# New Pythagoras

NEW PYTHAGORAS

By

Edward E. Rochon

Shakespir EDITION

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Edward E. Rochon on Shakespir

New Pythagoras

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Some Other Works by the Author

[Cubics: A Numbers Essay
Number Bases & Digits: An Essay
Contra Pantheism (Atheism): An Essay
Contra Nominalism: An Essay
God & Square Roots
God & Square Roots II
Jobmasters: An Essay
Pest Control: An Essay
Polygon Calculus]
Seven Month Pregnancy: An Essay

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Title Page

Preface

Chapter 1: Proportionality Proof

### Figures and Theorems

Fig1: Unit Side, Supposed Commensurable Overflow (x)

Preface

In previous work, I have offered several disproofs of irrational numbers with respect to the square root of 2. Here is another very forceful refutation of irrational numbers using a proportionality proof. This requires no higher mathematical background other than basic understanding of simple arithmetic and geometry.

We show that a diagonal of a unit square of sides equal to 1 can be given a fixed ratio of integer parts between the diagonal and the sides. This fixed ratio is in the form of line segments called commensurable segments. This ratio is a whole number ratio, even though we do not know exactly what those whole numbers are, but only the algebraic proof of their existence. Back to Table of Contents

Chapter 1: Proportionality Proof

We should be familiar with the Unit Square, I suppose, if you are reading this essay. I will not use up space drawing the diagonal line © and sides (s) of that famous isosceles right triangle. Write down the following constructions on a scrap of paper if you like while we go through the following compass and ruler constructions.

[Unit Side equals s = 1,
c = √2 (equals hypotenuse, diagonal of unit square),
i = (commensurable increment supposed),
I = (integer coefficient applied to (i) to equal line(s) lengths on L),
L = √2 + 1 + x (s and x constructed on to c),
x = (any supposed overflow of commensurable (i) upon line L)].

A commensurable theorem supposes an increment (i) will fit on to line L, such that a coefficient (I) times the increment (i) will equal line L: that is to say, Ii = L. Additionally, whole number multiples of (i) fit evenly into both the unit side and the diagonal.

Certain deceptive arguments have been used to claim this increment (i) does not exist for certain ratios along the line. Of course, any line can be divided by any integer divisor. All straight lines can be concatenated into one line. For our purpose, we must keep the line segments distinct, and then show this makes no difference. We can find an increment that fits into all the line segments without crossing an increment over (spanning) boundaries between segments. The stumbling block is that we are not exactly sure what the mantissa value of √2 is precisely (zeroed out on the last right digit.) My proof of the commensurable does not give that value, but we can still proceed using the concession method of argument. That is, let us concede that increment (i) does not fit into line L, while meeting the requirement that all segments are whole integer unit multiples of that increment. Now it is certain that for any coefficient (I) there is a value of increment (i) that is commensurable for side s = 1. Supposing this (i) does not fit into the √2 portion of L, we supply an overflow extension to line L to account for this as illustrated below:

Fig 1: √2, Unit Side, Supposed Commensurable Overflow (x)

We can also state that (c/s) = (√2/1) = (c’/s’). The last term (c’/s’) affirms that there are any number of ratios that can fit the proportion equal to (√2/1) , (√8/2) for example. We can also state that any numerical value can be equally divided into equal measures or parts by any integer value (I).

This is how the proportionality proof works; we will break (i) into (I-1) parts, then distribute these parts to each increment (i) in front of the overflow increment that makes (x) exist, assuming that (x) does exist. For if (i) is actually the commensurable term, there would be no overflow to produce (x). In effect, we fatten, thicken increment (i) to equal: )), and (I) is also the number of increments already superimposed upon L, including that portion of the supposed overflow, that part of the last increment at the endpoint of that segment equaling (√2/1).

By thickening each commensurable in this manner we grow or extend L until it reaches the overflow value endpoint at the end of line segment (x) without the last increment that we suppose causes the overflow (x). This must be so as this last increment is evenly divided into all the previous increments, fattening them to reach the endpoint of (x). Both parts of the line segments of (s) and © were proportionately increased, such that the ratio is still equal to (√2/1) = (c’/s’). NOTE: We also guarantee that the increment is evenly divided into (s) = 1, leaving any incremental overflow at the endpoint of ©. Rationally, we can certainly do this, then adjust for any discrepancy between the two line segments.

Remember, we deliberately insured that all the increments in the unit side (s) were fattened to extend the length of (s) by that proportion. Then we fattened all the increments within the diagonal © to keep all in proportion, all the way up to the endpoint of the supposed (x). We must therefore see that the following is true.

c/s = (c + x )/(s + (s(i + (i/(I-1))))) = c’/s’

NOTE: Just to make sure of the visualization, think of the overflow of the last increment at the endpoint of the square root segment being filled up by our fattened increments.

The preceding process clearly shows that there is a commensurable for the ratio of (c/s) = (c’/s’). Now you may suspect something is wrong because the absolute values of (√2/1), are different from the numerator and denominator for (c’/s’), a bit larger than (c/s). Just to reinforce the argument, the fundamental proof of all arithmetic is to reverse the process and come up with the original terms. We can certainly take out the extra padding that we added to increment (i), such that it shrinks back to the original absolute value of (c/s). We do this proportionately for the square root segment and the unit segment.

Do you say, “Wait a minute! We had an overflow that was incommensurable, but upon reversing the process, the incommensurable state is now commensurable. This is a paradox. It cannot be.” No, there is no paradox and this is why. Neither you nor I know what the actual numeric value of increment (i) is in proportion to the unit 1 and the square root of 2. You led me to assume that there was an overflow, because you claim no such commensurable exists. Your assumption is not proof of the assumption. I used the Zeno of Elea tactic of assuming your assumption and proving that assumption leads to contradiction. Rid yourself of your assumption and you will rid yourself of the paradox. Moreover, since the proportions of the ratio are the same, and one ratio is commensurable, it is certain that all such ratios are commensurable by that proportionality. QED.

ONE LAST VISUALIZATION:
Pick out a conveniently small increment that fits into side (s) = 1.
Fill up the diagonal © = √2 with the same increment (i) until we reach the endpoint, either that point or an overflow endpoint on (x).
Divide the last increment by (I-1), I the coefficient needed to sum the increments to the line (L).
We do not use the last increment, as this might overflow by some value (x).
The fattened increments will then reach that (x) endpoint of the line. We fatten (s) and © proportionately.
Both (s) and © are now longer by fattening, they are now (s’) and (c’).
Yet (c’/s’) is in proportion to (c/s).
We can shrink (c’/s’) proportionately down to (c/s) to prove the commensurable.
Yes, if one proportion is commensurable, all of them are: [√8/2, etc.
__]This is the Commensurable Theorem of Proportionality visualization.

Both the diagonal and sides of the unit square are definite values. They cannot string on forever. An infinite series would lead to an infinite extension, and reducing the increment to zero to get around that, produces a contradiction as well. A line cannot be composed of points, as points have no extensions, are merely relationships on the line. Aristotle was absolutely right on this point. Can we not put an end to this nonsense of trying to cram the infinite into the finite world of mathematics? And yet when it suited Aristotle’s purposes, he put infinity back into mathematical refutations of the paradoxes of Zeno, confounding his own assertions that an infinite series cannot exist and that an extension cannot be composed of mere points. Newton and Leibniz and Cantor and Euclid and Hippasus all did the same thing. It does not matter what their reputations amount to, they are all wrong. I am right. Pythagoras was right about all line segments being commensurable. Please, let us move on. Back to Table of Contents

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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