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GOD & SQUARE ROOTS

By

Edward E. Rochon

Shakespir EDITION

- * * * *

PUBLISHED BY:

Edward E. Rochon on Shakespir

God & Square Roots

Copyright © 2016 by Edward E. Rochon

Thank you for downloading this eBook. This book may not be reproduced, copied and distributed for non-commercial purposes, unless prior permission is given by the author.

Your support and respect for the property of this author is appreciated.

This book is a work of non-fiction and refers to no specific living persons, or current places, or current locales. The historic personages mentioned are described from historical information that may or may not be accurate, but whose description seems likely or at least possible to the author.

Some Other Works by the Author

[*Cubics: A Numbers Essay*

*Number Bases & Digits: An Essay*

*Contra Pantheism (Atheism): An Essay*

*Contra Nominalism: An Essay*

*Jobmasters: An Essay*

*Pest Control: An Essay*]

*Seven Month Pregnancy: An Essay*

Reading Material

*****

**Table of Contents**

**Title Page**

**Preface**

**Chapter 1: Projected Mirroring**

**Chapter 2: Previous Disproofs**

**Figure 1: Unit Triangle Projected on to Square Root Triangle**

**Figure 2: Unit Square Illustration**

**Commensurable Theorem**

**About the Author**

**Preface**

*In a previous work, I have disproven the arguments used to support irrational numbers, particularly, square root 2. I have included a commensurable theorem and arguments refuting Hippasus and Euclid : even and odd numbers with respect to Hippasus, and unbalanced prime factors with respect to Euclid. The Pythagorean attempt to bring infinity and God into mathematics is at the root of this irrational nonsense. Their problems with finding theorems to support commensurables for all lines stemmed from this. They brought their own problems down on themselves. The back draft of this hellish nonsense continues to the present day.*

*I have thought up another way to attack irrational numbers using the proportionality of triangles, that clearly demonstrates the validity of commensurable values in all cases, disproving the irrational and transcendental number position. Since these irrational mathematicians, these shamans of numbers, brought God and the infinite into this matter, I classify this work as primarily theological, an apology and affirmation of a transcendent God in contradistinction to the materialist mumbo-jumbo opposing it. For whether the idealism of George Berkeley, the materialism of David Hume, or the mystical materialism/idealism of Pythagoras, these are but variations on the same pagan nonsense, the idolatry of the cosmos, numbers and finite things.* Back to Table of Contents

**Chapter 1: Projected Mirroring**

We can speak of a trivial solution to the ratio between the square root of 2 and the unit side. This would be:

√[*2 / 1 = (Ax + Bx) / 1*

__

*A = 1/x (coefficient)*

*B = m/x (coefficient)*

*x = 1/A (commensurable of the sides)*

*m = mantissa of √2*]

We can see in the expansion of the right hand side of the equation that:

*(1/A) + ((m/(1/A))(1/A) = 1 + m = √2*

Our value *(x)* is the commensurable that we require. We give it algebraic value. We use *(x*) to divide the unit side into *(A*) parts. To reverse the process, *(A)* times *(x)* gives the unit value *(Ax) = 1*. Supposing *(x)* is also a commensurable of the diagonal line, the goal to be achieved, the mantissa of root 2 is divided by *(x*) as well: *(m/x)* to obtain coefficient *(B)*. And *(x) = (1 + m) =√**2*. But the sceptic will say this commensurable value *(x)* does not exist. Is there some other way we can compare the unit side to the diagonal to see how they relate to each other? We can do this by noting that all equilateral triangles are similar. This means they are proportional in all aspects.

We stack or superimpose a unit equilateral triangle (sides of 1) into or over an equilateral triangle of side *(√**2)*. This looks like this below:

[**F**

**igure 1: Unit Triangle Projected on to Square Root Triangle**]

We see that the unit side is proportional when projected onto the *(√**2)* side within the equilateral triangle. We note that the line segment extensions of the unit triangle (m) are proportional and equal to each other, congruent except for their diverging angles. Equilateral triangles are symmetric. We can construct another side, parallel to the unit side and to the *(√**2)* side from proportional points anywhere between the two line segments of *(m)*, the mantissa portion of *(√**2)*.

Returning to our trivial solution for the commensurable of the unit side to its diagonal, noting that any line can be divided by whole number counts into commensurable parts of any number and size, we continue to compare them.

Projecting a mirrored side, upwards from the unit side toward the diagonal side *(√**2)*, we can definitively conclude the following premise is true. For any commensurable subdivision of the unit side, producing *(Ax) = 1, (x)* the commensurable and A the coefficient needed to give the sum of the line length *(1)*, we can add the commensurable *(x)* once on either side of the unit side to create a new projection between *(1)* and *(√**2)*. This will be in the form: *Ax + 2x = s’*. We can also say that any line can be commensurably subdivided any number of times to give: *x’, x’‘, x’‘’….etc*., either upward or downward to produce a projected side *s’, s’‘, s’‘’, etc*. within any equilateral triangle. In our case, we will limit these commensurable divisions to within the distance between the square root side and the unit side. For example, we will not use a commensurable of *(.5)*, as this would exceed the length of *(√**2)* in the equation: *Ax + 2x = s’*.

We must certainly conclude that if the unit side is commensurable by *(x)*, then side *(s’)* is also commensurable to the unit side, it have precisely 2 extra units of the commensurable *(x)*. So the unit side *(s)* and *(s’)* are commensurable. This is also true, must be true, for all other sides so constructed.

The proponents of irrational numbers claim that the mantissa of *(√**2)* is endless. If this is so, it must also be true for the line segments *(m)* on either side of the equilateral triangle projection. Segment *(m)* is congruent with the mantissa of *(√**2)* except for direction. Since it is clear that I can commensurably slide a projection of the unit side up the equilateral triangle as far as I may please, can you tell me why this process somehow mysteriously ceases when I reach the *(√**2)* side, that is and must be a projected side, proportional and similar in every way to all other projected sides? I can answer for you. You cannot.

Moreover, inversely, we can make commensurables on (√2), and use a formula: Ax – 2x = s’ to inversely project (√2) onto the unit side (1*).*

We see that if the mantissa of so called irrational numbers are endless and continuous, we must be able to project similar sides up and down the line segments *(m)* of an equilateral triangle. So this endless assumption does not support irrational numbers. The other option is to deny this endless mantissa. In that case, your irrational number assumption is refuted. The mantissa must be capable of being zeroed out, and a commensurable value must exist between the unit side and the diagonal of the unit square.

Can we not put an end to this heathen nonsense? Infinity has no place in mathematical operations other than to show by logical exclusion that infinity must exist of necessity and cannot be mathematically operated upon. Back to Table of Contents

**Chapter 2: Previous Disproofs**

[Abstract from: *Cubics: A Numbers Essay*]

The diagonal line in relation to orthogonal coordinates has been a problem going back to antiquity. The Pythagoreans knew that there must be a commensurable increment that could fit evenly into the side of a rectangle or rectangular volume but were unable to develop a theorem to find one. A disciple named Hippasus came up with an algebraic argument that reputedly proved that no such commensurable existed between the diagonals and sides of a unit square and related geometric objects. From this developed the notion of irrational numbers whose fractional part (mantissa) went on endlessly without any apparent pattern. Such a mantissa is inconsistent with the **Fundamental Proof of Arithmetic** that clearly shows such an infinite progression is, as the name implies, irrational, undefined and untrue. The geometer Euclid added a variation to the argument of Hippasus using prime number theory. The arguments are as follows:

The argument of Hippasus and Euclid for irrational numbers:

*1 ^{2}*

Substitute p/q for the square root of two √2:

*1 ^{2}*

Factoring and isolating p:

*p ^{2}*

Then p^{2} is even. And powers of even numbers are even.

Substituting p = 2r:

^{2}*= 2q ^{2}*

Factoring and simplifying:

*2r ^{2}*

Then both p and q are even contradicting that (p/q) is relatively prime (having only 1 as a common factor [no primes in common]) and this is claimed proves that the square root of two is incommensurable (no common whole number multiple for the side and hypotenuse). More to the point of Euclid’s objection is that the two squared terms ® and (q) must be even with respect to their prime values. All numbers have a unique set of primes that can be multiplied to equal the value of the number. The number of primes can be an odd or an even number of terms. But when you square any number, you simply double the number of primes that make up the squared number. Since both ® and (q) are squared above, they both must have an even number of primes (Any odd number when doubled becomes even). But the coefficient 2 on the left is also a prime number so that we have an odd number of primes on the left and an even number on the right.

Fundamental Proof of Arithmetic: All arithmetic operations are reversible.

Here are the problems with this use of algebra:

*Irrational numbers do not exist because they require an infinitely long mantissa.*

*It is possible for a number to be both even and odd. Integers can only be odd or even, but rational numbers are all evenly divisible by 2 making all rational numbers even (.5 + .5 = 1; .5 is a rational number.) All integers are rational numbers, so all odd numbers are both even and odd when mixing integers and rational numbers. Since the diagonal of the square is evidently rational (fractional part) with respect to the unit side, one is compelled to use rational numbers in resolving the matter irrespective of whether the ratio (p/q) is made of integers or not.*

*Euclid arbitrarily excludes the number 1 from prime terms in the product of a number even though 1 is clearly a prime number. Therefore the 2 on the left side of the equation includes 1 in its sequence making the number of primes on the left even. In general, every prime number is a product of itself and 1. For a list of primes equal to n it is necessary to add to the sequence the number of 1’s equal to n and this is the same as 1 ^{n}*

Now I will offer my theorem proving that the commensurable that Pythagoras was looking for does exist.

**Figure 2: Unit Square Illustration**

Figure 1 shows one side of the 2 triangles created by the diagonal bisector of the unit square, so called because its sides are 1 unit in length (here inches). This problem has been difficult to resolve because it requires a polymetric solution. Polymetric means the use of two different metric scales. For example, a problem that involved the use of both inches and feet would be a polymetric problem. You would need to convert back and forth between the metric measurements to solve the various aspects of the problem. The Pythagoreans needed a common metric for the following reason:

Divide the side of line AC into even increments of whatever size. These are usually denoted by vertical risers or lines hatched across the line. If you were to extend these hatch lines through the diagonal, the lines would divide the diagonal into the same number of increments but proportionately longer since the diagonal is a longer line. If you were to rotate AC onto AB you would see that the increments did not match up to the diagonal hatch mark crossing points before rotation. However if you continued rotating AC onto BC the hatches would divide BC in the same proportions. This would be true even if BC were longer than AC. The orthogonal lines have a common metric but not with the diagonal. This requires a commensurable increment that would be evenly divisible into the diagonal and the sides of the square. Here is the proof that this commensurable exists with explanatory comments:

**Commensurable Theorem**

**For the unit square, rectangle or lines:**

**n, n*_{1}, etc. (place value digit for each position of base number system)*

**p = hypotenuse, q = side**

**Axiom 1: Any integer is the sum of an addend string of ones (1’s) (zero excluded)**

**Axiom 2: For all numbers {n.n _{1}***n*

* For p = n.n_{1}*n*_{2}*n*_{3~…n[~n}]*,

**(p/q) = ((nn _{1}***n*

*[*Then, p and q are commensurable by a metric unit r.]*

****Metric unit r (or metric r) refers only to the measure of the commensurable. It has no other connotation for the word metric. An inch is a metric unit as is a foot. The foot/pound system is a metric system just as the meter/gram system is a metric system. Any system of measure is a metric system when the word metric is used in its original sense. Do not confuse it with a common use of the word metric in higher mathematics. The basic idea here is that you shift the decimal point of the ratio of hypotenuse to side all the way to the right until you have two integers that are commensurable by the unit digit.*

Even if detractors still quibble about all rational numbers being even or that 1 must be included in the significant prime numbers for a given term (even if only implicitly), it is manifestly impossible for a mantissa sequence to be formed where the decimal point cannot rationally be moved to the right to form an integer. Whether the terms of the ratio are in relatively prime form or not is irrelevant. Two integers are always commensurable. Reader note that the theorem does not supply the required value of the integer multiple needed to make (p/q) commensurable in metric r. Back to Table of Contents

**#####**

**Other Works by the Author**

[**(*]Available online[**)*]

Elements of Physics: Matter

Elements of Physics: Space

Elements of Physics: Time

Unified Field Theory: An Essay

Liar Enigma: An Essay

Golden Age Essays

Golden Age Essays II

Golden Age Essays III

Golden Age Essays IV

Golden Age Essays V

**About the Author**

My current biography and contact links are posted at Shakespir.com/profile/view/EdRochon. My writings include essays, poetry and dramatic work. Though I write poetry, my main interest is essays about the panoply of human experience and knowledge. This includes philosophy, science and the liberal arts. Comments, reviews and critiques of my work are welcome. Thank you for reading my book.

Back to Title Page

A brief preface explains the purpose of the essay. Chapter 1 starts by creating a trivial solution to the ratio of √2/1 in the form (Ax + Bx)/1 = √2 with (x) the commensurable, A the coefficient multiplier needed to sum (x) to 1, B the coefficient needed to obtain the product equal to the mantissa of √2 by (x). I then project an equilateral unit triangle (sides = 1) on to the √2 equilateral triangle (sides = √2). I note that line segments (m) equal to the mantissa extend from the unit side to the root side. We note that all individual lines are divisible into commensurable parts through any number of divisions by integer values. We note that any commensurable (x) can be added to the unit line, one on either side, such that (Ax + 2x = s' and (s') is commensurable to (s) (unit length) by definition. Providing we remain within acceptable value limits between √2 and 1 for the commensurable, we can find lines across the triangle anywhere we please that are proportional and similar to 1 and √2 sides. We ask, why should we not be able to slide up to √2, this being so? The answer is that there is no reason, and we prove the commensurable hypothesis. Inversely, starting from √2, a formula for any value of (x) within limits can find a line between √2 and 1 by: Ax - 2x = s'. Chapter 2 restates my Commensurable Theorem and refutes the argument of Hippasus re: relative primes and both sides of the derived equation even numbered, and unbalanced prime factors re: Euclid's argument. They are both erroneous.

- Author: Edward E. Rochon
- Published: 2016-09-15 22:50:08
- Words: 2631