Not to mention, that keeping notes publicly does make wanders to being neat in thoughtful way, so – lets begin.

Suppose we hear the description of a man: “He is strong and with a loud strong but assuring voice”.

Who is this man, do we want to guess – is he a librarian or a military drill sergeant?

Is it 50%/50 chance, that given description of a man, is either librarian or sergeant?

We would more likely to guess that he is sergeant, because it is a way of more stereotypical sergeant, than librarian, at least one can make intuitively such a guess. Which is of course wrong.

To make **correct guess, one have to observe here actual ratio of librarians/sergeants** in our set of

It is important to observe that, even though description sounds much more like sergeant, if there is 1 sergeant in our bunch, and 99 of librarians, then even given the description, it is still much more probably that we are talking about strong and loud librarian, than anyone else.

So, lets say that probability of meeting sergeant in our bunch is , or in other words – for each sergeant we have four librarians.

Excellent, we’ve got our , now lets invent some values for likelihood of sergeants and librarians being with above mentioned properties.

Since those values are synthetic, lets invent that and

We are naturally here interested in* C=sarge,* and we require one more value, , which means that given any random guy we meet, chances of fitting the description is 29%, in case of geogebra app below this value is calculated from values mentioned above.

Now we have everything required, lets cook answer to the question: **“We’ve met guy from our bunch, he is strong and loud. What chance he is army sergeant?”, **or mathematically speaking

Estimate can be found by Bayes formula:

Or in our case

Application below tries to visualize behavior of event. Ie meeting *C*, after *x* is true.

Application has three handle to operate, those would be dots on top, left and right of the square.

Top one divides our set into *C* and *not C, *dot on the left chooses chance of meeting *x *in *C*, and dot on the right chooses the chance of meeting *x* in *not C* bunch.

Horizontal dotted line is our , chance of strong and loud member in our group – automatically generated.

And most important – vertical dotted line, which implies , chance of loud and strong being sergeant.

Следуюший, вполне логичный шаг, после создание видео-вопроса.

См. следующую статью

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In other words, we could literally assign natural number indexes to only one of each member of the set.

So, we have with cardinality , and with cardinality

Then cartersian product (set of all possible pairs from A and from B) – is also infinite, but countable. I mean, for , we can assign single natural number for each member of P.

Lets prove it. Suppose we injections *f* and *g, *so that and

Lets define function this way:

Then for and , because of unique division into primes and *f,g* injectivity **(a,b)=(c,d)**.

Therefore there is injection from *p(a,b)* to natural numbers, and cartesian product is infinitely countable.

– Then from naturals, lets construct

– Then from integers, lets construct

– And finally from rationals –

Mathematician Guiseppe Peano proposed next axiomatic definition at 1889, based on naive set theory(set, member of the set, and equivalence relation).

In this case, he also proposed such concept as *follower, *and thus Peano’s system (N,0,*) can be described with next axioms:

- is set, and
- For each member there is only one
*follower*n’ - Member 0 is not follower of any member in
- If and m’=n’, follows that m=n
- If , and if , then

It should be noticed, that we can define *less*(<) relation with natural numbers:

Lets define ~ equicalence relation on natural numbers in next maner:

Let

Therefore

Each rational x can be presented in a form x=a/b where and

And in a same way, like we received integers from naturals, we can

Therefore

Most interesting case in my opinion, first we can observe, that equation with integers 2 and 7 has no integer solution for x: 2x=7

However if then there is problem no more. x=7/2=3.5, which is rational numbers.

But sometimes rational numbers aren’t enough. For exampe x²=2 has no x in rational numbers.

To solve this problem, we define new number, lets call its

We can use it with rationals to get more numbers, etc

Thus we got extended rationals –

But that doesn’t give is solution to x²=3, x³=3,x³=9 etc. List is long,but we can define set of algebraic number (well, its not all algebraic numbers, we need complex numbers too).

Algebrain numbers are solution of all polynomials with integer coefficients

Thus we got real alreabric numbers –

Did we get all real numbers? No. Actually real algebraic numbers are countable set(all n-tuples {for finite} n of countable set is also countable set), and real numbers is not.

Except real alrebraic numbers, there are also transcendentals, here are two of them

π=3.1415.., e=2.7182

Those are exotic examples, but in reality – there is much, much more transcendental numbers than algebraic numbers.

Let us meet, what is called **Dedekind cut**.

Let there be a rational number set, graphically speaking:

But, our line doesn’t contain anything, except rational numbers!

I.e. 1,-4,⅗ are there, but *π *isn’t there. Neither √*2.*

So, lets make two sets out of rationals. Let

and Let

Ok, what if x=√*2?*

But.. but.. there is no , so

So, there is hole in rationals. It will, of course, happen with all irrational numbers. √*2,√3 *etc

Which makes bijection between all real numbers and all possible Dedekind cuts of rational numbers.

To makes things easier, way we define upper and lower sets, it is possible to refer to a cut using only lower set.

Dedekind cuts introduce relations to real numbers, which is ordered set.

So for *x* and *y* in real numbers,one thing is true: either *x<y*,*x>y* or *x=y*

Elementary cellular automata is essentially string of two symbols, lets call them 0 and 1, equipped with rules (and starting configuration, of course)

Rule takes symbol in the position, as well as left and right neighbors. Ie – when calculating symbol of the cell, we should consider 3 cells, cell to the left, cell itself and cell to the right.

Ie – suppose we have set of rules(*):*111 - 0**110 - 1**101 - 0**100 - 1**011 - 1**010 - 0**001 - 1**000 - 0*

Or graphically:

So if we have **10011** for starting configuration, calculation goes like this(from top to bottom):

1 |
0 |
0 |
1 |
1 |

(0) |
1 |
1 |
1 |
(1) |

(1) |
1 |
0 |
0 |
(1) |

*(italics)* means that we dont know which symbol is to the left and right to the starting configuration, for simplicity I consider here absent symbols to be *0*s (ie …*00000***10011***0000*…)

Now, what is 90, which is called *Wolfram rule*.

It is obvious, that 3 cells define state, and since we have two symbols, all possible combinations are , and since we have two possible symbols for each combination – rule of CA can be expressed as (0,2⁸=256) number.

90 in binary would be (*), which is enough to describe elementary ca behavior.

]]>Roughly translated: “*Define real axis ℝ Lebesque outermeasure on ℝ² surface*“

Lets get into it. Bit intuitively first.. Suppose we have segment of a line* (a,b)*

segment (*s*) itself can be expressed as

Actually, lets forget about *s*, lets consider without segments, ie

Then outer measure can be expressed as union of open sets:

for all

Where intervals are:

Thus length of intervals are from one integer to another – ,.. all integers

Height of intervals are

Ie , where biggest height is , which is quater of epsilon.

Therefore

or shorter: and for all

It is obvious from here

This article was inspired by **Margarita**.

I think I’d better come back to good old opiskelu.org, instead of spamming facebook page.

Oh yea, it all seems to work now.. kinda enjoy new editor for WordPress. So – here are plans for the future:

- Lebesque integral and measure, few notes regarding course and exam (english/finnish)
- Cellular Automata (english)
- Bitcoin shop with tor in mind (english)
- Немножко политики/russian politics (russian/english)

There. Lebesque integral notes&measure is coming today.

CA – on this weekend, based on university lecture.

Bitshop – next week.

Russian politics – when feel like it.

I believe, it would be quite useful to write out my research, here.

Ok, lets roll with project name [olleh], which later could be found in Github.

Soo.. all in all, let us begin.

`mkdir olleh`

`cd olleh`

Then, for following convenience, virtual enviroment is required:

`python3.6 -m venv venv`

Lets also make git repository

`git init`

Lets add out VENV to the git

`git add venv/`

I’d like to visualize storyline from my perspective operating on barber paradox, and why it is somewhat nonsense question.

Paradox states: The barber is the “one who shaves all those, and those only, who do not shave themselves.” The question is, does the barber shave himself?^{[Reference Wikipedia]}

But wait a minute, what is “all those, and those only, who do not shave themselves”?

It sounds like a group, or set. Many objects, which by some condition can be seen as a single entity.

Now if to think, that barber is inside redbox, and it is possible to “draw” him from the superset “ALL” by predicate “those who doesnt have themselves”, that indeed leads to quite amusing but useless paradox.

ZFC theory avoids fallacy here, and points that there is not enough reason to consider barber as possible candidate for “all those” set.

Giving it a little bit formal approach, one should be careful defining set and deriving subsets. It is quite easy to imagine self set of all real things, ie universe. However when one considers notion of , it is very obvious that this concept is far from reality. After all, “all” implies “all real” things.

]]>1,1,2,3,5,8,13,21…

Ne ovat Fibonacci luvut, niitä saadan seuraavalla tavalla.

Ensimmäiset kaksi ovat 1 ja 1.

Jokainen seuraava saadan laskemalla yhteen kaksi edellistä:

1,1

**1,1**,2(1,1,**1+1**)

1,**1,2**,3(1,1,2,**1+2**)

1,1,**2,3**,5(1,1,2,3,**2+3**)

1,1,2,**3,5**,8(1,1,2,3,5,**3+5**)

1,1,2,3,**5,8**,13(1,1,2,3,5,8,**5+8**)

Jne. Mutta entä jos piippu olisi tosi korkea, jos me halutaan vaikka 100, or tuhannes luku siitä jonosta. Laskemaan kaikki edelliset luvut olisi paljon työtä.

Voidaanko me keksiä joku näppärä keino saada Fibonacci luku ilman että laskisimme edelliset yhteen?

Formaali esitys meidän rekursio yhtälöön olisi: , kun

Entä jos me muutetaan täällä tavalla:

Voidaan poista n: ,

josta , siis

Hyvä. Siis

Nyt olisi hyvä löytää sopivia

Tiedämme että , josta johtuu kaksi yhtälöä:

Josta

Ja näin meidän

Tai sitten lyhytempi

Vaikka kaavassa on irrationaali se supistuu pois jos

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