, such proofs will look like

Prove $P(1)$.
Prove $P(1) \implies P(2)$.
Prove $P(2) \implies P(3)$.
Prove $P(3) \implies P(4)$.
Prove $P(4) \implies P(5)$.
Conclude $P(2)$, then $P(3)$, then $P(4)$, then $P(5)$ by modus ponens.

It will be able to prove any given $P(n)$, and the meta-theorem tells us that it will be able to prove any given $P(n)$. Without the axiom of induction, it will not be able to prove $\forall n\ge 1\, P(n)$.

Induction will hold an any set $X$ (possibly finite) with $x_0\in X$ (the "first" element) and $f: X \to X$ (the "successor" function) such that:
$~~~~~~X=\{ x_0, f(x_0), f(f(x_0)), \cdots \} $
We have:
$~~~~~~\forall P\subset X: [x_0\in P \land \forall a\in P:[ f(a)\in P] \implies P=X]$


Square root of a matrix not differentiable if $A$ is singular Consider the map
$$\phi:M_n(\mathbb{R}) \to M_n(\mathbb{R}), \qquad A \mapsto \sqrt{A^T A}$$
I claim that this map is differentiable if and only if $A$ is invertible.
Just from the inverse function theorem, it can be seen that the map $\phi$ is differentiable at any invertible matrix. Now I need to show that this map is not differentiable at singular matrices. I tried to do the singular value decomposition but it did not work. Any help or reference will be appreciated.
Thanks :)

Let $f(A)= \sqrt{A^TA}$. When $A$ is singular, let $A=U\pmatrix{S\\ &0}V^T$ be a singular value decomposition of $A$ and let $H=U\pmatrix{0\\ &I}V^T$. Then the directional derivative of $f$ along the direction $H$, i.e.
$$
\lim_{t\to0}\frac{f(A+tH)-f(A)}{t}=\lim_{t\to0}\frac{|t|}{t}V\pmatrix{0\\ &I}V^T,
$$
does not exist. Hence $f$ is not differentiable at $A$.

In the easiest case $n=1$ we have $ \phi(A)=\sqrt{A^2}=|A|.$
$\phi$ is not differentiable at $A=0.$


Prove that $mn  0$ and $n  0$. Prove that $mn < 0$ if and only if $m > 0$ and $n < 0$ or $m < 0$ and $n > 0$.
$m,n$ element of integers

Just starting out teaching myself discrete math still really bad at proofs, any help/advice on how to think/go about this would be greatly appreciated.

We need to make a few assumptions about how you have integers defined, but a very safe assumption is that you have defined $(\Bbb Z, +, \times, \leq)$ as an ordered commutative ring with identity constructed from the natural numbers in the usual way.

In particular, as part of the very definition of $(\Bbb Z,+,\times,\leq)$, we have the following (among many more properties, these are just the important ones necessary for my proof):

(In all of the following, $a,b,c$ and $x$ are arbitrary integers)

Properties of $+$

$0+a=a$ for any $a$
If $a$ and $b$ are integers, then $a+b$ is an integer

Properties of $\leq$ (and $<$)

If $a\leq b$ then $a+c\leq b+c$
If $0\leq a$ and $0\leq b$ then $0\leq a\times b$
$0\leq 1$

Properties of $-$

$(-x)$ is the unique number such that $(-x)+x=0$
$-(-x) = x$

Properties of $\times$

$a\times b = b\times a$
$(-x)=(-1)\times x$
$a\times b = 0$ if and only if $a=0$ or $b=0$
If $a$ and $b$ is an integer, then $a\times b$ is an integer

Starting the desired proof, we approach via contrapositive for the forward implication.  That is, we wish to prove that if ($0\leq m$ and $0\leq n$) or ($m\leq 0$ and $n\leq 0$) then $0\leq m\times n$.  In doing so, we will have proven that $m\times n<0$ implies $m<0$ and $0<n$ or $0<m$ and $n<0$.

Suppose $0\leq m$ and $0\leq n$.  Then by definition $0\leq m\times n$, so we are done.

Suppose $m\leq 0$ and $n\leq 0$.  Then $m+(-m)\leq 0+(-m)$ and therefore $0\leq -m$.  Similarly $0\leq -n$.  Then $0\leq (-m)\times (-n) = ((-1)\times m)\times ((-1)\times n) = (-1)\times (-1)\times m\times n = m\times n$.

Thus if $m\leq 0$ and $n\leq 0$ or if $0\leq m$ and $0\leq n$, we have $0\leq m\times n$

For the remaining direction for the proof, we wish to show that if $0<m$ and $n<0$ or if $m<0$ and $0<n$ then $m\times n < 0$

Suppose $0<m$ and $n<0$.  Then $n+(-n)<0+(-n)=-n$.  Thus $0<m\times (-n)$.  Thus $0+(m\times n) < m\times (-n)+(m\times n) = m\times (-1)\times n + m\times n = -(m\times n)+(m\times n) = 0$.  Therefore $m\times n<0$

The proof for $m<0$ and $0<n$ is identical by relabeling which was $m$ and which was $n$.

1) If m < 0 and n >0 or m>0 and n < 0,then mn < 0.

2) If mn < 0, suppose m > 0 and n > 0,this gives mn > 0 which is contradiction.Hence ,one of them should definitely be less than 0.If both are,then contradiction arises again as product would be greater than 0


Evaluate a limit in terms of derivative If $f$ is differentiable at $x = a$, with $a > 0$, evaluate the following limit in terms of $f'(a)$

$$\lim_{x \to a}\frac{f(x) - f(a)}{\sqrt{x} - \sqrt{a}}$$

HINT

Use that

$$\lim_{x \to a}\frac{f(x) - f(a)}{\sqrt{x} - \sqrt{a}}=\lim_{x \to a}\left(\frac{f(x) - f(a)}{x-a}\cdot \frac{x-a}{\sqrt{x} - \sqrt{a}}\right)$$

Take the derivative (l’hopital rule) to get: ( since it is zero divided by zero)

$f’(x)(2)x^{0.5}$

This gives

$f’(a)(2)a^{0.5}$


Any divisor of $(n + 1)^p - n^p$ is $1 \textrm{ mod } p$ I'm trying to work my way through the following problem.

Let $n$ be a natural number, $p$ a prime, and $d$ a divisor of $(n + 1)^p - n^p$.
Show that $d \equiv 1 \textrm{ (mod } p \textrm{)}$.

I'm not sure where to start on this. I can see $(n + 1)^p - n^p \equiv 1 \textrm{ (mod } p)$ and ofcourse $(n + 1)^p - n^p \equiv 0 \textrm{ (mod } d)$. So definitely $p, d$ are coprime. Also $(n + 1)^p - n^p \equiv 1 \textrm{ (mod } n)$. This feels like it shouldn't be too hard a problem but I'm not entirely sure what I'm missing.

Let $q$ be a prime divisor of $(n+1)^p-n^p$.

Shall show $q\equiv 1\ mod \ p$

Clearly $(q,n)=(q,n+1)=1$ and hence we have in the field $\mathbb Z_q$
$$(n+1)^p=n^p \ in  \ \ \mathbb Z_q$$
$$\implies ((n+1)n^{-1})^p=1 \ in \ \mathbb Z_q$$

SO elementary group theory says $order \  ((n+1)n^{-1}))=p \ in \ \ \mathbb Z_q^*$

So $p|q-1$ and hence any divisor $\equiv 1 \ mod \ p$

Well, the binomial theorem gives
$$(n+1)^p - n^p = \sum_{i=0}^p {p\choose i} n^i - n^p = \sum_{i=0}^{p-1}{p\choose i} n^i.$$
But $p$ divides ${p\choose i}$ for each $i=1,\ldots,p-1$ and so taken modulo $p$ there is only one summand left $(i=0)$, which is $1$.


How to find the solution of the differential equation Find the solution of the differential equation
  $$\frac{dy}{dx}=-\frac{x(x^2+y^2-10)}{y(x^2+y^2+5)}, y(0)=1$$

Trial: $$\begin{align} \frac{dy}{dx}=-\frac{x(x^2+y^2-10)}{y(x^2+y^2+5)} \\ \implies  \frac{dy}{dx}=-\frac{1+(y/x)^2-10/x^2}{(y/x)(1+(y/x)^2+5/x^2)} \\  \implies   v+x\frac{dv}{dx}=-\frac{1+v^2-10/x^2}{v(1+v^2+5/x^2)}  \end{align}$$ I can't seperate $x$ and $v$.

Hint: It is an exact equation. Assume there is an differentiable function $f(x,y)$ such that $$f_x=x^3+xy^2-10x,~~~f_y=x^2y+y^3+5y$$ and then find the function. The solution is as $$f(x,y)=C$$

Your differential equation can be written as
$$ (x^{3}+x y^{2}-10 x)dx+(x^{2}y+y^{3}+5y)dy=0$$
So it is of the form $M(x,y)dx+N(x,y)dy=0$
Now, $$M_{y}=2xy$$ and $$N_{x}=2xy.$$
$\therefore$ The given differential equation is exact.
So its general solution is
$$\int M dx+\int (\text{Terms in $N$ which does not contain $x$})dy=C$$
$$\Rightarrow \int (x^{3}+x y^{2}-10 x)dx+\int 0 dy=C$$
$$\Rightarrow \frac{x^{4}}{4}+\frac{x^{2}y^{2}}{2}-5x^{2}=c$$
Now $y(0)=1$, we have $C=0$.

$\therefore$ The solution of given differential equation is
$\displaystyle \frac{x^{4}}{4}+\frac{x^{2}y^{2}}{2}-5x^{2}=0$.


Existence of algebraically closed and algebraic field over another field Let $k$ be a field. There exists an extension $k^a$ which is algebraic over $k$ and algebraically closed.

Proof: Let $E$ be an extension of $k$ which is algebraically closed and let $$k^a:=\bigcup N$$ where the union taken over $N$ such that $k\subset N\subset E$, $N$ is algebraic over $k$. Note that this union is not empty because $k$ belongs to this family.

1) Let's show that $k^a$ is algebraic over $k$. Take $\alpha\in k^a$ then $\alpha\in N$ for some $N$ which is algebraic over $k$. Hence $\alpha$ is algebraic over $k$. Right?

2) Let's show that $k^a$ is algebraically closed. Take any $f\in k^a[X]$ with degree $\geq 1$ then we can consider it as polynomial over $E$, since $E$ is algebraically closed then $f$ has root $\beta\in E$. Hence $\beta$ is algebraic over $k^a$.

Question 1: How to show that $\beta\in k^a$?

Question 2: Also it's not obvious to me but why $k^a$ is a field?

Would be very grateful for detailed answer, please.

EDIT (This is the proof for the theorem which professor Lubin gave): We want to show that $k^a$ is algebraically closed. Take any monic polynomial $f(X)\in k^a[X]$, namely $f(X)=X^m+z_{m-1}X^{m-1}+\dots+z_1X+z_0$ then we can consider $f$ as an element of $E[X]$ but $E$ being algebraically closed means that $f(X)$ has root $\beta\in E$. Also $f(X)\in k(z_0,z_1,\dots,z_{m-1})$. Since $\beta$ is algebraic over $k(z_0,z_1,\dots,z_{m-1})$ then $[k(z_0,z_1,\dots,z_{m-1},\beta):k(z_0,z_1,\dots,z_{m-1})]<\infty$. Since each $z_i\in k^a$ then each $z_i$ is algebraic over $k$ and hence $[k(z_0,z_1,\dots,z_{m-1}):k]<\infty$.

Thus $[k(z_0,z_1,\dots,z_{m-1},\beta):k]<\infty$. Note that $$[k(z_0,z_1,\dots,z_{m-1},\beta):k]=[k(z_0,z_1,\dots,z_{m-1},\beta):k(\beta)][k(\beta):k].$$ It follows that $[k(\beta):k]<\infty$ and it follows that $\beta\in k^a$. We are done.

I would use a slightly different strategy, defining $k^a$ to be the union of all finite extensions of $k$ in the universe $E$. (And you know that finite implies algebraic.) Now, the union of two fields $F_1$ and $F_2$ is not ordinarily a field, but this set is contained the compositum $F_1\vee F_2$, which you may have seen defined as the smallest field containing both $F_1$ and $F_2$, or equivalently as $F_1(F_2)$. In any case, $F_1\vee F_2$ is obviously finite over $k$.

Under this definition, $k^a$ is a field, ’cause if you have elements $z,w\in k^a$, each is in a finite extension, and $z+w, zw\in k(z,w)$ (you can fill in the rest), and every element is algebraic over $k$, so the field itself is algebraic over $k$.

Algebraically closed? Take a monic polynomial $f(X)=X^m+z_{m-1}X^{m-1}+\cdots+z_1X+z_0\in k^a$, in fact $f\in k(z_0,\cdots,z_{m-1})[X]$, so that every root $\beta$ of $f$ has $[k(\beta):k]<\infty$, and thus $\beta\in k^a$.

If $f\in k^a[X]$, then $f\in N[x]$ for some algebraic extension $N$ of $k$. Then $f$ has a root $\beta$ in $E$. But then by definition, $\beta$ is algebraic over $N$, and thus algebraic over $k$. It follows that $N[\beta]$ is algebraic over $k$, so that $N[\beta]\subset k^a$ and thus $\beta\in k^a$.

For your second question, if $\alpha\ne 0$ is in $k^a$, choose some algebraic extension $N/k$ with $alpha\in N$; then $\alpha^{-1}\in N\subset k^a$. The others are similar.


Why not the domain is just $ ( - \infty , -3) $ of $ \sqrt{2-x} - \frac{1}{9-x^2} $ Consider

$$ f(x) = \sqrt{2-x} - \frac{1}{9-x^2} $$

Now for the radical to be definable,

  $$ 2-x \ge 0 $$

And for the fraction to be definable,

  $$ 9-x^2> 0 $$

SO , the number line of solution set looks like:

Where red line is for radical's definability & orange for fraction's.

Now,

The result from above stuff is

  $$ (- \infty , -3) $$

EDIT:
I got that the second inequality's solution set was wrong.

Its

  $$ (-3,3)$$

So final solution plot:

Which is  $ \color{red}{wrong} $

Why?

The fraction is defined if and only if $9-x^2\ne 0\Leftrightarrow x\ne\pm3.$

However, you have done the square root part correctly: $\sqrt{2-x}$ is defined if and only if $x\le 2$.

Because of this, we can conclude that $f(x)$ is defined if and only if $x\le 2$ and $x\ne -3$.

In one line:

$ x \le 2$ AND $(x \not = -3$ OR $x \not =3)$.

Hence $x \le 2$ AND $x \not = -3.$

$D= (-\infty, -3)\cup (-3,2]$.


Solve a linear function How do I solve this homework assignment?

For a linear function $y=f(x)$, $f(-3) = 25$ and $f(3) = 11$.

Determine $f(-20)$.

I know that with the values $f(-3) = 25$ and $f(3) = 11$ I am suppose to derive a formula and the plug in $f(-20)$ but how do I derive a formula from the values given?

Thanks!!

Hint: You don't have to derive the formula. Use the fact that the gradient of a linear function is always constant.

$$\frac{f(-20) - f(3)}{-20 - 3} = \frac{f(3) - f(-3)}{3 - (-3)}$$

Of course, if there is a need, you can derive similarly the formula of $f$. This is, by the way, my favourite method of deriving the equation of a linear function defined by two given points.

$$\frac{f(x) - f(3)}{x - 3} = \frac{f(3) - f(-3)}{3 - (-3)}$$

For a linear function, if you know its value at two points, you can fully determine the function. Do you know that general formula for a straight line in the plane? Write that down, and plug in (-3,25) and (3,11) to solve for the coefficients.


3D Camera transformation versus Object transformation A 3D camera is located at position $(eye_x,eye_y,eye_z)$ and looking at some object located at position $(0,0,0)$.

How can I achieve the same visual result but rotating the object (and having a static camera) instead of translating the camera around?

For example, if the camera eye is located at position $(2,2,2)$ looking at $(0,0,0)$, I guess I should rotate at least the object around Y axis 45° degrees. However, I don't know what calculations starting from the camera vectors (eye,at) should I apply in order to get there.

This is the camera transformation:

This is the desired transformation:

Prepare yourself for long and detailed explanation.

Let W be the transformation of the target object and V is the view matrix obtained from the transformation of the camera:

$
W=\begin{bmatrix}
mR_x&mR_y&mR_z&0\\
mU_x&mU_y&mU_z&0\\
mL_x&ML_y&mL_z&0\\
mT_x&mT_y&mT_z&1\\
\end{bmatrix} \ \ \ \ , \ \ \ \ \
V=\begin{bmatrix}
cR_x&cU_x&cL_x&0\\
cR_y&cU_y&cL_y&0\\
cR_z&cU_z&cL_z&0\\
-(cR\cdot cT)&-(cU\cdot cT)&-(cL\cdot cT)&1\\
\end{bmatrix}
$

where

mR, mU and mL are the local right, up and look vectors (axis), mT is the translation (at, in your case) vector of the target object;
cR, cU and cL are the local right, up and look vectors (axis), cT is the translation (eye, in your case) vector of the camera.

Since the camera always looks at the the object, the following should be used to obtain the cR, cU and cL vectors of the camera:

$ vU = (0, 1, 0)\\
cL = normalize(mT-cT) \\
cR = normalize(vU \times cL) \\
cU=normalize(cL\times cR)$

NOTE: "$ \times$" indicates a cross-product and "$\cdot$" indicates a dot-product.

To get the same image after projection, the multiplication of W (world) and V (view) matrices should remain the same (i.e. $W_0 \times V_0 = W_n \times V_n $). Note that I assumed the projection matrix does not change.

So you have initial world and view matrices, i.e. $W_0$ and $V_0$. Thus you also have $W_0 \times V_0$.

Then you moved the camera to somewhere else (i.e. cT is changed) but the target object is still at the origin. So you need to re-build the V matrix as I explained above. Let's name it as $V_n$.

Thus,

$W_n \times V_n=W_0 \times V_0 \\
\Rightarrow W_n=W_0 \times V_0 \times V_n^{-1}$

where $V_n^{-1}$ is the inverse of $V_n$; and $W_n$ is the new world matrix of the target object to guarantee the same image after projection.

If you extract the local axis vectors from $W_n$ then you can find the rotation angles compared to the initial state:

$
mR_n \cdot mR_0 = |mR_n| \ |mR_0| \ \cos\alpha\\
mU_n \cdot mU_0 = |mU_n| \ |mU_0| \ \cos\beta\\
mL_n \cdot mL_0 = |mL_n| \ |mL_0| \ \cos\theta
$

Note that the distance between camera and the target object should be the same if you want to only rotate the target object. Or else you'll also need to re-position it. The updated translation vector can be extracted from $W_n$ as well.

After struggling with the problem I found a solution that may not be optimal but at least solves the use case.

Here are the steps to rotate an object instead of moving the camera and have the same result:

Get the matrix of the camera:

Matrix.setLookAtM(cameraMatrix, 0, camera.xPos, camera.yPos, camera.zPos, camera.xView, camera.yView,
            camera.zView, camera.xUp, camera.yUp, camera.zUp);
Extract the rotation matrix:

float[] rotationMatrix = extractRotationMatrix(cameraMatrix);
// here it's possible to extract axis and angle from rotationMatrix
Apply transformation to object:

object.applyRotation(rotationMatrix);
// ModelMatrix=TranslationMatrix X ScaleMatrix X RotationMatrix


Complex conjugate and tensor product Let $V$ be a real vector space and $f : V \rightarrow V$ a linear endomorphism. Also, let $\sigma : \mathbb{C} \rightarrow \mathbb{C}$ be complex conjugation.

If $A$ is a real matrix, then it is clear how to interpret $A$ as a complex matrix and what the the claim $\sigma(A) = A$ means. I am confused how to write this for the abstract $f$.

The complex interpretation of $f$ is $f \otimes \mathrm{id} : V \otimes_{\mathbb{R}} \mathbb{C} \rightarrow V \otimes_{\mathbb{R}} \mathbb{C}.$ I would like to understand what the complex interpretation of "$\sigma(f) = f$" is. My thought was that applying $\sigma$ to $f \otimes \mathrm{id}$ should give us $f \otimes \sigma$, but I don't think this is true after looking at $f = \mathrm{id}_V$ (for example).

Sorry if my question is unclear.

The maps $\mathbb{C}\rightarrow\mathbb{C}$ are just multiplication by complex numbers: $z\overset{w}{\mapsto}wz$. Then $\sigma$ sends $f\otimes w$ to $f\otimes \sigma(w)$.


Intuition behind general solution of linear differential equation? If $y_1$, $y_2$ and so on till $y_n$ are solutions of a linear differential equation of order $n$, it is a theorem that $c_1y_1+ c_2y_2+\ldots +c_ny_n$ is the general solution for it. I have seen this in many places but I don't get how or why this is true. Would someone please give me some intuition in this result?

The obvious thing is that a linear combination of solutions is a solution. This is pretty much what it means for the equation to be linear and homogeneous: by assumption $L \left ( \sum_i c_i y_i \right ) = \sum_i c_i L(y_i)$ and the equation is $Ly=0$ so if each $L(y_i)=0$ then $L(y)=0$. Note that a linear inhomogeneous equation doesn't have this property, but it can be addressed in a similar way by summing up the general solution to the homogeneous problem with any particular solution to the inhomogeneous problem.

The less obvious thing is that the solution space has dimension $n$, so that if your $y_k$ are linearly independent then you have the general solution. To me the intuition for this is to parametrize the solutions by initial conditions at some arbitrary point; if you assume uniqueness of solutions to IVPs then it is apparent that the space of initial conditions has dimension $n$ so the same is true of the space of solutions. The problem then reduces to showing that a linear ODE of order $n$ with sufficiently nice coefficients and forcing has unique solutions to its IVPs.

A homogeneous linear differential equation of order $n$ looks like:
$$
\sum_{k = 0}^n c_ky^{(k)} = 0
$$
where $y^{(k)}$ is the $k$-th derivative of $y$ and $y^{(0)} = y$ by convention.
Observe that we may recursively define $u_0 = y$ and $u_k = u_{k - 1}'$ so that $u_k = y^{(k)}$. With this definition, the original equation becomes
$$
\sum_{k = 0}^n c_k u_k = 0
$$
Rearranging, we have
$$
u_{n} = -\sum_{k = 0}^{n-1}  \frac{c_k}{c_n} u_k
$$
Note that $c_n \neq 0$ by assumption; otherwise, the original equation would have had at most order $n - 1$. Defining $d_k = -\frac{c_k}{c_n}$ we can further rewrite the above as
$$
u_{n - 1}' = \sum_{k = 0}^{n-1} d_k u_k
$$
Meanwhile, we have the $n - 1$ other equations
$$
u_{k - 1}' = u_k
$$
with $1 \leq k \leq n - 1$. This system of equations can be transformed to the matrix equation
$$
\begin{bmatrix}
u_0' \\
\vdots\\
u_{k - 1}'\\
\vdots\\
u_{n - 1}'
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & \dots & 0 & 0 & 0 & \dots&0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & \dots & 0 & 1 & 0 & \dots & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
d_0 & d_1 & d_2 & \dots & d_{k - 1} & d_k & d_{k + 1} & \dots & d_{n - 1}
\end{bmatrix}
\begin{bmatrix}
u_0 \\
\vdots \\
u_k \\
\vdots \\
u_{n - 1}
\end{bmatrix}
$$
Define $\bf{u} = \begin{bmatrix} u_0 & \dots & u_{n - 1} \end{bmatrix}^T$ and $\bf{u}' = \begin{bmatrix}u_0' & \ldots & u_{n - 1}' \end{bmatrix}^T$. Let $A$ be the $n \times n$ matrix above, so that
$$
\bf{u}' = A\bf{u}
$$
Now $\bf{u}$ is a solution of the above equation with $u_0 = y$ if and only if $y$ is a solution of our original differential equation.
We will let $t$ be the variable with respect to which we have been differentiating. Thus, more explicitly, we have
$$
\bf{u}'(t) = A\bf{u}(t)
$$
Define
$$
\bf{v}(t) = e^{tA}\bf{v}(0)
$$
(See https://en.wikipedia.org/wiki/Matrix_exponential if the matrix exponential is unfamiliar). One can check that (the proof is identical to the proof that $\frac{d}{dt} e^{tx} = xe^{tx}$):
$$
\bf{v}'(t) = Ae^{tA} \bf{v}(0) = A\bf{v}(t)
$$
So $\bf{v}(t)$ as we have defined it is a solution to the matrix equation with initial condition $\bf{v}(0)$. Suppose $\bf{u}(t)$ is another solution to the matrix equation with $\bf{u}(0) = \bf{v}(0)$. Put $\bf{a}(t) = e^{-tA} \bf{u}(t)$. Then
\begin{align}
\bf{a}'(t) &= -Ae^{-tA} \bf{u}(t) + e^{-tA}\bf{u}'(t)
\\&= -Ae^{-tA} \bf{u}(t) + e^{-tA}A\bf{u}(t)
\\&= e^{-tA}(-A + A) \bf{u}(t)
\\&= 0
\end{align}
So it must be the case that $\bf{a}(t)$ is constant. But $\bf{a}(0) = e^{-0A}\bf{u}(0) = \bf{u}(0) = \bf{v}(0)$. Thus, $e^{-tA} \bf{u}(t) = \bf{a}(t) = \bf{v}(0)$, which shows $\bf{u}(t) = e^{tA} \bf{v}(0) = \bf{v}(t)$. Thus, $\bf{v}(t)$ is the unique solution to the matrix equation with initial condition $\bf{v}(0)$.
Now, if $y$ solves the original differential equation, then $\bf{u}$ with $u_0 = y$ solves the matrix equation. That means, by the above remarks, that $\bf{u}(t) = e^{tA} \bf{u}(0)$. Let $y_1(t) , \ldots , y_n(t)$ be the entries across the first row of $e^{tA}$. Then, reading off from $\bf{u}(t) = e^{tA}\bf{u}(0)$, we find
$$
y(t) = u_0(t) = \sum_{k = 0}^{n - 1} u_k(0) y_{k + 1}(t) = \sum_{k = 0}^{n - 1} y^{(k)}(0) y_{k + 1}(t)
$$
From this, we see that every solution to the differential equation is in the span of $y_1, \ldots, y_n$, with the coefficients determined by the initial conditions of $y$.
But it is worth noting that the $y_1, \ldots, y_n$ were not arbitrary -- they were just entries of $e^{tA}$. If we are handed an arbitrary set of solutions $f_1, \ldots, f_n$ how can we be sure that they span? Since the differential equation is linear (by definition) its solutions form a vector space. Because we showed that every solution is in the span of $n$ particular solutions, the dimension of the vector space can be no greater than $n$. Thus, any set of $n$ linearly independent (in the sense that no linear combination of the functions over $\mathbb{R}$ is identically $0$) must also span the solution space. This is just a restatement of the fact that in an $n$ dimensional vector space, a set of $n$ linearly independent vectors must span the space.
Concluding Remarks
This has been a long post, so let me summarize. We began with a homogeneous linear differential equation of order $n$. We transformed it to a matrix equation. We showed that the matrix equation has unique solutions for every choice of initial conditions. Using this solution of the matrix equation, we showed that any solution to the original differential equation had to live in the span of $n$ (not necessarily distinct!) solutions.
It all comes down to the fact that we have existence and uniqueness theorems for solutions of homogeneous linear differential equations and that the initial condition of a solution is specified by $n$ constants.


Measure Theoretic Definition of a Random Variable I am struggling a little with the definition of a RV: Let $(\Omega,F),(\Omega',F')$ be two event spaces. Then every mapping: $X:\Omega \to\Omega'$ is a RV provided $X^{-1}A' \in F,~~~ \forall A' \in F'$.

So my understanding of this: Let $T:X \to Y$ and $S\subseteq Y$. $T^{-1}(S)$ is the preimage of S under T. Now, $T(T^{-1}(S)\subseteq S $ and not necessarily:$T(T^{-1}(S)= S $. However, in the case of a random variable, we must have equality.

Is this the correct way to think about it?

The underlying idea ( in very practical terms) is simple. Suppose you are a statistician working with some survey. Lets suppose the survey has some questions about age, but only ask the respondent to identify his age in some given intervals, like $[0,18), [18, 25), [25,34), \dots $. Lets forget the other questions. This questionnaire defines an "event space", your $(\Omega,F)$. The sigma algebra $F$ codifies all information which can be obtained from the questionnaire, so, for the age question (and for now we ignore all other questions), it will contain the interval $[18,25)$ but not other intervals like $[20,30)$, since from the information obtained by the questionnaire we cannot answer question like: do the respondents age belong to this interval or not? More generally, a set is an event (belongs to $F$) if and only if we can decide if a sample point belongs to that set or not.

Now, let us define random variables with values in the second event space, $(\Omega', F')$. As an example, take this to be the real line with the usual (Borel) sigma-algebra. Then, an (un-interesting) function which is not a random variable is $f: $"respondents age is a prime number", coding this as 1 if age is prime, 0 else. No, $f^{-1}(1)$ do not belong to $F$, so $f$ is not a random variable. The reason is simple, we cannot decide from the information in the questionnaire if the respondents age is prime or not!  No you can make more interesting examples yourself.

I saw this kind of introduction first in the very good book by Peter Whittle "Probability via expectation" (Springer).


If $S$ is an affine independent subset of $R^n$ and $x \notin$ the affine span of $S$ then $S \cup \{x\}$ is affine independent. If $S$ is an affine independent subset of $R^n$ and $x \notin$ the affine span of $S$ then $S \cup \{x\}$ is affine independent.

I start by letting $S=\{a_0, \dots, a_m\}$ and therefore for by the definition of linear independence:
$$(\forall \{t_i\})[\sum_{i=1}^m t_i(a_i-a_0)=0 \Rightarrow t_i = 0, \forall i]$$  So then I need to show that $S \cup \{x\} = \{a_0, \dots , a_n, x\}$ is affine independent, or $$(\forall \{j_i\})[\sum_{i=1}^m j_i(a_i-a_0) + j_x(x-a_0)=0 \Rightarrow j_i = 0, \forall i]$$

But I'm a little confused as to how I'd continue from here after letting $\{j_i\}$ be arbitrary and assuming that $\sum_{i=1}^m j_i(a_i-a_0) + j_x(x-a_0)=0$.

Anyone have any ideas?

Let $x$ be outside the span of $S$. Suppose that $\alpha_1 v_1 + \dotsb + a_n v_n - \beta x = 0$ with $S = \{ v_1, \dotsc, v_n \}$ and $\alpha_1 + \dotsb + \alpha_n - \beta = 0$.

Suppose that $\beta \neq 0$, then $x = \frac{\alpha_1}{\beta} v_1 + \dotsb + \frac{\alpha_n}{\beta}$, and since we supposed $\alpha_1 + \dotsb + \alpha_n - \beta = 0$, we deduce that $\frac{\alpha_1}{\beta} + \dotsb + \frac{\alpha_n}{\beta} = 1$. This contradicts the hypothesis that $x$ is not in the span of $S$.

Therefore $\beta = 0$, we have $\alpha_1 v_1 + \dotsb + \alpha_n v_n = 0$ and $\alpha_1 + \dotsb + \alpha_n = 0$. The affine independence of $S$ implies that $\alpha_i = 0$ for all $i$. From this we deduce that $S \cup \{  x\}$ is affinely independent.

Starting from

$ \sum_{i=1}^m j_i(a_i-a_0) + j_x(x-a_0)=0  $, you could write $ \sum_{i=1}^m j_i(a_i-a_0) = - j_x(x-a_0)  $

Now, the left-side term is a combination of elements in $S$ and therefore is an element of $span(S)$. But, $x$ being not in $span(S)$, the right-hand term is not in $span(S)$ unless $j_x = 0$. Given that the two terms are equal, they should both belong to the same vector spaces; in particular, they should belong to $span(S)$.

Hence, the right hand term is $0$, and you can conclude that all $j_i$ are $0$ from here.