There Are No Irrational Numbers !

i’ve been sitting on this Rant for a few days now;

And i think that this is even more muddled that most of them.

But.

i really believe that there’s something here,

i’m just Not explaining it properly.

What got me really going on this idea that there are No Irrational Numbers, & also; The Concept of Infinity is being horrifically misused in Mathematics—

Started Quite A While ago;

But more recently; i was reading a proof that Cantor created to prove that there were a class of UnCountable Numbers, Notably, Irrational Numbers.

To do this; He first requires that a List of All Irrational Numbers be Created ( Hypothetically ) & even more annoyingly;

That is; More Annoying than Creating something that can’t even Hypothetically being Created without addressing The First Elemental Issue that Such a List would be impossible;

Is that he Completely Blows Off The First Step of this List by Suggesting that it Be Created using ‘Some Clever Method’.

If you were going to create a list of All Irrational Numbers;

How would you Begin ?

First Tier : All Irrational Numbers that begin with .000…

that continues for Infinity, but ends with a 1, then a 2, then a 3, & so on.

Second Tier : All Irrational Numbers that begin with .000…

that continues for Infinity, but ends with one additional 0, then 1, then a 2, then a 3 & so on…

Third Tier : All Irrational Numbers that begin with .000…

that continues for Infinity, But ends with 2 additional zeros,

…00, then a 1 ( …001 ), (…002 ), ( …003 ) & so on.

Certainly; This approach is Ridiculous.

What is worse; Is that by Introducing a Wondrously Elegant Jiggery Pokery Argument; He then goes on to Insist that This Hypothetical List of All Irrational Numbers ( Surprise Surprise ) Doesn’t Contain all Irrational Numbers.

You would think that any reasonably rational person would step back from that & say; ‘Hmmm, maybe your initial suggestion that it would be possible to create a list of all Irrational Numbers was seriously flawed, or maybe The Idea of Irrational Numbers itself is Flawed.

But no.

Instead; Mathematicians have universally accepted this proof & adopted The Idea that Some Infinities are Bigger than other Infinities.

i suspect that Mathematicians have so readily accepted this; Is because they are in complete denial or are just unaware that there is This ‘Alternative’ School of Logic that i Call Jiggery Pokery Logic.

The Kind of Logic that i call Fractional Propositional Logic can only be used to Prove True Things.

While Jiggery Pokery Logic can prove anything.

The Wondrous thing about Jiggery Pokery Logic is that it is Indistinguishable from Fractional Propositional Logic.

Just by Acknowledging that there is a Kind of Logic called Jiggery Pokery Logic & allowing that many Jiggery Pokery Logical Arguments are being passed off as Fractional Propositional Logical Arguments—

Proves that Logic is Bunk.

The Entire Purpose of Fractional Propositional Logic is to Prove things with Absolute Certainty; But this Simple & Direct Proof for Jiggery Pokery Logic Takes that away from Fractional Propositional Logic. You can Never be sure if your Argument is Really Fractional Propositional Logic or Jiggery Pokery Logic. You may believe that it’s Really Fractional Propositional Logic, But Mathematicians & Philosophers have been fooled many times in The Past.

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Fractional Propositional Logic ( btw )

Is just like Classical Propositional Logic, Except that it’s Not Restricted by Arbitrarily Asserting that some Observed Conditional is 100% True or 100% False. There’s a lot of times when you’re Not sure or are unable to determine The precise Truth Value of your Antecedent or Succedent.

The Operators for Fractional Proposition work like this : 

#    name       f.p.logic      binary logic

......

1    Null       0              0

2    Not Or     1-(Max(p,q))   p q OR ) NOT

3    JustBecuz  Min(p,(1-q)    p NOT ) q OR ) NOT

4    Not q      1-q            q NOT

5    ISaidSo    Min((1-p),q)   p ( q NOT ) OR ) NOT 

6    Not p      1-p            p NOT

7    ~Eql/XOr  ABS(p-q)        p q XOR

8    Not And    1-(Min(p,q))   p q AND )

9    And        Min(p,q)       p q AND

10   Eql/~XOr   1-(ABS(p-q))   p q XOR )

11   p          p              p

12   Because    Max(p,(1-q))   p ( q NOT ) OR

13   q          q              q

14   If Then    Max((1-p),q)   p NOT ) q OR

15   Or         Max(p,q)       p q OR

16   Tautology  1              1

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Returning to Irrational Numbers for a Moment :

i would suggest that it would be possible to Create a List of all Fractional Numbers using this series :

.0

.1

.2

.3

…

.9

.01

.11

.21

…

.99

.001

.101

.201

…

& So on.

This would Include All Fractional Values between Zero & One

Excluding One itself.

But they would all be Finite in Length.

It would Continue to Infinity, & Provide a Seamless Continuum of All Additional Points between any Two Specified Points. 

It would Exclude All Irrational Numbers.

They’re Not needed.

The Problem then is that The ‘Idea’ of Irrational Numbers is Sooooo Easy to ‘Imagine’. You’re initial response is; Of course they Exist, How could they Not Exist ?

Like π for instance; It has been proven that π is Irrational !

Really.

If you Take A Circle on a Sheet of Paper & Try to Measure The Diameter ( which is Defined as a Given Length ) & then try to Measure The Circumference, You’re going to find that it’s very Difficult to Get a Good Measurement.

So then; You think; Let’s just imagine that A Circle is a Polygon with Infinite Sides! We can Easily Calculate The Circumference of a Polygon to any Desired Degree of Accuracy—

If we hold that A Circle is a Polygon with Infinite Sides, Which it is Not ( ! )

On The Simplest Level; You have to define The Radius of The Polygon from The Center to The Vertice or Middle of A Side; Both of Which are Wrong.

 

Neither can you Calculate Both Polygons, One for The Inside Limit & one for The Outside Limit & Average them out; Because both are going to continue to Infinity, By Passing The Correct Value.

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This Brings up The Problem of Series’ that purport to Create an Irrational Number that Corresponds to A Labeled or Otherwise Defined Value. The Series is =Designed= by Definition to continue on forever & ever. It is Not Created to Suddenly Stop when it finds The Correct Measurement.

Even when you have a Sigma Series that Converges on a Fixed Value; The Series will never Reach it. It will get closer & closer & closer & closer, without end.

There are also very hinky Series’ that seem to converge on a fixed value, but if you wait long enough, allowing The calculation to repeat it’s itineration cycle enough times, it will slip past it’s previously assumed fixed value.

One of The Principle Warning Signs of Jiggery Pokery, is Excessively &/or Convoluted Steps to Prove (x.

The Wikipedia Page that Professes to Reveal The Proof for The Irrationality of π is page after page of Dense Calculus Gibberish.

It would have been very nice if The Author were to Express The Proof in plain English, A Step by Step Recounting for An Attentive 5th Grader to Follow. But i very much Suspect that if you were to ask someone that ostensibly understood this Explanation that was provided; They Would claim that such a Plain English Translation would be Impossible.

But what seems most obvious to me; What Justification is given that Relates these proofs to π ? You’ve got your π, you’ve got The Proofs, why do you think one corresponds to The other ?

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Another thing that occurred to me; If you going to allow that it is possible to ‘Define’ Irrational Numbers—

Would it be possible to Define an Irrational Number like this;

.000… …001

That is; We have an Number that goes on to Infinity, But it’s Last Digit is 1. Of Course; A Number like this that goes on forever, doesn’t have a last digit. But if A Definition like this is Impossible, Then would this one also be Impossible;

.000… …010… …000…

You’ve got a number that starts out as Zeros; Goes on for Infinity, Then somewhere in The Middle, has a 1, then goes on for infinity with more zeros.

Can you do that ?

Would that Number have a Unique & Different Value than:

.000… …001… …000…

?

or even

.000… …020… …000…

Would that be an entirely Different Number ?

Would it be Greater or Lesser than The Number with a 1 in The Middle?

The Problem would be that it’s Not ‘Really’ in The Middle.

You can’t have a Middle to Infinity.

Which brings up The Problem of A Universe that has Existed for Infinity. Sure; We’re in A Universe that ‘Started’ only a few Billion Years ago, But if We’re in a Larger ‘SuperVerse’, Then this Infinite Universe Hypothesis holds.

In which case; The Universe should have achieved Conscious Perfection after an Infinity of Existence, But We’re here now after an Infinity of Existence. Where is The Perfection ?

Is this The Best, Most Perfect of all Realities?

Is it more Perfect than yesterday ?

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The Fallacy of The Elegant Solution

i suspect that The Reason that Mathematicians believe that π is Irrational is that it doesn’t seem to have a simple elegant solution. Wouldn’t it be nice if gawd had provided this π ration with a nice simple fractional value ?

But it’s Not.

What if Numerator was a few billion digits long ?

There is No Way that you could ‘Measure’ that by any means.

You couldn’t even measure it if it were only a few dozen digits long.

So without an Elegant Solution Forthcoming; The Mathematicians created a New Kind of Elegant Solution, & Created all The Necessary Crazy Proofs to Prove it.

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Proof

Lettuce assume that a/b = √2

For this a/b to be The Simplest Terms

Both a or b may Not be Even or Divisible by a Common Factor

√2 = 1.4142…

√2 = a/b               : a = 7  b = 5 : a/b = 1.4

            a & b are Wildly Approximated

            So that we can see how The Algebraics are Working ( ? )

2 = a² / b²             : 2 = 7² / 5² = 1.96 ( Pretty Close ( ? ) )

a² = 2 · b²             : 7² = 2 · 5² : 49 = 50

Given then that b² is 5² = 25

And 25 x 2 is 50; An Even Number;

Forced to be Even by Multiplication by 2

Then a² is an Even Number

Our Approximations make a² = 49

But if a² were 50 : a = 7.0710

But we’re also asserting that a is axiomatically a Whole Number

So that if This were to work out so that a² were to be a Whole Number

& (a would also be a Whole Number,

Then (a would be Necessarily Even,

Since any Odd Number Squared is Odd.

e.g.; 3² = 9

So there’s something of a Problem with ±7.0710 being Even.

- - -

But Never Mind that —

Let us then Arbitrarily Replace (a with 2·k ( or 2k )

k would then be 3.5

But for this to ‘Work Out’; k would have to be a Whole Number

But by working with Symbolic Algebraics;

These Fractions are Swept under The Carpeting !

- -

Returning to :

2 = a² / b²             : 2 = 7² / 5² = 1.96

2 = (2k)² / b²        : 2 = (2 · 3.5)² / 5² = 1.96

(2k)² = 4 · k²

2 = 4 · k² / b²       : 2 = 4 · 3.5² / 5² = 1.96

2 · b² = 4 · k²        : 2 · 5² = 4 · 3.5² : 50 = 49

b² = 2 · k²             : 5² = 2 · 3.5² : 25 = 24.5

 

Which Means that b²  must be Even by The Same Logic Expressed above ( 2 · x ) must be an Even Number.

Now both a & b are Even according to this Juggling Act,

But this all Assumes that a & b Start out as Whole Numbers;

And all of The Permutations that they Endure allow their SubDivisions to Remain Whole Numbers too.

Which They don’t.

Even if you were to somehow allow that these Conditionals were Met;

The Jiggery Pokery here is Only Asserting that (a or (b are Not Odd or Even. It is somehow insisting that (a or (b are Outside The Realm of Whole Numbers.

It seems far more Reasonable to assume that this ‘Argument’ is A Paradox of The Zeno Type; And that while it seems Reasonable; It tacitly asserts things that it shouldn’t.

e.g.: That If a² is a Whole Even Number; (a must be a Whole Number as Well. It was assumed that (a was a Whole Number at The Beginning of The Argument; But then Craziness set in.

- - -

On a More Obvious Level; Doesn’t this Argument Structure assert that all Square Roots are Irrational; Which is Clearly Wrong.

√36 = 6

- - -

√9 = 3

√9 = a/b               : a = 3  b = 1 : a/b = 3

9 = a⁹ / b⁹             : 9 = 3⁹ / 1⁹ = 19683

- - -

What if it was supposed to be :

9 = a² / b²             : 9 = 3² / 1² = 9 / 1 = 9

- - -

Oh! So 2 works for Everything.

So that The 2 didn’t come from The 2 in √2

It came from The Square of (x : √x

- - -

So if we try a = 6 & b = 2 : 6/2 = 3

It should still work ( ? )

√9 = a/b

9 = a² / b²             : 9 = 6² / 2² = 36 / 4 = 9

It Still Works !

:*’``’*:-.,_,.-:*’``’*:-.,_,.-:*’``’*:-.,_,.-:*’``’*:-.,_,.-:*’``’*

√16 = a/b = 4

a = 12 : b = 3

16 = a² / b²           : 16 = 12² / 3² = 144 / 9 = 16

Which Still Works.

- -

So shouldn’t this mean that √16 is Irrational ?

Lettuce Continue :

a² = 16 · b²           : 12² = 16 · 3² : 144 = 16 · 9 : 144 = 144 

Then a² is an Even Number

144 is an Even Number.

a = 12

Which is also an Even Number.

- - -

Let us then Arbitrarily Replace (a with 2·k ( or 2k )

k would then be 6 : 2 · 6 = 12 : 12² = 144

- -

Returning to :

16 = a² / b²           : 16 = 12² / 3² = 144 / 9 = 16

16 = (2k)² / b²      : 16 = (2 · 6)² / 3² = 16

16 · b² = (2k)²      : 16 · 3² = (2 · 6)² : 144 = 144

16 · b² = 4 · k²      : 16 · 3² = 4 · 36 : 144 = 144

Divide Both Sides by 16 to Free up b²

b² = .25 · k²          : 3² = .25 · 36 : 9 = 9

But here; b² is supposed to be proven to be Even;

But .25 · k² doesn’t prove that.

What Happened ?

In The Original Proof for √2

a & b were supposed to be Whole Numbers that Weren’t Both Even,

But in our ReProof with √16—

Instead of Using The Simplest Fraction for 4, Which would have been 4/1; We Used 12/3, Because if we Used 1, when you use 1² in an Expression; Instead of getting a ‘responsible’ Answer; You’ll get 1.

If we Rework this with a = 4 & b = 1

We Get :

√16 = a/b = 4

a = 4 : b = 1

16 = a² / b²           : 16 = 4² / 1² = 16 / 1 = 16

a² = 16 · b²           : 4² = 16 · 1² : 15 = 16 · 1 : 16 = 16

Let us then Arbitrarily Replace (a with 2·k ( or 2k )

k would then be 2 : 2 · 2 = 4 : 4² = 16

- -

Returning to :

16 = a² / b²           : 16 = 4² / 1² = 16 / 1 = 16

16 = (2k)² / b²      : 16 = (2 · 2)² / 1² = 16

16 · b² = (2k)²      : 16 · 1² = (2 · 2)² : 16 = 16

16 · b² = 4 · k²      : 16 · 1² = 4 · 4 : 16 = 16

Divide Both Sides by 16 to free up b²

b² = .25 · k²          : 1² = .25 · 4 : 1 = 1

So here again; b² is supposed to be proven to be Even; And It’s Not.

The Original Proof is A Jiggery Pokery Argument that is Founded on The Confusion of The √2 which means √2 = (x

Which means that (x · (x = (x² = 2.

The Confusion is; Where did The 2 in (x² come from ?

Also; By Using Symbolic a’s & b’s; We’re never able to see if a or b is actually Even or Odd. The Algebraics tell us to believe if a or b is Even or Odd, When they are Not ( ! )