Second Proof 1=0
Posted: Tue Sep 03, 2019 2:12 am
A) 1 --> (1,1)=2 ---> (1,1,1)=3 ---> (1,1,1,1)=4 ---> (.....) ----> (1n ---> infinity)
***this can be observed as a standard 1 dimensional number line.
B) (1 <---> 1) <---> ((1--->1)&(1<---1))
C) (1--->1) <---> ((1--->2)--->3)...)
D) (1--->1) = (1n ---> infinity)
E) (1n ---> infinity) = (1--->1)◇
****"◇" means change or changing...wrong symbol but will do for now.
F) ((1--->1)&(1--->1)) contains as an element (1--->1)◇ therefore ((1--->1)&(1--->1))◇
G) (1<--->1) <----> ((1--->1)&(1<---1))◇ therefore ((1n ---> infinity) ---> 1)
H) ((1--->(1n--->infinity))&((1n--->infinity)--->1)
J) 1 <---> (1n ---> infinity)
K) 1 = 1◇
L) 1◇ <---> (1--->1)
M) (1--->1) ---> (1--->0) because 1---> (1n-->infinity)=
***1 approaching infinity always requires one manifesting fractions (or fractals on the number line) thus 1 approach to requires 1.1, 1.2, 1.3, etc. As 1.11, 1.12, 1.13, etc.
N) 1◇ <---> ((1--->1) & (1--->0))
O) 1◇ <---> (1 & 0)
P) 1◇ = (1 & 0)
Q) 1 = 1◇ therefore 1 = 0
***this can be observed as a standard 1 dimensional number line.
B) (1 <---> 1) <---> ((1--->1)&(1<---1))
C) (1--->1) <---> ((1--->2)--->3)...)
D) (1--->1) = (1n ---> infinity)
E) (1n ---> infinity) = (1--->1)◇
****"◇" means change or changing...wrong symbol but will do for now.
F) ((1--->1)&(1--->1)) contains as an element (1--->1)◇ therefore ((1--->1)&(1--->1))◇
G) (1<--->1) <----> ((1--->1)&(1<---1))◇ therefore ((1n ---> infinity) ---> 1)
H) ((1--->(1n--->infinity))&((1n--->infinity)--->1)
J) 1 <---> (1n ---> infinity)
K) 1 = 1◇
L) 1◇ <---> (1--->1)
M) (1--->1) ---> (1--->0) because 1---> (1n-->infinity)=
***1 approaching infinity always requires one manifesting fractions (or fractals on the number line) thus 1 approach to requires 1.1, 1.2, 1.3, etc. As 1.11, 1.12, 1.13, etc.
N) 1◇ <---> ((1--->1) & (1--->0))
O) 1◇ <---> (1 & 0)
P) 1◇ = (1 & 0)
Q) 1 = 1◇ therefore 1 = 0