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From: Archimedes Plutonium on 7 Mar 2010 02:38

Now how do we have a square root of 9999....9999?
Well using the AP-adics means of operators the square root of
9999....99999
would go like this:


sqrt9 = 3
sqrt99 = 9.94
sqrt999 = 31.60
sqrt9999 = 99.99
sqrt99999 = 316.22
sqrt999999 = 999.99
sqrt9999999 = 3162.27

Alright the above is a preliminary. I am going to make one last stab
at this in the
closing pages of this book. You can see that square root has two
alternatives depending
on whether the string of 9s is even or odd.

So according to this book, that 999....9999 is the same as -1. And I
said that the Euler
Identity makes e and pi special numbers as transcendental numbers
because those
two numbers allow us to tell whether 9999.....9999 at
infinity=negativenumbers whether
that number of its 9s is even or odd and what the square root of
9999.....99999 actually is.

In other words, this book predicts that there are two and only two
transcendental numbers and they are pi and e. And the reason they are
transcendental is because they fix what the
number 99999.....99999 and the square root of 99999.....99999 are.

In earlier editions of my three books I kept harping that pi and e are
transcendental because
they are growing numbers and not fixed numbers. That they depend on
time itself in the Cosmos as to what they are going to be a minute
later whereas a number like 1/3 is fixed
in all its digits as 0.3333..... a number like pi is growing and
unfixed in its digits. Well, I
no longer accept that. With finite number being 10^500 or less and
where infinity is not
endlessness but rather only negative numbers, I now feel that
Transcendental means
two special numbers that can "fix" the place value of 9999....9999 and
the square root
of 99999.....9999 = -1.

So the number 3162....... is about 31% of the distance of one complete
circuit around a
line of longitude on a sphere. And where 9999.....99999 is only 1 step
removed of a complete
circuit.

So now in the Euler Identity e^pi*i=-1 where we approx e as 3 and pi
as 3
we have 3^3*3162..... = 9999.....99999

This begins to look like 3^9999....9999 = 9999....99999

So the question becomes, is it ever the case that e^999...9999 is a
string
of 9s?

Now I know in binary of pi and e that 3 is 11 and 2 is 10.

And that e is closer to 3 than to 2 so that e is more like 11 rather
than 10 in
binary.

Now in decimal or binary when we have 111111.....111111 x
11111.....111111
we have something like this:

11 x 11 =
111 x 111=
1111 x 1111 =
11111 x 11111 =

And the answer in all those cases has the front digit and rear digit
always a
1. And because of the various digits in the middle of the 1 at the
rearend and
the 1 at the frontend are all cancelled and converted into 1s digits
in the middle
because of the pi and e variable digits. In pi of 3.14.... and e of
2.71.....
it is the 4 in pi and the 1 in e digits; and the 1 in pi and the 7 in
e and the 3 in
pi and the 2 in e in multiplication that all the digits end up looking
like

111111.......111111 in binary or 99999.....99999 in decimal.

In binary the number 99999.....9999 is of course 1111.....11111

So the use of pi and e, in the Euler Identity is that these are the
only two
numbers in all of mathematics that can fix the place value of
9999.....9999
and tell us whether the square root is 9999..... or 3162......

So the meaning of transcendental numbers in mathematics is that of
there are
two and only two such numbers and that they fix the place value of -1,
the infinite integer
of 9999.....99999. These two numbers are never algebraic because you
need these
two numbers to fix the place-value of Algebra.


Archimedes Plutonium
http://www.iw.net/~a_plutonium/
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
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