# Measure of 1 = 0

Ian Goddard (IAN@GODDARD.NET)
Fri, 10 Jul 1998 19:48:32 -0400

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

VISIT Ian Williams Goddard --------> http://Ian.Goddard.net