Thread: Mathematics - a new basis

Originally Posted by topsquark

I've looked at the paper a bit and even though I cannot understand much of what you are doing it appears that it makes some sort of sense. However you are not making sense in this thread as you have repeatedly ignored my requests for answers and merely posted snippets of information. You still haven't defined, for example, what a dynamic number is.

Since you are talking about a paper you have written this should properly belong to the Peer Math Review forum. Please post your thread there if you wish to continue the discussion.

-Dan

I will respond with pictures-dynamic number, R-positive real numbers , (d,d1,d2)-continuation of numbers (as a variable, so called dynamic numbers)
2Rd(3)

5.3_{3Rd1 (6) d2 (7)} +3 = 7 , 3_{3Rd1 (6) d2 (7)} +3 = 6 ,3_{3Rd1 (6) d2 (7)} +3 = 0
3Rd1 (6) d2 (7)

--------------
2.4 Mobile Number "2.2,2.3"
Theorem-Natural numbers can be specified and other numerical
point other than the point numeric 0th
Proof - Value (length) numeric point (0) and numeric point (2)
the number 2

Ratio (length) numeric point (1) and the numerical point of (3) is the number 2

Ratio (length) numerical point (2) and the numerical point of (4) is the number 2

...
A set of mobile numbers Nn = {[n]N}

2.5 Gap numbers "2.2,2.3,2.4"
Theorem - Natural number and mobile number of no contact,
(natural number and mobile number no contact) and have no contact mobile number, ..., in numeric longer.

EVIDENCE - natural number 2 and mobile number 2 no contact, you get the number of gaps 2/.1/2.

natural number 2 and mobile number 2 no contact, you get the number of gaps 2/.2/2.

natural number 2 and mobile number 2 no contact, you get the number of gaps 2/.3/2.

...
(natural number 2 and mobile number 2 no contact) and mobile number 1 no contact , you get the number of gaps
2/.1/2/.1/1

...
Set gap number G_N={a|/.b_n/c_n|(a,b_n,c_n) ,b_n>0}
a/.b₁/c₁
a/.b₁/c₁/.b₂/c₂
a/.b₁/c₁/.b₂/c₂/.b₃/c₃
a/.b₁/c₁/.b₂/c₂/.b₃/c₃/.b₄/c₄
...

Look, I am reading through some of your paper. Please do two things for me:
1. Please put this into the Peer Math forum.

2. If you want to post something then make it as a response to a question. This is a "Help" forum where we help people with their questions about Mathematics. This site is not a vehicle designed to get your ideas out in the open. Or whatever your goal is.

-Dan

Originally Posted by topsquark

I've looked at the paper a bit and even though I cannot understand much of what you are doing it appears that it makes some sort of sense.

It fails at the first hurdle:

Is just gobbledygook, to me at least.

I think the OP should try posting this to a maths site that uses his native language, and let them try to disentangle what he means (assuming it makes any sense at all, and I would not bet on that)

.

2.6. Mobile gap number "2.2,2.5"
Theorem-Gap numbers can be entered and the second numerical
point other than the point numeric 0th

EVIDENCE-ratio (length) numeric point (0) and the numerical point of (4) is
2/.1/1 number of gap.



ratio (length) numeric point (1) and the numerical point of (5) is
2/.1/1 number of gap.



ratio (length) numeric point (2) and the numerical point of (6) is
2/.1/1 number of gap.

...
A set of mobile numbers gap G_Nn={[n]G_N}

2.7. Points the number of "2.2,2.3,2.5"
Theorem - Number (N,G_N) has extended the numeric point, they
can write the opposite.

EVIDENCE - Number 5 has a point: [0], [1], [2], [3], [4], [5]. Opposite may
write: [.0], [.1], [.2], [.3], [.4], [.5].



Gaps has a number 2/.3/1 points: [0], [1], [2], [3], [4], [5], [6]. They can be
otherwise write: [.0], [.1], [.2], [.3], [.4], [.5), [.6].

2.8. The opposite number "2.2,2.3,2.5,2.7"
Theorem - Numbers (N, G_N) that have the same number of points
number, length becomes gap and rotation.
EVIDENCE - 4 0/.4/0 , 4s. = {4, 0/.4/0 } or 0/.4/0s.= {0/.4/0,4}.



1/.1/3 0/.1/1/.3/0 , 1/.1/3s. = {1/.1/3, 0/.1/1/.3/0} or
0/.1/1/.3/0s. ={ 0/.1/1/.3/0, 1/.1/3}



The general form of a b. a_s.= {a, b} or b_s. = {b, a}.

A set of numbers opposing S. = {(a, b) (N, G_N)}, S.n = {(a, b) (Nn, G_Nn)}

Though I have become loathe about this thread I will give you one, count them, one last chance. No. I've warned you before and you haven't followed instructions. I'd like to introduce you to a couple of my family members:
David Banner
Bruce Banner
topsquark Banner

-Dan