I have played the game "Ori and the Blind Forest definitive edition" on PC.
In easy mode, it took me 22 hours, and in normal mode (on the second lap), 12 hours.
(all skills are unlocked, 87% of maps is cleared, and as the last event finished, the end staffs roll was displayed. )
The game has an old classical 2D action style.
(same as Super Mario, but the motion of "Ori" is quicker than Mario. )
It has very mournful and good BGM and beautiful graphics.
The controls by which "Ori" defeats the enemies, escapes from the traps, and gets to the each goals are much difficult, because a little "Ori" is weak and the trap is not almost able to avoid if you have not already known.
(there are no puzzles, for each goals there is only one way which "Ori" have to choose. )
However, the difficult controls give me a high sense of accomplishment when the quests are completed, and I can feel my game play becomes better and my feeling becomes high and hot.
It is recommended on snowy cold day in winter vacation.
This is the last post in 2016.
I wish you a Merry Christmas and a happy New Year !!
2016/12/20
2016/12/06
axiomatic sets 28 (the number system)
We have seen the system of numbers or the structure of numbers,
where basic elements are the empty set and the axiom of ordered pair.
(1)Natural numbers $\mathbb{N}$ : $\phi,\left\{\phi \right\},\left\{\phi,\left\{\phi \right\} \right\},\cdots$ means $0,1,2,\cdots$ .
(2)Integers $\mathbb{Z}$ : $[(\phi,\phi)],[(\left\{\phi \right\},\phi)],[(\phi,\left\{\phi \right\})],[(\left\{\phi,\left\{\phi \right\} \right\},\phi)],[(\phi,\left\{\phi,\left\{\phi \right\} \right\})],\cdots$ means $0,1,-1,2,-2,\cdots$ .
(3)Rational numbers $\mathbb{Q}$ :
$[([(\phi,\phi)],[(\left\{\phi \right\},\phi)])]=0$ ,
$[([(\left\{\phi \right\},\phi)],[(\left\{\phi \right\},\phi)])]=1$ ,
$[([(\phi,\left\{\phi \right\})],[(\left\{\phi \right\},\phi)])]=-1$ ,
$[([(\left\{\phi,\left\{\phi \right\} \right\},\phi)],[(\left\{\phi \right\},\phi)])]=2$ ,
$[([(\phi,\left\{\phi,\left\{\phi \right\} \right\})],[(\left\{\phi \right\},\phi)])]=-2$ , $\cdots$ .
(4)Real numbers $\mathbb{R} $: the set of all "Dedekind cuts" ($\left\{x\in\mathbb{Q} : x\lt a,a\in\mathbb{Q} \right\}$)
We are not conscious for the making of numbers or the system of numbers.
However, have you ever had a little bit of doubts something like why $\frac{-1}{3}=\frac{1}{-3}$ or the definition of irrational numbers is numbers which is not rational numbers, etc, etc ?
The number system will answer some questions and you will see the rigorousness exists in every fields of mathematics.
It makes us very comfortable and gives much confidence.
*
In these definitions, the relation of $\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}$ will require one to one correspondence.
where basic elements are the empty set and the axiom of ordered pair.
(1)Natural numbers $\mathbb{N}$ : $\phi,\left\{\phi \right\},\left\{\phi,\left\{\phi \right\} \right\},\cdots$ means $0,1,2,\cdots$ .
(2)Integers $\mathbb{Z}$ : $[(\phi,\phi)],[(\left\{\phi \right\},\phi)],[(\phi,\left\{\phi \right\})],[(\left\{\phi,\left\{\phi \right\} \right\},\phi)],[(\phi,\left\{\phi,\left\{\phi \right\} \right\})],\cdots$ means $0,1,-1,2,-2,\cdots$ .
(3)Rational numbers $\mathbb{Q}$ :
$[([(\phi,\phi)],[(\left\{\phi \right\},\phi)])]=0$ ,
$[([(\left\{\phi \right\},\phi)],[(\left\{\phi \right\},\phi)])]=1$ ,
$[([(\phi,\left\{\phi \right\})],[(\left\{\phi \right\},\phi)])]=-1$ ,
$[([(\left\{\phi,\left\{\phi \right\} \right\},\phi)],[(\left\{\phi \right\},\phi)])]=2$ ,
$[([(\phi,\left\{\phi,\left\{\phi \right\} \right\})],[(\left\{\phi \right\},\phi)])]=-2$ , $\cdots$ .
(4)Real numbers $\mathbb{R} $: the set of all "Dedekind cuts" ($\left\{x\in\mathbb{Q} : x\lt a,a\in\mathbb{Q} \right\}$)
We are not conscious for the making of numbers or the system of numbers.
However, have you ever had a little bit of doubts something like why $\frac{-1}{3}=\frac{1}{-3}$ or the definition of irrational numbers is numbers which is not rational numbers, etc, etc ?
The number system will answer some questions and you will see the rigorousness exists in every fields of mathematics.
It makes us very comfortable and gives much confidence.
*
In these definitions, the relation of $\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}$ will require one to one correspondence.
2016/11/22
axiomatic sets 27 (real numbers 3)
In preceding posts, we have seen the properties of real numbers.
(1)arithmetics of addition
(2)arithmetics of multiplication
(3)distributive law
(4)ordered relation
(5)the completeness
Let us look at one of 'Dedekind cut' $\left\{ x\in\mathbb{Q} : x\lt a \right\}$ as one number $a$ ,and the set of all 'Dedekind cuts' as a set of numbers.
(you may remember the natural number $a+1=\left\{ a\cup\left\{ a \right\} \right\}$ . )
The set of all 'Dedekind cuts' $\mathcal{C}$ will satisfy from (1) to (4). (do you agree?)
Axiom for completeness (5) means that if a set of numbers is bounded above,
then it has a supremum. (a supremum must be in all numbers.)
We can not express $\sqrt{2}$ as a rational number and there is not $\sqrt{2}$ in $\mathcal{C}$ .
However, in $\mathcal{C}$ we are able to get a rational number close to $\sqrt{2}$
as much as you want. Because in $\mathcal{C}$ there is all rational numbers.
$1.4=\frac{14}{10}\in \mathcal{C}$
$1.41=\frac{141}{100}\in \mathcal{C}$
$1.414=\frac{1414}{1000}\in \mathcal{C}$
$1.4142=\frac{14142}{10000}\in \mathcal{C}$
$\cdots\cdots\cdots$
$1.41421356237=\frac{141421356237}{10000000000}\in \mathcal{C}$
$\cdots\cdots\cdots$
Here, we want you to remember that $0.99999\cdots=1$ in the preceding posts.
That is, the rational number series which becomes $1.41421356237\cdots$ will go to $\sqrt{2}$ .
Althogh $\sqrt{2}$ is not a rational number, there exists a 'Dedekind cut' $\left\{x\in\mathbb{Q} : x\lt\sqrt{2} \right\}$ .
Thus, $\mathcal{C}$ is almost enough for real numbers.
The problem is that there are many numbers which are not able to be expressed by rational numbers.
We call the numbers which are not rational numbers irrational numbers.
Thus, real numbers are constructed rational numbers and irrational numbers.
(Correctly, you must understand, all above is needed proofs. )
(1)arithmetics of addition
(2)arithmetics of multiplication
(3)distributive law
(4)ordered relation
(5)the completeness
Let us look at one of 'Dedekind cut' $\left\{ x\in\mathbb{Q} : x\lt a \right\}$ as one number $a$ ,and the set of all 'Dedekind cuts' as a set of numbers.
(you may remember the natural number $a+1=\left\{ a\cup\left\{ a \right\} \right\}$ . )
The set of all 'Dedekind cuts' $\mathcal{C}$ will satisfy from (1) to (4). (do you agree?)
Axiom for completeness (5) means that if a set of numbers is bounded above,
then it has a supremum. (a supremum must be in all numbers.)
We can not express $\sqrt{2}$ as a rational number and there is not $\sqrt{2}$ in $\mathcal{C}$ .
However, in $\mathcal{C}$ we are able to get a rational number close to $\sqrt{2}$
as much as you want. Because in $\mathcal{C}$ there is all rational numbers.
$1.4=\frac{14}{10}\in \mathcal{C}$
$1.41=\frac{141}{100}\in \mathcal{C}$
$1.414=\frac{1414}{1000}\in \mathcal{C}$
$1.4142=\frac{14142}{10000}\in \mathcal{C}$
$\cdots\cdots\cdots$
$1.41421356237=\frac{141421356237}{10000000000}\in \mathcal{C}$
$\cdots\cdots\cdots$
Here, we want you to remember that $0.99999\cdots=1$ in the preceding posts.
That is, the rational number series which becomes $1.41421356237\cdots$ will go to $\sqrt{2}$ .
Althogh $\sqrt{2}$ is not a rational number, there exists a 'Dedekind cut' $\left\{x\in\mathbb{Q} : x\lt\sqrt{2} \right\}$ .
Thus, $\mathcal{C}$ is almost enough for real numbers.
The problem is that there are many numbers which are not able to be expressed by rational numbers.
We call the numbers which are not rational numbers irrational numbers.
Thus, real numbers are constructed rational numbers and irrational numbers.
(Correctly, you must understand, all above is needed proofs. )
2016/11/08
coffee break 13 (the graphic card)
I bought a new graphic card nvidia geforce gtx1070 recently.
I have used amd radeon r9-280x.
Unfortunately on r9-280x Batman arkham knight crashes almost surely,
if I make the graphic level of my display monitor cheapest.
Many people says different types of the error reason on web and
my pc does not have a newest hardware configuration.
I thought one of reasons was a lack of memories in the graphic card.
(the memories of r9-280x is 3gb and my PC is 16gb.)
Despite I worried about errors caused by the graphic driver (changing from amd to nvidia),
the change of the card went successfully.
The new card gtx1070 gives me playing the game comfortably.
(the memories of gtx1070 is 8gb.)
I remember that a few years ago, when I could not play TorchLight 2 by intel HD graphics on mother board, I bought the garaphic card radeon HD7750.
***
I have heard that gtx1070 being configured micron memories has OC issues.
By GPU-Z, my gtx1070 has micron memories.(orz)
I have used amd radeon r9-280x.
Unfortunately on r9-280x Batman arkham knight crashes almost surely,
if I make the graphic level of my display monitor cheapest.
Many people says different types of the error reason on web and
my pc does not have a newest hardware configuration.
I thought one of reasons was a lack of memories in the graphic card.
(the memories of r9-280x is 3gb and my PC is 16gb.)
Despite I worried about errors caused by the graphic driver (changing from amd to nvidia),
the change of the card went successfully.
The new card gtx1070 gives me playing the game comfortably.
(the memories of gtx1070 is 8gb.)
I remember that a few years ago, when I could not play TorchLight 2 by intel HD graphics on mother board, I bought the garaphic card radeon HD7750.
***
I have heard that gtx1070 being configured micron memories has OC issues.
By GPU-Z, my gtx1070 has micron memories.(orz)
2016/10/12
axiomatic sets 26 (real numbers 2)
We can not give a construction of real numbers from rational numbers.
Because it's proof is very long and a little bit tedious for blogs.
We will see a brief outline of the construction.
Let us make a subset $\alpha$ in rational numbers $\mathbb{Q}$ as follow.
(1)$\alpha\ne \phi,\mathbb{Q}$ .
(2)$\alpha$ is, for a number $a$, $\left\{x\in\mathbb{Q} : x\lt a \right\}$ .
(3)There is not a maximum number in $\alpha$ .
You must remember that all rational numbers is completely ordered "$\lt$" .
Therefore, if $y\lt a$ ,then $y\in \alpha$ and if $y\ge a$ ,then $y\notin\alpha$ .
As rational numbers are dense, if $y\lt a$ , then there must exists a $z$ such that $y\lt z\lt a$ .
Thus, $\alpha$ has not a maximum number.
By the number $a$ , rational numbers $\mathbb{Q}$ are separated to two sets $\alpha$ and $\alpha^c$ .
(You may prefer to think the both-sides open interval $\alpha=(-\infty, a)$ in the line of $\mathbb{Q}$ .
Please note that $\alpha$ does not have a point $a$ . )
It was called the 'Dedekind cut' $\alpha$ in the preceding post.
When we give many rational numbers to $a$ , there can be many 'Dedekind cuts' $\alpha$ .
We will express the set of all 'Dedekind cuts' $\mathcal{C}$ .
That is, $\mathcal{C}$ has the all intervals in which the left-side is $-\infty$
and the right-side is a rational numbers.
It is an astonishing fact that the element of the set $\mathcal{C}$ is a real number.
(We have to recognize the term "all" is much useful. )
Because it's proof is very long and a little bit tedious for blogs.
We will see a brief outline of the construction.
Let us make a subset $\alpha$ in rational numbers $\mathbb{Q}$ as follow.
(1)$\alpha\ne \phi,\mathbb{Q}$ .
(2)$\alpha$ is, for a number $a$, $\left\{x\in\mathbb{Q} : x\lt a \right\}$ .
(3)There is not a maximum number in $\alpha$ .
You must remember that all rational numbers is completely ordered "$\lt$" .
Therefore, if $y\lt a$ ,then $y\in \alpha$ and if $y\ge a$ ,then $y\notin\alpha$ .
As rational numbers are dense, if $y\lt a$ , then there must exists a $z$ such that $y\lt z\lt a$ .
Thus, $\alpha$ has not a maximum number.
By the number $a$ , rational numbers $\mathbb{Q}$ are separated to two sets $\alpha$ and $\alpha^c$ .
(You may prefer to think the both-sides open interval $\alpha=(-\infty, a)$ in the line of $\mathbb{Q}$ .
Please note that $\alpha$ does not have a point $a$ . )
It was called the 'Dedekind cut' $\alpha$ in the preceding post.
When we give many rational numbers to $a$ , there can be many 'Dedekind cuts' $\alpha$ .
We will express the set of all 'Dedekind cuts' $\mathcal{C}$ .
That is, $\mathcal{C}$ has the all intervals in which the left-side is $-\infty$
and the right-side is a rational numbers.
It is an astonishing fact that the element of the set $\mathcal{C}$ is a real number.
(We have to recognize the term "all" is much useful. )
2016/10/03
axiomatic sets 25 (real numbers)
We have constructed rational numbers by using the ordered pairs of two integers.
Unfortunately irrational numbers cannot be explicitly constructed from integers by such the way.
But we know there are many irrational numbers. For example, $\sqrt{2},\sqrt{3},\pi,e,\cdots$ .
We are enough in living using rational numbers. Almost people will be satisfied by the number "3.14" as $\pi$ .
No one requires the numbers which has infinite digits.
However, in rigorous thought as mathematics numbers which can not be represented by rational numbers are needed.
Area of circle must be $\pi r^2$ , not $3.14r^2$ , and
the answer of the equation $x^2-2=0$ must be $\pm\sqrt{2}$ , not $\pm 1.4142$ .
Irrational numbers may be defined by numbers which cannot be represented by a rational number
(that is, one ordered pair of two integers. )
As you know well, real numbers are made by rational numbers and irrational numbers.
In other words, real numbers are made by rational numbers and numbers which are not rational numbers.
Such a definition will give us much frustration. I will also agree with it.
We have already known complex numbers or quaternion and these are neither rational
nor irrational numbers.
We want to know how to make real numbers by rational numbers and irrational numbers.
Thus, what is real number was one of much important issues around the end of 19century.
The answer was given by R.Dedekind or G.Cantor, etc.
(In the preceding posts, we have seen the properties of real numbers.)
Unfortunately irrational numbers cannot be explicitly constructed from integers by such the way.
But we know there are many irrational numbers. For example, $\sqrt{2},\sqrt{3},\pi,e,\cdots$ .
We are enough in living using rational numbers. Almost people will be satisfied by the number "3.14" as $\pi$ .
No one requires the numbers which has infinite digits.
However, in rigorous thought as mathematics numbers which can not be represented by rational numbers are needed.
Area of circle must be $\pi r^2$ , not $3.14r^2$ , and
the answer of the equation $x^2-2=0$ must be $\pm\sqrt{2}$ , not $\pm 1.4142$ .
Irrational numbers may be defined by numbers which cannot be represented by a rational number
(that is, one ordered pair of two integers. )
As you know well, real numbers are made by rational numbers and irrational numbers.
In other words, real numbers are made by rational numbers and numbers which are not rational numbers.
Such a definition will give us much frustration. I will also agree with it.
We have already known complex numbers or quaternion and these are neither rational
nor irrational numbers.
We want to know how to make real numbers by rational numbers and irrational numbers.
Thus, what is real number was one of much important issues around the end of 19century.
The answer was given by R.Dedekind or G.Cantor, etc.
(In the preceding posts, we have seen the properties of real numbers.)
2016/09/21
axiomatic sets 24 (arithmetics in rational numbers)
Arithmetics in rational numbers are defined as follow.
Given two rational numbers $x=(a,b)$ and $y=(m,n)$ ,where $a,b,m,n$ are integers and $b,n$ are not zero.
The addition is
\[ x+y=(a,b)+(m,n)=(an+bm,bn) . \]
The multiplication is
\[ xy=(a,b)(m,n)=(am,bn) \]
,where "+" is the addition and "$an$" means the multiplication of integers "$a$" and "$n$".
The magnitude relations "$\gt$", "$=$", and "$\lt$" are
\[ x\gt y \quad \mbox{if and only if }\quad an\gt bm , \]
\[ x= y \quad \mbox{if and only if }\quad an= bm \]
,and
\[ x\lt y \quad \mbox{if and only if }\quad an\lt bm . \]
We say that $x$ is from above "greater than", "equivalent to" ,and "less than" $y$ .
The exponents are;
\[ x^0=1 \]
,where "1" is the rational number "1", (i.e. $1=[(1,1)]$ .)
\[ x^1=x=(a,b) , \]
\[ x^2=xx=(aa,bb) , \]
\[ x^3=(xx)x=((aa)a,(bb)b) \]
,and so on.
Immediately we will get $x^rx^s=x^{r+s}$ .
If $a$ and $b$ are not zero ,then
\[ x^{-1}=(a,b)^{-1}=(b,a) . \]
This is an inverse element of multiplication.
That is,
\[ xx^{-1}=(a,b)(b,a)=(ab,ab)=1 \]
,where "1" is the rational number "1".
We will ordinarily denote $x^{-1}=\frac{1}{x}$ .
More over, $x^{-2}=\frac{1}{x^2}$ , $x^{-3}=\frac{1}{x^3}$ and so on.
More generally, we will write $x^{-1}y=yx^{-1}=\frac{y}{x}$ .
(It is called the quotient. )
The definition of $x^{c}\quad (0\lt c\lt 1)$ is complicated and not easy.
Given a rational number $w(\gt 0)$ and $w^2=ww=z$ ,
then we will define and denote $z^{\frac{1}{2}}=w$ .
In same way, if $w^3=z$, then $z^{\frac{1}{3}}=w$ ,and so on.
However, for any rational number $z$ , we know that there does not exist a rational number $z^{\frac{1}{2}}$ .
(in the preceding post, we have proven $2^{\frac{1}{2}}=\sqrt{2}$ is not a rational number. )
Unfortunately, we can not always assert $x^{c}\quad (0\lt c\lt 1)$ is in rational numbers.
(Although it is necessary to prove, almost numbers $x^{c}$ are not rational numbers. )
In addition, we have known many numbers such $\pi$ and $e$ are not rational numbers.
Thus irrational numbers and real numbers will be needed.
Given two rational numbers $x=(a,b)$ and $y=(m,n)$ ,where $a,b,m,n$ are integers and $b,n$ are not zero.
The addition is
\[ x+y=(a,b)+(m,n)=(an+bm,bn) . \]
The multiplication is
\[ xy=(a,b)(m,n)=(am,bn) \]
,where "+" is the addition and "$an$" means the multiplication of integers "$a$" and "$n$".
The magnitude relations "$\gt$", "$=$", and "$\lt$" are
\[ x\gt y \quad \mbox{if and only if }\quad an\gt bm , \]
\[ x= y \quad \mbox{if and only if }\quad an= bm \]
,and
\[ x\lt y \quad \mbox{if and only if }\quad an\lt bm . \]
We say that $x$ is from above "greater than", "equivalent to" ,and "less than" $y$ .
The exponents are;
\[ x^0=1 \]
,where "1" is the rational number "1", (i.e. $1=[(1,1)]$ .)
\[ x^1=x=(a,b) , \]
\[ x^2=xx=(aa,bb) , \]
\[ x^3=(xx)x=((aa)a,(bb)b) \]
,and so on.
Immediately we will get $x^rx^s=x^{r+s}$ .
If $a$ and $b$ are not zero ,then
\[ x^{-1}=(a,b)^{-1}=(b,a) . \]
This is an inverse element of multiplication.
That is,
\[ xx^{-1}=(a,b)(b,a)=(ab,ab)=1 \]
,where "1" is the rational number "1".
We will ordinarily denote $x^{-1}=\frac{1}{x}$ .
More over, $x^{-2}=\frac{1}{x^2}$ , $x^{-3}=\frac{1}{x^3}$ and so on.
More generally, we will write $x^{-1}y=yx^{-1}=\frac{y}{x}$ .
(It is called the quotient. )
The definition of $x^{c}\quad (0\lt c\lt 1)$ is complicated and not easy.
Given a rational number $w(\gt 0)$ and $w^2=ww=z$ ,
then we will define and denote $z^{\frac{1}{2}}=w$ .
In same way, if $w^3=z$, then $z^{\frac{1}{3}}=w$ ,and so on.
However, for any rational number $z$ , we know that there does not exist a rational number $z^{\frac{1}{2}}$ .
(in the preceding post, we have proven $2^{\frac{1}{2}}=\sqrt{2}$ is not a rational number. )
Unfortunately, we can not always assert $x^{c}\quad (0\lt c\lt 1)$ is in rational numbers.
(Although it is necessary to prove, almost numbers $x^{c}$ are not rational numbers. )
In addition, we have known many numbers such $\pi$ and $e$ are not rational numbers.
Thus irrational numbers and real numbers will be needed.
2016/09/02
axiomatic sets 23 (types of numbers)
There are some kinds of numbers. Let us list these samples up.
The empty set means the number "0", that is, $\phi=\left\{ \right\}=0$ .
We will see that "0" is $0_N$ in natural numbers.
The natural number "1" is the set whose member is only the empty set, $1_N=\left\{\phi\right\}$ .
The natural number "2" is the set whose members are natural numbers "0" and "1",
$2_N=\left\{0_N,1_N \right\}=\left\{\phi,\left\{\phi\right\} \right\}$ .
The integer "0" is the ordered pair class $0_Z=[(0_N,0_N)]$ ,
but it's class has only one element $(0_N,0_N)$ .
The non negative integer "1" is $[(1,0)]$ ,where "1" is the natural number $1_N$, and "[ ]" means a equivalent class. Therefore,
\[ 1_Z=[(1_N,0_N)]=[(\left\{\phi\right\},\phi)]=[(\left\{\left\{ \right\}\right\},\left\{ \right\})] \]
As the equivalent relation is for two integers $(a_N,b_N),(c_N,d_N)$ ,
$a_N+d_N=b_N+c_N$ must be satisfied,
you have to note that $(1_N,0_N)\sim (2_N,1_N)\sim (3_N,2_N)\sim\cdots$ .
Those are in $1_Z$ ,and we use every elements as $1_Z$ .
The non negative integer "2" is also the ordered pair class $[(2_N,0_N)]=2_Z$ ,
The negative integer "-2" is $[(0_N,2_N)]=-2_Z$ .
The rational number "0" is also the ordered pair class $[(0,a)]$ ,where $a$ is an arbitrary integer, but not zero, and "[ ]" means a equivalent class.
\[ 0_Q=[(0_Z,a_Z)]=[([(0_N,0_N)],[(a_N,0_N)])] . \]
The equivalent relation on rational numbers is for two rational nummbers $(p_Z,q_Z),(r_Z,s_Z)$
(and $q_Z,s_Z\ne 0)$, $p_Zs_Z=q_Zr_Z$ must be satisfied.
The non negative rational number "1" is $[(1,1)]$ ,where "1" is the integer number "$1_Z$".
Therefore,
\[ 1_Q=[(1_Z,1_Z)]=[([(\left\{\phi\right\},\phi)],[(\left\{\phi\right\},\phi)])] , \]
and for $x\in 1_Q$ ,
\[ x\sim (1_Z,1_Z)\sim (-1_Z,-1_Z)\sim (2_Z,2_Z)\sim (-2_Z,-2_Z)\sim (3_Z,3_Z)\sim\cdots \]
By same way, $2_Q=[(2_Z,1_Z)]=\left\{(2_Z,1_Z),(-2_Z,-1_Z),(4_Z,2_Z),(-4_Z,-2_Z),(6_Z,3_Z),\cdots \right\}$ .
You will find these kinds of numbers are constructed by $\phi$ and axioms of sets.
Please try to make some types of numbers.
The empty set means the number "0", that is, $\phi=\left\{ \right\}=0$ .
We will see that "0" is $0_N$ in natural numbers.
The natural number "1" is the set whose member is only the empty set, $1_N=\left\{\phi\right\}$ .
The natural number "2" is the set whose members are natural numbers "0" and "1",
$2_N=\left\{0_N,1_N \right\}=\left\{\phi,\left\{\phi\right\} \right\}$ .
The integer "0" is the ordered pair class $0_Z=[(0_N,0_N)]$ ,
but it's class has only one element $(0_N,0_N)$ .
The non negative integer "1" is $[(1,0)]$ ,where "1" is the natural number $1_N$, and "[ ]" means a equivalent class. Therefore,
\[ 1_Z=[(1_N,0_N)]=[(\left\{\phi\right\},\phi)]=[(\left\{\left\{ \right\}\right\},\left\{ \right\})] \]
As the equivalent relation is for two integers $(a_N,b_N),(c_N,d_N)$ ,
$a_N+d_N=b_N+c_N$ must be satisfied,
you have to note that $(1_N,0_N)\sim (2_N,1_N)\sim (3_N,2_N)\sim\cdots$ .
Those are in $1_Z$ ,and we use every elements as $1_Z$ .
The non negative integer "2" is also the ordered pair class $[(2_N,0_N)]=2_Z$ ,
The negative integer "-2" is $[(0_N,2_N)]=-2_Z$ .
The rational number "0" is also the ordered pair class $[(0,a)]$ ,where $a$ is an arbitrary integer, but not zero, and "[ ]" means a equivalent class.
\[ 0_Q=[(0_Z,a_Z)]=[([(0_N,0_N)],[(a_N,0_N)])] . \]
The equivalent relation on rational numbers is for two rational nummbers $(p_Z,q_Z),(r_Z,s_Z)$
(and $q_Z,s_Z\ne 0)$, $p_Zs_Z=q_Zr_Z$ must be satisfied.
The non negative rational number "1" is $[(1,1)]$ ,where "1" is the integer number "$1_Z$".
Therefore,
\[ 1_Q=[(1_Z,1_Z)]=[([(\left\{\phi\right\},\phi)],[(\left\{\phi\right\},\phi)])] , \]
and for $x\in 1_Q$ ,
\[ x\sim (1_Z,1_Z)\sim (-1_Z,-1_Z)\sim (2_Z,2_Z)\sim (-2_Z,-2_Z)\sim (3_Z,3_Z)\sim\cdots \]
By same way, $2_Q=[(2_Z,1_Z)]=\left\{(2_Z,1_Z),(-2_Z,-1_Z),(4_Z,2_Z),(-4_Z,-2_Z),(6_Z,3_Z),\cdots \right\}$ .
You will find these kinds of numbers are constructed by $\phi$ and axioms of sets.
Please try to make some types of numbers.
2016/08/25
axiomatic sets 22 (rational numbers)
We saw the two kinds of numbers would be created.
Natural numbers $\mathbb{N}$ are constructed by $\phi$ and successor sets.
Integers $\mathbb{Z}$ are constructed by the equivalent classes $[(a,b)]$ of an ordered pair of two natural numbers.
In this post, we shall define rational numbers $\mathbb{Q}$ .
Rational numbers will be defined by the same way of integers.
These are constructed by the equivalent classes $[(p,q)]$ of an ordered pair of two integers.
Suppose an arbitrary ordered pair $(p,q)$ of two integers, where $q\ne 0$ .
(you can see it $\frac{p}{q}$ . )
The equivalent relation "$\sim$" means, on two ordered pairs $(p,q)$ and $(r,s)$ (where $q\ne 0$ and $s\ne 0$),
\[ (p,q)\sim (r,s)\quad \leftrightarrow\quad ps=qr . \]
We make $\mathbb{Q}$ denote the set of all equivalent classes with respect to "$\sim$" .
The elements of $\mathbb{Q}$ will be called rational numbers $[(p,q)]$.
It is very simple.
Natural numbers $\mathbb{N}$ are constructed by $\phi$ and successor sets.
Integers $\mathbb{Z}$ are constructed by the equivalent classes $[(a,b)]$ of an ordered pair of two natural numbers.
In this post, we shall define rational numbers $\mathbb{Q}$ .
Rational numbers will be defined by the same way of integers.
These are constructed by the equivalent classes $[(p,q)]$ of an ordered pair of two integers.
Suppose an arbitrary ordered pair $(p,q)$ of two integers, where $q\ne 0$ .
(you can see it $\frac{p}{q}$ . )
The equivalent relation "$\sim$" means, on two ordered pairs $(p,q)$ and $(r,s)$ (where $q\ne 0$ and $s\ne 0$),
\[ (p,q)\sim (r,s)\quad \leftrightarrow\quad ps=qr . \]
We make $\mathbb{Q}$ denote the set of all equivalent classes with respect to "$\sim$" .
The elements of $\mathbb{Q}$ will be called rational numbers $[(p,q)]$.
It is very simple.
2016/08/03
axiomatic sets 21 (arithmetics in integers)
The addition of two integers is defined as follow.
Given two integers $[(a,b)],[(c,d)]\quad (a,b,c,d\in\mathbb{N})$ ,
\[ [(a,b)]+[(c,d)]=[(a+c,b+d)] . \]
You will not have any questions.
For example,
\[ [(2,1)]+[(3,1)]=[(5,2)] . \]
This means $1+2=3$ .
\[ [(4,1)]+[(1,4)]=[(5,5)] . \]
This means $3+(-3)=0$
\[ [(1,3)]+[(1,4)]=[(2,7)] . \]
This means $(-2)+(-3)=-5$ and equal to
\[ [(21,23)]+[(41,44)]=[(62,67)] . \]
The multiplication of integers is defined same as addition.
\[ [(a,b)]\times [(c,d)]=[(ac+bd,ad+bc)] . \]
Intuitively, as $(a-b)\times(c-d)=(ac+bd)-(ad+bc)$ is true, it will be also true.
You will be able to see the integers are closed under addition (subtraction) and multiplication.
The integers have been constructed on natural numbers $\mathbb{N}$ and some axioms.
(On this definition, unfortunately, $\mathbb{N}\subset\mathbb{Z}$ is not true as a matter of form. )
Given two integers $[(a,b)],[(c,d)]\quad (a,b,c,d\in\mathbb{N})$ ,
\[ [(a,b)]+[(c,d)]=[(a+c,b+d)] . \]
You will not have any questions.
For example,
\[ [(2,1)]+[(3,1)]=[(5,2)] . \]
This means $1+2=3$ .
\[ [(4,1)]+[(1,4)]=[(5,5)] . \]
This means $3+(-3)=0$
\[ [(1,3)]+[(1,4)]=[(2,7)] . \]
This means $(-2)+(-3)=-5$ and equal to
\[ [(21,23)]+[(41,44)]=[(62,67)] . \]
The multiplication of integers is defined same as addition.
\[ [(a,b)]\times [(c,d)]=[(ac+bd,ad+bc)] . \]
Intuitively, as $(a-b)\times(c-d)=(ac+bd)-(ad+bc)$ is true, it will be also true.
You will be able to see the integers are closed under addition (subtraction) and multiplication.
The integers have been constructed on natural numbers $\mathbb{N}$ and some axioms.
(On this definition, unfortunately, $\mathbb{N}\subset\mathbb{Z}$ is not true as a matter of form. )
2016/07/21
axiomatic sets 20 (integers)
In preceding post, we have defined negative numbers using ordered pairs of $0$ and a natural number $n$.
\[ \mathbb{N}^-=\left\{(0,n) : n\in\mathbb{N},n\ne 0 \right\} \]
Therefore, integers $\mathbb{Z}$ will be simply constructed by $\mathbb{N}\cup\mathbb{N}^-$ .
However, such integers are complicated as we always have to concern the difference of positive numbers and negative numbers. Because integers are made by two kinds of sets (0, positive numbers $1,2,3,\cdots$ and negative numbers $(0,1),(0,2),(0,3),\cdots$) .
We shall introduce an equivalent relation "$\sim$" to any ordered pair of two natural numbers.
For any natural numbers $k,h,n,m\in\mathbb{N}$ , $(k,h)\sim (n,m)$ if and only if $k+m=h+n$ .
That is,
\[ (1,0)\sim (2,1)\sim (3,2)\sim (4,3)\sim\cdots , \]
\[ (2,0)\sim (3,1)\sim (4,2)\sim (5,3)\sim\cdots , \]
\[ (0,0)\sim (1,1)\sim (2,2)\sim (3,3)\sim\cdots , \]
\[ (0,1)\sim (1,2)\sim (2,3)\sim (3,4)\sim\cdots , \]
$\cdots,\cdots$ .
We will write $[(k,h)]$ as meaning
\[ [(k,h)]=\left\{(n,m) : n,m\in\mathbb{N}, (k,h)\sim(n,m) \right\} . \]
For given $k,h$ , $[(k,h)]$ becomes a class of sets.
For example,
\[ [(1,0)]=\left\{(1,0),(2,1),(3,2),(4,3),\cdots \right\} , \]
\[ [(2,0)]=\left\{(2,0),(3,1),(4,2),(5,3),\cdots \right\} , \]
\[ [(0,0)]=\left\{(0,0),(1,1),(2,2),(3,3),\cdots \right\} , \]
\[ [(0,1)]=\left\{(0,1),(1,2),(2,3),(3,4),\cdots \right\} , \]
$\cdots,\cdots$ .
As a matter of course, for example,
\[ [(1,0)]=[(2,1)]=[(3,2)]=[(4,3)]=\cdots . \]
You will have to find $[(1,0)]$ as the intuitive natural number $1$ and
$[(2,0)]$ as $2$, $[(0,0)]$ as $0$ , $[(0,1)]$ as $-1$ , $\cdots\cdots$ .
We make $\mathbb{Z}$ denote the set of all equivalent classes with respect to $\sim$ .
The elements of $\mathbb{Z}$ will be called integers.
You will see intuitively it means
\[ \mathbb{Z}=\left\{ 0,1,-1,2,-2,\cdots \right\} . \]
Also, you can do define
\[ \mathbb{Z}=\left\{ [(0,0)],[(2,1)],[(2,3)],[(4,2)],[(3,5)],\cdots \right\} . \]
\[ \mathbb{N}^-=\left\{(0,n) : n\in\mathbb{N},n\ne 0 \right\} \]
Therefore, integers $\mathbb{Z}$ will be simply constructed by $\mathbb{N}\cup\mathbb{N}^-$ .
However, such integers are complicated as we always have to concern the difference of positive numbers and negative numbers. Because integers are made by two kinds of sets (0, positive numbers $1,2,3,\cdots$ and negative numbers $(0,1),(0,2),(0,3),\cdots$) .
We shall introduce an equivalent relation "$\sim$" to any ordered pair of two natural numbers.
For any natural numbers $k,h,n,m\in\mathbb{N}$ , $(k,h)\sim (n,m)$ if and only if $k+m=h+n$ .
That is,
\[ (1,0)\sim (2,1)\sim (3,2)\sim (4,3)\sim\cdots , \]
\[ (2,0)\sim (3,1)\sim (4,2)\sim (5,3)\sim\cdots , \]
\[ (0,0)\sim (1,1)\sim (2,2)\sim (3,3)\sim\cdots , \]
\[ (0,1)\sim (1,2)\sim (2,3)\sim (3,4)\sim\cdots , \]
$\cdots,\cdots$ .
We will write $[(k,h)]$ as meaning
\[ [(k,h)]=\left\{(n,m) : n,m\in\mathbb{N}, (k,h)\sim(n,m) \right\} . \]
For given $k,h$ , $[(k,h)]$ becomes a class of sets.
For example,
\[ [(1,0)]=\left\{(1,0),(2,1),(3,2),(4,3),\cdots \right\} , \]
\[ [(2,0)]=\left\{(2,0),(3,1),(4,2),(5,3),\cdots \right\} , \]
\[ [(0,0)]=\left\{(0,0),(1,1),(2,2),(3,3),\cdots \right\} , \]
\[ [(0,1)]=\left\{(0,1),(1,2),(2,3),(3,4),\cdots \right\} , \]
$\cdots,\cdots$ .
As a matter of course, for example,
\[ [(1,0)]=[(2,1)]=[(3,2)]=[(4,3)]=\cdots . \]
You will have to find $[(1,0)]$ as the intuitive natural number $1$ and
$[(2,0)]$ as $2$, $[(0,0)]$ as $0$ , $[(0,1)]$ as $-1$ , $\cdots\cdots$ .
We make $\mathbb{Z}$ denote the set of all equivalent classes with respect to $\sim$ .
The elements of $\mathbb{Z}$ will be called integers.
\[ \mathbb{Z}=\left\{ [(0,0)],[(1,0)],[(0,1)],[(2,0)],[(0,2)],\cdots \right\} \]
You will see intuitively it means
\[ \mathbb{Z}=\left\{ 0,1,-1,2,-2,\cdots \right\} . \]
Also, you can do define
\[ \mathbb{Z}=\left\{ [(0,0)],[(2,1)],[(2,3)],[(4,2)],[(3,5)],\cdots \right\} . \]
2016/07/05
axiomatic sets 19 (negative numbers)
We want to extend natural numbers to integers.
Integers, as you know well, are constructed by natural numbers (which are positive numbers and 0) and negative numbers.
In axiomtaic set theory, all objects must be sets which are collections of defined something.
A set can not become negative and always be positive because those exist here and there.
Thus, we will identify some sort of sets same as negative numbers.
Most simple way for negative numbers is the ordered pair $(0,n)$ ,where $n$ is a natural number and not $0$ .
That is to say, $(0,1)$ means $-1$ , and $(0,2)$ is $-2$ ,･･･, $(0,n)$ is $-n$, and so on.
For this definition, we have to accept the following lemma. But it is very natural.
If natural numbers $m,n$ are $m\lt n$ ,then there is a natural number $i$ such that $m+i=n$ .
We will state an "$i$" "$n-m$" .
Next, we want to use this rule in the case when $m\ge n$ ,
because it is very useful and we do not need to think about the magnitude relation of $m,n$ .
Especially, if $n=0$ ,then $i=-m$ omitting $n$ .
Then, we will define the ordered pair $(0,m)$ as $-m$. ( $(0,0)=0$ and If $n=m$ ,then $i=0$ )
This is one of definitions of negative numbers.
Integers, as you know well, are constructed by natural numbers (which are positive numbers and 0) and negative numbers.
In axiomtaic set theory, all objects must be sets which are collections of defined something.
A set can not become negative and always be positive because those exist here and there.
Thus, we will identify some sort of sets same as negative numbers.
Most simple way for negative numbers is the ordered pair $(0,n)$ ,where $n$ is a natural number and not $0$ .
That is to say, $(0,1)$ means $-1$ , and $(0,2)$ is $-2$ ,･･･, $(0,n)$ is $-n$, and so on.
For this definition, we have to accept the following lemma. But it is very natural.
If natural numbers $m,n$ are $m\lt n$ ,then there is a natural number $i$ such that $m+i=n$ .
We will state an "$i$" "$n-m$" .
Next, we want to use this rule in the case when $m\ge n$ ,
because it is very useful and we do not need to think about the magnitude relation of $m,n$ .
Especially, if $n=0$ ,then $i=-m$ omitting $n$ .
Then, we will define the ordered pair $(0,m)$ as $-m$. ( $(0,0)=0$ and If $n=m$ ,then $i=0$ )
This is one of definitions of negative numbers.
(Might you have any questions? OK, I will continue.)
2016/06/14
axiomatic sets 18 (arithmetic in natural numbers 2)
The multiplication of natural numbers is defined same as that of the addition.
The operation $g$ of the multiplication is a function satisfying following conditions.
(You must remember the function $g$ is also a set.)
(1)$g(a,0)=0$
(2)$g(a,b^+)=g(a,b)+a$
,where $a,b\in\mathbb{N}$ and $b^+$ is the successor set of $b$ . ($b^+=b\cup\left\{b\right\}$)
These mean simply ;
$a\times 0=0$ ,
$a\times (b+1)=(a\times b)+a$ .
For example,
$g(2,3)=2\times(2+1)=(2\times 2)+2=((2\times 1)+2)+2$
$=(((2\times 0)+2)+2)+2=0+2+2+2=6$ .
In addition, the exponent of natural numbers is extended.
By using the multiplication, the function $e$ of the exponent satisfies following conditions.
(1)$e(a,0)=1$ ,
(2)$e(a,b^+)=g(e(a,b),a)$ .
That is,
$a^0=1$ ,
$a^{b+1}=(a^b)\times a$ .
If $a=2,b^+=3$ ,then
$e(2,3)=g(e(2,2),2)=g(g(e(2,1),2),2)=g(g(g(e(2,0),2),2),2)=1\times 2\times 2\times 2$ .
These are called a finite recursion formula.
The operation $g$ of the multiplication is a function satisfying following conditions.
(You must remember the function $g$ is also a set.)
(1)$g(a,0)=0$
(2)$g(a,b^+)=g(a,b)+a$
,where $a,b\in\mathbb{N}$ and $b^+$ is the successor set of $b$ . ($b^+=b\cup\left\{b\right\}$)
These mean simply ;
$a\times 0=0$ ,
$a\times (b+1)=(a\times b)+a$ .
For example,
$g(2,3)=2\times(2+1)=(2\times 2)+2=((2\times 1)+2)+2$
$=(((2\times 0)+2)+2)+2=0+2+2+2=6$ .
In addition, the exponent of natural numbers is extended.
By using the multiplication, the function $e$ of the exponent satisfies following conditions.
(1)$e(a,0)=1$ ,
(2)$e(a,b^+)=g(e(a,b),a)$ .
That is,
$a^0=1$ ,
$a^{b+1}=(a^b)\times a$ .
If $a=2,b^+=3$ ,then
$e(2,3)=g(e(2,2),2)=g(g(e(2,1),2),2)=g(g(g(e(2,0),2),2),2)=1\times 2\times 2\times 2$ .
These are called a finite recursion formula.
2016/06/06
axiomatic sets 17 (arithmetic in natural numbers)
The set of all natural numbers is closed under addition (and multiplication).
We have already defined one equation of addition.
\[ n+1=n\cup\left\{ n \right\} \qquad (n\in\mathbb{N}, 0\in\mathbb{N},1=\left\{ 0 \right\}) \]
Since, for any natural number $n$ , there must be an $n\cup\left\{ n \right\}$ by axiom of infinity,
and for any natural number $m\ne 0$ , there must be an $n$ such that $n\cup\left\{ n \right\}=m$ ,
this equation is true.
We want to extend this to the definition of "$a+b$".
For this purpose the function "$f$" is needed.
A collection of a pair of two natural numbers is a set.
(i.e. Cartesian Product of two $\mathbb{N}$s is a set. )
\[ \mathbb{N}\times\mathbb{N}=\left\{\left\{ a,b \right\}: a\in\mathbb{N},b\in\mathbb{N} \right\} \]
In axiomatic set theory, as the function $f$ is also a set,
\[ f:\mathbb{N}\times\mathbb{N}\rightarrow \mathbb{N} , \]
and
\[ \forall a\forall b\exists f[f(\left\{a,b \right\})\in\mathbb{N}] , \]
and $f$ has to be satisfied the following two conditions;
(1)$f(\left\{a,0\right\})=a$
(2)$f(\left\{a,b^+\right\})=(f(\left\{a,b\right\}))^+$
,where $b^+$ is the successor set of $b$ . Namely, $b^+=b\cup\left\{ b \right\} $ .
You may see it is not easy to understand. However, its mean, simply,
(1)$a+0=a$ ,
(2)$a+(b+1)=(a+b)+1$ .
If $b$ is not $0$, there exists a $c$ such that $c^+=c+1=b$ .
Thus,
\[ a+(b+1)=(a+b)+1=(a+c^+)+1=(a+(c+1))+1=((a+c)+1)+1 . \]
Same operations take us to
\[ a+b=(\cdots(a+0)+1)+1)\cdots )+1 . \]
,where $(\cdots (0+1)+1)\cdots )+1=b$ .
This is the definition of an arithmetic of addition in axiomatic set theory.
We have already defined one equation of addition.
\[ n+1=n\cup\left\{ n \right\} \qquad (n\in\mathbb{N}, 0\in\mathbb{N},1=\left\{ 0 \right\}) \]
Since, for any natural number $n$ , there must be an $n\cup\left\{ n \right\}$ by axiom of infinity,
and for any natural number $m\ne 0$ , there must be an $n$ such that $n\cup\left\{ n \right\}=m$ ,
this equation is true.
We want to extend this to the definition of "$a+b$".
For this purpose the function "$f$" is needed.
A collection of a pair of two natural numbers is a set.
(i.e. Cartesian Product of two $\mathbb{N}$s is a set. )
\[ \mathbb{N}\times\mathbb{N}=\left\{\left\{ a,b \right\}: a\in\mathbb{N},b\in\mathbb{N} \right\} \]
In axiomatic set theory, as the function $f$ is also a set,
\[ f:\mathbb{N}\times\mathbb{N}\rightarrow \mathbb{N} , \]
and
\[ \forall a\forall b\exists f[f(\left\{a,b \right\})\in\mathbb{N}] , \]
and $f$ has to be satisfied the following two conditions;
(1)$f(\left\{a,0\right\})=a$
(2)$f(\left\{a,b^+\right\})=(f(\left\{a,b\right\}))^+$
,where $b^+$ is the successor set of $b$ . Namely, $b^+=b\cup\left\{ b \right\} $ .
You may see it is not easy to understand. However, its mean, simply,
(1)$a+0=a$ ,
(2)$a+(b+1)=(a+b)+1$ .
If $b$ is not $0$, there exists a $c$ such that $c^+=c+1=b$ .
Thus,
\[ a+(b+1)=(a+b)+1=(a+c^+)+1=(a+(c+1))+1=((a+c)+1)+1 . \]
Same operations take us to
\[ a+b=(\cdots(a+0)+1)+1)\cdots )+1 . \]
,where $(\cdots (0+1)+1)\cdots )+1=b$ .
This is the definition of an arithmetic of addition in axiomatic set theory.
2016/05/23
axiomatic sets 16 (Peano's axioms)
By an axiom of infinity, the collection of all natural numbers was gotten.
By Peano's axioms, all natural numbers $\mathbb{N}$ has been defined.
$w$ which satisfies the following conditions (1)-(5) is $\mathbb{N}$ ;
(1)$0\in w$
(2)if any $n\in w$ , then $n\cup\left\{ n \right\}\in w$
(3)for each $n\in w$ , $n\cup\left\{ n \right\}\ne 0$
(4)if $x$ is a subset of $w$ such that $0\in x$ and if $n\in x$ , then $n\cup\left\{ n \right\}\in x$ ,
then $x=w$
(5)if $n,m\in w$ and $n\cup\left\{ n \right\}=m\cup\left\{ m \right\}$ , then $n=m$
Please remember that
\[ n\cup\left\{ n \right\}=n+1 . \]
By using preceding axioms, these can be proved.
$n\cup\left\{ n \right\}$ is called a successor set of $n$ .
Please note that (4) means the principle of "mathematical induction".
It may seem same as an axiom of infinity.
By Peano's axioms, all natural numbers $\mathbb{N}$ has been defined.
$w$ which satisfies the following conditions (1)-(5) is $\mathbb{N}$ ;
(1)$0\in w$
(2)if any $n\in w$ , then $n\cup\left\{ n \right\}\in w$
(3)for each $n\in w$ , $n\cup\left\{ n \right\}\ne 0$
(4)if $x$ is a subset of $w$ such that $0\in x$ and if $n\in x$ , then $n\cup\left\{ n \right\}\in x$ ,
then $x=w$
(5)if $n,m\in w$ and $n\cup\left\{ n \right\}=m\cup\left\{ m \right\}$ , then $n=m$
Please remember that
\[ n\cup\left\{ n \right\}=n+1 . \]
By using preceding axioms, these can be proved.
$n\cup\left\{ n \right\}$ is called a successor set of $n$ .
Please note that (4) means the principle of "mathematical induction".
It may seem same as an axiom of infinity.
2016/05/09
axiomatic sets 15 (the set of all Natural numbers)
We got all natural numbers.
\[ 0=\left\{ \right\}, \quad n+1=n\cup\left\{n\right\}\quad (n\ge 0) \]
You are able to know any natural numbers which you desire, increasing one by one.
However, it is not a set or collection of all natural numbers.
In the set $\mathbb{N}=\left\{0,1,2,\cdots \right\}$ ,
'･･･' does not mean any natural number.
No matter how a large number, it does not mean '･･･'.
Hence, the axiom is needed.
\[ \exists x \forall y[y\in x\rightarrow (y\cup\left\{ y \right\})\in x] \]
It is called an axiom of infinity.
There can be exists a set whose elements are infinite.
Let $y$ be $0$ , the set $x$ is the set of all natural numbers.
\[ 0=\left\{ \right\}, \quad n+1=n\cup\left\{n\right\}\quad (n\ge 0) \]
You are able to know any natural numbers which you desire, increasing one by one.
However, it is not a set or collection of all natural numbers.
In the set $\mathbb{N}=\left\{0,1,2,\cdots \right\}$ ,
'･･･' does not mean any natural number.
No matter how a large number, it does not mean '･･･'.
Hence, the axiom is needed.
\[ \exists x \forall y[y\in x\rightarrow (y\cup\left\{ y \right\})\in x] \]
It is called an axiom of infinity.
There can be exists a set whose elements are infinite.
Let $y$ be $0$ , the set $x$ is the set of all natural numbers.
2016/04/15
axiomatic sets 14 (a natural number)
Natural number is an ordinal.
You must know well,
the first=1,
the second=1+1=2,
the third=2+1=3,
the fourth=3+1=4,
･･････
In addition, '0' is also a natural number.
In axiomatic set theory, those must be sets.
We have already defined
\[ 0=\phi=\left\{ \right\} . \]
As, if 'n' is a natural number, then 'n+1' is also a natural number and an ordinal,
we will define
\[ n+1=n\cup\left\{ n\right\}=\left\{ n, \left\{ n\right\} \right\} . \]
That is to say,
$1=0\cup\left\{ 0\right\}=\phi\cup\left\{ 0\right\}=\left\{ 0\right\}=\left\{ \phi \right\}=\left\{ \left\{ \right\}\right\}$ ,
$2=1\cup\left\{ 1\right\}=\left\{ 0\right\}\cup\left\{ \left\{ 0\right\} \right\}=\left\{ 0,\left\{ 0\right\}\right\}$ ,
$3=2\cup\left\{ 2\right\}=\left\{ 0,\left\{ 0\right\},\left\{ 0,\left\{ 0\right\}\right\} \right\}$ ,
$4=3\cup\left\{ 3\right\}=\left\{ 0,\left\{ 0\right\},\left\{ 0,\left\{ 0\right\}\right\},\left\{ 0,\left\{ 0\right\},\left\{ 0,\left\{ 0\right\}\right\} \right\} \right\}$ ,
･･････
You will understand
$1=\left\{0 \right\}$ ,
$2=\left\{0,1 \right\}$ ,
$3=\left\{0,1,2 \right\}$ ,
$4=\left\{0,1,2,3 \right\}$ ,
･･････
We are able to get all natural numbers.
You must know well,
the first=1,
the second=1+1=2,
the third=2+1=3,
the fourth=3+1=4,
･･････
In addition, '0' is also a natural number.
In axiomatic set theory, those must be sets.
We have already defined
\[ 0=\phi=\left\{ \right\} . \]
As, if 'n' is a natural number, then 'n+1' is also a natural number and an ordinal,
we will define
\[ n+1=n\cup\left\{ n\right\}=\left\{ n, \left\{ n\right\} \right\} . \]
That is to say,
$1=0\cup\left\{ 0\right\}=\phi\cup\left\{ 0\right\}=\left\{ 0\right\}=\left\{ \phi \right\}=\left\{ \left\{ \right\}\right\}$ ,
$2=1\cup\left\{ 1\right\}=\left\{ 0\right\}\cup\left\{ \left\{ 0\right\} \right\}=\left\{ 0,\left\{ 0\right\}\right\}$ ,
$3=2\cup\left\{ 2\right\}=\left\{ 0,\left\{ 0\right\},\left\{ 0,\left\{ 0\right\}\right\} \right\}$ ,
$4=3\cup\left\{ 3\right\}=\left\{ 0,\left\{ 0\right\},\left\{ 0,\left\{ 0\right\}\right\},\left\{ 0,\left\{ 0\right\},\left\{ 0,\left\{ 0\right\}\right\} \right\} \right\}$ ,
･･････
You will understand
$1=\left\{0 \right\}$ ,
$2=\left\{0,1 \right\}$ ,
$3=\left\{0,1,2 \right\}$ ,
$4=\left\{0,1,2,3 \right\}$ ,
･･････
We are able to get all natural numbers.
2016/04/05
axiomatic sets 13 (functions)
We can get a definition of a function,
adding conditons to the definition of a relation.
As a relation is a set, a function is a subset of a relation.
'$F$' is a function if and only if '$F$' is a relation and
\[ \forall x\forall y\forall z[xFy\wedge xFz\rightarrow y=z] \]
It means there exists only one $y$ in a function $xFy$ for giving $x$.
This definition of a function is equivalent to
\[ \forall z[z\in F \rightarrow E!(y)[z=<x,y>\rightarrow xFy]] \]
,where $<x,y>$ is an ordered pair.
In general, the realtion $\lt(\gt)$ are not a function
because when $x\lt(\gt) y$ and $x\lt(\gt) z$ , we are not able to say $y=z$ .
You must know that in standard notations we write $xFy$ $y=F(x)$ ,
namely $y$ is $F$ of $x$ .
We are able to define a funtion as
\[ F(x)=y\leftrightarrow [E!z[xFz]\wedge xFy]\vee [\neg(E!z[xFz])\wedge y=0] . \]
In this case, for example, if
\[ F=\left\{<x,y> ; <1,2>,<1,3>,<3,4>,<4,4> \right\} \]
,then
\[ F(1)=0,\qquad F(2)=0,\qquad F(3)=4,\qquad F(4)=4 . \]
You know $\left\{ \right\}=0 $ .
adding conditons to the definition of a relation.
As a relation is a set, a function is a subset of a relation.
'$F$' is a function if and only if '$F$' is a relation and
\[ \forall x\forall y\forall z[xFy\wedge xFz\rightarrow y=z] \]
It means there exists only one $y$ in a function $xFy$ for giving $x$.
This definition of a function is equivalent to
\[ \forall z[z\in F \rightarrow E!(y)[z=<x,y>\rightarrow xFy]] \]
,where $<x,y>$ is an ordered pair.
In general, the realtion $\lt(\gt)$ are not a function
because when $x\lt(\gt) y$ and $x\lt(\gt) z$ , we are not able to say $y=z$ .
You must know that in standard notations we write $xFy$ $y=F(x)$ ,
namely $y$ is $F$ of $x$ .
We are able to define a funtion as
\[ F(x)=y\leftrightarrow [E!z[xFz]\wedge xFy]\vee [\neg(E!z[xFz])\wedge y=0] . \]
In this case, for example, if
\[ F=\left\{<x,y> ; <1,2>,<1,3>,<3,4>,<4,4> \right\} \]
,then
\[ F(1)=0,\qquad F(2)=0,\qquad F(3)=4,\qquad F(4)=4 . \]
You know $\left\{ \right\}=0 $ .
2016/03/30
coffee break 12 (movies)
Recently, I watched several movies below at a time.
Nodame Cantabile (She plays 'cantabile')
Minna ESPER dayo (Everyone is ESPERs)
Umimachi diary (Our little sister)
Biri gal（With flying colors）
Himizu (Mole)
Mamma mia
American Sniper
Star Wars ep.7 The force awakens
Watashi no otoko (My man)
Jigoku de naze warui (Why don't you play in hell?)
All is good and very interesting.
No.1 is 'Our little sister'.
The film was awarded the Japan Academy Award.
Its original story is the comic book and not finished.
The contents of the movie have about 5/7 on the published comic books.
'Umimachi' means a town nearby the shore in japanese language.
The movie makes me feel the atmosphere of the town and
the sea breeze comfortably.
Star Wars ep.7 disappointed me a little than old films.
Nodame Cantabile (She plays 'cantabile')
Minna ESPER dayo (Everyone is ESPERs)
Umimachi diary (Our little sister)
Biri gal（With flying colors）
Himizu (Mole)
Mamma mia
American Sniper
Star Wars ep.7 The force awakens
Watashi no otoko (My man)
Jigoku de naze warui (Why don't you play in hell?)
All is good and very interesting.
No.1 is 'Our little sister'.
The film was awarded the Japan Academy Award.
Its original story is the comic book and not finished.
The contents of the movie have about 5/7 on the published comic books.
'Umimachi' means a town nearby the shore in japanese language.
The movie makes me feel the atmosphere of the town and
the sea breeze comfortably.
Star Wars ep.7 disappointed me a little than old films.
2016/03/02
axiomatic sets 12 (relations)
In axiomatic set theory, a relation is also a set.
Ordinally we will look at what is a relation between the element $x$ and $y$ .
For instance, in real numbers an ordered relation is satisfied.
Namely, in an arbitrary pair $\lt x,y\gt$ of two elements of real numbers
only one of the three relations
\[ x=y,\quad x\lt y,\quad x\gt y \]
must occure.
However, in axiomatic set theory we are interested in
the set $R$ which all $\left\{ \lt x,y\gt \right\}$ belongs to.
Please note that there is a set before a relation.
If $R$ is a binary relation,
\[ \forall z[z\in R\rightarrow \exists x\exists y[z=\lt x,y\gt ] ] . \]
,where $\lt x,y\gt$ is an ordered pair, which has been explained in the preceding post.
We will state it $xRy$ . $ xRy$ if and only if $\lt x,y\gt\in R$ .
If $R$ is a ternary relation,
\[ \forall z [z\in R\rightarrow \exists x\exists y\exists w[z=\lt\lt x,y\gt,w\gt ] ] . \]
In addition, well-known properties of a relation are as follow;
$R$ is reflexive in $A$ if and only if $\forall z[z\in A\rightarrow zRz ]$ .
$R$ is symmetric in $A$ if and only if $\forall x\forall y[x,y\in A , xRy \rightarrow yRx] $
$R$ is transitive in $A$ if and only if $\forall x\forall y\forall z[x,y,z\in A , xRy , yRz\rightarrow xRz] $
You must know the relation which satisfies above three conditions is an equivalence relation "=" on the set.
(This is not an axiom.)
Ordinally we will look at what is a relation between the element $x$ and $y$ .
For instance, in real numbers an ordered relation is satisfied.
Namely, in an arbitrary pair $\lt x,y\gt$ of two elements of real numbers
only one of the three relations
\[ x=y,\quad x\lt y,\quad x\gt y \]
must occure.
However, in axiomatic set theory we are interested in
the set $R$ which all $\left\{ \lt x,y\gt \right\}$ belongs to.
Please note that there is a set before a relation.
If $R$ is a binary relation,
\[ \forall z[z\in R\rightarrow \exists x\exists y[z=\lt x,y\gt ] ] . \]
,where $\lt x,y\gt$ is an ordered pair, which has been explained in the preceding post.
We will state it $xRy$ . $ xRy$ if and only if $\lt x,y\gt\in R$ .
If $R$ is a ternary relation,
\[ \forall z [z\in R\rightarrow \exists x\exists y\exists w[z=\lt\lt x,y\gt,w\gt ] ] . \]
In addition, well-known properties of a relation are as follow;
$R$ is reflexive in $A$ if and only if $\forall z[z\in A\rightarrow zRz ]$ .
$R$ is symmetric in $A$ if and only if $\forall x\forall y[x,y\in A , xRy \rightarrow yRx] $
$R$ is transitive in $A$ if and only if $\forall x\forall y\forall z[x,y,z\in A , xRy , yRz\rightarrow xRz] $
You must know the relation which satisfies above three conditions is an equivalence relation "=" on the set.
(This is not an axiom.)
2016/02/18
axiomatic sets 11 (making a subset)
Making a subset $y$ of a set $x$ , we will use a formula $P(z)$ in the preceding post .
It is a statement of properties which all elements $z$ of the subset $y$ has.
\[ \forall x\exists y \forall z[z\in y \leftrightarrow z\in x\wedge P(z)] \]
That is to say,
\[ z\in y=\left\{z\in x : P(z) \right\} \]
It is called axiom schema of separation, abstruction or subset.
It is not an axiom because there can be many statements $P(z)$ .
Although $P(z)$ must satisfy the promise, as we can not write all $P(z)$ of subsets,
Axiom schema has been used.
As a simple example, let $P(z)$ be $z\in w$ . Then,
\[ y=\left\{ z\in x : z\in w \right\} . \]
We call $y$ the intersection of $x$ and $w$ , and
we write $y=x\cap w$ .
Therefore, $x\cap w=w\cap x$ .
You must note that $\left\{z\in x : P(z) \right\}$ is different to $\left\{z : P(z) \right\}$ .
Axiom schema of separation is very important for avoiding the paradoxes.
It is a statement of properties which all elements $z$ of the subset $y$ has.
\[ \forall x\exists y \forall z[z\in y \leftrightarrow z\in x\wedge P(z)] \]
That is to say,
\[ z\in y=\left\{z\in x : P(z) \right\} \]
It is called axiom schema of separation, abstruction or subset.
It is not an axiom because there can be many statements $P(z)$ .
Although $P(z)$ must satisfy the promise, as we can not write all $P(z)$ of subsets,
Axiom schema has been used.
As a simple example, let $P(z)$ be $z\in w$ . Then,
\[ y=\left\{ z\in x : z\in w \right\} . \]
We call $y$ the intersection of $x$ and $w$ , and
we write $y=x\cap w$ .
Therefore, $x\cap w=w\cap x$ .
You must note that $\left\{z\in x : P(z) \right\}$ is different to $\left\{z : P(z) \right\}$ .
Axiom schema of separation is very important for avoiding the paradoxes.
（ This is the 100th post. ）
2016/02/04
axiomatic sets 10 (a formula)
In naive set theory, we wrote the set of real numbers in the interval $[0,1]$
like $\left\{x\in\mathcal{R} | x\in [0,1] \right\}$ .
The formula $x\in [0,1]$ is the property or the condition which the elements $x$ have.
A set is a collection of some objects.
(there are also some cases in which a collection is not a set and becomes a class. )
We have to state various kinds of properties which elements of a set have for defining the set.
It is a formula $P(x)$ .
IF $P(x)$ is a formula (a statement of properties of $x$ ),
then logical notations below can be only accepted in $P(x)$ ;
belongs to : $\in$
or : $\vee$
and : $\wedge$
not : $\neg$
If then : $\rightarrow$
for all : $\forall$
for any : $\exists$
Occasionally, $E!(A)$ means " there exists only one A" .
Of course, new notations which have been derived by above are possible.
For example, $\subset$ , $\cup$ ,and so on.
We can also write $\neg(x\in a)$ $x\notin a$ .
This is not an axiom. It is a promise.
like $\left\{x\in\mathcal{R} | x\in [0,1] \right\}$ .
The formula $x\in [0,1]$ is the property or the condition which the elements $x$ have.
A set is a collection of some objects.
(there are also some cases in which a collection is not a set and becomes a class. )
We have to state various kinds of properties which elements of a set have for defining the set.
It is a formula $P(x)$ .
IF $P(x)$ is a formula (a statement of properties of $x$ ),
then logical notations below can be only accepted in $P(x)$ ;
belongs to : $\in$
or : $\vee$
and : $\wedge$
not : $\neg$
If then : $\rightarrow$
for all : $\forall$
for any : $\exists$
Occasionally, $E!(A)$ means " there exists only one A" .
Of course, new notations which have been derived by above are possible.
For example, $\subset$ , $\cup$ ,and so on.
We can also write $\neg(x\in a)$ $x\notin a$ .
This is not an axiom. It is a promise.
2016/01/12
axiomatic sets 9 (power set)
The definition of a subset has be explained in the preceding post. Here, we shall use the definition.
If a set have many elements, there will be many subsets of the set.
Given a set $A$ which is not empty, a set whose elements are all subsets of $A$ is called the power set of $A$.
It is a set and expressed $\mathcal{A}$ .
\[ \forall x\exists z \forall w[w\in z\leftrightarrow w\subset x] \]
It is axiom of power set. If you do not want to use the symbol $\subset$ ,
\[ \forall x\exists z \forall w[w\in z\leftrightarrow \forall y[y\in w\rightarrow y\in x ]] . \]
If $A=\left\{ x,y \right\}$ , then all subsets of $a$ is $\left\{ x \right\}, \left\{ y \right\},$ and $\left\{ x,y \right\}$ .
Therefore, $\mathcal{A}=\left\{ \phi, \left\{ x \right\}, \left\{ y \right\}, \left\{ x,y \right\} \right\}$ .
As you see, if the number of elements of a set is $n$ , the number of the elements of the power set is $2^n$ .
(including empty set $\phi$.)
As $\phi=0$ , the power set of $\left\{ 0,1 \right\}$ is $\left\{ 0,\left\{ 0,1 \right\} \right\}=\left\{ 0, \left\{ 1 \right\} \right\}$ .
The power set of $\phi$ is $\left\{ 0 \right\}$ ,
and the power set of the power set of $\phi$ is $\left\{ 0, \left\{ 0 \right\} \right\}=\left\{ 0, 1 \right\}$ .
If a set have many elements, there will be many subsets of the set.
Given a set $A$ which is not empty, a set whose elements are all subsets of $A$ is called the power set of $A$.
It is a set and expressed $\mathcal{A}$ .
\[ \forall x\exists z \forall w[w\in z\leftrightarrow w\subset x] \]
It is axiom of power set. If you do not want to use the symbol $\subset$ ,
\[ \forall x\exists z \forall w[w\in z\leftrightarrow \forall y[y\in w\rightarrow y\in x ]] . \]
If $A=\left\{ x,y \right\}$ , then all subsets of $a$ is $\left\{ x \right\}, \left\{ y \right\},$ and $\left\{ x,y \right\}$ .
Therefore, $\mathcal{A}=\left\{ \phi, \left\{ x \right\}, \left\{ y \right\}, \left\{ x,y \right\} \right\}$ .
As you see, if the number of elements of a set is $n$ , the number of the elements of the power set is $2^n$ .
(including empty set $\phi$.)
As $\phi=0$ , the power set of $\left\{ 0,1 \right\}$ is $\left\{ 0,\left\{ 0,1 \right\} \right\}=\left\{ 0, \left\{ 1 \right\} \right\}$ .
The power set of $\phi$ is $\left\{ 0 \right\}$ ,
and the power set of the power set of $\phi$ is $\left\{ 0, \left\{ 0 \right\} \right\}=\left\{ 0, 1 \right\}$ .
2016/01/07
coffee break 11 (happy new year)
Happy new year
Recently I am playing middle-earth:shadow of mordor goty on steam.
The game is action RPG with open world. However, RPG elements are a little bit.
The action style is based on combats of Batman and wall climbing and stealth kills of Assassin Creed
These are very fun and feel good.
Player's character death is usually occurred in an action game.
Retrying the scene of the game, as player's skill becomes good, players can advance the game.
Then, there are normally no death penalties in action games.
However, there is a death penalty in this game and it is very interesting.
When player's character was slain, the enemies become level up and stronger.
Therefore, player's character can not afford to die.
Now as my character has been dead many times, I am afraid the clear of the game.
Recently I am playing middle-earth:shadow of mordor goty on steam.
The game is action RPG with open world. However, RPG elements are a little bit.
The action style is based on combats of Batman and wall climbing and stealth kills of Assassin Creed
These are very fun and feel good.
Player's character death is usually occurred in an action game.
Retrying the scene of the game, as player's skill becomes good, players can advance the game.
Then, there are normally no death penalties in action games.
However, there is a death penalty in this game and it is very interesting.
When player's character was slain, the enemies become level up and stronger.
Therefore, player's character can not afford to die.
Now as my character has been dead many times, I am afraid the clear of the game.