Paradoxica Amelioratora: not a bullet poem: extensis

Initial thought:

0 + 1 should be 0 not 1 (else -1 + 0 = 1)

(that when 0 is not a-axial with one, it must be axial)

AND

questioning 0 + 1 - 0 = 1 because 0 is less than 1

(that 1 is neither negative nor positive)

qualifics

--Eucaleh

QUALIFIC OPERATORS 'QUALIFICS' / DIFFERING SYSTEMS AND COUNTERPROOFS [EXTENDED]

Consider the basis for mathematics, a notion of one and zero, being and nothing, value or void;

Although operations with the analogical one and zero are basic to mathematics, the use of non-numerical signs implies

that numbers are subset within a systemic context; although we need not conclude anything about this pre-condition,

one may go so far as to say that there is a slipperiness in operators:

For example, we may simultaneously posit that:

1 + 0 = 1 not 0 AND

1 - 0 = 1 not 0

[10 = the Scales, thus 10 is not 1,0 ; to posit gain is to posit complex gain in numbers]

In the context of the digital, comparing the set 1, 0 to the number 10 creates a schism;

the first is pared until it requires systemic referrant, while in the second case the system

becomes accretively both ambiguous and complex

Questioning the view that 0 + 1 - 0 = 1 because for example 0 is less than 1 creates the idea that

earlier statements might be redefined (dangerously) with unexpected results. One might look also

at two other related expressions:

One might then claim that 0 - 1 = 0 not -1 (else 1 - 0 = 1/2)

Evidence might be cited in the equivalent expressions [(1 + 1) - ( 0 + 0)] = 2 not 1 yet [(1 - 0) - (1 - 0)] = 0 not 1,

the first is said to equal 2 and the second zero; simultaneously it is held that two positives amount to positive,

and two negatives amount to positive, however in this case that isn't true. If it is not negatives

operating, how may one explain that one expression is greater than the other with approximately identical terms?

(e.g. how is the extension [(1 + 1) - (0 + 0)] + [(1 - 0) - (1 - 0)] > [(1 - 0) - (1 - 0)] + [(1 - 0) - (1 - 0)] + 1 ? ad infinitum?)

Also, 0 + 1 should be 0 not 1 (else -1 + 0 = 1)

Responding to this, notation may be altercated such that:

1 balanced = 0 [1--> + <--1 = <--0-->] AND

1 unbalanced = 0 balanced [1--> + 1--> = 0--> or 1--0-->]

This view, although not elaborated, meets greater adequacy in geometry, for example that the digit two is not

more axial than one, but rather an extension of number through zero

Considering briefly what such a mathematics would entail, one might suppose that

any additional one is a separate entity, thus by identity all ones are opposed unless via unifying;

hence here non-unified 1 + 1 + 1 = 0{3 (this might be equivalent to space divided in three portions)

also, unified ones do nothing to the strength of 0

hence (1 + 1 + 1)--> = 3{0 (this might be equivalent to a non-spacial qualitative value of 3)

the expression x{y might be compared to 'x root y' or 'x stem y' superficially but doesn't entail any operation specifically so much as a context-point within which operations may occur

saying that 0{3 is 1/3 would be wrong, for its implicit here that 3 is a subset of 0 that still has for lack of a better term quantitative value

(the best conclusion may be that there is an implicit non-spacial geometry depending on concepts of distinction-as-variable)

also, saying that 3}0 is 1 or 3 in a traditional sense is misleading; the distinction between 2{0 and

3{0 may be seen in the same way as 2}0 + 1}0 = 3}0 in the sense of piling coins, however it is more accurate to say that systemically 0{0 is of greatest importance,

while qualifically 99{0 is like a mobile operator and 0{99 a matter of many dimensions as yet unqualified

Standard Arithmatic

1 + 0 = 1 not 0

1 - 0 = 1 not 0

so logically 0 - 1 should be 0 not -1

else 1 - 0 = 1/2

proof: [ (1 + 1) - ( 0 + 0 ) ] = 2 not 1

however, [ ( 1 - 0 ) - ( 1 - 0 ) ] = 0 not 1

if 1 - 0 = 1/2 and 0 - 1 = 0, then

[(1 + 1) - ( 0 + 0 )] = 2 and

[( 1 - 0 ) - ( 1 - 0 )] = 0

in the first case 1 + 1 = 1/2 + 1/2 (unity)

and 0 + 0 = [( 0 - 1 ) + ( 0 - 1 )]

however, we already know that

-1 -1 = -2 not 0, else

[ (1 + 1) - (0 + 0) ] should = 1

in either case 0 + 0 = 1

and 1 + 1 = 1/2 + 1/2

it has the fortunate conclusion of solving the dilemma of

0 - 1 / 1 - 0 = [ ( 1 + 1 ) - ( 0+ 0 ) ] / [ ( 1 - 0 ) - ( 1 - 0) ]

-1 / 1 = (0 - 1) / (1 - 0) = ( 1 + 1) - (0 + 0) / (1 - 0) - (1 - 0) = 1 / 1

else (1 + 1) - (0 + 0) / (0 - 1) + (0 - 1) = 2 / -2

since

(0 - 1) + (0 - 1)

if = 0 not= -2

while (1 + 1) + (0 + 0) = 2

yet -1 + -1 = [(0 - 1) + (0 - 1)]

even if -1 + -1 = -2

furthermore a comparison may be made where

(1 + 1) - (0 + 0) / (0 - 1) + (0 - 1) not= -1 / 1

because [ (1 + 1) - (0 + 0)] not = [(1 - 0) - (1 - 0)]

in the context of the premises 1 + 0 = 1 and 1 - 0 = 1

thus either

(1 + 1) - (0 + 0) / (1 - 0) - (1 - 0) not= (1 + 1) - (0 + 0) / (1 - 0) - (1 - 0)

or -1 / 1 = 2 / -2

or -1 / 1 = 1 / 1

[note: for sanity's sake, perhaps geometry precedes #, I suspect this]

since the last equation seems completely false ( -1 not= 1)

we must approach the idea that -1 / 1 = 2 / -2

Deliberating over previous equations, the conclusion is then that in the first place

(-1) + (-1) = -2

However, this means (0 - 1) + (0 - 1) = -2

Thus (1 - 0) - (1 - 0) not= 0

because logically

1 - 0 / 0 - 1 = 0

However, if 1 - 1 not= 0

this flies in the face of traditional arithmatics

The conclusion then is that 1 - 0 / 0 - 1 not= 0

OR

[(1 + 1) - (0 + 0)] = [(1 - 0) - (1 - 0)]

However, in the view that

1 + 0 = 1 and 1 - 0 = 1

we know that

(1 + 1) - (0 + 0) = 2

and

(1 - 0) - (1 - 0) = 0

although these seem reasonable,

we know either

[(1 + 1) - (0 + 0)] = [(1 - 0) - (1 - 0)]

or

1 - 0 / 0 - 1 not= 0

if (1 + 1) - (0 + 0) = 2, it is

not= (1 - 0) - (1 - 0) = 0

Thus, either we may question the premises

1 + 0 = 1 and 1 - 0 = 1

OR

1 - 0 / 0 - 1 not= 0

according to (1 + 1) - {(1 - 0 / 0 - 1)+( 1 - 0 / 0 - 1)} = 0

[equivalent to (1 + 1) - (0 + 0) = (1 - 0) - (1 - 0)]

2 - 0 = 0 if 0 + 0 = 0

AND 1 - 0 / 0 - 1 = 0, however

Since we know either 1 - 0 / 0 - 1 not= 0

or fundamental premises are false,

2 - 0 is not 0, rather

(1 - 0 / 0 - 1) + (1 - 0 / 0 - 1) = 2

or the premises are false [here begins a second notebook titled MXC3 arbitrarily

it might be considered whether this system extended confutes unknowability / unsolvability vis. Godel]

Since (1 - 0 / 0 - 1) + (1 - 0 / 0 - 1) = 2

or 1 + 0 not= 1 and/or 1 - 0 not= 1

The conclusion according to traditional arithmatic is that

1 - 0 / 0 - 1 not= 1 - 0 / 0 - 1

(else 1 - 0 / 0 - 1 not= 1 - 0 / 0 - 1)

However the conclusion here is that 1 / -1 = 1

and 1 - 0 = 1

if 1 / - 1 = 1 then

(1 / - 1) + (1 / -1) = 2

( 0 - 1 ) + ( 0 - 1) = -2 or -1 not= 1 (according to previously)

then (if premises hold), 1 / -1 + 1/ -1 + [(0 - 1) + (0 - 1)] = 0

if (1 / -1 + 1 / -1) + [(0 - 1) + (0 - 1)] = 0

Then according to traditional basics, 2 - 2 = 0

[However, according to the logic followed as a basis for comparison _ _ _]

since

[(1 + 1) - (0 + 0) not= (1 - 0) - (1 - 0)]

the 2nd equation may be translated as:

1 / -1 + 1 / -1 not= not= 0

since we know by (1 + 1) - (0 + 0) = 2

that also 2 not= (1 - 0) - (1 - 0) = 0

so that

[(1 + 1) - (0 + 0)] not= [ 2 = 0 ]

if 1 / -1 + 1 / -1 not= not= 0 [NOT AN ERROR: EMPHASIS]

then 1 / -1 + 1 / -1 = 0

or equivalent

if so, then

(1 / -1) + (1 / -1) not= 2

thus 1 / -1 not= 1

However, according to previous arguments, the only conclusion flies in the face of traditional arithmatics:

posing 1 - 0 / 0 - 1 not= 1

then either 1 - 0 / 0 - 1 not= 1 - 0 / 0 - 1

OR

1 - 0 / 0 - 1 not= 0 AND (1 - 0 / 0 - 1) + (1 - 0 / 0 - 1) not= 2

Since it runs agianst all rationality to claim a statement is not itself,

The conclusion is that since 1 - 0 / 0 - 1 not= 0

AND

1 - 0 / 0 - 1 + 1 - 0 / 0 - 1 not= 2

thus

1 + 0 not= 1 AND/OR 1 - 0 not= 1

Since this flies in the face of fundamental mathematics, the most rational thing is to refigure at the primary equations in a manner suiting to resolution_ _ _

The initial assumption was that 1 + 0 = 1 and not 0

and 1 - 0 = 1 and not 0

YET 0 - 1 is -1

AND 0 + 1 is 1 and not 0

Considering on the surface these four entities (surrounding a zero point),

the quadrants are (counterclockwise) 1 + 0 = 1, 0 + 1 = 1, 0 - 1 = -1, and 1 - 0 = 1

The total is 3 and not 0, although they are situated axially around a point of no value--this in itself ought to bring criticism.

(0 - 1 = -1 is the only negative box of four).

Reconsidering, it cannot be concluded that 1 + 0 is other than 1, if zero is actually nothing, nor may it be concluded that one

minus 0 is anything other than 1, if 0 is merely a loss of nothing

Then if the premises may be questioned, it is only in two statements, namely

if 0 - 1 = -1 and

if 0 + 1 = 1

(essentially in my belief these were popularized by Descartes and the Cartesian diagram)

If 1 - 0 / 0 - 1 not= 0

AND

1 - 0 / 0 - 1 + 1 - 0 / 0 - 1 not= 2

if 0 - 1 = - 1 and

1- 0 holds as 1

then (1 / -1) + (1 / -1) not= 0 or 2

However, since 1 / 1 and -1 / -1 are conceived as = 1

and 0 / 0, 1 / 0, and 0 / 1 may be equal to 0

the statement that

1 / 1 - 1 / 1 [equiv. to (1 / 1) + (-1 / 1)]

- [(-1 / 1) + (1 / 1)] = 0

seems to vindicate that two terms, technically opposite, amount to zero

While this isn't startling,

it now comes to the fore that

-1 / 1 + 1 / -1 = 0

[since otherwise opposite terms would not cancel]

Thus saying that

[(1 / 1) + (1 / 1)] - [ (-1 / 1) + (-1 / 1)] = [(-1 / 1) - (-1 / 1)] - [(-1 / 1) - (-1 / 1)]

isn't valid under the arguments

1 - 0 / 0 - 1 + 1 - 0 / 0 - 1 not= 2

and

1 - 0 / 0 - 1 not= 1

[

if (1 / -1) + (-1 / 1) not= 0 then according to arguments [ [(1 - 0 ) / (0 - 1)] + [(1 - 0) / (0 - 1)] / 2] not= (1 - 0) / (0 - 1)

THUS

[(1 + 1) - (0 + 0) = (1 - 0) - (1 - 0)] is true

since the two halves of the basis [(1 + 1) - (0 + 0) = (1 - 0) - (1 - 0)]

are equivalent to

1 + 0 = 1 and 1 - 0 = 1 respectively,

and (1 - 0) / (0 - 1) not= 1

and (1 - 0) / (0 - 1) not= 0

[and [(1 - 0) / (0 - 1)] + [(1 - 0) / (0 - 1)] not= 2]

the avg. [ [(1 - 0) / (0 - 1)] + [(1 - 0) / (0 - 1)] / 2 ] in equivalence to 1 - 0 / 0 - 1

substantiates the following: if 1 / -1 = 0

then (tentatively)

1 - 0 / 0 - 1 + 1 - 0 / 0 - 1 = 1/2

and

1 - 0 / 0 - 1 = 2

However, the clear conclusion here upon simplifying terms is that

1 / -1 + 1 / -1 = 1/2

and

1 / -1 = 2

Although easily reproofed under conventions, under the earlier arguments these serve to substantiate

criticism not of the first premises ( 1 + 0 = 1 and 1 - 0 = 1), instead the 2nd two

[1] That 0 - 1 should be 0

and not -1

(else 1 - 0 = 1/2

since (1 + 1) - (0 + 0) = 2

not 1 yet (1 - 0) - (1 - 0) is 0 not 1)

(1 - 0) + (0 - 1) = 1/2 or (1 + 0) + ( 0 - 1) = 0 in spite of 0 + 0 + 1 + 1 = 2 and 0 doing no operation :: vis. the earlier example of equivalent terms

yet (1 + 1) - ( 0 + 0) not= (1 + 0) - (1 + 0)

AND

[2] 0 + 1 should be 0 not 1

(since -0 + 1 = 1)

else -1 + 0 = 1

(cancellation of signs)

else pending argument of the validity of positive or negative numbers since the view of zero as a canceller is promoted in these methods, yet also in the concept that -0 + 1 = 1 in the context of cancelling operators [EMPHASIS]

This altered view, although utterly insane by most standards, suggests an alternate approach to mathematics in which an attenuated proportion within the traditional graduated system (e.g. for example integral numbers redefined in terms of fractions or geometry) provides a foundation for a calculative method

Maybe sometime it will be determined for a childhood like mine, that calculus was too diabolical, we needed a diabolical calculus

[18th 5 X 7" page in a day, 6/11/2008]

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1 month ago

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