(d)0 = i
d + 0 = d
i > 0 < d

</> is a form of multiplication that incorporates addition, it results in a logarithm which helps to explain;
a) 0 is a constant.
b) There is a margin of error which expands upon ascending the numerical system.
c) That there are 'multiples' of 0.
d) If we shift one decimal position we discover that 0 is perceived as not a constant.
e) (i) exists.

My aim is to prove that 0 is closer to 1 than 1 is to 2.

1. (q) </> (w) = (qw)' (w')(q') = (qw = y' = e) (w' = r) (q' = t) = (e)(r)(t)

Or, with '</>' in a different 'grammatical' position;

2. (q)(w) </> (qw)' (w')(q') = (qw = y' = e) (w' = r) (q' = t) = (e)(r)(t)

12 </> 13 = (12 x 13)' (12') (13') = (156)' (3) (4) = 144
12 x 13 </> (12 x 13)' (12') (13') = (156)' (3) (4) = 144

if q = 12, w = 13 then (e)(r)(t) = 144

1. 12 </> 13 = 144
Or;
2. 12 x 13 </> 144

12 </> 13 = (12 x 13)' (12') (13') = (156)' (3) (4) = 144

if q = 13, w = 14 then (e)(r)(t) = 220
13 </> 14 = 220
__________________________________________________

0 </> 0 = 0
1 </> 0 = 1
2 </> 0 = 2
3 </> 0 = 3
4 </> 0 = 4
5 </> 0 = 5
6 </> 0 = 6
7 </> 0 = 7
8 </> 0 = 8
9 </> 0 = 9
?
10 </> 0 = 1
11 </> 0 = 2
12 </> 0 = 3
13 </> 0 = 4
14 </> 0 = 5
15 </> 0 = 6
16 </> 0 = 7
17 </> 0 = 8
18 </> 0 = 9
19 </> 0 = 10
20 </> 0 = 2
21 </> 0 = 3
22 </> 0 = 4
23 </> 0 = 5
24 </> 0 = 6
25 </> 0 = 7
26 </> 0 = 8
27 </> 0 = 9
28 </> 0 = 10
29 </> 0 = 11
30 </> 0 = 3
31 </> 0 = 4
32 </> 0 = 5
33 </> 0 = 6
34 </> 0 = 7
35 </> 0 = 8
36 </> 0 = 9
37 </> 0 = 10
38 </> 0 = 11
39 </> 0 = 12
40 </> 0 = 3
41 </> 0 = 5
42 </> 0 = 6
43 </> 0 = 7
44 </> 0 = 8
45 </> 0 = 9
46 </> 0 = 10
47 </> 0 = 11
48 </> 0 = 12
49 </> 0 = 13
50 </> 0 = 5

So one can infer that
10 x 19 x 29 x 39 x 49 = 10 x 20 x 30 x 40 x 50
'margin of error' </> i
So,
10 x 20 x 30 x 40 x 50 = 12 000 000 - i

? TW Stacey 14.10.2009