Vyacheslav Telnin. Search for New Numbers.


 Cordelia Anthony
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1 Vycheslv Telnin Serch for New Numbers. 1
2 CHAPTER I 2
3 I.1 Introduction. In 1984, in the first issue for tht yer of the Science nd Life mgzine, I red the rticle "NonStndrd Anlysis" by V. Uspensky, in which the notion of hyperrel numbers hd been introduced. And, lredy somewhere in June 1984, I thought "There ws time when they were not known bout. And now lerned. So mybe there re other numbers tht we don't know now, nd then we'll find out. And is it possible not to wit for this, but somehow to find out bout these unknown now numbers by myself?. I.2 The help of dilectics. Dilectics sys, tht the subject should be studied in development. In this cse, the subject is numbers. And their development cn be trced bck to my own experience  wht numbers, nd in wht order I studied t school. I.3 Numbers, studied t school. First we studied numbers: 1, 2, 3,...  they re clled nturl. Then zero nd negtive numbers. Nturl, zero nd negtive re clled integers. Then there were frctionl numbers. Integers nd frctionl numbers re clled rtionl numbers. After them we studied the irrtionl numbers. Rtionl numbers together with irrtionl numbers re clled rel numbers. They were followed by imginry numbers. Rel numbers together with imginry numbers re clled complex numbers. I.4 The direct opertions, studied in school. 3
4 There re mny opertions, studied in the school. But the min ones re ddition, multipliction, exponentition. They will be clled "direct opertions". We introduce the following nottion: [1] ddition, [2] multipliction, [3]  exponentition. Also, we will introduce the conventionl symbols of these three "direct opertions": [1] "+", [2] "*", [3] "^". I.5 The first direct opertion. Tke ny two nturl numbers:, b. And pply to them the first direct opertion. The result is lso nturl number c: [1] b = c ( + b = c). (I.5.1) [1] b c Tble I.1 I.6 The second direct opertion. Tke ny two integers:, b. And pply to them the second direct opertion. The result will lso be n integer c: [2] b = c ( * b = c). (I.6.1) [2] b c Tble I.2 I.7 The third direct opertion. 4
5 Tke ny two nturl numbers:, b. And pply to them the third direct opertion. The result is lso nturl number  c: [3] b = c ( ^ b = c). (I.7.1) [3] b c Tble I.3 I.8 The definition of inverse opertions. Let direct opertion n be defined on some set of numbers: [n] b = c; (I.8.1) (Here, b re numbers from this set. They re clled opernds.) If in (I.8.1) we rerrnge the first opernd "" with the result "c", we get: c [n]1] b = ; (I.8.2) This is the definition of the first inverse opertion for n direct opertion [n]1]. If in (I.8.1) we rerrnge the second opernd "b" with the result "c", we get: [n]2] c = b; (I.8.3) This is the definition of the second inverse opertion for n direct opertion [n]2]. I.9 The first inverse opertion for the first direct opertion. Now for [1]1] we build Tble I.4 from Tble I.1 (using (I.8.1) nd (I.8.2)): 5
6 [1]1] b c Tble I.4 We see from Tble I.4, tht [1]1] coincides with "". Tht is: = c [1]1] b = c  b; (I.9.1) But not ll the cells in Tble I.4 re filled in yet. Let's try to find such nturl numbers с nd b, so tht too ws nturl number. And tht's wht we get: [1]1] b c Tble I.5 There re no numbers in the upper right corner of Tble I.5 mong nturl numbers. These numbers: 22, 33, 44, 23, 34, 24. Tht is, opertion [1]1] is not equl to opertion [1] on the set of nturl numbers. To chieve equlity with opertion [1], it is necessry to move to wider rnge of numbers thn nturl 6
7 numbers. To integers. These re nturl numbers, zero, nd negtive numbers. Then ny pir of integers c nd b for opertion [1]1] gives n integer. And for opertion [1] ny pir of integers nd b lso gives n integer. And then opertions [1]nd [1]1] become equl. And empty cells in Tble 5 cn be filled with new numbers: 0, 1, 2. Then we get Tble I.6: [1]1] b c Tble I.6 I.10 The second inverse opertion for the first direct opertion. Now, for [1]2], let's build the Tble I.7 from the Tble I.1 (using (I.8.1) nd (I.8.3)): [1]2] c b Tble I.7 From (I.8.2), (I.8.3), Tble I.4, nd Tble I.7, we see tht b = [1]2] c = c [1]1] = c  ; (I.10.1) 7
8 To populte the Tble I.7 completely, you hve numbers, c be tken from the integers. Then b will lso be n integer. And opertion [1]2] becomes equl with opertion [1]. And then we get Tble I.8: [1]2] c b Tble I.8 We now hve: x [1] y = x + y; (I.10.2) x [1]1] y = x  y; (I.10.3) x [1]2] y = y  x; (I.10.4) I.11 Connection of opertion [1] with opertion [2]. The opertion of ddition is tught t school first. Then move on to the study of the second opertion  multipliction. Let's look t some exmples of these opertions = 3 * 2 = 6; (I.11.1) = 3 * 3 = 9; (I.11.2) = 2 * 4 = 8; (I.11.3) We rewrite these exmples in the new nottion: 3 [1] 3 = 3 [2] 2 = 6; (I.11.4) 3 [1] 3 [1] 3 = 3 [2] 3 = 9; (I.11.5) 8
9 2 [1] 2 [1] 2 [1] 2 = 2 [2] 4; (I.11.6) The multipliction opertion is formed s follows: to the left of the multipliction opertion, the summnd is written, nd to the right of it, the number of these summnds is written. I.12 The first inverse opertion for the second direct opertion. Second direct opertion [2]: [2] b = c; (I.12.1) From (I.8.2) for n = 2 we obtin definition for [2]1]: c [2]1] b = ; (I.12.2) Using Tble I.2, we obtin Tble I.9 for [2]1]: [2]1] b c Tble I.9 Tble I.9 shows tht [2]1] is division / : = c [2]1] b = c / b; (I.12.3) But not ll the cells in Tble I.9 re filled in yet. Let's look for c nd b mong integers so tht is lso integer. And we get Tble I.10: 9
10 [2]1] b c Tble I.10 For empty cells in Tble I.10 there were no numbers mong integers. Tht is, opertion [2]1] is not equl to opertion [2] on the set of integers. Equlity of these opertions cn be chieved by introducing new frctionl numbers. Then ll empty cells in Tble I.10 cn be filled with new frctionl numbers. And we get Tble I.11: [2]1] b 1 1 1/2 1/ / / / /2 5/ /2 7/ / /2 3 c Tble I.11 Integers together with frctionl numbers re clled rtionl numbers. And for ny pir of rtionl numbers, b in (I.12.1) there 10
11 will be rtionl number c. Also for ny pir of rtionl numbers c nd b in (I.12.2) there will be rtionl number. Tht is, opertions [2] nd [2]1] becme equl on the set of rtionl numbers. I.13 The second inverse opertion for the second direct opertion. Second direct opertion [2] of (I.8.1) t n = 2: [2] b = c; (I.13.1) From (I.8.3) for n = 2 we obtin definition for [2]2]: [2]2] c = b; (I.13.2) Using Tble I.2, we obtin Tble I.12 for [2]2]: [2]2] c b Tble I.12 This tble shows tht b = c / ; (I.13.3) But not ll the cells in Tble I.12 re filled in yet. Let's look for c nd mong integers so tht b is lso integer. And we get tble I.13: [2]2] c b Tble I.13 11
12 For empty cells in Tble I.13 there were no numbers mong integers. Tht is, opertion [2]2] is not equl to opertion [2] on the set of integers. Equlity of these opertions cn be chieved by introducing new frctionl numbers. Then ll empty cells in Tble I.13 cn be filled with new frctionl numbers. And we get Tble I.14: [2]2] c /2 1 3/2 2 5/2 3 7/2 4 9/2 3 1/3 2/3 1 4/3 5/3 2 7/3 8/3 3 b Tble I.14 Integers together with frctionl numbers re clled rtionl numbers. And for ny pir of rtionl numbers, b in (I.13.1) there will be rtionl number c. Also for ny pir of rtionl numbers nd c in (I.13.2) there will be rtionl number b. Tht is, opertions [2] nd [2]2] becme equl on the set of rtionl numbers. Given (I.12.3), (I.13.2), (I.13.3) we obtin: b = [2]2] c = c / = c [2]1] ; (I.13.4) We now hve: x [2] y = x * y; (I.13.5) x [2]1] y = x / y; (I.13.6) x [2]2] y = y / x; (I.13.7) I.14 Connection of opertion [2] with opertion [3]. 12
13 The opertion of exponentition  [3] is studied in school fter multipliction [2]. These opertions re closely relted to ech other. Here re some exmples of this connection: 3 * 3 = 3 ^ 2 = 9; (I.14.1) 3 * 3 * 3 = 3 ^ 3 = 27; (I.14.2) 2 * 2 * 2 * 2 = 2 ^ 4 = 16; (I.14.3) We rewrite these exmples in the new nottion: 3 [2] 3 = 3 [3] 2 = 9; (I.14.4) 3 [2] 3 [2] 3 = 3 [3] 3 = 27; (I.14.5) 2 [2] 2 [2] 2 [2] 2 = 2 [3] 4 = 16; (I.14.6) The exponentition opertion is formed s follows: the fctor is written to the left of the exponentition opertion sign, nd the number of these fctors is written to the right of this sign. I.15 The first reverse opertion for the third direct opertion. Now we will build for [3]1] Tble I.15 from the Tble I.3 (using (I.8.1) nd (I.8.2)): [3]1] b c Tble I.15 13
14 Tble I.15 shows tht [3]1] coincides with the root: = root of degree b of c; (I.15.1) Not ll cells in Tble I.15 re occupied. Let's try to occupy them with rtionl numbers. Then we get Tble I.16: [3]1] b c Tble I.16 Not ll the cells of the Tble I.16 were filled with rtionl numbers. Tht is, opertion [3]1] is not equl to opertion [3] on the set of rtionl numbers. We hve to introduce new numbers here the squre root of 2, the root of the third degree of 2, the squre root of 3, the root of the third degree of 3, nd so on. These numbers re clled irrtionl. They together with rtionl numbers re clled rel numbers. And now ll the empty cells in Tble I.16 re filled. And now we will build Tble I.17: [3]1] b c Tble I.17 One cell in the top row of Tble I.17 could not be filled not only with rtionl, but even with rel number. Tht is, opertion 14
15 [3]1] is still unequl to opertion [3] even on the set of rel numbers. Let's introduce nother number  the root of the second degree from 1. It is clled "imginry unit", nd is usully denoted by the letter i: 1 [3]1] 2 = i; (I.15.2) i ^ 2 = 1 < 0; (I.15.3) If the imginry unit is multiplied by the rel number b, you get nother new imginry number. Tht is, if it is rised to the second degree, it will lso be negtive number: (i * b) ^ 2 =  (b * b) < 0; (I.15.4) If rel numbers re dded with imginry numbers, then complex numbers will be obtined. Now we cn fill n empty cell in the top row of Tble I.17 with the imginry unit i nd get Tble I.18: [3]1] b 11 i c Tble I.18 And now fter the introduction of irrtionl numbers, nd imginry numbers opertion [3]1] becme equl to opertion [3]: For ny complex numbers c, b for opertion [3]1] there is lwys complex number. And for ny complex, b for opertion [3] there is lwys complex number c. 15
16 I.16 The second reverse opertion for the third direct opertion. From (I.8.1), (I.8.3) for n = 3 we hve definitions of [3] nd [3]2] opertions: [3] b = c; (I.16.1) [3]2] c = b; (I.16.2) Bsed on these definitions nd Tble I.3, we construct Tble I.19: [3] c b Tble I.19 Tble I.19 shows tht opertion [3]2] coincides with the logrithm: b = logrithm (c) for bse ; (I.16.3) Let's try to fill in the empty cells in Tble I.19 using complex numbers. The result is presented in Tble I.20: [3] c 1 1?????? b Tble I.20 The "+" sign mrks the cells, where the known numbers re locted the rel numbers, only the cells re too smll to fit these numbers. But zero fit, nd it's stnding in two cells. 16
17 Those cells, where no known complex number cn stnd, re mrked with sing "?". It is possible, tht there is new clss of numbers. I.17 Numbers with?. We introduce into considertion the new numbers through Tble I.21: [3] c 1 1?2?3?4?8?9? b Tble I.21 Numbers?2,?3,?4, nd so on  re defined through opertion [3] s follows: 1 [3]?2 = 1 ^?2 = 2; (I.17.1) 1 [3]?3 = 1 ^?3 = 3; (I.17.2) 1 [3]?4 = 1 ^?4 = 4; (I.17.3) 1 [3]? = 1 ^? = ; (I.17.4) Here is ny complex number. And the number? will be clled "question number". I.18 Opertion [1] for the two question numbers. Let's define the ddition opertion [1] for ny two question numbers?,?b:? +?b =?c; (I.18.1) 17
18 1 ^ (? +?b) = (1 ^?) * (1 ^?b) = * b = 1 ^?( * b); (I.18.2)? +?b =?( * b); (I.18.3)? [1]?b =?( [2] b) =?c; (I.18.4) According to this formul, we build Tble I.22: [1]?1?2?3 b?1?1?2?3?2?2?4?6?3?3?6?9 c Tble I.22 I.19 The first inverse opertion [1]1] for the first direct opertion [1] on the set of question numbers. For the question numbers?,?b,?c for opertion [1] in Tble I.22, construct opertion [1]1 nd disply it in Tble I.23: [1]1]?1?2?3 b?1?1?2?2?1?3?3?1?4?2?6?3?2?9?3 c Tble I.23 Let's define the subtrction opertion "  " [1]1] for ny two question numbers?c,?b: 1 ^ (?c ?b) = (1 ^?c) * (1 ^ (?b)) = c / (1 ^?b) = c / b = = 1 ^?(c / b); (I.19.1) 18
19 ?c ?b =?(c /b); (I.19.2)?c [1]1]?b =?c ?b =?(c / b); (I.19.3) According to this formul for opertion "  " [1]1], we fill in the empty cells in Tble I.23 nd get Tble I.24: [1]1]?1?2?3 b?1?1?(1/2)?(1/3)?2?2?1?(2/3)?3?3?(3/2)?1?4?4?2?(4/3)?6?6?3?2?9?9?(9/2)?3 c Tble I.24 I.20 The second inverse opertion [1]2] for the first direct opertion [1] on the set of question numbers. According to Tble I.22, moving b nd c, we get Tble I.25: [1]2]?1?2?3?4?6?9 c?1?1?2?3?2?1?2?3?3?1?2?3 b Tble I.25 From the comprison of Tble I.25 with Tble I.23 (where the difference is:?c ?b =?) Tble I.25 shows the difference:?c ? =?b; (I.20.1) 19
20 From (I.19.2) fter replcement?b on?  we get:?c ? =?{c / ); (I.20.2)?b =? [1]2]?c =?c [1]1]? =?c ? =?(c / ); (I.20.3) And now we cn fill in the Tble I.25 empty cells nd get the Tble I.26: [1]2]?1?2?3?4?6?9?c?1?1?2?3?4?6?9?2?(1/2)?1?(3/2)?2?3?(9/2)?3?(1/3)?(2/3)?1?(4/3)?2?3??b Tble I.26 So we hve:?x [1]?y =?(x * y); (I.20.4)?x [1]1]?y =?(x / y); (I.20.5)?x [1]2]?y =?(y / x); (I.20.6) I.21 Opertion [2] for the two question numbers. We define the multipliction opertion [2] for two ny question numbers?,?b:? [2]?b =? *?b =?c; (I.21.1) 1 ^ (? [2]?b) = (1 ^?) ^?b = ^?b = 1 ^?( ^?b) = 1 ^?c; (I.21.2)?c =?( ^?b); (I.21.3) c = ^?b; (I.21.4)? [2]?b =?( [3]?b); (I.21.5) According to the formul (I.21.5), we build Tble I.27: 20
21 [2]?1?2?3?b?1?1?2?3?2?(2 [3]?1)?(2 [3]?2)?(2 [3]?3)?3?(3 [3]?1)?(3 [3]?2)?(3 [3]?3)??c Tble I.27?2 increses 1: 1 [3]?2 = 2;?(1/2) reduces 1: 1 [3]?(1/2) = 1/2;?1 sves 1: 1 [3]?1 = 1; Wht follows?1 = 1; (I.21.6) With this in mind, the Tble I.27 becomes the Tble I.28: [2]?1?2?3?b?1?1?2?3?2?2?(2 [3]?2)?(2 [3]?3)?3?3?(3 [3]?2)?(3 [3]?3)??c Tble I.28 Let's introduce b: b = ^?b; (I.21.7) b = [3]?b; (I.21.8) After tht, the Tble I.28 will tke the form of the Tble I.29: [2]?1?2?3?b?1?1?2?3?2?2?(2 2)?(2 3)?3?3?(3 2)?(3 3)??c Tble I.29 21
22 Here's useful formul: (?) * b =?( ^ b); (I.21.9) (?) [2] b =?( [3] b); (I.21.10) We'll get it out.: 1 ^ ((?) * b) = (1 ^?) ^ b = ^ b = 1 ^?( ^ b); (I.21.11) Got. We'll be using it soon. In the mentime, build logrithm: = b ^ log (b ); (I.21.12) = b ^ {log () bse b} For complex b, d we hve: d * (?b) = (?b) * d; (I.21.13) But this formul (I.21.13) we hve not checked for correctness. And here is nother useful formul: b = b ; (I.21.14) We will del with its conclusion: (21.7) + (21.12) + (21.13) + (21.9) = (I.21.14) b = ^?b = b ^ {(Log(b )) *?b} = (I.21.13) = = b ^ { (?b * Log(b ) } = (I.21.9) = = b ^?(b ^ Log(b )) = = b ^? = b ; Apply to Tble I.29 eqution (I.21.14), nd we obtin Tble I.30: 22
23 [2]?1?2?3?b?1?1?2?3?2?2?(2 2)?(2 3)?3?3?(2 3)?(3 3)??c Tble I.30 I.22 The first inverse opertion [2] 1] for the second direct opertion [2] on the set of question numbers. (I.21.1):? [2]?b =?c;? =?c [2]1]?b; (I.22.1) (I.21.4): c = ^?b; From here we hve: = root of degree?b from c ; (I.22.2) (I.22.1) + (I.22.2) = (I.22.3):? =?c [2]1]?b =?(root of degree?b from c); (I.22.3) Let's switch plces in Tble I.30? nd?c nd get Tble I.31: [2]1]?1?2?3?b?1?1?2?2?1?3?3?1?(2 2)?2?(2 3)?3?2?(3 3)?3?c? Tble I.31 Next, consider the three empty cells (?b =?1 = 1): 23
24 ? =?(root of degree 1 from 2 2); (I.22.4)? =?(root of degree 1 from 2 3); (I.22.5)? =?(root of degree 1 from 3 3); (I.22.6) Tble I.31 tkes the form of Tble I.32: [2]1]?1?2?3?b?1 1?2?2 1?3?3 1?(2 2)?(2 2)?2?(2 3)?(2 3)?3?2?(3 3)?(3 3)?3?c? Tble 32 (I.22.3):? =?c [2]1]?b =?(root of degree?b from c); Substitute (I.22.3) in Tble I.32 nd get Tble I.33: [2]1]?1?2?3?b? ?2?2 1 3+?3?3 4+ 1?(2 2)?(2 2)?2 5+?(2 3)?(2 3)?3?2?(3 3)?(3 3) 6+?3?с? Tble I.33 Becuse of the closeness I hd to enter the symbols: 1+ =?(root of degree?2 of 1); (I.22.4) 2+ =?(root of degree?2 of 1); (I.22.5) 24
25 3+ =?(root of degree?2 of 2); (I.22.6) 4+ =?(root of degree?2 of 3); (I.22.7) 5+ =?(root of degree?2 of 2 2); (I.22.8) 6+ =?(root of degree?2 of 3 3); (I.22.9) I.23 The second inverse opertion [2]2] for the second direct opertion [2] on the set of question numbers. (I.21.1):? [2]?b =?c;?b =? [2]2]?c; (I.23.1)?b = log (c) on the bsis of ; (I.23.2)? [2]2]?c = log (c) on the bsis of =?b; (I.23.3) Let's move?b nd?c in Tble I.30. Then we get Tble I.34: [2]2]?1?2?3?(2 2)?(2 3)?(3 3)?c?1?1?2?3?2?1?2?3?3?1?2?3??b Tble I.34 Let's look t the three empty cells in the top row of Tble I.34: Substitute in (I.23.3): = 1; c = 2 2; nd we get:?b =?(2 2); (I.23.4) To verify the correctness of this, rise 1 to the power (I.23.3) nd use (I.23.4): 25
26 1 ^ (log (2 2) bse 1) = 1 ^?(2 2) = 2 2; = 1; c = 2 3; nd we get:?b =?(2 3); (I.23.5) = 1; c = 3 3; nd we get:?b =?(3 3); (I.23.6) Substituting this into the Tble I.34 to get Tble I.35: [2]2]?1?2?3?(2 2)?(2 3)?(3 3)?c?1?1?2?3?(2 2)?(2 3)?(3 3)?2?1?2?3?3?1?2?3??b Tble I.35 The remining six empty cells in Tble I.35 re filled with numbers using (I.23.3), nd we get Tble I.36: [2]2]?1?2?3?2 2?2 3?3 3?c?1 1?2?3?2 2?2 3?3 3? ?2?3 6+? ?2?3??b Tble I.36 Here, the numbers tht do not fit in these six cells re indicted by the numbers with the plus sign to the right of the number:? [2]2]?c = log (c) on the bsis of =?b; (I.23.3) 1+ = logrithm (1) for bse 2 = 0; (I.23.7) 2+ = logrithm (1) for bse 3 = 0; (I.23.8) 3+ = logrithm (2) for bse 3; (I.23.9) 4+ = logrithm (3) for bse 2; (I.23.10) 26
27 5+ = logrithm (2/2) for bse 3; (I.23.11) 6+ = logrithm (3/3) on bse 2; (I.23.12) Given (I.23.7) nd (I.23.8) we obtin Tble I.37: [2]2]?1?2?3?2 2?2 3?3 3?c?1 1?2?3?2 2?2 3?3 3? ?2?3 6+? ?2?3??b Tble I.37 I.24 Opertion [3] for the two question numbers.? [3]?b = c; (I.24.1) По формуле (I.24.1), с учётом?1 = 1, построим Таблицу I.38: [3]?1?2?3?b? ?2?2?2 [3]?2?2 [3]?3?3?3?3 [3]?2?3 [3]?3? c Tble I.38 I.25 More bout question numbers. Let us return to the definition of question numbers. If is ny number (complex or question), 1 [3]? = 1 ^? = : (I.25.1)?() = log () on the bsis of 1; (I.25.2) 27
28 It cn be seen, tht the question mrk is the opertor:?() = log () on the bsis of 1 ; (I.25.3) It cn be pplied to the sme number multiple times: b =?(?()); c =?(?(?(b))); (I.25.4) And more from (I.25.1) it follows tht t > 1,? is one of infinitely lrge numbers. And t < 1,? is one of infinitely smll numbers. Question numbers cn be used in formuls long with complex numbers. They cn be multiplied by the imginry unit. Here is n exmple when the question number is represented s the sum of the imginry question, nd the ctul question summnds:?(2) =?(e) * i * PI +?(2); (I.25.5) Check this formul by rising 1 to the power of (I.25.5): 1 [3]?(2) = 1 [3] {?(e) * i * PI +?2 } 2 = { (1 [3]?e) [3] (i * PI) } * (1 [3]?2) 2 = { e [3] (i * PI) } * 22 = { 1 } * 22 = 2 Here PI = 3.14, nd i is the imginry unit, nd e = 2.7. I.26 Building new direct opertions. 28
29 So we trced the process of the emergence of new numbers, strting with nturl to complex. And found tht they rose t the introduction of reverse opertions to direct opertions. The introduction of subtrction gve zero nd negtive numbers. Introduction of division gve the frctionl numbers. The introduction of the root gve irrtionl nd imginry numbers. Their ddition gve complex numbers. This pttern ws confirmed for ll numbers. And if someone wnts to build new numbers, you cn use this pttern. But the fct is tht ll reverse opertions hve lredy been used. And ll direct opertions hve lredy been used. If we were given new direct opertion, we would build pir of inverse opertions for it, nd then pply the found pttern to build new numbers. Let us turn to dilectics gin. It sys tht the subject should be studied in development. In this cse, the subject is direct opertions. And their development cn be trced bck to our own experience  wht direct opertions, nd in wht order were studied t school. At the very beginning, we hve lredy mentioned ll the currently known direct opertions, nd numbered them in the following order: ddition [1], multipliction [2], exponentition [3]. In this order they hve been studied in school. We re now interested in the rules of building new direct opertions on the bsis of previous direct opertions. In point I.11 the construction of multipliction opertion  [2] by ddition opertion [1] is considered. And t the end of this prgrph, this rule is summrized. Here it is: "The multipliction opertion is formed s follows: to the left of the multipliction opertion, the summnd is written, nd to the right of it, the number of these summnds is written.» 29
30 In prgrph 14 we consider the construction of the opertion of exponentition [3] by opertion of multipliction [2]. And t the end of this prgrph, this rule is summrized. Here it is: "The exponentition opertion is formed s follows: the cofctor is written to the left of the exponentition opertion sign, nd the number of these cofctors is written to the right of this sign.» These two definitions cn be summrized s follows: Opertion [n+1] is formed by opertion [n] s follows: opernd for opertion [n] is written to the left of [n+1], nd the number of these opernds is written to the right of [n+1]. I.27 Construction of the fourth direct opertion  [4]. Let's build some exmples to obtin [4] by [3]: 2 [4] 2 = 2 [3] 2 = 4; (I.27.1) 3 [4] 2 = 3 [3] 3 = 27; (I.27.2) 2 [4] 4 = 2 [3] 2 [3] 2 [3] 2 = X; (I.27.3) In the lst formul there re two options the ccount in the righthnd side of the eqution to led lefttoright (get 256), or from right to left (get ). How to choose the right option? Shll we use the question number?2. Is it n infinitely lrge number ( 1 [3]?2 = 2 ): 2 [4]?2 = 2 [3] 2 [3] 2 [3] ; (I.27.4) In this formul, there is only one option from left to right. The second option (from right to left) is not possible becuse the number of opernds on the right side (I.27.4) is infinite nd there 30
31 is no right opernd. Of course, this is specil exmple, but the right option should be universl. Therefore, the eqution (I.27.3) tkes the form: 2 [4] 4 = (( 2 [3] 2 ) [3] 2 ) [3] 2 = 256; (I.27.3) Now the fourth direct opertion [4] is uniquely determined. And we'll be filling out the Tble I.39 for it.: [4] b c Tble I.39 In this tble we will put the formul: [4] b = c; (I.27.4) 1 [4] 1 =1; (I.27.5) To the right of [4] is 1, which mens, tht fter the equl sign there is only one opernd 1 nd no opertion [3]. 2 [4] 1 = 2; (I.27.6) 3 [4] 1 = 3; (I.27.7) 1 [4] 2 = 1 [3] 1 = 1; (I.27.8) 1 [4] 3 = (1 [3] 1) [3] 1 = 1 [3] 1 = 1; (I.27.9) 2 [4] 2 = 2 [3] 2 = 4; (I.27.10) 2 [4] 3 = (2 [3] 2) [3] 2 = 4 [3] 2 = 16; (I.27.11) 3 [4] 2 = 3 [3] 3 = 27; (I.27.12) 31
32 3 [4] 3 = (3 [3] 3) [3] 3 = 27 [3] 3 = ; (I.27.13) [4] b c Tble I.40 So we filled in the Tble I.39. And now it is clled the Tble I.40. And now we show shortened method of recording the fourth direct opertion [4] through the third direct opertion [3]. Tke the formul (I.27.4): [4] b = c; And let b = 5. By definition of the fourth opertion, we write this formul s: c = [4] 5 = ((( [3] ) [3] ) [3] ) [3] ; (I.27.14) Since the properties [3] nd [2] we hve: ( [3] ) [3] = [3] ( [2] ); (I27.15) Then (27.5) will look like this: c = [4] 5 = ( [3] ( [2] ) [3] ) [3] = = ( [3] ( [2] [2] )) [3] = = [3] ( [2] [2] [2] ) = [3] ( [3] 4); (I.27.16) This exmple shows the following formul: [4] b = [3] ( [3] (b 1)); (I.27.17) 32
33 This conclusion is suitble for nturl b. But perhps the finl formul (I.27.17) is lso true for wider clss of numbers. I.28 The first inverse opertion [4]1] for the fourth direct opertion [4]. Let's look t item I.8. The formul (I.8.1) for n = 4 hs the form: [4] b = c; (I.28.1) And the formul (I.8.2) for n = 4 hs the form: c [4]1] b = ; (I.28.2) Let us rerrnge in Tble I.40 nd c, nd we get Tble I.41: [4]1] b c Tble I.41 Using (I.28.1) nd (I.28.2) we will try to fill in the empty cells in Tble I.41: c = 4; b = 1; from (28.1) we obtin: = 4; (I.28.3) c = 16; b = 1; from (28.1) we obtin: = 16; (I.28.4) c = 27; b = 1; from (28.1) we obtin: = 27; (I.28.5) c = ; b = 1; from (28.1) we obtin: = ; (I.28.6) 33
34 And then the equtions for re not included in the cells of the tble, nd we will indicte these equtions by number with plus (we will use (I.27.17)): 1+ = ; c= 2; b = 2; [4] 2 = [3] = c = 2; (I.28.7) 2+ = ; c = 2; b = 3; [4] 3 = [3] ( [3] 2) = c = 2; (I.28.8) 3+ = ; c = 3; b = 2; [4] 2 = [3] = c = 3; (I.28.9) 4+ = ; c = 3; b = 3; [4] 3 = [3] ( [3] 2) = c = 3; (I.28.10) 5+ = ; c = 4; b = 3; [4] 3 = [3] ( [3] 2) = c = 4; (I.28.11) 6+ = ; c = 16; b = 2; [4] 2 = [3] = c = 16; (I.28.12) 7+ = ; c = 27; b = 3; [4] 3 = [3] ( [3] 2) = c = 27; (I.28.13) 8+ = ; c = ; b = 2; [4] 2 = [3] = c = ; (I.28.14) The completed Tble I.41 is now clled Tble I.42: [4]1] b c Tble I.42 I.29 The second inverse opertion [4]2] for the fourth direct opertion [4]. Formul (I. 8.1) t n = 4 gives: 34
35 [4] b = c; (I.29.1) Tke the formul (I. 8.3) t n = 4: [4]2] c = b; (I.29.2) Let's chnge b nd c in the Tble I.40, nd get the Tble. 43: [4]2] c 1 1, 2, b Tble I.43 Here we re gin fced with new type of numbers: numbers strings. In cell b, where = 1, c = 1 of Tble I.43 re just 3 numbers: 1, 2, 3. And, if some formul will meet this number b, it is uncler which of its three vlues (1, 2, 3) to tke for clcultions. We hve seen this in Tble I.19 nd in Tble I.20. But there we were silent bout this feture. The only thing we cn do now is to develop specil designtion for such numbersstrings, so tht it is entirely plced in one cell. If the first vlue of the numberline denote  "m", the number of vlues in the numberline denote "N", nd dditive to move to the next vlue denote "n", nd s seprtor of these symbols to tke verticl line" ", we get such record: m N n (I.29.3) In our exmple (1, 2, 3) it will be written s: (I.29.4) Now fill in the empty cells in Tble I.43 nd get Tble I.44: [4]2] c 1 1, 2, Tble I b
36 = 2; c = 1; 2 [4] b = 1; 2 [4]2] 1 = b = 1+; (I.29.5) = 3; c = 1; 3 [4] b = 1; 3 [4]3] 1 = b = 2+; (I.29.6) In equtions 1+ nd 2+, the vlues b reduce 2 nd 3 to 1. Tht mens they re infinitely smll. = 1; c = 2; 1 [4] b = 2; 1 [4]2] 2 = b = 3+; (I.29.7) = 1; c = 3; 1 [4] b = 3; 1 [4]2] 3 = b = 4+; (I.29.8) = 1; c = 4; 1 [4] b = 4; 1 [4]2] 4 = b = 5+; (I.29.9) = 1; c = 16; 1 [4] b = 16; 1 [4]2] 16 = b = 6+; (I.29.10) = 1: c = 27: 1 [4] b = 27; 1 [4]2] 27 = b = 7+; (I.29.11) = 1; c = ; 1 [4] b = ; 1 [4]2] = b = 8+; (I.29.12) In equtions from 3+ to 8+, the vlues b enlrge the 1 to 2, or to even lrger numbers. So the vlues b re infinitely lrge. = 2; c = 3; 2 [4] b =3; 2 [4]2] 3 = b = 9+; (I.29.13) = 2; c = 27; 2 [4] b = 27; 2 [4]2] 27 = b = 10+; (I.29.14) = 2; c = ; 2 [4] b = ; 2 [4]2] = b = 11+; (I.29.15) = 3; c = 2: 3 [4] b = 2; 3 [4]2] 2 = b = 12+; (I.29.16) = 3; c = 4; 3 [4] b = 4: 3 [4]2] 4 = b = 13+; (I.29.17) =3; c = 16; 3 [4] b = 16; 3 [4]2] 16 = b = 14+; (I.29.18) In equtions from 9+ to 14+ vlues increse or decrese the originl vlues, but neither the originl nor the finl vlues re equl to 1. So the vlues b in these equtions re lso finite. I.30 Question numbers. Extension of the concept to new direct opertions. 36
37 Remember the definition of question numbers in prgrph I.17: [3]? = 1 ^? = ; (I.17.4) Here is ny complex number. And the number? will be clled "question number" Now tht the fourth direct opertion hs ppered, this definition cn be generlized to ny direct opertion with number n: 1 [n] n? = ; (I.30.1) Here is ny complex or question number. And number n? will be clled "question number". The connection of the old "question numbers"? nd new "question numbers" n? is this: 3? =?; (I.30.2) 1 [3] 3? = ; (I.30.2) And for the fourth direct opertion [4] definition (I.30.1) gives: 1 [4] 4? = ; (I.30.3) For = 2 we obtin the eqution: 1 [4] 4?2 = 2; (I.30.4) Here the number 4?2 increses 1 to 2. So this is n infinitely lrge number. Number 4?3 it turns out, too, endlessly lrge number. Also infinitely lrge numbers re 4?4, 4?16, 4?27, 4?19683, nd so on. For = 1/2, we obtin the eqution: 1 [4] 4?1/2 = 1/2; (I.30.5) 37
38 In this exmple, the number 4?1/2 reduce 1 to 1/2. So it's n infinitely smll number. Number 4?1/3 is lso n infinitely smll number. Also infinitesiml numbers re 4?2/5, 4?3/7, 4?1/12 ll, which hve <1. And number 4?1 gives the eqution: 1 [4] 4?1 = 1; (I.30.6) It stores 1 nd thus is itself equl to 1: 4?1 = 1; (I.30.7) I.31 Opertion [4] for question numbers? ( 3? ). From point I.27 we tke the formul: [4] b = [3] ( [3] (b1)); (I.27.17) For the brevity of the formuls, we will use the old nottion. Recll tht?1 = 1.?1 [4]?1 =?1 [3] (?1 [3] (?1 1)) = 1 [3] (1 [3] (1 1)) = = 1 [3] (1 [3] 0) = 1 [3] 1 = 1; (I.31.1)?1 [4]?2 =?1 [3] (?1 [3] (?2 1)) = 1 [3] (1 [3] (?2 1)) = = 1 [3] ((1 [3]?2) [2] (1 [3] (1)) = 1 [3] (2 [2] 1) = = 1 [3] 2 = 1; (I.31.2)?1 [4]?3 = 1 [3] (1 [3] (?3 1)) = 1 [3] ((1 [3]?3) [2] (1 [3] (1)) = = 1 [3] (3 [2] 1) = 1 [3] 3 = 1; (I.31.3)?2 [4]?1 =?2 [3] (?2 [3] (?1 1)) =?2 [3] (?2 [3] (1 1)) = =?2 [3] ((?2 [3] 1) / (?2 [3] 1)) =?2 [3] (?2 /?2) =?2 [3] 1 = =?2; (I.31.4) 38
39 ?3 [4]?1 =?3 [3] (?3 [3] (?1 1)) =?3 [3] (?3 [3] (1 1)) = =?3 [3] ((?3 [3] 1) / (?3 [3] 1)) =?3 [3] (?3 /?3) = =?3 [3] 1 =?3; (I.31.5)?2 [4]?2 =?2 [3] (?2 [3] (?2 1)) = (По Таблице I.24:?21 =?2 [1]1]?1 =?2) =?2 [3] (?2 [3]?2); (I.31.6)?2 [4]?3 =?2 [3] (?2 [3] (?3 1)) = =?2 [3] (?2 [3]?3); (I.31.7)?3 [4]?2 =?3 [3] (?3 [3] (?2 1)) = =?3 [3] (?3 [3]?2); (I.31.8)?3 [4]?3 =?3 [3] (?3 [3] (?3 1)) = =?3 [3] (?3 [3]?3); (I.31.9) These results re summrized in Tble 45: [4]?1?2?3 b? ?2?2?2 [3] (?2 [3]?2)?2 [3] (?2 [3]?3)?3?3?3 [3] (?3 [3]?2)?3 [3] (?3 [3]?3) c Tble I.45 Tble I.45 contins the formul: [4] b = c; (I.31.10) A little note. From (I.30.4) we hve: 1 [4] 4?2 = 2; And from (I.31.2) should (given 3?2 =?2): 1 [4] 3?2 = 1; Tht mens 4?2 > 3?2; (I.31.11) 39
40 I.32 The first inverse opertion [4]1] for the fourth direct opertion [4] for question numbers? ( 3? ). Let's look t item I.8. The formul (I.8.1) for n = 4 hs the form: [4] b = c; (I.32.1) And the formul (I.8.2) for n = 4 hs the form: c [4]1] b = ; (I.32.2) We introduce the following nottions:?2 [3] (?2 [3]?2) = "222";?2 [3] (?2 [3]?3) = "223"; (I.32.3)?3 [3] (?3 [3]?2) = "332";?3 [3] (?3 [3]?3) = "333"; (I.32.4) If we rerrnge "" nd "c" in Tble I.45, we get Tble I.46 for opertion [4]1] pplied to the numbers? (3?): [4]1]?1?2?3 b ?2?2?3?3 222?2 223?2 332?3 333?3 c Tble I.46 Using the formul (I.32.1) set b = 1 nd obtin = c. Tble I.46 tkes the form of Tble I.47: 40
41 [4]1]?1?2?3 b ?2?2?3? ? ? ? ?3 c Tble I.47 Fill the empty cells in the Tble I.47 with icons 1+, 2+,..., 7+, 8+  nd get Tble I.48: [4]1]?1?2?3 b ?2? ?3? ? ? ? ?3 c Tble I.48 [4] b = c; Give the icons their corresponding vlues: c =?2; b =?2; [4]?2 =?2; 1+ = ; (I.32.5) c =?2; b =?3; [4]?3 =?2; 2+ = ; (I.32.6) c =?3; b =?2; [4]?2 =?3; 3+ = ; (I.32.7) c =?3; b =?3; [4]?3 =?3; 4+ = ; (I.32.8) 41
42 c = 222; b =?3; [4]?3 = 222; 5+ = ; (I.32.9) c = 223; b =?2; [4]?2 = 223; 6+ = ; (I.32.10) c = 332; b =?3; [4]?3 = 332; 7+ = ; (I.32.11) c = 333; b =?2; [4]?2 = 333; 8+ = ; (I.32.12) I.33 The second inverse opertion [4]2] for the fourth direct opertion [4] for question numbers? ( 3? ). The formul (I.8.1) for n = 4 gives: [4] b = c; (I.33.1) Tke the formul (I.8.3) for n = 4: [4]2] c = b; (I.33.2) We introduce the following nottions:?2 [3] (?2 [3]?2) = "222";?2 [3] (?2 [3]?3) = "223"; (I.33.3)?3 [3] (?3 [3]?2) = "332";?3 [3] (?3 [3]?3) = "333"; (I.33.4) Let's rerrnge b nd c in Tble I.45 nd get Tble I.49: [4]2] 1?2? c?1 1+?2 1?2?3?3 1?2?3 b Tble I.49 Here 1+ is numberstring: 1+ = 1,?2,?3; (I.33.5) Fill the remining empty cells of the Tble I.49 with icons 2+, 3+,..., 14+, 15+ nd get the Tble I.50: 42
43 [4]2] 1?2? c ? ?2? ? ?2?3 b Tble I.50 Give the vlues for the new icons in the Tble I.50: ( [4] b = c ) ( [4]2] c = b ) =?2; c = 1;?2 [4] 2+ = 1; (I.33.6) =?3; c = 1;?3 [4] 3+ = 1; (I.33.7) = 1; c =?2; 1 [4] 4+ =?2; (I.33.8) = 1; c =?3; 1 [4] 5+ =?3; (I.33.9) = 1; c = 222; 1 [4] 6+ = 222; (I.33.10) = 1; c = 223; 1 [4] 7+ = 223; (I.33.11) = 1; c = 332; 1 [4] 8+ = 332; (I.33.12) = 1; c = 333; 1 [4] 9+ = 333; (I.33.13) =?3; c =?2;?3 [4] 10+ =?2; (I.33.14) =?2; c =?3;?2 [4] 11+ =?3; (I.33.15) =?3; c = 222;?3 [4] 12+ = 222; (I.33.16) =?3; c = 223;?3 [4] 13+ = 223; (I.33.17) =?2; c = 332;?2 [4] 14+ = 332; (I.33.18) =?2; c = 333;?2 [4] 15+ = 333; (I.33.19) 43
44 I.34 Opertion [4] for question numbers 4?. [4] b = c; (I.34.1) From (I.30.7) tke the eqution: 4?1 = 1; Then we will get: 4?1 [4] 4?1 = 1; (I.34.2) 4?1 [4] 4?2 = 2; (I.34.3) 4?1 [4] 4?3 = 3; (I.34.4) 4?2 [4] 4?1 = 4?2; (I.34.5) 4?3 [4] 4?1 = 4?3; (I.34.6) For the formul (I.34.1) will be the Tble I.51: [4] 4?1 4?2 4?3 b 4? ?2 4?2 4?2 [4] 4?2 4?2 [4] 4?3 4?3 4?3 4?3 [4] 4?2 4?3 [4] 4?3 c Tble I.51 I.35 The first inverse opertion [4]1] to the fourth direct opertion [4] on the set of question numbers 4?. To obtin the first inverse opertion [4]1] to the fourth direct opertion [4] it is necessry to rerrnge "" nd "c" in Tble I.51: c [4]1] b = ; (I.35.1) And then we get Tble I.52: 44
45 [4]1] 4?1 4?2 4?3 b ?2 4?2 4?3 4?3 4?2 [4] 4?2 4?2 4?2 [4] 4?3 4?2 4?3 [4] 4?2 4?3 4?3 [4] 4?3 4?3 c Tble I.52 Using the formuls (I.32.1), (I.30.7) nd the Tble I.52 ( [4] 1 = c), we fill in few more cells, nd get Tble I.53: [4]1] 4?1 4?2 4?3 b ?2 4?2 4?3 4?3 4?2 [4] 4?2 4?2 [4] 4?2 4?2 4?2 [4] 4?3 4?2 [4] 4?3 4?2 4?3 [4] 4?2 4?3 [4] 4?2 4?3 4?3 [4] 4?3 4?3 [4] 4?3 4?3 c Tble 53 In the remining empty cells full expressions will not fit. So we'll fill them with short icons nd then mtch the icons to the expressions. Tble with icons will be clled Tble I.54: 45
46 [4]1] 4?1 4?2 4?3 b ?2 4? ?3 4? ?2 [4] 4?2 4?2 [4] 4?2 4?2 9+ 4?2 [4] 4?3 4?2 [4] 4? ?2 4?3 [4] 4?2 4?3 [4] 4?2 4? ?3 [4] 4?3 4?3 [4] 4? ?3 c Tble I [4] 4?2 = 1; (I.35.2) 2+ [4] 4?3 = 1; (I.35.3) 3+ [4] 4?3 = 2; (I.35.4) 4+ [4] 4?2 = 3; (I.35.5) 5+ [4] 4?2 = 4?2; (I.35.6) 6+ [4] 4?3 = 4?2; (I.35.7) 7+ [4] 4?2 = 4?3; (I.35.8) 8+ [4] 4?3 = 4?3; (I.35.9) 9+ [4] 4?3 = 4?2 [4] 4?2; (I.35.10) 10+ [4] 4?2 = 4?2 [4] 4?3; (I.35.11) 11+ [4] 4?3 = 4?3 [4] 4?2; (I.35.12) 12+ [4] 4?2 = 4?3 [4] 4?3; (I.35.13) 46
47 I.36 The second inverse opertion [4]2] to the fourth direct opertion [4] on the set of question numbers 4?. To obtin the second inverse opertion [4]2] to the fourth direct opertion [4] it is necessry to rerrnge "b" nd "c" in Tble I.51, nd we get the Tble I.55: [4]2] ?2 4?3 22? 23? 32? 33? c 4?1 1 4?2 4?3 4?2 4?1 4?2 4?3 4?3 4?1 4?2 4?3 b Tble I.55 [4] b = c; (I.36.1) [4]2] c = b; (I.36.2) In Tble 55 there re pplied the reductions: "22?"= "4?2 [4] 4?2"; "23?"= "4?2 [4] 4?3"; (I.36.3) "32?"= "4?3 [4] 4?2"; "33?"= "4?3 [4] 4?3"; (I.36.4) From (I.30.7) it is tken the eqution: 4?1 = 1; Let's set "" = 4?1 = 1 nd we will substitute in (I.36.1) different "c". We will get different equtions for "b : c = 4?2; 1 [4] b = 4?2; b = 4?(4?2); b1 = b; (I.36.5) c = 4?3; 1 [4] b = 4?3; b = 4?(4?3); b2 = b; (I.36.6) c = 22?; 1 [4] b = 22?; b = 4?(22?); b3 = b; (I.36.7) c = 23?; 1 [4] b = 23?; b = 4?(23?); b4 = b; (I.36.8) c = 32?; 1 [4] b = 32?; b = 4?(32?); b5 = b; (I.36.9) 47
48 c = 33?; 1 [4] b = 33?; b = 4?(33?); b6 = b; (I.36.10) Let's put into Tble I.55 the vlues: b1, b2, b3, b4, b5, b6 in pproprite plces, nd get Tble I.56: [4]2] ?2 4?3 22? 23? 32? 33? c 4?1 1 4?2 4?3 b1 b2 b3 b4 b5 b6 4?2 1 4?2 4?3 4?3 1 4?2 4?3 b Tble I.56 Let's fill the remining empty cells by indexes: 1+, 2+,..., 11+, 12+ nd get Tble I.57: [4]2] ?2 4?3 22? 23? 32? 33? c 4?1 1 4?2 4?3 b1 b2 b3 b4 b5 b6 4? ?2 4? ? ?2 4?3 b Tble I.57 Now let's compre its vlue to ech index: ( [4] b = c ) ( [4]2] c = b ) = 4?2; c = 1; 4?2 [4] b = 1; 1+ = b; (I.36.11) = 4?3; c = 1; 4?3 [4] b = 1; 2+ = b; (I.36.12) = 4?2; c = 2; 4?2 [4] b = 2; 3+ = b; (I.36.13) = 4?3; c = 2; 4?3 [4] b = 2; 4+ = b; (I.36.14) = 4?2; c = 3; 4?2 [4] b = 3; 5+ = b; (I.36.15) = 4?3; c = 3; 4?3 [4] b = 3; 6+ = b; (I.36.16) 48
49 In these six formuls the indexes 1+, 2+, 3+, 4+, 5+, 6+ re infinitely smll. = 4?3; c = 4?2; 4?3 [4] b = 4?2; 7+ = b; (I.36.17) = 4?2; c = 4?3; 4?2 [4] b = 4?3; 8+ = b; (I.36.18) = 4?3; c = 22?; 4?3 [4] b = 22?; 9+ = b; (I.36.19) = 4?3; c = 23?; 4?3 [4] b = 23?; 10+ = b; (I.36.20) = 4?2; c = 32?; 4?2 [4] b = 32?; 11+ = b; (I.36.21) = 4?2; c = 33?; 4?2 [4] b = 33?; 12+ = b; (I.36.22) In the lst four formuls the indexes 9+, 10+, 11+, 12+ re infinitely lrge. I.37 The construction of the fifth direct opertion  [5]. Prgrph I.26 describes the generl rule of constructing [n + 1] direct opertion by [n] direct opertion. Here it is: Opertion [n+1] is formed by opertion [n] s follows: the opernd for opertion [n] is written to the left of [n+1], nd the number of these opernds is written to the right of [n+1] You cn look t prgrph I.27  it describes the construction of the fourth direct opertion [4]. It obtins generl rule of prenthesis rrngements so, tht the clcultions re performed from left to right. In our cse, n = 4. Let's look t some exmples: 49
50 [5] b = c; (I.37.1) 2 [5] 2 = 2 [4] 2 = 4; (I.37.2) 2 [5] 3 = (2 [4] 2) [4] 2 = 4 [4] 2 = 4 [3] 4 = 256; (I.37.3) 3 [5] 2 = 3 [4] 3 = ; (I.37.4) 3 [5] 3 = (3 [4] 3) [4] 3 = [4] 3; (I.37.5) 1 [5] 1 = 1; 1 [5] 2= 1; 1 [5] 3 = 1; (I.37.6) 2 [5] 1 = 2; 3 [5] 1 = 3; (I.37.7) Let us summrize them in Tble I.58: [5] b [4] 3 c Tble I.58 Then you should ct s in the cse of the fourth opertion. And then build the sixth opertion, nd so on. This concludes Chpter I
51 CHAPTER II 51
52 II.1 Summry of Chpter I. In 1984, I red in the journl "Science nd life" rticle V. Ouspensky "nonstndrd nlysis". The hyperrel numbers were introduced in considertion in it. And, lredy somewhere in June 1984, I thought " There ws time when they did not know bout. And now lerned. So mybe there re other numbers tht we don't know now, nd then we'll find out. And is it possible not to wit for this, but somehow to find out bout these unknown numbers now?". Dilectics sys tht the subject should be studied in development. In this cse, the subject is numbers. And their development cn be trced from my own experiencewht numbers, nd in wht order I studied t school. First, we studied the numbers: 1, 2, 3,...  they re clled nturl. Then zero nd negtive numbers. Nturl, zero nd negtive re clled integers. Then there were frctionl numbers. Integers nd frctionl numbers re clled rtionl numbers. Then irrtionl numbers were studied. Rtionl numbers together with irrtionl numbers re clled rel numbers. They were followed by imginry numbers. Rel numbers together with imginry numbers re clled complex numbers. The opertions, studied in school, re mny. But the min ones re ddition, multipliction, exponentition. They will be clled  "direct opertions". We introduce the following nottions: [1] ddition, [2] multipliction, [3] exponentition. Also, we will introduce the conventionl symbols of these three "direct opertions": [1] "+", [2] "*", [3] "^". 52
53 Let direct opertion n be defined on some set of numbers: [n] b = c; (I.8.1) (Here, b re numbers from this set. They re clled opernds.) If in (8.1) we rerrnge the first opernd "" with the result "c", we get: c [n]1] b = ; (I.8.2) This is the definition of the first reverse opertion [n]1] to the direct opertion n. If in (8.1) we rerrnge the second opernd "b" with the result "c", we get: [n]2] c = b; (I.8.3) This is the definition of the second inverse opertion [n]2] for n direct opertion. So we trced the process of the emergence of new numbers, strting from nturl to complex. And found tht they occurred with the introduction of reverse opertions to direct opertions. The introduction of subtrction gve zero nd negtive numbers. The introduction of division gve the frctionl numbers. The introduction of the root gve irrtionl nd imginry numbers. Their ddition gve complex numbers. This pttern ws confirmed for ll numbers. And, if someone wnts to build new numbers, you cn use this pttern. But the fct is tht ll reverse opertions hve lredy been used. And ll direct opertions hve lredy been used. If we were given new direct opertion, we would build pir of inverse opertions for it, nd then pply the found regulrity to construct new numbers. Let us turn to dilectics gin. It sid tht 53
54 the subject should be studied in development. In this cse, the subject is the direct opertions. And their development cn be trced bck to our own experience  wht direct opertions, nd in wht order studied t school. At the very beginning, we hve lredy mentioned ll the currently known direct opertions, nd numbered them in the following order: ddition [1], multipliction [2], exponentition [3]. In this order they hve been studied in school. We re now interested in the rules of building new direct opertions on the bsis of previous direct opertions. The multipliction opertion is formed s follows: to the left of the multipliction opertion sign, the summnd is written, nd to the right of the sign the number of these summnds is written. The exponentition opertion is formed s follows: the cofctor is written to the left of the exponentition opertion sign, nd the number of these cofctors is written to the right of this sign. These two definitions cn be summrized s follows: Opertion [n+1] is formed by opertion [n] s follows: the opernd for opertion [n] is written to the left of [n+1], nd the number of these opernds is written to the right of [n+1]. 2 [4] 4 = 2 [3] 2 [3] 2 [3] 2 = X; (I.27.3) In the lst formul there re two options the ccount in the righthnd side of the eqution to led lefttoright (get 256), or from right to left (get ). How to choose the right option? We shll use the question number?2. It Is n infinitely lrge number (1 [3]?2 = 2): 54
55 2 [4]?2 = 2 [3] 2 [3] 2 [3]...; (I.27.4) In this formul there is only one option from left to right. The second option (from right to left) is not possible becuse the number of opernds on the right side of (I.27.4) is infinite nd there is no right opernd. Of course, this is prticulr exmple, but the right option should be universl. Therefore, the eqution (I.27.3) tkes the form: 2 [4] 4 = (( 2 [3] 2 ) [3] 2 ) [3] 2 = 256; (I.27.3) Recll the definition of question numbers in prgrph I.17: [3]? = 1 ^? = ; (I.17.4) Here is ny complex number. And the number? will be clled "question number" Now tht the fourth direct opertion hs ppered, this definition cn be generlized to ny direct opertion with number n: 1 [n] n? = ; (I.30.1) Here is ny complex or question number. And the number n? will be clled "question number". Connection of old "question numbers"? nd the new "question numbers" n? is this: 3? =?; (I.30.2) 1 [3] 3? = ; (I.30.2) 55
56 And now we show shortened method of recording the fourth direct opertion [4] through the third direct opertion [3]. [4] b = [3] ( [3] (b 1)); (I.27.17) This conclusion is suitble for nturl b. But perhps the finl formul (I.27.17) is lso true for wider clss of numbers. [4]2] c 1 1, 2, b Tble I.43 Here we re gin fced with new type of numbers: numbers strings. In cell b, where = 1, c = 1 of Tble I.43 re just 3 numbers: 1, 2, 3. And, if some formul will meet this number b, it is uncler which of its three vlues (1, 2, 3) to tke for clcultions. We hve lredy seen this in Tble I.19 nd Tble I.20. But there we were silent bout this feture. The only thing we cn do now is to develop specil designtion for such numbersstrings, so tht it fits entirely in one cell. If the first vlue of the numberline we denote "m", the number of vlues in the numberline denote "N", nd dditive to move to the next vlue denote "n", nd s seprtor of these symbols tke verticl line " ", we get such record: m N n (I.29.3) In our exmple (1, 2, 3) it will be written s: (I.29.4) 56
57 II.2 Determintion of direct zero [0] nd direct negtive [m] opertions. Chpter I estblishes connection between direct opertion [n] nd direct opertion [n + 1]. Here re couple of exmples of this connection: 3 [n + 1] 2 = 3 [n] 3; (II.2.1) 2 [n + 1] 3 = (2 [n] 2) [n] 2; (II.2.2) If we tke n = 0, we cn determine by the known direct opertion [1] (ddition) new direct opertion [0]. And then, by direct opertion [0] build new direct opertion [1]. Thus we cn construct ll direct opertions with negtive numbers. So the number of the direct opertion cn tke the vlue of ny integer. II.3 Zero direct opertion [0]. Tke n = 0. Use lso: 2 [1] 2 = = 4; (II.3.1) In the smple (II.2.1) replce 3 with 2, nd then we get the eqution for 2 [0] 2: 4 = 2 [1] 2 = 2 [0] 2; (II.3.2) Further 2 [1] 3 = = 5; (II.3.3) Apply the smple (II.2.2) nd obtin the eqution for 4 [0] 2: 5 = 2 [1] 3 = (2 [0] 2) [0] 2 = 4 [0] 2; (II.3.4) Now, considering (II.3.4), we construct the eqution for 5 [0] 2: 6 = 2 [1] 4 = ((2 [0] 2) [0] 2) [0] 2 = (4 [0] 2) [0] 2 = 5 [0] 2; (II.3.5) And now, considering (II.3.5), the eqution for 6 [0] 2: 57
58 7 = 2 [1] 5 = (5 [0] 2) [0] 2 = 6 [0] 2; (II.3.6) Next, we get couple of equtions: 8 = 2 [1] 6 = (6 [0] 2) [0] 2 = 7 [0] 2; (II.3.7) 9 = 2 [1] 7 = (7 [0] 2) [0] 2 = 8 [0] 2; (II. 3.8) The equtions obtined re generlized by the formul: [0] b = c; (II.3.9) Now let's build five more equtions, but with b = 3: 5 = 3 [1] 2 = 3 [0] 3; (II.3.10) 6 = 3 [1] 3 = (3 [0] 3) [0] 3 = 5 [0] 3; (II.3.11) 7 = 3 [1] 4 = 6 [0] 3; (II.3.12) 8 = 3 [1] 5 = 7 [0] 3; (II.3.13) 9 = 3 [1] 6 = 8 [0] 3; (II.3.14) And now four more equtions with b = 4: 6 = 4 [1] 2 = 4 [0] 4; (II.3.15) 7 = 4 [1] 3 = (4 [0] 4) [0] 4 = 6 [0] 4; (II.3.16) 8 = 4 [1] 4 = 7 [0] 4; (II.3.17) 9 = 4 [1] 5 = 8 [0] 4; (II.3.18) We construct three equtions with b = 5: 7 = 5 [1] 2 = 5 [0] 5; (II.3.19) 8 = 5 [1] 3 = (5 [0] 5) [0] 5 = 7 [0] 5; (II.3.20) 9 = 5 [1] 4 = 8 [0] 5; (II.3.21) Now let's build two equtions with b = 6: 8 = 6 [1] 2 = 6 [0] 6; (II.3.22) 9 = 6 [1] 3 = (6 [0] 6) [0] 6 = 8 [0] 6; (II.3.23) 58
59 And finlly, the lst eqution (with b = 7): 9 = 7 [1] 2 = 7 [0] 7; (II.3.24) [0] b = c; (II.3.9) All these equtions re presented in the form of Tble II.1: [0] b c Tble II.1 It is impossible to mke formuls for empty cells. It remins only to speculte. If you try to mke it logicl, blnced nd beutiful, you will get the Tble II.2: [0] b c Tble II.2 Here is n extended tble for [0]: 59
60 [0] b c Tble II.2.1 II.4 The first inverse opertion [0]1] to the zero direct opertion [0]. Permute nd c in Tble II.2 nd get Tble II.3 for the first inverse to zero: c [0]1] b = ; (II.4.1) [0]1] b c Tble II.3 60
61 The contents of the empty cells conjectures, logicl nd simple. The result is in the Tble II.3.1: [0]1] b c Tble II.3.1 II.5 The second inverse opertion [0]2] to the zero direct opertion [0]. Prt 1. Swp the c nd b in Tble II.2 nd obtin the Tble II.4: [0]2] c = b; (II.5.1) c b Tble II.4 [0]2] 61
62 1+ = 4, 5, 6, 7; 2+ = 5, 6, 7; 4+ = 6, 7; (II.5.2) 3+ = 2, 3; 5+ = 2, 3, 4; 6+ = 2, 3, 4, 5; (II.5.3) In this tble we gin met the numberlines: 1+,..., 6+. We will write them down in the designtions defined in prgrph 29: If the first vlue of the numberline denote "m", the number of vlues in the numberline denote "N", nd dditive to move to the next vlue denote "n", nd s seprtor of these symbols tke verticl line " ", we get such record: m N n (29.3) First, do this with the formuls (II.5.2): 1+ = 4 4 1; 2+ = 5 3 1; 4+ = 6 2 1; (II.5.4) Let's look t Tble II.2 gin: [0] b c Tble II.2 When = 2 fter the number of 4.5 nd up to the border of the Tble II.2 there is continuous row of fives. The first of these is bordered by number of 4.5 nd hs b = 4. So in 1+ for m we get b = 4. The number of these fives from 4.5 to the right border of the tble = 4. Therefore, for N we took 4. The first five hs b = 4. The second five hs b = 5. Additive = 1. The third five hs b = 6. 62
63 Supplement, too, = 1. The fourth five hs b = 7. Supplement, too, = 1. Therefore, n = 1. And we hve for series of four fives: 1+ = m N n = 4 4 1; (II.5.5) For series of three sixes t = 3 we hve fter the number 5.5 the first six with b = 5. So in 2+ for m, we get b = 5. The number of these sixes from the number 5.5 to the right border of the tble = 3. Therefore, for N we took 3. The first six hs b = 5. The second six hs b = 6. Addition to the first six = 1. The third six hs b =7. Additive to the second six = 1. So n = 1. And we hve for series of three sixes: 2+ = m N n = 5 3 1; (II.5.6) For series of two sevens t = 4 we hve the number 6.5 on the left s the boundry. The first seven to the right of 6.5 hs b = 6. Therefore, in 4+ we took m = 6. The number of these sevens from 6.5 on the left, nd to the right border of Tble II.2 is equl to 2. Therefore, N = 2. The first seven hs b = 6. The second seven hs b = 7. Addition to the first seven = 1. Therefore, n = 1. And for series of two sevens we hve: 4+ = m N n = 6 2 1; (II.5.7) For Tble II.2 we hve tken for its right boundry b = 7. If it were not for the size limit of Tble II.2 for the zero direct opertion [0] of the pge frmes in this book, then from the generl view of Tble II.2 it is cler tht the row of fives t = 2 continues to the right endlessly. Tht is, N =?2. (Question number?2 coincides with the number of nturl numbers.) And then for 1+ we hve: 1+ = 4?2 1; (II.5.8) The sme resoning is true for the row of three sixes. And we hve for 2+: 63
64 2+ = 5?2 1; (II.5.9) The sme thing hppens with row of two sevens t = 4: 4+ = 6?2 1; (II.5.10) Now let's move on to the formuls (II.5.3). Let's strt with 3+ in Tble II.4. There re two sixes t = 5 in Tble II.2 to the left of the number 6.5. The left border of the Tble II.2 is locted to the left of them. And to the left of it hides n endless series of sixes. Therefore, it is possible to strt the numbering of sixes in the numberline 3+ only from the right with the number b = 3: m = 3. The number of sixes is infinite: N =?2 (number of positive integers). The six to the left of the first hs b = 2. So the ddition is 1, n = 1. And then we hve: 3+ = m N n = 3?21; (II.5.11) Continue 5+ in Tble II.4. There re three sevens t = 6 in Tble II.2 to the left of 7.5. The left border of the Tble II.2 is locted to the left of them. And to the left of this border hides n infinite number of sevens. Therefore, it is possible to strt the numbering of sevens only from the right with seven hving b = 4: m = 4. N =?2. The second seven in the numberline 5+ hs b = 3. So the ddition is 1, n = 1. And then we hve: 5+ = m N n = 4?21; (II.5.12) Now is the turn to the 6+ in Tble II.4. There re four eights t = 7 in Tble II.2 to the left of 8.5. To the left of them out of border of the Tble II.2 is n infinite number of eights. Then N =?2. And the rightmost eight hs b = 5: m =5. The next eight to the left of it hs b = 4. Then increment = 1, n = 1. And we now hve: 6+ = m N n = 5?21; (II.5.13) 64
65 Let us introduce in Tble II.4 four more numbersstrings: 7+ = 2?21; 8+ = 1?21; 9+ = 7?2 1; 0+ = 8?2 1; (II.5.14) And get the Tble II.5: c b Tble II.5 [0]2] Now let's explin  where from did these four numberslines ppered. The single five is plced in the Tble II.2 with = 4, t the left border. And on the other side of the border there is n endless row of fives. And the beginning of ll these fives is the right five with b = 2: m = 2. N =?2. To the left of it (just brod) is five with b = 1. Then increment = 1, n = 1. Here is the result: 7+ = m N = 2?21; (II.5.15) In Tble II.2 t = 3 nd b = 1 (just beyond the left border) is 4. To the right of it there is the number 4.5, nd to the left of it there is n infinite number of fours. Tht is, m = b = 1 nd N =?2. The fours extend to the left in descending order of b. Additive = 1: n = 1. And we got: 8+ = m N n = 1?21; (II.5.16) In Tble II.2, with = 5 nd b = 7 is 8. To its left is the number 7.5. To its right is the right border of Tble II.2, nd just over the border  n endless series of figure eights. So m = b = 7. N =?2. Eights re 65
66 numbered in scending order of b. It mens, tht supplement = 1. And we hve: 9+ = m N n = 7?2 1; (II.5.17) In Tble II.2 for = 6 nd b = 8, 9 is locted (to the right of 8.5 just beyond the right border of Tble II.2). It is followed to the right by n infinite number of nines with n increment of 1: n = 1. Also we hve m = b = 8. N =?2. And s result: 0+ = m N n = 8?2 1; (II.5.18) II.6 The second inverse opertion [0]2] to the zero direct opertion [0]. Prt 2. Floor 1st. Let us reproduce the finl results obtined in Prt 1: c b Tble II.5 [0]2] 1+ = 4?2 1; (II.5.8) 2+ = 5?2 1; (II.5.9) 4+ = 6?2 1; (II.5.10) 9+ = m N n = 7?2 1; (II.5.17) 0+ = m N n = 8?2 1; (II.5.18) 66
67 8+ = m N n = 1?21; (II.5.16) 7+ = m N n = 2?21; (II.5.15) 3+ = m N n = 3?21; (II.5.11) 5+ = m N n = 4?21; (II.5.12) 6+ = m N n = 5?21; (II.5.13) The numberlines. Wht to do with them? In wht order to extrct numbers from them? Let's ssume tht ll initil vlues of the row numbers re in the plne of Tble II.5. And ll the others re bove them. Ech consecutive number from the numberline is on its floor. All the second numbersrows from 1+ to 6+  re on the second floor of the threedimensionl Tble II.5. All third numbers of ll row numbers re on the third floor of Tble II.5. And so on. Let's look t the first floor of Tble II.5: c b Tble II.5 [0]2] 1st floor This turned out to be the "skeleton" of the tble. This is wht is obtined from formuls relting opertions [0] nd [1]. We must to drem (most likely) nd build on this "skeleton" everything else. If you look closely t the" skeleton", you cn see this feture of the" skeleton": horizontl numbers re prts of integer series 67
68 they cn be extended to the left nd to the right of the" skeletl " middle. If you look t this "skeleton" gin, you cn see its second feture: verticl numbers re prts of integer series they cn be extended up nd down from the "skeletl" middle. There is lso third feture: the horizontl ddition of integers does not interfere with the verticl ddition of integers. And verticl dditions of integers lso do not interfere with horizontl dditions of integers. After these observtions grow on the "skeleton" "met" nd get Tble II.6: c b Tble II.6 [0]2] 1st floor II.7 The second inverse opertion [0]2] to the zero direct opertion [0]. Prt 2. Floor 2nd. Now let's go to the second floor. Insert in the cells of Tble II.5 the second numbers of the corresponding numbersrows nd get the "skeleton" of Tble II.7: 68
69 c b Tble II.7 [0]2] Floor 2nd If you look t this "skeleton", you cn see tht it hs no horizontl ordering. At lest there is no horizontl integer order. A verticl ordering my be present. Where "c" is semiinteger, integer ordering is possible. And where" c " is n integer, n even or odd ordering is possible. Nothing else comes to mind, so let's increse our verticl "met" on this "skeleton" nd get the Tble II.8: c b Tble II.8 [0]2] Floor 2nd II.8 The second inverse opertion [0]2] to the zero direct opertion [0]. Prt 2. The 3rd floor. 69
70 Now let's go to the third floor. Insert in the cells of Tble II.5 the third numbers of the corresponding numbersrows nd get the Tble II.9: c b Tble II.9 [0]2] The 3rd floor There is no horizontl ordering in this "skeleton". While the verticl ordering is present. Where "c" is semiinteger, integer ordering is possible. And where "c" is n integer, the numbers re locted with step 3. Let's build on this "skeleton" "met" nd get Tble II.10: Tble II.10 [0]2] The 3rd floor II.9 The second inverse opertion [0]2] to the zero direct opertion [0]. Prt 2. Floor 4th nd bove. 70
71 Now it is cler wht the ordering is on the 4th floor nd bove. Where "c" is hlfinteger, there the integer verticl lignment is possible. And where "c" is n integer, the numbers verticlly go in increments equl to the floor number. II.10 Zero direct opertion [0] on the set of question numbers?. We will now fill in Tble II.11:? [0]?b = c; (II.10.1) [0]?1?2?3?4?b?1?2?3?4? c Tble II.11?1 [0]?1 = 1 [0] 1 = 1 [1] 2 = = 3; (II.10.2)?2 [0]?2 =?2 [1] 2 =?2 [1] (1 [1] 1) = (?2 [1] 1) [1] 1 = =?2 [1] 1 =?2; (II.10.3)?3 [0]?3 =?3 [1] 2 =?3 [1] (1 [1] 1) = (?3 [1] 1) [1] 1 = =?3 [1] 1 =?3; (II.10.4)?4 [0]?4 =?4 [1] 2 =?4 [1] (1 [1] 1) = (?4 [1] 1) [1] 1 = =?4 [1] 1 =?4; (II.10.5)?2 [0]?1 =?2 [0] 1 =?2; (II.10.6)?3 [0]?1 =?3 [0] 1 =?3; (II.10.7) 71
72 ?4 [0]?1 =?4 [0] 1 =?4; (II.10.8)?1 [0]?2 = 1 [0]?2 = (((1 [1] 1) [1] 1) [1] 1) [1] 1 = = (((1 [0] 2) [1] 1) [1] 1) [1] 1 = Use Tble II.2.1 (expnsion of Tble II.2) = ((3.5 [1] 1) [1] 1) [1] 1 = Use Tble II.26: = (4 [1] 1) [1] 1 = Use Tble II.26 nd (II.13.76)?2 times: = 4 [1] 1... = 4 [1] 1... = 4; (II.10.9) Tht is:?1 [0]?2 = 4; (II.10.9) Similrly (?3 times):?1 [0]?3 = 1 [0]?3 = (((1 [1] 1) [1] 1) [1] 1) [1] 1 = = (4 [0] 1) [0] 1 = 4 [0] 1 = 4; (II.10.10) Also (?4 times):?1 [0]?4 = 4; (II.10.11) We did the Tble II.12: [0]?1?2?3?4?b?1 3?2?3?4?2?2?2?3?3?3?4?4?4? c Tble II.12 72
73 Let's continue filling the cells in Tble II.12:?2 [0]?3 = (((?2 [1]?2) [1]?2) [1]?2) [1]?2 = = (((?2 [0] 2) [1]?2) [1]?2) [1]?2 = Remember the definition of?1: 1 [3]?1 = 1; Here s?1 ny nturl number is good. Including 2. Advntge of this: = (((?2 [0]?1) [1]?2) [1]?2) [1]?2 = And now we hve the right to replce?1 on 1 since s?1 ny nturl number is good. Including 1. = (((?2 [0] 1) [1]?2) [1]?2) [1]?2 = = (((?2 [1]) [1]?2) [1]?2) [1]?2 = = ((?2 [1]?2) [1]?2) [1]?2 = We took one step. Do?3 steps nd get: =?2; (II.10.12) Tht wy?2 [0]?3 =?2; (II.10.12) In the sme wy but mking lredy?4 steps we will get:?2 [0]?4 =?2; (II.10.13) And, by doing?2 nd?4 steps, we get couple more formuls:?3 [0]?2 =?3; (II.10.14)?3 [0]?4 =?3; (II.10.15) And the result is Tble II.13: 73
74 [0]?1?2?3?b? ?2?2?2?2?3?3?3?3? c Tble II.13? [0]?b =?; ( = 2, 3, 4, ; b = 1, 2, 3, 4, ); (II.10.16) Looking t the expnded Tble II.2.1 for [0] we will see (?1 = 1): 1 [0] 0 = 2.5; (II.10.17) 0 [0] 0 = 2; (II.10.18) 0 [0] 1 = 2.5; (II.10.19) 0 [0] 2 = 3; (II.10.20) 0 [0] 3 = 3; (II.10.21) 0 [0] 3 = (0 [1] 0) [1] 0 = (0 [0] 2) [1] 0 = 3 [1] 0; (II.10.22) (II.10.21) + (II.10.22) = (II.10.23): 3 [1] 0 = 3; (II.10.23) 0 [0]?2 = ((0 [1] 0) [1] 0) [1] 0 = ((0 [0] 2) [1] 0) [1] 0 = = (3 [1] 0) [1] 0... = 3 [1] 0... =...?2 times nd = 3; (II.10.24) Tht is 0 [0]?2 = 3; (II.10.24) 0 [0]?3 =...?3 times nd = 3; (II.10.25) Insert these vlues into Tble II.13 nd get Tble II.13.1: 74
75 [0] 0?1?2?3?b ? ?2?2?2?2?3?3?3?3? c Tble II.13.1 Two cells re not cler how to fill, so let's limit ourselves to Tble II.13. II.11 The first inverse opertion [0]1] to the zero direct opertion [0] on the set of question numbers?. Rerrnge "?" nd "c" in Tble II.13 nd get the Tble II.14: [0]1]?1?2?3?b 3?1 4?1?1?2?2?2?2?3?3?3?3 c? Tble II.14? [0]?1 =? [0] 1 = 4; (II.11.1) If you look t Tble II.2.1 (expnded to Tble II.2) then we'll see tht? = 3. Tht is 4 [0]1]?1 = 3; (II.11.2)? [0]?2 = 3; ccording to Tble II.13.1 see:? = 0; (II.11.3) 3 [0]1]?2 = 0; (II.11.4)? [0]?3 = 3; ccording to Tble II.13.1 see:? = 0; (II.11.5) 3 [0]1]?3 = 0; (II.11.6) 75
76 Tble II.14 will become Tble II.15: [0]1]?1?2?3?b ?2?2?2?2?3?3?3?3 c? Tble II.15 And s result  for the first inverse opertion [0]1] to the zero direct opertion [0] for numbers? we hve Tble II.15. II.12 The second inverse opertion [0]2] to the zero direct opertion [0] on the set of question numbers?. Let's see in detil how numbersstrings re born. Consider the first row in Tble II.13 nd leve it in Tble II.16: [0]?1?2?3?b? c?2?3? Tble II.16 Rerrnge the "c" nd "?b". Get Tble II.16.1: [0]2] c?1?1?2?3?b?2?3? Tble II
77 The column hedings should be different, so let's combine the columns with the sme hedings in Tble II.16.2: [0]2] 3 4 c?1?1?2,?3?b?2?3? Tble II.16.2 Consider the second row in Tble II.13 nd leve it in Tble II.16.3: [0]?1?2?3?b?1?2?2?2?2 c?3? Tble II.16.3 Rerrnge the "c" nd "?b". Get Tble II.16.4: [0]2]?2?2?2 c?1?2?1?2?3?b?3? Tble II.16.4 Agin, the sme column hedings. Agin, combine the contents of these columns under common nme nd get Tble II.16.5: [0]2]?2 c?1?2?1,?2,?3?b?3? Tble II
78 Consider the third row in Tble II.13 nd leve it in Tble II.16.6: [0]?1?2?3?b?1?2?3?3?3?3 c? Tble II.16.6 Rerrnge the "c" nd "?b". Get Tble II.16.7: [0]2]?3?3?3 c?1?2?3?1?2?3?b? Tble II.16.7 Agin, the sme column hedings. Agin, combine the contents of these columns under common nme nd get Tble II.16.8: [0]2]?3 c?1?2?3?1,?2,?3?b? Tble II.16.8 Combine the Tbles: II.16.2, II.16.5, II.16.8 nd get Tble II.20: [0]2] 3 4?2?3 c?1?1?2,?3?2?1,?2,?3?3?1,?2,?3??b Tble II.20 78
79 Numberslines denote s follows: 1+ =?2,?3; 2+ =?1,?2,?3; 3+ =?1,?2,?3; (II.12.1) Tble II.20 will tke the form of Tble II.20.1: [0]2] 3 4?2?3 c?1?1 1+?2 2+?3 3+??b Tble II.20.1 For numberslines, there is still vrint of record: 1+ =?2 2?+1; (II.12.2) 2+ =?1 3?+1; (II.12.3) 3+ =?1 3?+1; (II.12.4) Before the first verticl line  " " the first number in the numberline is specified. It is followed by the number of numbers in the numberline. Then comes the second verticl line. It is followed by the step between the numbers in the numberline. In this cse, for quntities beginning with the sign "?", specified index step following sign "?". We introduce the floors bove the tble. The very first vlue in ll numberslines is locted on the first floor in Tble II.21: [0]2] 3 4?2?3 c?1?1?2?2?1?3?1??b Tble II.21 1st floor On the second floor of opertion [0]2] for? ech numberline contins its second vlue. This is shown in Tble II.22: 79
80 [0]2] 3 4?2?3 c?1?1?3?2?2?3?2??b Tble II.22 2nd floor The third floor contins the third vlue of numberslines. Once in numberline 1+ only two vlues, then the lst it s vlue ?3 is tken. The result is presented in Tble II.23: [0]2] 3 4?2?3 c?1?1?3?2?3?3?3??b Tble II.23 3rd floor nd bove The fourth floor contins in ech cell the fourth vlue of numberslines. In those cells, where the number of vlues in the numberline is less thn four, the most recent vlues of these numberslines re locted. Get gin Tble II.23. It is lso true for higher floors. We got three "skeletl" Tbles: II.21, II.22, II.23. Let's try to increse them "met". It is not cler how to fill in the first two Tbles II.21 nd II.22, nd here is the third Tble  II.23, which forms itself in Tble II.24: [0]2] 3 4?2?3 c?1?1?3?5?7?2?(1)?1?3?5?3?(3)?(1)?1?3??b Tble II.24 3rd floor nd bove 80
81 II.13 Direct opertion [1]. In Chpter II, prgrph 2, it is written, how to define n unknown opertion [1] by known opertion [0]: 3 [n + 1] 2 = 3 n] 3; (II.2.1) 2 [n + 1] 3 = (2 [n] 2) [n] 2; (II.2.2) At n = 1 we obtin: 3 [0] 2 = 3 [1] 3; (II.13.1) 2 [0] 3 = (2 [1] 2) [1] 2; (II.13.2) Vlues for opertion [0] will be tken from Tble II.2. If the desired vlue lies outside the boundry of Tble II.2, then you cn push the boundries of Tble II.1 s fr s you need, by entering new rows nd new columns, nd filling them with new vlues, nd then on the principle of symmetry fill the remining empty cells. In further clcultions, the vlues for opertion [0] re tken from such n extended Tble II = 2 [0] 2 = 2 [1] 2; (II.13.3) 1 = 2 [0] 3 = (2 [1] 2) [1] 2 = 1 [1] 2; (II.13.4) 1 = 2 [0] 4 = 1 [1] 2; (II.13.5) The repets re endless, so 1+ = 1?2 0; in cell 1[1] 2; (II.13.6) 2 = 1 [0] 2 = 1 [1] 1; (II.13.7) 2 = 1 [0] 3 = (1 [1] 1) [1] 1 = 2 [1] 1; (II.13.8) 81
82 2 = 1 [0] 4 = 2 [1] 1; (II.13.9) The repets re endless, so 2+ = 2?2 0; in cell 2 [1] 1; (II.13.10) 3 = 0 [0] 2 = 0 [1] 0; (II.13.11) 3 = 0 [0] 3 = (0 [1] 0) [1] 0 = 3 [1] 0; (II.13.12) 3 = 0 [0] 4 = 3 [1] 0; (II.13.13) The repets re endless, so 3+ = 3?2 0; in cell 3 [1] 0; (II.13.14) 3.5 = 1 [0] 2 = 1 [1] 1; (II.13.15) 4 = 1 [0] 3 = (1 [1] 1) [1] 1 = 3.5 [1] 1; (II.13.16) 4 = 1 [0] 4 = 4 [1] 1; (II.13.17) 4 = 1 [0] 5 = 4 [1] 1; (II.13.18) 4 = 1 [0] 6 = 4 [1] 1; (II.13.19) The repets re endless, so 4+ = 4?2 0; in cell 4 [1] 1; (II.13.20) 4 = 2 [0] 2 = 2 [1] 2; (II.13.21) 4.5 = 2 [0] 3 = (2 [1] 2) [1] 2 = 4 [1] 2; (II.13.22) 5 = 2 [0] 4 = ((2 [1] 2) [1] 2) [1] 2 = 4.5 [1] 2; (II.13.23) 82
83 5 = 2 [0] 5 = 5 [1] 2; (II.13.24) 5 = 2 [0] 6 = 5 [1] 2; (II.13.25) 5 = 2 [0] 7 = 5 [1] 2; (II.13.26) The repets re endless, so 5+ = 5?2 0; in cell 5 [1] 2; (II.13.27) 4.5 = 3 [0] 2 = 3 [1] 3; (II.13.28) 5 = 3 [0] 3 = (3 [1] 3) [1] 3 = 4.5 [1] 3; (II.13.29) 5.5 = 3 [0] 4 = 5 [1] 3; (II.13.30) 6 = 3 [0] 5 = 5.5 [1] 3; (II.13.31) 6 = 3 [0] 6 = 6 [1] 3; (II.13.32) 6 = 3 [0] 7 = 6 [1] 3; (II.13.33) The repets re endless, so 6+ = 6?2 0; in cell 6 [1] 3; (II.13.34) 5 = 4 [0] 2 = 4 [1] 4; (II.13.35) 5.5 = 4 [0] 3 = 5 [1] 4; (II.13.36) 6 = 4 [0] 4 = 5.5 [1] 4; (II.13.37) 6.5 = 4 [0] 5 = 6 [1] 4; (II.13.38) 7 = 4 [0] 6 = 6.5 [1] 4; (II.13.39) 7 = 4 [0] 7 = 7 [1] 4; (II.13.40) 83
84 7 = 4 [0] 8 = 7 [1] 4; (II.13.41) The repets re endless, so 7+ = 7?2 0; in cell 7 [1] 4; (II.13.42) 6 = 5 [0] 2 = 5 [1] 5; (II.13.43) 6 = 5 [0] 3 = 6 [1] 5; (II.13.44) 6.5 = 5 [0] 4 = 6 [1] 5; (II.13.45) 11+ = 6, 6.5; in cell 6 [1] 5; (II.13.46) 7 = 5 [0] 5 = 6.5 [1] 5; (II.13.47) 7.5 = 5 [0] 6 = 7 [1] 5; (II.13.48) 8 = 5 [0] 7 = 7.5 [1] 5; (II.13.49) 8 = 5 [0] 8 = 8 [1] 5; (II.13.50) 8 = 5 [0] 9 = 8 [1] 5; (II.13.51) The repets re endless, so 8+ = 8?2 0; in cell 8 [1] 5; (II.13.52) 7 = 6 [0] 2 = 6 [1] 6; (II.13.53) 7 = 6 [0] 3 = 7 [1] 6; (II.13.54) 7 = 6 [0] 4 = 7 [1] 6; (II.13.55) 7.5 = 6 [0] 5 = 7 [1] 6; (II.13.56) 10+ = 7, 7, 7.5; in cell 7 [1] 6; (II.13.57) 84
85 8 = 6 [0] 6 = 7.5 [1] 6; (II.13.58) 8.5 = 6 [0] 7 = 8 [1] 6; (II.13.59) 8 = 7 [0] 2 = 7 [1] 7; (II.13.60) 8 = 7 [0] 3 = 8 [1] 7; (II.13.61) 8 = 7 [0] 4 = 8 [1] 7; (II.13.62) 8 = 7 [0] 5 = 8 [1] 7; (II.13.63) 8.5 = 7 [0] 6 = 8 [1] 7; (II.13.64) 9+ = 8, 8, 8, 8.5; in cell 8 [1] 7; (II.13.65) All the bove 65 equtions re considered nd the results re summrized in Tble II.26 ( [1] b = c): [1] b c Tble II.26 In the Tble II.26 there re the following numberstrings: 85
86 1+ = 1?2 0; 2+ = 2?2 0; (II.13.75) 3+ = 3?2 0; 4+ = 4?2 0; (II.13.76) 5+ = 5?2 0; 6+ = 6?2 0; 7+ = 7?2 0; 8+ = 8?2 0; (II.13.77) 9+ = 8, 8, 8, 8.5; 10+ = 7, 7, 7.5; 11+ = 6, 6.5; (II.13.78) For simplicity, let us leve in the Tble II.26 only lines with integer hedings. Get Tble II.27: [1] b c Tble II.27 Now consider the "skeleton" for the first floor. To do this, tke the first vlue from ech numberline nd plce it in the pproprite cell. Insted of 11+ will be the number 6. Insted of 10+ will be 7. Insted of 9+ will stnd 8. And for the rest of the numberslines it will be enough to remove the sign"+". And we get the Tble II.28: 86
87 [1] b c Tble II.28 1st floor skeleton Now, fter thinking, let's increse the "met" on this "skeleton". And we get the Tble II.29: [1] b c Tble II.29 1st floor Now consider the "skeleton" for the second floor. To do this, from ech numberline tke the second vlue from the left nd put it in the pproprite cell. Insted of 11+ will stnd the number
88 Insted of 10+ will be 7. Insted of 9+ will stnd 8. And for the rest of the numberslines, it will be enough to remove the sign"+". After ll, they get the second vlue by dding 0 to the first vlue. And we get the Tble II.30: [1] b c Tble II.30 2nd floor skeleton [1] b c Tble II.31 2nd floor 88
89 Now consider the "skeleton" for the third floor. It coincides with the skeleton for the second floor, but insted of 10+ is not 7, but 7.5 nd we get the Tble II.32 nd Tble II.33: [1] b c Tble II.32 3rd floor skeleton [1] b c Tble II.33 3rd floor 89
90 Now consider the "skeleton" for the 4th floor. It coincides with the skeleton for the 3rd floor, but insted of 9+ is not 8, but 8.5, nd we get Tble II.34 nd Tble II.35: [1] b c Tble II.34 4th floor nd bove skeleton [1] b c Tble II.35 4th floor nd bove Tble II.34 nd II.35 re correct nd for ll overlying floors. 90
91 II.14 Direct opertion [1] for question numbers?. Let's fill the digonl in Tble II.40: [1]?1?2?3?4?b?1?2?3?4? c Tble II.40 In the mentime, let's discuss one rule of computtion with right 1 opernd. We my do so in the course of filling in the Tble II.40. Proceed:?1 [1]?1 = 1 [1] 1 = There re two possible wys: = 1 [0] 2 = 3.5; (II.14.1) = 1 [2] = 1; (II.14.2) It is difficult to justify logiclly my choice, but from my prctice of computing I choose the first wy. Resume:?2 [1]?2 =?2 [0] 2 = Remember the definition of?1: 1 [3]?1 = 1; (II.14.3) Here s?1 ny nturl number is good. Including the number 2. Let us use this nd the Tble II.13: =?2 [0]?1 =?2; (II.14.4) 91
92 ?3 [1]?3 =?3 [0] 2 =?3 [0]?1 =?3; (II.14.5)?4 [1]?4 =?4 [0] 2 =?4 [0]?1 =?4; (II.14.6) The digonl is full, nd we got Tble II.41: [1]?1?2?3?4?b?1 3.5?2?2?3?3?4?4? c Tble II.41?2 [1]?1 =?2 [1} 1 =?2 [2] =?2; (II.14.7)?3 [1]?1 =?3 [1] 1 =?3 [2] =?3; (II.14.8)?4 [1]?1 =?4 [1] 1 =?4 [2] =?4; (II.14.9) Now we hve Tble II.42: [1]?1?2?3?4?b?1 3.5?2?2?2?3?3?3?4?4?4? c Tble II.42 Let's continue our clcultions:?1 [1]?2 = 1 {1]?2 = ((((1 [1] 1) [1] 1) [1]1) [1] 1) [1] 1... = Apply the formul (II.14.1) = (((3.5 [1] 1) [1] 1) [1] 1) [1] 1... = Apply Tble II.26: 92
93 3.5 [1] 1 = 4; (II.14.10) = ((4 [1] 1) [1] 1) [1] 1... = Apply Tble II.26 nd (II.13.76): 4 [1] 1 = 4; (II.14.11) = (4 [1] 1) [1] 1... = Agin pply (II.14.11): = 4 [1] 1... =... =... = We will pply (II.14.11)?2 times nd get: = 4; (II.14.12) Tht is:?1 [1]?2 = 4; (II.14.12) Further?1 [1]?3 = 1 [1]?3 = ((((1 [1] 1) [1] 1) [1]1) [1] 1) [1] 1... = Apply (II.14.1): = (((3.5 [1] 1) [1] 1) [1] 1) [1] 1... = Apply (II.14.10): = ((4 [1] 1) [1] 1) [1] 1... = Apply (II.14.11): = (4 [1] 1) [1] 1... = Agin pply (II.14.11): = 4 [1] 1... =... =... = We will pply (II.14.11)?3 times nd get: 93
94 = 4; (II.14.13) Tht is:?1 [1]?3 = 4; (II.14.13) Similrly, we obtin:?1 [1]?4 = 4; (II.14.14) And now we hve filled three more boxes in the Tble II.42, nd got the Tble II.43: [1]?1?2?3?4?b? ?2?2?2?3?3?3?4?4?4? c Tble II.43 Let's continue our clcultions:?2 [1]?3 = ((?2 [2]?2) [2]?2) [2]?2... = = ((?2 [1] 2) [2]?2) [2]?2... = Apply the words following (II.14.3): = ((?2 [1]?1) [2]?2) [2]?2... = Apply here (II.14.7):?2 [1]?1 =?2; (II.14.7) = (?2 [2]?2) [2]?2... = By doing?3 such steps, we will get: =?2; (II.14.15) Tht is:?2 [1]?3 =?2; (II.14.15) Similrly, we obtin:?2 [1]?4 =?2; (II.14.16)?3 [1]?2 = ((?3 [2]?3) [2]?3) [2]?3... = = ((?3 [1] 2) [2]?3) [2]?3... = 94
95 = ((?3 [1]?1) [2]?3) [2]?3... = Apply (II.14.8)?3 [1]?1 =?3 [1] 1 =?3 [2] =?3; (II.14.8) = (?3 [2]?3) [2]?3... = By doing?2 such steps, we will get: =?3; (II.14.17) Tht is:?3 [1]?2 =?3; (II.14.17) Similrly, we obtin:?3 [1]?4 =?3; (II.14.18) Similrly, we obtin couple of formuls:?4 [1]?2 =?4; (II.14.19)?4 [1]?3 =?4; (II.14.20) So we filled in the empty cells in the Tble II.43 nd received the Tble II.44: [1]?1?2?3?4?b? ?2?2?2?2?2?3?3?3?3?3?4?4?4?4?4? c Tble II.44 Recll tht Tble II.44 represents the direct opertion [1] for the question numbers?. 95
96 CHAPTER III 96
97 III.1 Summry of Chpter II. In prgrph II.1 there is summry of Chpter I. In prgrph II.2 the principle of construction of zero nd negtive direct opertions is described: Chpter I estblishes the connection between direct opertion [n] nd direct opertion [n + 1]. Here re couple of exmples of this connection: 3 [n + 1] 2 = 3 [n] 3; (II.2.1) 2 [n + 1] 3 = (2 [n] 2) [n] 2; (II.2.2) If we tke n = 0, we cn determine new direct opertion [0] by the known direct opertion [1] (ddition). And then (t n = 1), by direct opertion [0] construct new direct opertion [1]. Thus, we cn construct ll direct opertions with negtive numbers. So the number of the direct opertion cn tke the vlue of ny integer. In prgrph II.3 by the known opertion [1] ddition built "skeleton" of the zero direct opertion [0]. This is wht ws possible to clculte by the formuls of connection of these opertions: All these equtions re presented in the form of Tble II.1: [0] b c Tble II.1 97
98 It is impossible to mke formuls for empty cells. It remins only to speculte. If you try to mke it logicl, blnced nd beutiful, you get Tble II.2: [0] b c Tble II.2 In prgrph II.4 it is considered the reverse opertion [0]1] for the zero direct opertion [0]. Rerrnge nd c in Tble II.2 nd get Tble II.3 for the first reverse opertion to zero opertion: c [0]1] b = ; (II.4.1) [0]1] b c Tble II.3 98
99 The contents of empty cells think logiclly nd simply. Result in Tble II.3.1: [0]1] b c Tble II.3.1 In prgrphs from II.5 to II.9 the second inverse opertion [0]2] to the zero direct opertion [0] is considered. A significnt prt is the description of the work with numbers strings. The concept of "floors" is introduced. In prgrph II.10 we consider the zero direct opertion [0] on the set of question numbers?. And the result is Tble II.13: [0]?1?2?3?b? ?2?2?2?2?3?3?3?3? c Tble II.13 99
100 In prgrph II.11 the first inverse opertion [0]1] to the zero direct opertion [0] on the set of question numbers? is considered. Rerrnge "?" nd " c " in Tble II.13 nd get Tble II.14: [0]1]?1?2?3?b 3?1 4?1?1?2?2?2?2?3?3?3?3 c? Tble II.14? [0]?1 =? [0] 1 = 4; (II.11.1) If you look t Tble II.2.1 (expnded to Tble II.2) then we'll see tht? = 3. Tht is 4 [0]1]?1 = 3; (II.11.2)? [0]?2 = 3; ccording to Tble II.13.1 see:? = 0; (II.11.3) 3 [0]1]?2 = 0; (II.11.4)? [0]?3 = 3; ccording to Tble II.13.1 see:? = 0; (II.11.5) 3 [0]1]?3 = 0; (II.11.6) Tble II.14 will become Tble II.15: [0]1]?1?2?3?b ?2?2?2?2?3?3?3?3 c? Tble II
101 And s result  for the first inverse opertion [0]1] to the zero direct opertion [0] for numbers?  we hve Tble II.15. In prgrph II.12 the second inverse opertion [0]2] to the zero direct opertion [0] on the set of question numbers? is considered. The birth of numberslines nd floors is considered in detil. In prgrph II.13 the direct opertion [1] is considered. All the bove 65 equtions re considered nd the results re summrized in Tble II.26 ( [1] b = c): [1] b c Tble II.26 There re numbersstrings in Tble II.26: 1+ = 1?2 0; 2+ = 2?2 0; (II.13.75) 101
102 3+ = 3?2 0; 4+ = 4?2 0; (II.13.76) 5+ = 5?2 0; 6+ = 6?2 0; 7+ = 7?2 0; 8+ = 8?2 0; (II.13.77) 9+ = 8, 8, 8, 8.5; 10+ = 7, 7, 7.5; 11+ = 6, 6.5; (II.13.78) For the first four floors of Tble II.26 "skeletl" tbles were built, nd "met" ws built on them. In prgrph II.14 the direct opertion [1] on the set of question numbers? is considered: So we filled in the empty cells in Tble II.43 nd derived Tble II.44: [1]?1?2?3?4?b? ?2?2?2?2?2?3?3?3?3?3?4?4?4?4?4? c Tble II.44 III.2 Numbers N1 nd n1. Bck in 1984, I tried to extrct new numbers from new equtions. Also I tried the eqution: x + 1 = x; (III.2.1) But then I couldn't find use for it. And it wsn't until few yers lter, tht I relized, tht this eqution describes n infinitely lrge number. And designted it s N1: N1 + 1 = N1; (III.2.2) 102
103 From this definition it follows: N1  N1 = 1; (III.2.3) N1 * (1 1) = 1; (III.2.4) We introduce new vlue n1 s follows: N1 * n1 = 1; (III.2.5) It follows: n1 = 1 / N1; (III.2.6) Tht is, n1 is n infinitely smll number. (III.2.5) + (III.2.4) = (III.2.7) n1 = 1 1; (III.2.7) III.3 Numbers N2 nd n2. Using the number N1 we construct new infinitely lrge number N2: N2 = 2 ^ N1 = 2 [3] N1; (III.3.1) N2 > N1; (III.3.2) We construct new number n2 s follows: n2 = 1 / N2; (III.3.3) It is n infinitesiml number, nd it is less thn n1 n2 < n1; (III.3.4) This follows from (III.2.6), (III.3.3), (III.3.2) III.4 Numbers N3 nd n3. Using the number N2 we construct new infinitely lrge number N3: N3 = 2 ^ N2 = 2 [3] N2; (III.4.1) N3 > N2; (III.4.2) 103
104 We construct new number n3 s follows: n3 = 1 / N3; (III.4.3) It is n infinitesiml number, nd it is less thn n2 n3 < n2; (III.4.4) This follows from (III.4.3), (III.3.3), (III.4.2) III.5 Numbers NA nd na. Let's tke series of integers A (... 3, 2, 1, 0, 1, 2, 3,...) nd build series of infinitely lrge numbers NA (bsed on the lredy constructed numbers N1, N2, N3): N (A + 1) = 2 ^ NA = 2 [3] NA; (III.5.1) N (A + 1) > NA; (III.5.2) When A = 3, we obtin definition for N4: N4 = 2 [3] N3; (III.5.3) For A = 4, we obtin definition for N5 through N4: N5 = 2 [3] N4; (III.5.4) When tking incresing indices A, incresing numbers NA re obtined. Now let's tke A = 0 nd get: N1 = 2 [3] N0; (III.5.5) This eqution defines N0 by N1. Tke A = 1 nd get: N0 = 2 [3] N (1); (III.5.6) Here N (1) is defined by N0. 104
105 When A = 2, we obtin the definition of N (2) by N (1): N (1) = 2 [3] N (2); (III.5.7) Tking A = 3, 4, 5,... will produce new positive infinitely lrge numbers N (3), N (4), N (5),... nd ll of them will fit in the intervl from N1 to 1. Now let's build series of infinitesiml numbers na: na = 1 / NA; (III.5.8) n (A + 1) < na; (III.5.9) III.6 Topology of the numericl line for Nnumbers. The representtion of number line is usully given s follows: Figure 1 Given the infinitely lrge nd the infinitely smll NA nd na numbers give the following two figures: Figure 2 105
106 Figure 3 From (III.(2.7) should:  n1.=  (1 1) = = 1 1 = n1; (III.6.1)  n1 = n1; (III.6.2) Physiclly, n1 nd n1 lie on different number lines, but topologiclly they coincide. In other words, these two numeric lines intersect t this point. The result is in Figure 4. Figure 4 106
107 (III.2.5): N1 * n1 = 1; (III.6.2):  n1 = n1; ( N1) * n1 =  (N1 * n1) = N1 * (n1) = N1 * n1; (III.6.3) It follows:  N1 = N1; (III.6.4) So these two very lrge numbers is lso topologiclly identicl. In some multidimensionl spce, these two numericl "lines" bend to intersect t this point. Result in figure 5: Figure 5 107
108 ((III.3.1): N2 = 2 ^ N1; (III.3.3): n2 = 1 / N2; (III.6.4):  N1 = N1; N2 = 2 ^ N1 = 2 ^ ( N1) = 1 / (2 ^ N1) = 1 / N2 = n2; (III.6.5) N2 = n2; (III.6.6) So n infinitely lrge number coincides with n infinitely smll number! (opposites coincide  the min lw of dilectics). The numbers N2 nd n2 lie on the sme stright line of positive numbers, but in different plces. And so they coincide topologiclly, this "direct" line need to curve in some multidimensionl spce tht they re superimposed on ech other. The result is figure 6. Figure 6 108
109 If the eqution (III.6.6) is multiplied by 1, we get:  N2 =  n2; (III.6.7) This pir of numbers ( N2 nd n2) lies on one "direct line" of negtive numbers t its different ends. But they re topologiclly the sme. To do this, the "direct line" of negtive numbers should be curved in specil multidimensionl spce so tht it hs selfintersection. The result is in figure 7. Figure 7 It is possible to obtin new pirs of topologiclly coinciding different numbers infinitely. Ech new pir is bsed on the previous pir. But it is incresingly difficult to disply these 109
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