From: Ian Goddard (IAN@GODDARD.NET)
Date: Fri Jul 10 1998 - 17:48:32 MDT
This pretty much defines the shortfall in Dan's
argument that the two observers measured the other
as smaller free from reference to something larger,
and thus (stated in error) a zero-sum isn't found.
Every thing, A, is measured by its deviation from zero A.
zero A = no A
Every measured unit of A is a displacement from 0 A:
0A--->1A--->2A
The positive displacement of 1 from 0 is symmetrical to
the negative displacement of 0 from 1, and the measurement
taking place is equally both displacements, one displacement
is not a more valid representation of the measure taking place:
0--->1 = (+1) = (1 is 1 more than 0)
0<---1 = (-1) = (0 is 1 less than 1)
The degree to which 1 is larger (+) than 0 is equal and opposite
to the degree that 0 is smaller (-) than 1. Indeed, the statement
"1 is 1 unit more than 0" tells us the same thing as the statement
"0 is 1 unit less than 1." This equality expresses the equality
of the degree to which both of the symmetrical displacements shown
above between 0 and 1 are valid representations of the measure of
difference between 1 and 0. We can express the equal validity of
each side of the 0,1 difference as such, where "v" = "validity":
v( 1 is 1 more than 0 ) = v( 0 is 1 less than 1 )
Those two equally valid statements are inverse measurements of
the defined difference, and as such are identical to inverse
subtractions, which are also measures of difference, thus:
(1 is 1 more (+) than 0) = (1 - 0) = +1
(0 is 1 less (-) than 1) = (0 - 1) = -1
Therefore, the difference between 0 and 1 is equally both 1 and
-1, not 1 more than -1, since both statements are equally valid
expressions of the measurement of 1 from 0. It clearly follows
therefore that the sum of all valid measurements derived from
the differentiation of 1 from 0 will = 0, since 1 + (-1) = 0.
Negative Numbers Are "In" Positive Numbers
Negative 1 is internally "nested" within positive 1 due to the
fact that both -1 and 1 are equally and simultaneously valid
expressions of the 0,1 relation that defines 1 as 1. All the
negative numbers are "in" the positive numbers, since each
positive number is its difference/displacement from 0 and
that difference is symmetrical (n + (-n)) and sums to 0.
Whatever we measure, 1 unit of X is derived by its deviation
from 0 X, or "no X." The displacement from 0 is symmetrical
such that 1X is as much (more)X (+1) as 0X is (less)X (-1)
and the net displacement (+1 + (-1)) always equals 0.
0 1 2 3
____________
0 | 0 1 2 3 |
| |
1 |-1 0 1 2 |
| |
2 |-2 -1 0 1 |
| |
3 |-3 -2 -1 0 |
--------------
Every number has a discrete relation with 0, and every number
IS its difference from 0: 1 is 1 more (+) than 0, 2 is 2 more
(+) than 0, 3 is 3 more (+) than zero, and so forth... At the
same time, 0 is 1 less (-) than 1, 2 less (-) than 2, and 3
less (-) than 3. The relation of n to 0 is simultaneously
expressed by equal yet opposite values that sum to zero.
Both of those symmetrical values are valid at the same time
just as the statement " 0 < 1 " says both "0 is less than 1"
and "1 is more than 0" at the same time. And the sum of this
"more-less" difference always = 0, and thus the sum of all
difference must always = 0, as the matrix above proves.
But moreover: the sum of all valid measurements always = 0
because the measurement of 1 unit of X is composed of two
symmetrically different components that sum to 0. So all
measurement, being founded upon the symmetrical difference
between 0 and 1, can never actually exceed a net sum of zero.
The conservation of measure is the conservation of difference.
The conservation of difference is the conservation of identity.
Identity conservation: http://www.erols.com/igoddard/matrix.htm
(c) 1998 Ian Williams Goddard
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VISIT Ian Williams Goddard --------> http://Ian.Goddard.net
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