Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.



CHANGING THE DIAGONAL!
Posted:
Dec 28, 2012 7:16 PM


+>  0. 542..  0. 983..  0. 143..  0. 543..  ... v OK  THINK  don't back explain to me. You run down the Diagonal 5 8 3 ... IN YOUR MIND 
[1] you change each digit ONE AT A TIME 0.694... but this process NEVER STOPS
[2] so you NEVER CONSTRUCT A NEW DIGIT SEQUENCE!
[1] [1]>[2] [2]
******* *PROOF* *******
AD METHOD (binary version) Choose the number 0.a_1a_2a_3...., where a_i = 1 if the ith number in your list had zero in its iposition, a_i = 0 otherwise.
LIST R1= < <314><15><926><535><8979><323> ... > R2= < <27><18281828><459045><235360> ... > R3= < <333><333><333><333><333><333> ... > R4= < <888888888888888888888><8><88> ... > R5= < <0123456789><0123456789><01234 ... > R6= < <1><414><21356><2373095><0488> ... > ....
By breaking each infinite expansion into arbitrary finite length segments
[3] The antiDiagonal never produces a unique segment (all finite segments are computable)
[4] The antiDiagonal never produces a unique sequence of segments (all segment sequences are computable)
CONCLUSION: Changing the diagonal just changes the permutation, every digit change is accommodated into the same set.
G Cooper (BInfTech)



