Algebraic Applications in Creating Imaginary Abecedaries

7/28/2019 Algebraic Applications in Creating Imaginary Abecedaries

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Algebraic applications in creating imaginaryabecedaries

by/per Alejandro Ochoa G.

Where/Ubi:Guadalajara, Jalisco,Mexico.20.702085 degrees North latitude,103.32535 degrees West longitude.

20 42 07.506 N, 103 19 31.26 W.

Gvadalaxara, Xaliscum, Mexicum.Arrica, Nova Gallciam, Nova Hispaniam.

In fact, the author does not know Latin; instead, he writes some words and paragraphs in fake

Latin. If you find errors, please, you may report them, in theComments area.Nam auctor nescire Latine, sed et in locis et fictus quaedam latine scripsit. Si aliquam invenire

erroribus potest referre Commentarii area.

What/Quid:Algebraic applications in creating imaginaryabecedaries

Ex algebrae app l icat ionis creat ionemabecedriicommentcia

When/Quando:June, 2013. VI (Inius), MMXIII.


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Copyright 2013 by Alejandro Hector Ochoa-Gonzalez.Guadalajara, Jalisco, Mexico.

2013 by

All rights reserved under International and Pan American Copyright Conventions, and under The Digital MillenniumCopyright Act (DMCA) and the World Intellectual Property Organization Copyright Treaty (WIPO Copyright Treaty).No part of this e-book may be used, reproduced or transmitted in any manner or by any means whatsoever without

written or electronic permission from the publisher, except in the case of brief or long quotations, embodied in criticalreviews and articles.

Of course, it is authorized, from now and on, the copying of all or any parts of this article, provided there are no profit

purposes; only for study, personal use, reproduction with no monetary interests, duplication to give it as a gift, et cetera.

This writing has been translated from Spanish into English mainly by usingGoogle Translator, so please excuse me for any spelling and grammar errors.Spanish is my mother tongue.

It is intended that the following writing constitute, in the distant future(due to lack of time, for now), a partial basis for an Introduction of areverse dictionary in which the words that constitute the entries arewritten from right to left, but the order or alphabetical indexing will be


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from left to right according to the order of the letters that make thewords resulting from the reversions; such as:

yrreb berry

yrrebkcalb blackberry

yrrebnarc cranberry

yrrebpsar raspberry

yrrebwarts strawberry

yrrehc cherry,

et cetera.

It is not about discovering warm water.

Reverse dictionaries already exist, for example, one by John

Walker, The rhyming dictionary of the English language: in which thewhole language is arranged according to its terminations...,

Routledge & Kegan Paul, 1983.

Imaginary abecedariesAbecedrii commentcia1. Sometimes, grammar and mathematics agree on some points, or

may reveal certain relationships, or the human brain can make, or

imagine, that certain elements, fragments, procedures, routes, et

cetera, harmonize, coincide or tune in some places, subdivisions,

times, systems... under certain conditions.

All sciences, disciplines and specialties have certain relations

to each other. A truism.


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This paper will match certain mechanisms of juxtapositions of

letters to form words (or groups of letters), with some values and

formulas of algebra.

English Grammar and Babylonian-Persian-Indo-Arab Algebrawill come together, but algebrathat very important discipline, a

branch or division of the mathematical science this time will be

serving Grammar as an auxiliary.

In 1586, William Bullokar printed the first English grammar,Bref Grammar for English.

By the 1930s American mathematicians Paul K. Rees andFred W. Sparks released theirCollege Algebra, published byMcGraw-Hill Book Company, Incorporated.

This book, expanded, was still printed in the 1990s.

1.1. Note 1: in chapter number7 there is a summaryShould you want to skip lengthy lucubrations and

constructions.

1.2. Note 2: in chapter number 8 there are exposed twotheories resulting from an observation exercise* (inductivemethod). Please see item 7.2.3.

*This one.

2. Symbols used.

In this paper, three main symbols will be used:

B basep powerR result.

2.1. In this algebraic-grammarian exercise, the number of letters ineach imaginary abecedary that we will create will constitute amathematical base (which is represented by the symbol B).


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2.2. The number of letters contained in the words that we will beable to build, will constitute the exponent or power,represented by the symbol p, written as a superscriptcharacter, an indicator of the power that the bases cited in 2.1.

will have to be raised to; that is, B

p

.

2.2.1. Furthermore, the different values that p will be acquiring,depending on the number of letters in each word we will be able tobuild, will indicate three things:

2.2.1.1. The number of terms each algebraic formula will have, andthe highest exponential value of each monomial, binomial, trinomial,quadrinomial, pentanomial, et cetera. The highest exponential valueof each nomialwill match the number of terms of such nomial, as wewill discuss below:

B this formula has one term, and this monomial is the simplestformula of all; here, p equals one, id est: Bp = B1 = B.

B(B-(p-1)) =B(B-(2-1)) =B(B-1) =B

2-B this formula has two terms, and in this binomial the highest

exponential value of p is two.

B(B-(p-2))(B-(p-1)) =B(B-(3-2))(B-(3-1)) =B(B-1)(B-2) =(B

2-B)(B-2) =

B3-B2-2B2+2B =B3-3B2+2B this formula has three terms, and in this trinomial thehighest exponential value of p is three.

B(B-(p-3))(B-(p-2))(B-(p-1)) =B(B-(4-3))(B-(4-2))(B-(4-1)) =B(B-1)(B-2)(B-3) =(B2-B)(B-2)(B-3) =(B3-B2-2B2+2B)(B-3) =(B3-3B2+2B)(B-3) =B

4-3B

3+2B

2-3B

3+9B

2-6B =

B4-6B3+11B2-6B this formula has fourterms, and in this

quadrinomial the highest exponential value of p is four.


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B(B-(p-4))(B-(p-3))(B-(p-2))(B-(p-1)) =B(B-(5-4))(B-(5-3))(B-(5-2))(B-(5-1)) =B(B-1)(B-2)(B-3)(B-4) =(B2-B)(B-2)(B-3)(B-4) =(B

3

-B

2

-2B

2

+2B)(B-3)(B-4) =

(B3-3B

2+2B)(B-3)(B-4) =

(B4-3B

3+2B

2-3B

3+9B

2-6B)(B-4) =

(B4-6B3+11B2-6B)(B-4) =B5-6B4+11B3-6B2-4B4+24B3-44B2+24B =B5-10B4+35B3-50B2+24B this formula has five terms, and in thispentanomial the highest exponential value of p is five.

You can create more formulas (please see the fifth paragraph ofsection 6.6.), but I think that with the above examples it will besuffice.

2.2.1.2. In the algebraic pre-formulas, within algebraic factors thathave two or more elements, the successive values of the symbol pwill serve as the bases for the construction of certain subtrahends tobe subtracted from B, for example:

(The different nomials, in these five pre-formulas are separated bysemicolons.)

B; monomial

B(B-(p-1)); binomial

B(B-(p-2))(B-(p-1)); trinomial

B(B-(p-3))(B-(p-2))(B-(p-1)); quadrinomial

B(B-(p-4))(B-(p-3))(B-(p-2))(B-(p-1)); pentanomial.

Note: In the above examples, in the first nomialthe symbol p has avalue of 1; in the second nomialthe symbol p has a value of 2; in thethird nomialthe symbol p has a value of 3; in the fourth nomialthesymbol p has a value of 4; and in the fifth nomialthe symbol p has avalue of 5.

After being substituted the different values of p, math operations will

generate the following pre-formulas:


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B;

B(B-(2-1));

B(B-(3-2))(B-(3-1));

B(B-(4-3))(B-(4-2))(B-(4-1));

B(B-(5-4))(B-(5-3))(B-(5-2))(B-(5-1));

B;

B(B-1);

B(B-1)(B-2);

B(B-1)(B-2)(B-3);

B(B-1)(B-2)(B-3)(B-4);

According to the algebraic multiplications made in 2.2.1.1., weobtain the following five nomials, which are examples and nothingmore, because the possibilities are endless:

B

B2-BB3-3B2+2BB

4-6B

3+11B

2-6B

B5-10B4+35B3-50B2+24B2.2.1.3. In the algebraic formulas resulting from the algebraicmultiplications indicated by the pre-formulas, the symbol pindicates the highest exponential value of each nomial, andexponents or powers of the terms of each nomialgo down 1 by 1,

from left to right, up to the value of 1.


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The number of terms of each nomialequals the highestexponent or power, p, of such nomial, as it can be seen in the finalpart of 2.2.1.2.

2.3. The symbol R, which means: result, will be used rarely.

Note: in certain mathematical disciplines, as well as incomputer programming and software creation, some authors andengineers indicate or denote raising a number to a power, not bywriting a superscript number (exempli gratia, 23 = 8), but through thecollation of a caret, hood, or circumflex accent (which in Word 2003and Word 2007 can be written by pressing the following keys: ALT94 ^, or by using the Unicode key 005E ^; or another Unicode key,0302: ^), exempli gratia, 2 ^ 3 = 8. Both forms are correct; here, wewill be using predominantly the superscript character, but in tables ofchapter 7, we will use the hood or caret (or circumflex accent).

Moreover, the multiplication sign (by) can be obtained inWord 2003 and Word 2007 by pressing ALT 158: , or ALT0215, or by using the Unicode key 00D7: .If in some Windows or Unix platform you use, you cannot write

some symbols because the software does not support ALT orUnicode keys, or if your keyboard does not respond, I will suggestsomething: please firstly write in Word the symbols, signs, funny

letters, Greek letters, strange characters, rarely used characters, etcetera, and then select, copy and paste them in the Web space inwhich you are writing.

In another post of one of my other blogs, you can find anumber of ALT and Unicode keys to symbols, signs, Latin letters,Greek letters, et cetera:

http://algodatos.blogspot.mx/2013/05/algunos-codigos-ascii-y-unicodes-para.html

3. Lets imagine an ary(abecedary) with a single letter: a.

3.1. Imagine that we could form one-letter words each, only.

3.1.1. According to sections 2.1. and 2.2., and by virtue of thatin ourarythere is only oneletter, lets raise the base 1 to the firstpower: Bp = B1 = 11 = 1, which is the number of words that would

exista single word: a.


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3.3.2. Now, lets introduce a restriction equal to the previousone: that in each and every of the resulting words, we may notrepeat any letter.

Instead of raising the number of letters in our ary(1) to a

power equal to the number of letters that each and every word couldhave (3 in this example), under the restriction we apply the formula:B(B-1)(B-2) = (B

2-B)(B-2) = B

3-B

2-2B

2+2B = B

3-3B

2+2B.

Then, as in this case ouraryhas only one letter: the letter a,we apply the formula B3-3B2+2B, and substitute: 13-(3)(12)+2(1) = 1-3+2 = 0 (zero), which is the number of words we obtain, according toourrestriction: no word, in view that if ouraryhas only one letter(a), and we want to form three-letter words each, words in which noletter is repeated, that is impossible.

3.4. Further, imagine that we could form four-letter wordseach, only.

3.4.1. According to sections 2.1. and 2.2., and as in ourarythere is only one letter, we raise the base 1 to the fourth power: 14 =1 1 1 1 = 1, which is the number of words that would existasingle word: aaaa.

3.4.2. Now, lets introduce a restriction equal to the above:

that in each and every of the resulting words, we may not repeat anyletter.

Instead of raising the number of letters in our ary(1) to apower equal to the number of letters that each and every word couldhave (4 in this example), under the restriction we apply the formula:B(B-1)(B-2)(B-3) =(B2-B)(B-2)(B-3) =(B3-B2-2B2+2B)(B-3) =(B3-3B2+2B)(B-3) =B

4

-3B3

+2B2

-3B3

+9B2

-6B =B

4-6B

3+11B

2-6B.

Then, as in this case ouraryhas only one letter: the letter a,we apply the formula B4-6B3+11B2-6B, and substitute: 14-(6)(13)+(11)(12)-(6)(1) = 1-6+11-6 = 0 (zero), which is the number ofwords we obtain, according to our restriction: no word, because ifouraryhas only one letter (a), and we want to build words of fourletters each, words in which no letter is repeated, that is impossible.


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(An explanatory parentheses: as you can see, the formulas forthe restrictions come from simpler formulas, such as the onesshown above, in the second paragraph of item 3.2.2., and in thesecond paragraph of item 3.3.2 .)

3.5. Further, imagine that could form five-letter words eachonly.

3.5.1. According to sections 2.1. and 2.2., and as in ourarythere is only one letter, we raise the base 1 to the fifth power: 15 = 1 1 1 1 1 = 1, which is the number of words that would existasingle word: aaaaa.

3.6. And so on.

3.7. Now, lets imagine that we could form one-letter and two-letter words each, only.

There would be two words: a, aa (11

+ 12

= 1 + 1 = 2). This lastarithmetic operation is a result from the algebraic formula: Bp + Bp =R, where B is the base or number of letters in our ary(1), p is thepower or exponent that represents the number of letters the wordswe may form will have, and R is the result, in this case: 1

1+ 1

2= 1 +

1 = 2.

The above statement is according to what was set in sections2.1. and 2.2. of this algebraic-grammarian text.

3.8. Also, imagine that we could form one-letter, two-letter, andthree-letter words each, only.

There would be three words: a, aa, aaa (11

+ 12

+ 13

= 1 + 1 +1 = 3).

3.9. In addition, lets assume that we could form one-letter,

two-letter, three-letter, and four-letter words each, only.There would be four words: a, aa, aaa, aaaa (1

1+ 1

2+ 1

3+ 1

4

= 1 + 1 + 1 + 1 = 4).

3.9.1. And so on.

4. Now, lets imagine an abeary(abecedary) with two letters: a,b.



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4.1.1. According to sections 2.1. and 2.2., and as in ourabeary

there are only twoletters, lets raise the base 2 to the first power Bp= B1 = 21 = 2, which is the number of words that would existtwo

words: a, b.

4.1.2. Now lets introduce a restriction: that in each and everyof the resulting words, we may not repeat any letter.

Instead of raising the number of letters in our abeary(2) to apower equal to the number of letters that each and every word couldhave (1 in this example), under the restriction we apply the formula:B.

Then, as in this case ourabearyhas only two letters (a, b),and there is a restriction, we apply the formula B, and substitute: B= 2 (two), which is the number of words we obtain, according to ourrestriction: two words, because if ourabearyhas only two letters(a, b), and we want to form one-letter words each, words in which noletter is repeated, that gives as a result two words: a, b.

4.2. Then imagine that we could form two-letter words each,only.

4.2.1. According to sections 2.1. and 2.2., and as in ourabearythere are only twoletters, lets raise the base 2 to the second power(or squared) 22 = 2 2 = 4, which is the number of words that wouldexistfour words: aa, ab, ba, bb.

4.2.2. Now, lets introduce a restriction, as above: that in eachand every of the resulting words, we may not repeat any letter.

Instead of raising the number of letters in ourabeary(2 in this

example) to a power equal to the number of letters that each andevery word could have (2 in this example), under the restriction weapply the formula: B(B -1) = B2-B.

This formula is the same as the one that appears in thesecond paragraph of item 3.2.2.

Then, as in this case ourabearyhas only two letters (a, b), weapply the formula B2-B, and substitute: 22-2 = 4-2 = 2, which is thenumber of words that would exist, according to our restrictiontwowords: ab, ba.

4.3. Imagine that we could form three-letter words each only.


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have (4 in this example), under ourrestriction we apply theformula:B(B-1)(B-2)(B-3) =(B2-B)(B-2)(B-3) =(B

3

-B

2

-2B

2

+2B)(B-3) =

(B3-3B

2+2B)(B-3) =

B4-3B

3+2B

2-3B

3+9B

2-6B =

B4-6B3+11B2-6B.This formula is the same as the one that appears in the

second paragraph of item 3.4.2.

Then, we apply the formula B4-6B3+11B2-6B, and substitute:24-6(23)+11(22)-6B = 16-(6)(8)+(11)(4)-(6)(2) = 16-48+44-12 = 0(zero), which is the number of words we obtain, according to ourrestriction: no word, in view that if ourabearyhas only two letters(a, b), and we want to form four-letter words each, words in which noletter is repeated, that is impossible.

4.5. Imagine that we could form five-letter words each only.

4.5.1. According to sections 2.1. and 2.2., and as in ourabearythere are only two letters, lets raise the base 2 to the fifth power: 25

= 2 2 2 2 2 = 32, which is the number of words that wouldexistthirty two words:

(16 with initial a):

aaaaa (5 as)aaaab aaaba aabaa abaaa (4 as and 1 b)aaabb aabab aabba abaab ababa abbaa (3 as and 2 bs)aabbb ababb abbab abbba (2 as and 3 bs)abbbb (1 a and 4 bs)(16 with initial b):baaaa (1 b and 4 as)baaab baaba babaa bbaaa (2 bs and 3 as)baabb babab babba bbaab bbaba bbbaa (3 bs and 2 as)babbb bbabb bbbab bbbba (4 bs and 1 a)bbbbb (5 bs)


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4.6. And so on.

(In mathematics and computer programming, when it is not

possible to write a power or exponent in the form of a superscriptcharacter [as in: Bp] many individuals use the hood or caret or

circumflex accent, please read the Note under section 2.3.(Thus, it is true that: 25 = 2 ^ 5 = 32.(Bp = B ^ p)

4.7. Next, imagine that we could form one-letter, and two-letterwords each, only.

There would be six words: a, b, aa, ab, ba, bb (21 + 22 = 2 + 4= 6).

4.8. Then, imagine that we could form one-letter, two-letter,and three-letter words each, only.

There would be fourteen words:

a baa ab ba bbaaa

aab aba baaabb bab bba

bbb(21 + 22 + 23 = 2 + 4 + 8 = 14).

4.9. Then, imagine that we could form one-letter, two-letter,three-letter, and four-letter words each, only.

There would be 30 words:

a baa ab ba bbaaa aab aba baa


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abb bab bbabbb

aaaa aaab aaba abaa baaaaabb abab abba baab baba bbaa

abbb babb bbab bbba bbbb.

(21

+ 22

+ 23

+ 24

= 2 + 4 + 8 + 16 = 30).

4.10. Further, imagine that we could form one-letter, two-letter,three-letter, four-letter, and five-letter words each, only.


a baa ab ba bbaaa

aab aba baaabb bab bba

bbbaaaa aaab aaba abaa baaa

aabb abab abba baab baba bbaaabbb babb bbab bbba bbbb

aaaaaaaaab aaaba aabaa abaaaaaabb aabab aabba abaab ababa abbaaaabbb ababb abbab abbbaabbbbbaaaabaaab baaba babaa bbaaabaabb babab babba bbaab bbaba bbbaababbb bbabb bbbab bbbbabbbbb


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(21 + 22 + 23 + 24 + 25 = 2 + 4 + 8 + 16 + 32 = 62).

4.11. And so on.

5. Then, imagine an abeceary(abecedary) with three letters: a,b, c.


5.1.1. According to sections 2.1. and 2.2., and as in ourabecearythere are only threeletters, lets raise the base 3 to thefirst power Bp = B1 = 31 = 3, which is the number of words thatwould existthree words: a, b, c.

5.1.2. Now, lets introduce a restriction: that in each andevery of the resulting words, we may not repeat any letter.

Instead of raising the number of letters in our abeceary(3) to a

power equal to the number of letters that each and every word couldhave (1 in this example), under the restriction we apply the formula:B.

Then, as in this case ourabecearyhas only three letters (a, b,c), and there is a restriction, we apply the formula B, andsubstitute: B = 3 (three), which is the number of words we obtain,according to ourrestriction: three words, because if ourabecearyhas only three letters (a, b, c), and we want to form one-letter wordseach, words in which no letter is repeated, that gives as a result

three words: a, b, c.

Then, as in this case ourabecearyhas three letters (a, b, c), weapply the formula, B, and substitute: B = 3, which is the number ofwords that we agree with our restriction, which requires not repeatany letters in words: a, b, c.

5.2. Imagine that we could form two-letter words each only.

5.2.1. According to sections 2.1. and 2.2., and as in ourabecearythere are only three letters, lets raise the base 3 to the


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second power (or squared): 32 = 9, which is the number of wordsthat would existnine words: aa, ab, ac, ba, bb, bc, ca, cb, cc.

5.2.2. Now, lets introduce a restriction: that in each and

every of the resulting words, we may not repeat any letter.Instead of raising the number of letters in ourabeceary(3 inthis example) to a power equal to the number of letters that eachand every word could have (2 in this example), under the restrictionwe apply the formula: B(B -1) = B2-B.


Then, as in this case ourabecearyhas only three letters (a, b,c), we apply the formula B2-B, and substitute: 32-3 = 9-3 = 6, whichis the number of words that would exist, according our restrictionsix words: ab, ac, ba, bc, ca, cb.

5.3. Now imagine that we could form three-letter words eachonly.

5.3.1. In accordance with sections 2.1. and 2.2., and as in ourabecearythere are only three letters, lets raise the base 3 to thethird power (or cubed): 3

3= 3 3 3 = 27, which is the number of

words that would existtwenty-seven words:

By abecedaricorder:

aaa, aab, aac, aba, abb, abc, aca, acb, acc,(9 begin with an a)

baa, bab, bac, bba, bbb, bbc, bca, bcb, bcc,(9 start with a b)

caa, cab, cac, cba, cbb, cbc, cca, ccb, ccc(9 begin with a c).

Another way to arrange the same 27 words:

abc, acb, bac, bca, cab, cba, (with no repeated letters)

one letter repeated:aab, aac, aba, aca, baa, caa, (words with 2 as)

abb, bab, bba, bbc, bcb, cbb, (words with 2 bs)acc, cac, cca, bcc, cbc, ccb, (words with 2 cs)


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each letter appears three times:aaa, bbb, ccc (3 as, 3 bs, 3 cs).


Instead of raising the number of letters in ourabeceary(3) to apower equal to the number of letters that each and every word couldhave (3 in this example), under the restriction we apply the formula:

B(B-1)(B-2) = (B2-B)(B-2) = B

3-B

2-2B

2+2B = B

3-3B

2+2B.


Then, as in this case ourabecearyhas only three letters (a, b,c), we apply the formula B

3-3B

2+2B, and substitute: 3

3-(3)(3

2)+2(3) =

27-27+6 = 6 (six), which is the number of words we obtain,according to ourrestriction:six words: abc, acb, bac, bca, cab,cba.

5.4. Imagine that we could form four-letter words each only.

5.4.1. In accordance with sections 2.1. and 2.2., and as in ourabecearythere are only three letters, we raise the base 3 to thefourth power: 34 = 3 3 3 3 = 81, which is the number of wordsthat would exist:

(27 begin with an a):

aaaa, aaab, aaac,aaba, aabb, aabc, (9 begin with aa)aaca, aacb, aacc,abaa, abab, abac,abba, abbb, abbc, (9 begin with ab)abca, abcb, abcc,acaa, acab, acac,acba, acbb, acbc, (9 begin with ac)acca, accb, accc.


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(27 start with a b):baaa, baab, baac,baba, babb, babc, (9 start with ba)

baca, bacb, bacc,bbaa, bbab, bbac,bbba, bbbb, bbbc, (9 start with bb)bbca, bbcb, bbcc,bcaa, bcab, bcac,bcba, bcbb, bcbc, (9 start with bc).bcca, bccb, bccc.

(27 begin with a c)caaa, caab, caac,caba, cabb, cabc, (9 begin with ca)caca, cacb, cacc,cbaa, cbab, cbac,cbba, cbbb, cbbc, (9 begin with cb)cbca, cbcb, cbcc,ccaa, ccab, ccac,ccba, ccbb, ccbc, (9 begin with cc)ccca, cccb, cccc,5.4.2. Now, lets introduce a restriction: that in each and

every of the resulting words, we may not repeat any letter.

Instead of raising the number of letters in ourabeceary(3) to apower equal to the number of letters that each and every word couldhave (4 in this example), under the restriction we apply the formula:

B(B-1)(B-2)(B-3) =(B2-B)(B-2)(B-3) =(B3-B2-2B2+2B)(B-3) =(B

3-3B

2+2B)(B-3) =

B4-3B3+2B2-3B3+9B2-6B =B

4

-6B3

+11B2

-6B.


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We apply the formula B4-6B3+11B2-6B, and substitute:

3

4

-(6)(3

3

)+(11)(3

2

)-(6)(3) = 81-(6)(27)+(11)(9)-18 = 81-162+99-18 =0 (zero), which is the number of words we obtain, according to ourrestriction: no word, in view that if ourabecearyhas only threeletters (a, b, c), and we want to form four-letter words each, words inwhich no letter is repeated, that is impossible.

5.5. Now, imagine that we could form one-letter, and two-letterwords only, each.

There would be twelve words: a, b, c, aa, ab, ac, ba, bb, bc, ca,cb, cc.

(31 + 32 = 3 + 9 = 12).

5.6. Now imagine that we could form one-letter, two-letter, andthree-letter words each, only.


a, b, c,aa, ab, ac, ba, bb, bc, ca, cb, cc,

aaa, aab, aac, aba, abb, abc, aca, acb, acc,baa, bab, bac, bba, bbb, bbc, bca, bcb, bcc,caa, cab, cac, cba, cbb, cbc, cca, ccb, ccc.

(31

+ 32

+ 33

= 39).

5.7. Now imagine that we could form one-letter, two-letter, three-letter, and four-letter words each, only.


a, b, c,

aa, ab, ac, ba, bb, bc, ca, cb, cc,

aaa, aab, aac, aba, abb, abc, aca, acb, acc,baa, bab, bac, bba, bbb, bbc, bca, bcb, bcc,caa, cab, cac, cba, cbb, cbc, cca, ccb, ccc,

aaaa, aaab, aaac,aaba, aabb, aabc,


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aaca, aacb, aacc,abaa, abab, abac,abba, abbb, abbc,abca, abcb, abcc,

acaa, acab, acac,acba, acbb, acbc,acca, accb, accc.baaa, baab, baac,baba, babb, babc,baca, bacb, bacc,bbaa, bbab, bbac,bbba, bbbb, bbbc,bbca, bbcb, bbcc,bcaa, bcab, bcac,bcba, bcbb, bcbc,bcca, bccb, bccc,caaa, caab, caac,caba, cabb, cabc,caca, cacb, cacc,cbaa, cbab, cbac,cbba, cbbb, cbbc,cbca, cbcb, cbcc,ccaa, ccab, ccac,ccba, ccbb, ccbc,ccca, cccb, cccc.(31 + 32 + 33 + 34 = 3 + 9 + 27 + 81 = 120).

5.8. And so on.

6. Now, imagine an abecedarywith four letters: a, b, c, d.

6.1. Imagine that could form one-letter words each, only.


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6.1.1. According to sections 2.1. and 2.2. and as in ourabecedarythere are only four letters, we raise the base 4 to the firstpower 41 = 4, which is the number of words that would existfourwords: a, b, c, d.


Instead of raising the number of letters in our abecedary(4) toa power equal to the number of letters that each and every wordcould have (1 in this example), under the restriction we apply theformula: B.

Then, as in this case ourabecedaryhas only four letters (a, b,c, d), we apply the formula B, and substitute: B = 4 (four), which isthe number of words we obtain, according to ourrestriction: a, b, c,d.

6.2. Imagine that we could form two-letter words each, only.

6.2.1. According to sections 2.1. and 2.2., and as in ourabecedarythere are only four letters (a, b, c, d), we raise the base 4to the second power (or squared): 42 = 16, which is the number ofwords that would exist: aa, ab, ac, ad, ba, bb, bc, bd, ca, cb, cc, cd,da, db, dc, dd.


Instead of raising the number of letters in ourabecedary(4) toa power equal to the number of letters that each and every wordcould have (2 in this example), under the restriction we apply theformula: B2-B.

Then as in this case our abecedary has only four letters (a, b,c, d) we apply the formula, B2-B, and substitute: 42-4 = 16-4 = 12

(twelve), which is the number of words we obtain, according to ourrestriction: ab, ac, ad, ba, bc, bd, ca, cb, cd, da, db, dc.

6.3. Now, lets imagine that we could form three-letter wordseach, only.

6.3.1. In accordance with sections 2.1. and 2.2., and as in ourabecedarythere are four letters, we raise the base 4 to the thirdpower (or cubed): 4.3 = 4 4 4 = 64, which is the number of words

that would exist:


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By abecedaricorder:

aaa, aab, aac, aad, aba, abb, abc, abd, aca, acb, acc, acd,ada, adb, adc, add.

(16 begin with an a)

baa, bab, bac, bad, bba, bbb, bbc, bbd, bca, bcb, bcc, bcd,bda, bdb, bdc, bdd.

(16 begin with a b)

caa, cab, cac, cad, cba, cbb, cbc, cbd, cca, ccb, ccc, ccd, cda,cdb, cdc, cdd.

(16 begin with a c).

daa, dab, dac, dad, dba, dbb, dbc, dbd, dca, dcb, dcc, dcd,dda, ddb, ddc, ddd.

(16 begin with a d).

6.3.2. Now, introduce a restriction: that in each and every ofthe resulting words, we may not repeat any letter.

Instead of raising the number of letters in ourabecedary(4 inthis case) to a power equal to the number of letters that each andevery word could have (3 in this example), under the restriction we

apply the formula: B3-3B2+2B.Then as in this case ourabecedaryhas only four letters (a, b,

c, d) we apply the formula B3-3B2+2B, and substitute: 43-(3)(42)+(2)(4) = 64-(3)(16)+8 = 64-48+8 = 24 (twenty-four), which isthe number of words we obtain, according to our restriction:

abc, abd, acb, acd, adb, adc.(6 begin with an a)

bac, bad, bca, bcd, bda, bdc.(6 begin with a b)

cab, cad, cba, cbd, cda, cdb.(6 begin with a c).

dab, dac, dba, dbc, dca, dcb.(6 begin with d).

6.4. Now imagine that could form four-letter words each, only.


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6.4.1. In accordance with sections 2.1. and 2.2., and as in our

abecedarythere are four letters (a, b, c, d), we raise the base 4 tothe fourth power 44 = 4 4 4 4 = 256, which is the number of

words that would exist:

aaaa baaa caaa daaaaaab baab caab daabaaac baac caac daacaaad baad caad daadaaba baba caba dabaaabb babb cabb dabbaabc babc cabc dabcaabd babd cabd dabdaaca baca caca dacaaacb bacb cacb dacbaacc bacc cacc daccaacd bacd cacd dacdaada bada cada dadaaadb badb cadb dadbaadc badc cadc dadcaadd badd cadd daddabaa bbaa cbaa dbaaabab bbab cbab dbababac bbac cbac dbacabad bbad cbad dbadabba bbba cbba dbbaabbb bbbb cbbb dbbbabbc bbbc cbbc dbbcabbd bbbd cbbd dbbdabca bbca cbca dbcaabcb bbcb cbcb dbcbabcc bbcc cbcc dbccabcd bbcd cbcd dbcdabda bbda cbda dbdaabdb bbdb cbdb dbdbabdc bbdc cbdc dbdcabdd bbdd cbdd dbddacaa bcaa ccaa dcaaacab bcab ccab dcabacac bcac ccac dcacacad bcad ccad dcadacba bcba ccba dcba


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acbb bcbb ccbb dcbbacbc bcbc ccbc dcbcacbd bcbd ccbd dcbdacca bcca ccca dccaaccb bccb cccb dccb

accc bccc cccc dcccaccd bccd cccd dccdacda bcda ccda dcdaacdb bcdb ccdb dcdbacdc bcdc ccdc dcdcacdd bcdd ccdd dcddadaa bdaa cdaa ddaaadab bdab cdab ddabadac bdac cdac ddacadad bdad cdad ddadadba bdba cdba ddbaadbb bdbb cdbb ddbbadbc bdbc cdbc ddbcadbd bdbd cdbd ddbdadca bdca cdca ddcaadcb bdcb cdcb ddcbadcc bdcc cdcc ddccadcd bdcd cdcd ddcdadda bdda cdda dddaaddb bddb cddb dddbaddc bddc cddc dddcaddd bddd cddd dddd.

6.4.2. Now, lets introduce a restriction: that in each and everyof the resulting words, we may not repeat any letter.

Instead of raising the number of letters in ourabecedary(4 inthis case) to a power equal to the number of letters that each and

every word could have (4 in this example), under the restriction weapply the formula: B

4-6B

3+11B

2-6B, and substitute: 4

4-

(6)(43)+(11)(42)-(6)(4) = 256-(6)(64)+(11)(16)-(6)(4) = 256-384+176-24 = 24 (twenty-four), which is the number of words we obtainaccording to hour restriction:

abcd, abdc, acbd, acdb, adbc, adcb,bacd, badc, bcad, bcda, bdac, bdca,cabd, cadb, cbad, cbda, cdab, cdba,dabc, dacb, dbac, dbca, dcab, dcba.


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6.5. Now, lets imagine that we may form five-letter wordseach, only.

6.5.1. In accordance with sections 2.1. and 2.2., and as in our

abecedarythere are four letters (a, b, c, d), we raise the base 4 tothe fifth power: 45

= 4 4 4 4 4 = 1,024, which is the numberof words that exist:

aaaaa, aaaab, aaaac, aaaad...

and 1,020 more words in addition to the four written above assamples

6.5.2. Now, introduce a restriction: that in each and every ofthe resulting words, we may not repeat any letter.

Instead of raising the number of letters in our abecedary (4) toa power equal to the number of letters that each and every wordcould have (5 in this example), under the restriction we apply theformula: B5-10B4+35B3-50B2+24B.

Then, as in ourabecedarythere are four letters (a, b, c, d), weapply the formula:B5-10B4+35B3-50B2+24B,

and we substitute:45-(10)(44)+(35)(43)-(50)(42)+(24)(4) =1024-(10)(256)+(35)(64)-(50)(16)+(24)(4) =1024-2560+2240-800+96 = 0 (zero), which is the number of words

we obtain, according to our restriction: no word, in view that if ourabecedaryhas only four letters (a, b, c, d), and we want to form five-letter words each, words in which no letter is repeated, that isimpossible.

6.5.3. And so on.

6.5.4. Next, imagine that could form one-letter, and two-letterwords each, only.


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There would be 20 (twenty) words: a, b, c, d, aa, ab, ac, ad,ba, bb, bc, bd, ca, cb, cc, cd, da, db, dc, dd (41 + 42 = 4 + 16 = 20).

6.5.5. Then, imagine that we could form one-letter, two-letter,

and three-letter words each, only.


a, b, c, d,aa, ab, ac, ad, ba, bb, bc, bd, ca, cb, cc, cd, da, db, dc, dd,aaa, aab, aac, aad, aba, abb, abc, abd, aca, acb, acc, acd,

ada, adb, adc, add,baa, bab, bac, bad, bba, bbb, bbc, bbd, bca, bcb, bcc, bcd,

bda, bdb, bdc, bdd,caa, cab, cac, cad, cba, cbb, cbc, cbd, cca, ccb, ccc, ccd, cda,

cdb, cdc, cdd,daa, dab, dac, dad, dba, dbb, dbc, dbd, dca, dcb, dcc, dcd,

dda, ddb, ddc, ddd.

(41 + 42 + 43 = 4 + 16 + 64 = 84).6.5.6. And so on.

6.6. In order to obtain very many combinations of letters, youcould use the formula 26 ^ 26, id est 2626, that is twenty-six raised tothe twenty-sixth power, a mathematical operation that would result ina frightening, figure balmbica. Twenty-six is the number of lettersin the international Latin order also, of the English abecedary, etcetera.

a b c d e f g h I j k l m n o p q r s t u v w x y z.

[In a blog of mine,algodatos.blogspot.comin some posts ofJune, 2013, you will find many groups of letters similar to those ofitem 6.4.1. of this algebraic-grammatical writing, but those groupsare much larger, with the inconvenience for you English-speakingreaders, those groups of letters include the Spanish letter (soundsalways like ny in canyon).
http://www.algodatos.blogspot.com/http://www.algodatos.blogspot.com/http://www.algodatos.blogspot.com/http://www.algodatos.blogspot.com/


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[In some May 2013 posts, there are 27 posts with 729 three-letter groups each, which makes a total of 19,683 three-lettergroups.]

Now imagine formulas similar to those written at the end ofitem 2.2.1.2., Starting with: B26

-(an X number)B25

+(an Xnumber)B

24... and rather very lengthy formulas that would be

horrific.

Furthermore, the Hawaiian alphabet/abecedary has only 12letters: a, e, h, i, k, l, m, n, o, p, u, w.

Here, by applying 12 ^ 12 = 1212 = 8,916,100,448,256, or eighttrillion nine hundred and sixteen billion one hundred million fourhundred and forty-eight thousand two hundred fifty-six possiblewords (or groups of letters), a figure also somewhat scary, but notas high as the ones obtained when performing operations outlined inprevious paragraphs.

7. Summary.

7.1. Table 1.Table 1.NL 1L 1Lw 2L 2Lw 3L 3Lw 4L 4Lw 5L 5Lw 6L 6Lw1 1 1 1 0 1 0 1 0 1 0 1 02 2 2 4 2 8 0 16 0 32 0 64 03 3 3 9 6 27 6 81 0 243 0 729 04 4 4 16 12 64 24 256 24 1024 0 4096 05 5 5 25 20 125 60 625 120 3125 120 15625 06 6 6 36 30 216 120 1296 360 7776 720 46656 720

B I B^2 II B^3 III B^4 IV B^5 V B^6 VINote: B ^ 2 means B squared; B ^ 3, B cubed; B ^ 4, B raised tothe fourth power, et cetera.

Roman numerals indicate in Table 1, which formulas listed below,are applied in the respective columns of Table 1:

I) B.II) B2-B.III) B3-3B2+2B.


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IV) B4-6B

3+11B

2-6B.

V) B5-10B4+35B3-50B2+24B.VI) The author did not want to develop the formula VI.

Meanings of the acronyms or abbreviations that appear in the firstrow of Table 1:NL Number of letters in each imaginary abecedary.1L 1-letter words.1Lw 1-letter words without repeating any letters.2L 2-letter words.2Lw 2-letter words without repeating any letters.3L 3-letter words.3Lw 3-letter words without repeating any letters.4L 4-letter words.4Lw 4-letter words without repeating any letters.5L 5-letter words.5Lw 5-letter words without repeating any letters.6L 6-letter words.6Lw 6-letter words without repeating any letters.

7.2. Now, let's examine some points in Table 1:

7.2.1. First, a curiosity: we can see two parallel quasi-diagonal

broken lines, from the upper left to the lower right corner of Table 1,marked or constituted by the respective factorials of the followingnumbers: 1, 2, 3, 4, 5, and 6.

Let's see what is a factorial:The Merriam-Webster Dictionary defines the factorial: the

product of all the positive integers from 1 to n. Thesymbol of thefactorial is: ! (the same as an exclamation mark).

So, as an example, the factorial of 4 is 4! = 1 2 3 4 = 24.

Factorial of 1: 1! = 1 1 = 1.Factorial of 2: 2! = 1 2 = 2.Factorial of 3: 3! = 1 2 3 = 6.Factorial of 4: 4! = 1 2 3 4 = 24.Factorial of 5: 5! = 1 2 3 4 5 = 120.Factorial of 6: 6! = 1 2 3 4 5 6 = 720.


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So, in the Table 1, we can see these two parallel quasi-diagonal broken lines, each formed by the numbers 1, 2, 6, 24, 120,and 720.

These factorials are always in columns whose titles/heads endwith a lower case w, id est, in columns that do not allow therepetition of letters.

7.2.2. Second, something obvious: in the even-numbercolumns(second, fourth, sixth, eighth, tenth and twelfth columns) of Table 1,the exponents or powers (defined as p in 2.2. of this text) increasetheir values 1 by 1 power from left to right: B (or: B ^ 1), B ^ 2, B ^ 3,B ^ 4, B ^ 5, B ^ 6.

7.2.3. Third, and most importantly: it is observed that there

is a pattern in each row, from left to right, which justifies, andsimultaneously follows the pre-formula* B(B-1 ), that is, Bmultiplied by B-minus-one.

* Apparently, if step by step each element of the pattern has given rise toa pre-formula, then it has been utilized the inductive-synthetic method (fromparticulars to generals-from parts to wholes); a theoretical formula has beenconstructed, it is justified.

If the pre-formula is applied in a practical way and it can be applied inequal cases (or very similar cases), always giving equal results (or very similarresults, according to the respective instance), it has been used the deductive-analytical method (from generals to particulars-from wholes to parts): theformula is followed by us.

The first pre-formulas are simple, and give rise to morecomplex pre-formulas and formulas, such as those developed initem 2.2.1.2.

The explanation is as follows:If you look at the third column (a column whose title ends with

a lower case w, which means it does not support repetition ofletters, and at the bottom of such column there is a Roman numeralI, which corresponds to the formula B, the simplest), and yourepresent the value of each number with a letter B, this B is thefirst pre-formula or formula, and it is the simplest of all, as noted inthe second paragraph of 2.2.1.1. (The symbol B means base, asindicated in 2.1.) - You will be able to see that if you apply the pre-


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formula B(B-1), you get certain results, which were written in thefifth column of each respective row, in Table 1.

Lets verify

The pre-formula B(B-1) is equal to the formula B

2

-B.

First row: 1(1-1) = (1)(0) = 0.Or: 1(1-1) = 1-1 = 0.

Second row: 2(2-1) = (2)(1) = 2.Or: 2(2-1) = 4-2 = 2.

Third row: 3(3-1) = (3)(2) = 6.Or: 3(3-1) = 9-3 = 6.

Fourth row: 4(4-1) = (4)(3) = 12.Or: 4(4-1) = 16-4 = 12.

Fifth row: 5(5-1) = (5)(4) = 20.Or: 5(5-1) = 25-5 = 20.

Sixth row: 6(6-1) = (6)(5) = 30.Or: 6(6-1) = 36-6 = 30.

So far, we have the pre-formula B(B-1), which gives rise to theformula B2-B.

7.2.4. If we consider our resultsproduct of algebraic operationsperformed in item 7.2.3. when following the pre-formula B(B-1),values that have also been written in the fifth column of Table 1, asa result of an induced data capture (inductive-synthetic method),

and we take those data or values individually, and then multiplyeach one by (B-2), we obtain some results, which were written in theseventh column of Table 1:

First row: 0(B-2) = 0(1-2) = 0(-1) = 0.

Second row: 2 (B-2) = 2 (2-2) = 2 (0) = 0.

Third row: 6 (B-2) = 6 (3-2) = 6 (1) = 6.

Fourth row: 12 (B-2) = 12 (4-2) = 12 (2) = 24.


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Fifth row: 20 (B-2) = 20 (5-2) = 20 (3) = 60.

Sixth row: 30 (B-2) = 30 (6-2) = 30 (4) = 120.

Remember that the six values of the far left in the above operations,id est, 0, 2, 6, 12, 20, and 30, are the product of applying the pre-formula B(B-1) in item 7.2.3. and are written in the fifth column ofTable 1, and in turn the results of the algebraic operationsperformed in 7.2.4. (id est, in this item) of the extreme right, id est, 0,0, 6, 24, 60 and 120, are the result of applying the pre-formulaB(B-1)(B-2) and have been written in the seventh column of Table 1.

This pre-formula B(B-1)(B-2) has been shown in item 2.2.1.1(please see such item in this writing), and leads to the formula B3-3B2+2B), so that having used the inductive-synthetic method afteran empirical or experimental exercise, we can state that thealgebraic pre-formulas and formulas shown in item 2.2.1.1.have emerged from the pattern observed in Table 1 andmentioned in the opening lines of item 7.2.3., although Table 1has been placed many paragraphs below item 2.2.1.1., because theintention has been to take the reader first by the empirical road.

So far, we have the pre-formula B(B-1)(B-2) which leads to theformula B3-3B2+2B.

7.2.5. Now, lets consider our results of the algebraic operations initem 7.2.4., product of having followed the pre-formula B(B-1)(B-2)

or the formula B3-3B

2+2B, values that have also been written in

the seventh column of Table 1, as a result of an induced datacapture (inductive-synthetic method), and take those data or valuesin the individual, and then multiply each by (B-3), to obtain certain

results, which were written in the ninth column of Table 1:

First row: 0(B-3) = 0(1-3) = 0(-2) = 0.

Second row: 0(B-3) = 0(2-3) = 0(-1) = 0.

Third row: 6(B-3) = 6(3-3) = 6(0) = 0.

Fourth row: 24(B-3) = 24(4-3) = 24(1) = 24.

Fifth row: 60(B-3) = 60(5-3) = 60(2) = 120.


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Sixth row: 120(B-3) = 120(6-3) = 120(3) = 360.

So far, we have the pre-formula B(B-1)(B-2)(B-3), which leads to

the algebraic formula B

4

-6B

3

+11B

2

-6B.

7.2.6. Now, similarly to what has been performed in the immediatelypreceding paragraph, lets consider our results of the algebraicoperations in item 7.2.5., product of having followed the pre-formula B(B-1)(B-2)(B-3), values that have also been written in theninth column of Table 1, a result of an induced data capture(inductive-synthetic method), and lets take the data or values in theindividual, and then multiply each by (B-4) to obtain certain results,which have also been written in the eleventh column of Table 1:

First row: 0(B-4) = 0(1-4) = 0(-3) = 0.

Second row: 0(B-4) = 0(2-4) = 0(-2) = 0.

Third row: 0(B-4) = 0(3-4) = 0(-1) = 0.

Fourth row: 24(B-4) = 24(4-4) = 24(0) = 0.

Fifth row: 120(B-4) = 120(5-4) = 120(1) = 120.

Sixth row: 360(B-4) = 360(6-4) = 360(2) = 720.

So far, we have the pre-formula B(B-1)(B-2)(B-3)(B-4), which leadsto the algebraic formula B

5-10B

4+35B

3-50B

2+24B.

7.2.7. Next, and to come to an end in our task of illustrating byexamples, a similar maneuver to what was done in the preceding

item, lets consider the results of our algebraic operations in item7.2.6., product of having followed the pre-formula B(B-1)(B-2)(B-3)(B-4)or the formula B5-10B4+35B3-50B2+24B, values that havealso been written in the eleventh column of Table 1 as a result of aninduced data capture (inductive-synthetic method), and take thedata or values in the individual, and then multiply each by (B-5) toobtain certain results, which were written in the thirteenth column ofTable 1:

First row: 0(B-5) = 0(1-5) = 0(-4) = 0.


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Second row: 0(B-5) = 0(2-5) = 0(-3) = 0.

Third row: 0(B-5) = 0(3-5) = 0(-2) = 0.

Fourth row: 0(B-5) = 0(4-5) = 0(-1) = 0.

Fifth row: 120(B-5) = 120(5-5) = 120(0) = 0.

Sixth row: 720(B-5) = 720(6-5) = 720(1) = 720.

So far, we have the pre-formula B(B-1)(B-2)(B-3)(B-4)(B-5), whichleads to a long formula whose highest exponential value is 6, andthat will not be developed in this text.

The odd column titles, from the third column to the right, in Table 1,end with a lower case w (without repetition), which means thatthe resulting words, formed from corresponding formulas, do notsupport any letter repetition, whereby the formulas for the oddcolumns are much more complex than those of the even columns.

7.3. Table 1, a copy.

Below, Table 1 has been copied, in order to facilitate any possiblecomparison with Table 2 ifthe original Table 1 has been printedfaraway (above) from Table 2, in the text:

Table 1, a copyNL 1L 1Lw 2L 2Lw 3L 3Lw 4L 4Lw 5L 5Lw 6L 6Lw1 1 1 1 0 1 0 1 0 1 0 1 02 2 2 4 2 8 0 16 0 32 0 64 03 3 3 9 6 27 6 81 0 243 0 729 04

4

4

16

12

64

24

256

24

1024

0

4096

0

5 5 5 25 20 125 60 625 120 3125 120 15625 06 6 6 36 30 216 120 1296 360 7776 720 46656 720


Roman numerals indicate in Table 1 which formulas listed below,are applied in the respective columns of Table 1:


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I) B.II) B2-B.III) B3-3B2+2B.IV) B4-6B3+11B2-6B.V) B

5

-10B

4

+35B

3

-50B

2

+24B.

VI) The author did not want to develop the formula VI.

Meanings of the acronyms or abbreviations that appear in the firstrow of Table 1:

NL Number of letters in each imaginary abecedary.1L 1-letter words.1Lw 1-letter words without repeating any letters.2L 2-letter words.2Lw 2-letter words without repeating any letters.3L 3-letter words.3Lw 3-letter words without repeating any letters.4L 4-letter words.4Lw 4-letter words without repeating any letters.5L 5-letter words.5Lw 5-letter words without repeating any letters.6L 6-letter words.6Lw 6-letter words without repeating any letters.

7.4. Table 2.

Next, it is shown Table 2, very similar to Table 1, except that in thefourth, sixth, eighth, tenth and twelfth columns, the numbers havebeen expressed with a base (B) raised to the appropriate power.

Tabla 2.NL 1L 1Lw 2L 2Lw 3L 3Lw 4L 4Lw 5L 5Lw 6L 6Lw1 1 1 1^2 0 1^3 0 1^4 0 1^5 0 1^6 02 2 2 2^2 2 2^3 0 2^4 0 2^5 0 2^6 03 3 3 3^2 6 3^3 6 3^4 0 3^5 0 3^6 04 4 4 4^2 12 4^3 24 4^4 24 4^5 0 4^6 05 5 5 5^2 20 5^3 60 5^4 120 5^5 120 5^6 06 6 6 6^2 30 6^3 120 6^4 360 6^5 720 6^6 720



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Roman numerals indicate in Table 2 which formulas listed below,are applied in the respective columns of Table 2:

I) B.II) B

2

-B.

III) B3-3B

2+2B.

IV) B4-6B

3+11B

2-6B.

V) B5-10B4+35B3-50B2+24B.VI) The author did not want to develop the formula VI.

Meanings of the acronyms or abbreviations that appear in the firstrow of Table 2:

NL Number of letters in each imaginary abecedary.1L 1-letter words.1Lw 1-letter words without repeating any letters.2L 2-letter words.2Lw 2-letter words without repeating any letters.3L 3-letter words.3Lw 3-letter words without repeating any letters.4L 4-letter words.4Lw 4-letter words without repeating any letters.

5L 5-letter words.5Lw 5-letter words without repeating any letters.6L 6-letter words.6Lw 6-letter words without repeating any letters.

8. Theories.

In accordance with what was exposed in items 2.1. and 2.2., the

following theoretical statements are formulated:

8.1. Given an abecedary consisting of B number of letters, themaximum number of words of p letters that can be formed isobtained by applying the formula: Bp.

(The symbol B represents the number of letters in a givenabecedary, while the symbol p represents the number of letters tobe included in each word that will be formed.)


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The author asks that this theory, although obvious to almost anyteenager versed in algebra and grammar, be known as: Theory ofabecedaries and maximum number of words.

However, this name is a proposal, an outline, barely. In the futuresomeone may suggest, propose or even impose a better name.

In section 7.1., in Table 1 (there is a copy in section 7.3.), thenumber of words that can be formed by applying the above formulais, respectively, in the second, fourth, sixth, eighth, tenth, and twelfthcolumns, according to the number of letters in each abecedary, andthe maximum number of letters each word will be able to include.

8.2. Given an abecedary consisting of B number of letters, themaximum number of words of p letters that can be formedwithout repeating any letter within each word, is obtained byapplying the following formulas:

1) B, where p = 1.2) B(B-1) B2-B, where p = 2.3) B(B-1)(B-2) B3-3B2+2B, where p = 3.4) B(B-1)(B-2)(B-3) B4-6B3+11B2-6B, where p = 4.5) B(B-1)(B-2)(B-3) (B-4) B5-10B4+35B3-50B2+24B, where p =

5,et cetera.

(The symbol B represents the number of letters in a givenabecedary, while the symbol p represents the number of letters tobe included in each word that will be formed.)

The author asks that this theory be known as: Theory of

abecedaries and maximum number of words, with the restriction ofnot repeating any letters in words.

However, this name is a proposal, an outline, barely. In the futuresomeone may suggest, propose or even impose a better name.

In section 7.1., in Table 1 (there is a copy in item 7.3.), the numberof words that can be formed by applying the above formula is,respectively, in the third, fifth, seventh, ninth, eleventh and thirteenth

columns, according to the number of letters in each abecedary, and


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the maximum number of letters each word will be able to include,with their respective restraints.

9. Other possible methods.

The author of the above theories absolutely ignores if instead orbesides of using algebraic formulas of the types outlined above, idest:

1) B.2) B

2-B.

3) B3-3B2+2B.4) B4-6B3+11B2-6B.5) B5-10B4+35B3-50B2+24Bet cetera, there could be applied formulas, procedures or easierways to find answers, either by vectors or Cartesian graphs(invented by the father of French skepticism, le pre du scepticismefranais, philosopher and mathematician Ren Descartes, 1596-1650) or natural or napierian logarithms* (whose base is theirrational number e,** id est 2.718281828459 ...) or decimallogarithms (whose base is number 10), or calculus, invented by such

portentous scientists, German mathematician, logician, andphilosopher Gottfried Wilhelm Leibniz (1646-1716) and Englishmathematician and physicist Sir Isaac Newton (1643-1727).

* It should be noted that the Italian mathematician and Franciscan friar FraLuca Pacioli (1445-1517), a systematic analyzer of the accounting methodcalled double-entry accounting system and precursor of probability calculus,utilized logarithmic approximations, a century before John Napier did the same(Scottish mathematician, 1550-1617, a Presbyterian and furious anti-Catholic),excellent definer of natural or napierian logarithms.

** This is represented by the symbol e, in honor of Leonhard Euler (1707-1783), a bright Swiss mathematician, whose last name begins with an e.

10. Four useless virtual devices.

10.1. A sphere of characters.A computer file/program, not created yet, simulating a giant hollowsphere (virtual, of course), with many characters (letters, Greek

letters, accented vowels, numbers, symbols, punctuation, signs,spaces, et cetera) printed on its inside; of each character, there are


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at least two copies; of the most used characters, there are morecopies, about 50, for example, for the letter e, however, the x couldhave only four copies, and within the sphere a virtual mosquito flyingat the speed of light can stop (and in fact stops) before (or at)

numerous letters, numbers, symbols, one by one (for amillisecond, a microsecond or a nanosecond each) and form words,text, books, treatises, encyclopedias, et cetera.

It would have to be implemented an algorithm for the creation ofwords, one after another, and a repository which would keep thewords, paragraphs, and pages created, and also, perhaps, analgorithm that prevents the creation of groups of characters such asxah^vywh*tl~aezdd8ff, that would not make sense in anylanguage. Or maybe it would make sense in the so called machinelanguage, a concept which is widely used by computer engineersand programmers.

The mosquito could write a word in Croatian, then another inGerman, one in Spanish, one fourth word in English language, andso on, thus, someone has to create one or more intra-algorithms toclassify the words by language.

10.2. A virtual circumference of characters.

Imagine a circle, crossed by two diametrical straight lines, onehorizontal, and a second vertical line, perpendicular to the first one.

This circle would have 3,600 (three thousand six hundred)degrees (it would be divided into 3,600 parts), and in each grade itwould be a character (letter, number, symbol or sign).

For example, in the position of 0 degrees, at the right end ofthe horizontal line, it could be the value to space, in the grade 1, itcould be the number zero; in grade 2, it could be the 1; in the grade3, the number 2; in the grade 10, the number 9; in the grade 11, the

lower case letter a; in grade 12, the lowercase letter b, and so on.The position of 0 degrees (space) would be at the right end

of the horizontal diametrical straight line; the grade 1 would be onedivision above, at the degree 1 position (it would have the "printable"value of 0 [zero]); the grade 2 (with the "printable" value of number 1[one], would be two divisions above the position of 0 degrees,counterclockwise, et cetera.

Each character would be printed orengraved on thecircumference or immediately outside it, at least twice, the

characters that represent the sounds most used in the language(such as vowels e, a, i, o, u, and certain consonants) would have


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more representation (would be repeated more times in thecircumference, and placed not together, but scattered).

A central hand with a virtual rotary axis in the center of thecircle defined by the circumference, could move in any of two

directions (clockwise, and counterclockwise), in search of thenearest convenient character.The hand would stop for a millisecond, a microsecond or a

nanosecond before each characterchosen to be marked(selected), and a virtual printer would be printing (writing) on avirtual page or virtual repository, each character to form words,phrases, sentences, paragraphs, pages articles, treatises, books,and so on.

Similar to section 10.1., it would be needed a programmer orcomputer engineer, and a linguist or a philologist, to create asuitable program, algorithms, constraints, and so on.

10.3. The lock of characters.A combination lock, similar to those used to lock the closing of somesuitcases (called suitcase combination locks), which usually havethree swivel wheels with ten divisions each, numbered 0-9; eachwheel has a notch in a certain position. When you match threenotches, the suitcase can be opened. The owner of the suitcase canreprogram the opener combination, to be the one he wants, for

example, 842.In our case, our virtual wheels would not have ten notches or

divisions, but some 1,024 (one thousand twenty-four divisions); eachdivision would have a printed character, and three wheels would notbe enough, so our virtual combination lock would have somethinglike 35 gyrating wheels.

Each character (letter, number, symbol, sign, space, formula,et cetera) would be engraved or printed at least twice on eachwheel, and the most used characters would be represented (printed

or recorded) more times.No wheel would have printed or engraved characters in the

same order. There would be 35 ways to accommodate thecharacters, one way at each wheel.

Every millisecond, microsecond or nanosecond, the wheelswould stop (would rotate at different speeds and in oppositedirections alternately, or all swirling at the same speed, but inopposite directions alternately, and according to a specialalgorithm), and a reader orposition sensor would record or

would print (write) in a virtual page or in a virtual space the wordsor groups of letters that would be forming.


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As in previous cases, it is necessary to create one or morealgorithms to run this virtual machine.

10.4. Recombinant DNA language.

The structure of deoxyribonucleic acid (DNA) was determined anddefined in 1953 by English molecular biologist, biophysicist, andneuroscientist Francis Crick (1916-2004), American molecularbiologist James D. Watson (1928 -), and British physicist andmolecular biologist born in New Zealand, Maurice Wilkins (1916-2004).

Basic structure of DNA, deoxyribonucleic acid: a double helix withplenty of four nitrogenous bases molecules, linked by pairs:

adenine=thymine, A=T; and guanine cytosine, GC.

[In RNA, ribonucleic acid, there is no thymine; uracil occupies theplace of thymine: adenine=uracil, A=U; and guanine cytosine,GC]

Some argue that the three winners of the Nobel Prize in Physiologyor Medicine in 1962, Crick, Watson and Wilkins, had disparagedBritish biophysicist and X-ray crystallographer Rosalind Franklin

(1920-1958), and also say that scientifically she deserved morerecognition, since she was the one who took the first pictures ofX-ray diffraction of the double helix structure of DNA, who firstinterpreted them, and who showed that the support structure,phosphate, should be on the outside, and nitrogenous baseson the inside.

Although, as the Nobel Prizes are not given posthumously, herrelatives could not have received the Nobel Prize on her behalf in

1962.

Well, by now enough of the history of chemistry, biochemistry,biology, medicine, physiology, and so on.

In the fifth column of Table 1, shown in section 7.1. of this writing, itappears the formula II (two Roman numeral), which is: B2-B, and inthe row corresponding to an abecedaryof four letters to form wordsof two letters with the restriction set in item 6.2.2., in the sense that

no letter can be repeated in each word formed, the result of applyingthe formula was: B2-B = 16-4 = 12, meaning that out of 16 possible


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combinations of two letters, four should be discarded, in ordertwelve remain.

The twelve combinations which meet the restriction are:

ab, ac, ad, ba, bc, bd, ca, cb, cd, da, db, dc.

The four combinations discarded (because a letter is repeated) are:

aa, bb, cc, dd.

But here in this section where we are right now, 10.4., we have notan abecedaryof four letters, but an acegetary, consisting of fourletters: a, c, g, t, which are the initials of four nitrogenous bases:adenine, cytosine, guanine, and thymine.

Also here the restriction is broader than in 6.2.2.: besides notrepeating any letter (id est, a nitrogenous base may not bind toanother equal to itself, so the combinations aa, cc, gg , tt areimpossible ones), each nitrogenous base may bind to only one ofthe three other bases, to form pairs that are always producedbetween adenine and thymine: at (or thymine and adenine: ta) onone side; and cytosine and guanine : cg (or guanine and cytosine:gc) on the other.

(The author will repeat here paragraphs 2.1., and 2.2. of this writing:

2.1. In this algebraic-grammarian exercise, the number of letters in eachimaginary abecedary that we will create will constitute a mathematical base(which is represented by the symbol B).

2.2. The number of letters contained in the words that we will be able to build,will constitute the exponent or power, represented by the symbol p, written as asuperscript character, an indicator of the power that the bases cited in 2.1. will

have to be raised to; that is, B

p

.

We are faced here, in this case if item 10.4., that the value of base Bis 4 (the number of letters in ourabecedary, which in this case is anacegetary, which also fulfills the condition of 2.1.) And the power orexponent p is 2 as it will form two-letter words each.

Then, in view of a broader restriction in this case, we will apply adif ferent pre-formula: B(B-3), resulting in a formula that alsodiffers from all previous formulas: B2-3B.


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The pre-formula complies with the condition 2.1. (In this case, Bhas a value of 4, which is the number of letters in ouracegetary).

Furthermore, the pre-formula meets the broader restriction that

each of the four letters (a, c, g, t) can bind to only one of theremaining three, which may be expressed algebraically as follows:B-3 (operation that results in 1, in this case, and that meets thebroader restriction that each nitrogenous base may bind to only oneof the other three, and besides that the nitrogenous base it can bindto, may not be a base equal to itself). And if the pre-formula meetsthese conditions, the formula also meets them.

Finally, the formula B2-3B, meets the condition specified in 2.2.: inthis case the highest exponential (power) value of p in the formula is2: B2-3B.

By applying the formula and substituting, we have: B2-3B = 16-12 =

4, which is the number of combinations that we have, and these are:at, ta, cg, gc.

The twelve combinations discarded are: aa, cc, gg, tt, ac, ag, ca, ct,ga, gt, tc, tg.Now if here ouracegetaryhas only four letters, oursuper-abecedarycould have 256 or 512 or 1,024 characters.

That is, here the architecture or structure of combinations torepresent letters would be different from the points 3, 4, 5, and 6 ofthis paper. Please note that at this item 10.4. we are dealing withone of fouruselessvirtual devices, and there is more freedom toact.

To form chains of pairs of nitrogenous bases we can form randomcombinations structures almost ad infinitum (remember: the onlyvalid ones are the following four: at, ta, cg, gc), and arbitrarily havebeen sectioned lengthy chains consisting of billions of pairs ofnitrogenous bases, in order to form chains of eight pairs each.

Each of these chains formed by eight pairs or 16 nitrogenous basesrepresents one character belonging to oursuper-abecedaryof, letssay, 1,024 characters.

For example, the chain:


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attaatta

cggccggccould represent the letter a.

The chain:

attaatcgtagccggccould represent the letter b.

The chain:

attacgattagccggccould represent the letter c, and so on.

After a computer program (including certain algorithms, conditionsand restrictions) section off chains to form mini-chains of eight pairs,the latter can be arranged so that letters can form words lead to

real orreasonably possible to exist words (after the programhave been trashed all of the mini-chains of eight pairs that hinder or


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are nonexistent in our range of 1,024 mini-chains of eight pairs ofnitrogenous bases each) in any languages that use the Latinabecedary, a copy of the original very, very long (billionaire) chaincan be considered.

After having formed a considerable number of actual words, let'ssee a copy of our original chain.

In this copy, the left column moves up, lets say, n number ofplaces, and the right column either remains stationary or movesdown n oro number of places (according to algorithms) to formtrillions of new pairs; the impossible pairs (the twelve pairs discardedin the above paragraphs) are trashed, and the possible, survivingpairs (at, ta, cg, gc) are grouped into sections of eight pairs each, toform new groups representing different letters and characters of oursuper-abecedary, which will be forming new words.

It is likely that after the realignment of sequences, the program(software) thrash away more than 75 percent of the new base pairs(for not complying with the broaderrestriction), so it might bepossible to consider the introduction of an additional algorithm or anunder-algorithm orintra-algorithm inside the main algorithmwhich form groups of eight pairs each, from valid pairs (at, ta, cg, gc)

that are not necessarily contiguous, but several or many rows apart,but keeping the order they bring from the original sequence.

There may be millions (or billions?) of billions of possibilities, and thelargest computers of today may not have the capacity to handlemany rearrangements, as it happens with computers that areintended to simulate the multiple possible recombinations of the realDNA.

The TX5 Cray supercomputer price is about a million dollars, plus afurther 50 thousand dollar to maintain it in a controlled, dust-free,low-temperature room, an appropriate place where to locate it.Experts claim that this is an inflated price, that it should cost half, butas Cray has little competition...Too much box or continent and lesscontent (really useful hardware).

10.5. In the Wikipedia, you can read about the infinite monkey

theorem:


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http://en.wikipedia.org/wiki/Infinite_monkey_theorem

Also, you may want to check something very interesting: the Turing

machine:

http://aturingmachine.com

http://en.wikipedia.org/wiki/Turing_machine

The virtual Turing machine was created by outstanding Britishmathematician Alan Turing (1912-1954).

The author, Alejandro Hector Ochoa-Gonzalez, a guadalajarense ortapatio by birth,inheritance, residency, conviction and conveniencewith a Sonoran stamp dedicates thiswriting, with much gratitude and appreciation to his teachers of preschool and elementaryeducation at the Colegio Niebla, in Ciudad Obregon, State of Sonora 747* South SinaloaStreet: Cornelia Niebla, Dolores Niebla de Garca, Isabel Caloca de Velzquez, Teresa GarcaNiebla, Yolanda Duarte Castro, Gloria Ypiz Coronado, Magda Acosta, Evangelina Camacho(English teacher), and the principal, professor Rosario Niebla Lugo, and to my InductiveSpanish teacher (the author of the textbook, Espaol Inductivo was: Maurilio Barriga Gaona) athigh school in the Instituto La Salle, in Ciudad Obregon, Sonora (a school where basketball andbaseball are widely practiced), professor Gregorio Patrn, as well to my math teachers at highschool, same institute: engineer Btiz and Lasallian brother Ignacio Navarro Castaeda (whoused to smoke cigarettes of a brand named Fiesta [in English: Party]), and at college, sameinstitute: Alfonso Rodrguez Garca de Alba (Lasallian brother and principal of the school, whotold us, his disciples, that we had to bought the book College Algebra [translated into Spanish,of course], by Rees & Sparks, and also taught chemistry and sometimes English), engineerBrquez, and Victoriano Garca Angis (patron of my graduating class, the 10th GraduationClass, 1974-1977), who could write with his back toward the blackboard with either of bothhands while verbally exposing the class, and my literature teacher at College, Jorge HerreraChavarra. Some have died; for them, In memoriam.

* Unforgettable as for any former child your elementary school; in this case, also memorablebecause the house number matches the number of the Jumbo jet, the Boeing 747.
http://en.wikipedia.org/wiki/Infinite_monkey_theoremhttp://en.wikipedia.org/wiki/Infinite_monkey_theoremhttp://aturingmachine.com/http://aturingmachine.com/http://en.wikipedia.org/wiki/Turing_machinehttp://en.wikipedia.org/wiki/Turing_machinehttp://en.wikipedia.org/wiki/Turing_machinehttp://aturingmachine.com/http://en.wikipedia.org/wiki/Infinite_monkey_theorem

Algebraic Applications in Creating Imaginary Abecedaries

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