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Surreal Number. Tianruo Chen. Introduction. In mathematics system, the surreal number system is an arithmetic continuum containing the real number as infinite and infinitesimal numbers. Construction of surreal number.

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## Surreal Number

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**Surreal Number**Tianruo Chen**Introduction**• In mathematics system, the surreal number system is an arithmetic continuum containing the real number as infinite and infinitesimal numbers.**Construction of surreal number**• Surreal number is a pair of sets of previously created surreal number. • If L and R are two sets of numbers, and no member of L is ≥ any member of R, then we get a number {L|R} • We can construct all numbers in this way • For example, • { 0 | } = 1 • { 1 | } = 2 • { 0 | 1 } = 1/2 • { 0 | 1/2 } = 1/4**Convention**• If x={L|R}, we write xLfor the typical member of L and xRfor the typical member of R. • So we can write {xL|xR} to represent x itself.**Definition**• Definition 1 • Wesayx ≥yiffno xR≤ y and x≤ no yL • Definition 2 • x=y iff x ≥ y and y ≥ x • x>y iffx≥y and y is not more than or equal to x**Let’s construct the surreal numbers**• Every number has the for {L|R} based on the construction. • But what do we have at the beginning? Since initially there will be no earlier constructed number. • The answer is that there is a certain set of number named the empty set Ø. • So the earliest number can only be {L|R} where L=R=Ø. In the simplist notation { | }. We call this number 0.**Is the surreal number well-formed**• We have mentioned that no member of L is ≥ any member of R. • We call the number well-formed if it satisfies this requirement. • So are any members of the right set less than or equal to any members of the left set? • Since both the sets are empty for { | }. It doesn’t matter here.**The construction of -1 and 1**• We can create 3 new numbers now. • {0| }, { |0} and {0|0} • Since the last number {0|0} is not well-formed, because 0≤0. We only have 2 appropriate surreal number {0| } and { |0}. • Here we call 1={0| } and -1={ |0}. • We can prove that -1= -1, -1<0, -1<1, 0<1, 1=1 For example, Is -1≥ 1? -1≥1 iff no -1R ≤ 1 and -1≤ no 1L But 0≤1 and -1≤0 , So we don’t have -1≥1**The Construction of 2,½,-2,-½**• As we find before, -1<0<1 • And we have particular set • { }, {-1},{0},{1},{-1,0},{-1,1},{0,1},{-1,0,1} • We use it for constructing surreal number with L and R • { |R}. {L| }, {-1|0}, {-1|0,1}, {-1|1}, {0|1},{-1,0|1} • We define {1| }=2, {0 |1}=½ • And For number x={ 0,1| }, 0<x and 1<x, since 1<x already tells us 0<1<x, the entry 0 didn’t tell us anything indeed. So x={0,1| }={1| }=2**The Construction of 2,½,-2,-½**0={−1 | }={ | 1} ={-1| 1} 1={−1, 0 | } 2={0, 1 | } = {−1, 1 | } ={−1, 0, 1 | } -1={ | 0, 1} −2={ | − 1, 0} = { | − 1, 1} ={ | − 1, 0, 1} ½={−1, 0 | 1} -½={−1 | 0, 1}**Arithmetical operation**• Definition of x＋y • x+y={xL+ y,x+yL| xR+ y, x + yR} • Definition of –x • -x = { -xR | -xL } • Definition of xy. • xy= {xLy + xyL– xLyL, xRy + xyr –xRyR|xLy +xyR –xLyR,xRy + xyL-xRyL}**The number {Z| }**• Since • Because 0 is in Z, 1={0| } and -1={ |0} are also in Z. Therefore, all numbers born from these previous number set are in Z. Then we can create a new surreal number {Z| } • What is the value of it? • It is a number that greater than all integers. It’s value is infinity. We use Greek letter ω to denote it**Red-Blue HackenbushGame**• Rule: • There are two players named “Red” and “Blue” • Two players alternate moves, Red moves by cutting a red segment and Blue, by cutting a blue one • Whenaplayerisunabletomove,heloses. • A move consists of hacking away one of the segments, and removing that segment and all segments above it that are not connected to the ground.**Analyzing Games**• Everygamehastoendwithawinneroraloserandwheretherearefinitenumberofpossiblemovesandthegamemustendinfinitetime. • Let’sconsideraboutthefollowingHackenbushGame • AndweassumethatBluemakesthefirstmove，there are sevenpossiblemoveswecanreach.**Thetreeforthegame**• We cannowdrawthefollowingcompletetreeforthegame • Inthiscase,ifRedplaycorrectly,hecanalwayswinifBluehasthefirstmove.**SomeFractionalGames**• Let’sassumethatcomponentsof positionsaremade of entirely n blue segments, it will have a value of +n, and if there are n red segments, it will have the value of –n. In the picture (A), the blue has exactly 1 move, so it can be assigned the value of +1. However, in all other four diagrams, blue can win whether he starts first or not and Red has more and more options. So what is the value of the other four pictures?**SomeFractionalGames**• Let’s consider the picture (F), it has a value of 0, since whoever moves first will lose. And the red segments has the value of -1. This meant that two copies of picture(B) has the sum value of +1. So it the picture (B) has the value ½ . • And consider about the picture (G), we can get the picture(C) has the value ¼ .**Finding a Game’s Value**• Let’s consider the following game. • We can find that the value of the picture is +1. (three blue moves for +3 and 2 red moves for -2) • If blue need to move, the remaining picture will have values of 0,-1 and -2. If the red move first the remaining picture will have value of +2 and +3. • V ={ B1,B2,…,Bn|R1,R2,…,Rm}**Finding a Game’s Value**• For the sample game above,we can write the value as: {−2, −1, 0|2, 3}. • We can ignore the “bad” moves and all that really concerns us are the largest value on the left and the smallest value on the right: {−2, −1, 0|2, 3} = {0|2} = 1**Calculating a Game Value**• Let’s work out the value of the following HackenbushGame. The pair of games on the left show all the possibility that can be obtained with a blue move and the ones on the right are from a red move.**Calculating a Game’s Value**• From previous work, we know the game values of all the components except for the picture (B) • Repeating the previous steps,weget: • Weknowthatthevalueof(C),(D),(E)and(F)are+½,+¼,0and+1. • SowegetthevalueofB={+½,0|+1,+1}=+¾ • ValueofA={+¼,0|+¾,+½}=+⅜**Thankyouforlistening**Reference: • [Conway, 1976] Conway, J. H. (1976). On Numbers and Games. Academic Press, London, New York, San Francisco. [Elwyn R. Berlekamp, 1982] Elwyn R. Berlekamp, John H. Conway, R. K. G. (1982). Winning Ways, Volume 1: Games in • General. Academic Press, London, New York, Paris, San Diego, San Francisco, Sa ̃o Paulo, Sydney, Tokyo, Toronto. • [Knuth, 1974] Knuth, D. E. (1974). Surreal Numbers. Addison-Wesley, Reading, Massachusetts, Menlo Park, California, London, Amsterdam, Don Mills, Ontario, Sydney. • Hackenbush.Tom Davis http://www.geometer.org/mathcircles December 15, 2011

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