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.

: - - -
- - - - - - - - - - - - : o

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

: - - -
- - - - - - - - - - - - : o

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.

: - - -
- - - - - - - - - - - - : o

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 ?

: - - -
- - - - - - - - - - - - : o

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 ?

: - - -
- - - - - - - - - - - - : o

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.

: - - -
- - - - - - - - - - - - : o

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 = a

^{2}/ b^{2}: 2 = 7^{2}/ 5^{2}= 1.96*( Pretty Close ( ? ) )*
a

^{2}= 2 · b^{2}: 7^{2}= 2 · 5^{2}: 49 = 50
Given
then that b

^{2 }is 5^{2}= 25
And 25
x 2 is 50; An Even Number;

Forced
to be Even by Multiplication by 2

Then a

^{2 }is an Even Number
Our
Approximations make a

^{2}= 49
But if
a

^{2}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 a

^{2}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.; 3

^{2}= 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 = a

^{2}/ b^{2}: 2 = 7^{2}/ 5^{2}= 1.96
2 =
(2k)

^{2}/ b^{2}: 2 = (2 · 3.5)^{2}/ 5^{2}= 1.96
(2k)

^{2}= 4 · k^{2}
2 = 4 ·
k

^{2}/ b^{2}: 2 = 4 · 3.5^{2}/ 5^{2}= 1.96
2 · b

^{2 }= 4 · k^{2}: 2 · 5^{2 }= 4 · 3.5^{2}: 50 = 49
b

^{2 }= 2 · k^{2}: 5^{2 }= 2 · 3.5^{2}: 25 = 24.5
Which
Means that b

^{2}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 a

^{2 }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 = a

^{9}/ b^{9}: 9 = 3^{9}/ 1^{9}= 19683
- - -

What if
it was supposed to be :

9 = a

^{2}/ b^{2}: 9 = 3^{2}/ 1^{2}= 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 = a

^{2}/ b^{2}: 9 = 6^{2}/ 2^{2}= 36 / 4 = 9
It
Still Works !

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

√16 =
a/b = 4

a = 12
: b = 3

16 = a

^{2}/ b^{2}: 16 = 12^{2}/ 3^{2}= 144 / 9 = 16
Which
Still Works.

- -

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

Lettuce
Continue :

a

^{2}= 16 · b^{2}: 12^{2}= 16 · 3^{2}: 144 = 16 · 9 : 144 = 144
Then a

^{2 }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 : 12

^{2}= 144
- -

Returning
to :

16 = a

^{2}/ b^{2}: 16 = 12^{2}/ 3^{2}= 144 / 9 = 16
16 =
(2k)

^{2}/ b^{2}: 16 = (2 · 6)^{2}/ 3^{2}= 16
16 · b

^{2 }= (2k)^{2}: 16 · 3^{2 }= (2 · 6)^{2 }: 144 = 144
16 · b

^{2 }= 4 · k^{2}: 16 · 3^{2 }= 4 · 36 : 144 = 144
Divide
Both Sides by 16 to Free up b

^{2}
b

^{2 }= .25 · k^{2}: 3^{2 }= .25 · 36 : 9 = 9
But
here; b

^{2 }is supposed to be proven to be Even;
But .25
· k

^{2 }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 1

^{2 }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 = a

^{2}/ b^{2}: 16 = 4^{2}/ 1^{2}= 16 / 1 = 16
a

^{2}= 16 · b^{2}: 4^{2}= 16 · 1^{2}: 15 = 16 · 1 : 16 = 16
Let us
then Arbitrarily Replace (a with 2·k ( or 2k )

k would
then be 2 : 2 · 2 = 4 : 4

^{2}= 16
- -

Returning
to :

16 = a

^{2}/ b^{2}: 16 = 4^{2}/ 1^{2}= 16 / 1 = 16
16 =
(2k)

^{2}/ b^{2}: 16 = (2 · 2)^{2}/ 1^{2}= 16
16 · b

^{2 }= (2k)^{2}: 16 · 1^{2 }= (2 · 2)^{2 }: 16 = 16
16 · b

^{2 }= 4 · k^{2}: 16 · 1^{2 }= 4 · 4 : 16 = 16
Divide
Both Sides by 16 to free up b

^{2 }
b

^{2 }= .25 · k^{2}: 1^{2 }= .25 · 4 : 1 = 1
So here
again; b

^{2}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 = (x

^{2}= 2.
The
Confusion is; Where did The 2 in (x

^{2}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 ( ! )

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