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M.S.

I will present you a math composed of only two basis (natural and realistic basis)

Current mathematics (CM.)

Natural Base
-natural straight line the main axiom, its beginning or end point and natural straight line a defined length and with two points
NOTATION - natural straight line (lower case), points (capital letters or numbers (when specified point uploads metric (such as the number line)))

-natural gaps negation natural straight line , natural emptiness and emptiness is defined with two points
NOTATION - natural gaps (small underlined letter)

-basic rule merger - natural straight line and natural gaps are connected only points
-basic set - all possibilities defined theorem
(CM.)does not know the natural straight line , point is not defined, knows no natural gap, is not defined by basic set

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Reply 1
wtf?
Original post by point.ms
I will present you a math composed of only two basis (natural and realistic basis)

Current mathematics (CM.)

Natural Base
-natural straight line the main axiom, its beginning or end point and natural straight line a defined length and with two points
NOTATION - natural straight line (lower case), points (capital letters or numbers (when specified point uploads metric (such as the number line)))

-natural gaps negation natural straight line , natural emptiness and emptiness is defined with two points
NOTATION - natural gaps (small underlined letter)

-basic rule merger - natural straight line and natural gaps are connected only points
-basic set - all possibilities defined theorem
(CM.)does not know the natural straight line , point is not defined, knows no natural gap, is not defined by basic set


A *straight line* is defined as a function f:RRf: \mathbb{R} \to \mathbb{R} (more general fields are available) given by xax+bx \mapsto a x+b for some real constants a,ba, b. A *point* is defined as a 2-tuple of reals.
Wow, I have no idea what you're talking about.
Reply 4
Original post by Smaug123
A *straight line* is defined as a function f:RRf: \mathbb{R} \to \mathbb{R} (more general fields are available) given by xax+bx \mapsto a x+b for some real constants a,ba, b. A *point* is defined as a 2-tuple of reals.



you are given evidence from two axioms (line, line number)
and I gave two axioms (natural straight line and natural gaps)

to be the definition of the gaps in the current math?

Theorem - Natural straight line (natural gap) are connected in the direction of the points AB (0.1)
PROOF - straight line (gaps) b (b\underline{b}) -defined AC (0,2)

- straight line (gaps) c (c\underline{c}) -defined AD (0,3)

...
infinite one way straight line (oneway infinite gaps) (\underline{\infty}) defined A∞ (0, ∞)

(CM.) - straight line (not from the natural basis), not gaps, a one-way infinite straight line the (semi-line (not from natural base)), one-way infinite gaps does not exist
Original post by point.ms
to be the definition of the gaps in the current math?

A *gap* between points A=(r,s)A=(r, s) and B=(t,u)B=(t, u) is defined to be the line f:xsurtx+r(su)rt+sf: x \mapsto \frac{s-u}{r-t}x + \frac{-r(s-u)}{r-t}+s, restricted to the domain [r,t][r, t]. The definition must be suitably altered in the case r=tr=t, of course.

A *one-way infinite gap* from point A through point B is defined to be the gap between A and B, but where ff is restricted to the domain [r,)[r, \infty) (assuming wlog that r<tr < t).
Reply 6
Theorem - there is a relationship between the points 0 and all points one-way infinite straight line(one-way infinite gaps) including points 0


PROOF - relationship points 0 points 0 and the number 0



-relationship points 0 points 1 and the number 1 (1\underline{1})



-relationship points 0 points 2 and the number 2 (2\underline{2})


...

basic set of natural numbers No={0,1,2,3,4,5,...}N^o=\{0 , 1 , 2 ,3 ,4 ,5 ,...\}
basic set of natural numbers gaps Npo={0,1,2,3,4,5,...}N_p^o=\{0 , \underline{1} ,\underline{2} ,\underline{3} ,\underline{4} ,\underline{5} ,...\}


(CM.) - natural numbers are given as an axiom, there is no natural gaps numbers (there is this form, but do not call numbers ({0,0}{a,a}aN)(\{0,0\}\cup\{a,a\} a\in N)
Original post by point.ms
(CM.) - natural numbers are given as an axiom, there is no natural gaps numbers (there is this form, but do not call numbers ({0,0}{a,a}aN)(\{0,0\}\cup\{a,a\} a\in N)

The natural numbers are given as an axiom for you as well. Your system, so far, is entirely finite. You're going to need some kind of axiom of infinity, as well as some way of supporting induction, in order to assert the existence of infinitely many points, let alone natural numbers.
(edited 8 years ago)
Reply 8
Theorem - natural numbers and natural numbers gaps can be connected in the direction AB (0.1)


PROOF - Number 1 and number 1\underline {1} receives the combined number of 111\underline {1} or dup (duž , praznina )



-Number 1\underline {1} and number 1 receives the combined number of 11\underline {1}1 or dup



-Number 1 and number 2\underline {2} receives the combined number of 121\underline {2} or dup

...
- A basic set of combined natural numbers Ko=(an,bn,anNo,bnNpo,(an,bn)>0)K^o=(a_n,\underline{b}_n,a_n\in{N^o},\underline{b}_n\in{N_p^o},(a_n,\underline{b}_n)>0)


a1b1a_1\underline{b}_1
b1a1\underline{b}_1a_1
a1b1a2a_1\underline{b}_1a_2
b1a1b2\underline{b}_1a_1\underline{b}_2

...

(CM.) - Dup do not know, not know the combined numbers (there is this form, but not numbers {0,a}{c,c},{0,0}{a,b},{0,b}{c,d},{0,0}{a,b}{c,c},...\{0,a\}\cup\{c,c\},\{0,0\}\cup\{a,b\},\{0,b\}\cup\{c,d\},\{0,0\}\cup\{a,b\}\cup\{c,c\},... )

Smaug123 - to current mathematics defined dup ?
Original post by point.ms
Theorem - natural numbers and natural numbers gaps can be connected in the direction AB (0.1)


PROOF - Number 1 and number 1\underline {1} receives the combined number of 111\underline {1} or dup (duž , praznina )



-Number 1\underline {1} and number 1 receives the combined number of 11\underline {1}1 or dup



-Number 1 and number 2\underline {2} receives the combined number of 121\underline {2} or dup

...
- A basic set of combined natural numbers Ko=(an,bn,anNo,bnNpo,(an,bn)>0)K^o=(a_n,\underline{b}_n,a_n\in{N^o},\underline{b}_n\in{N_p^o},(a_n,\underline{b}_n)>0)


a1b1a_1\underline{b}_1
b1a1\underline{b}_1a_1
a1b1a2a_1\underline{b}_1a_2
b1a1b2\underline{b}_1a_1\underline{b}_2

...

(CM.) - Dup do not know, not know the combined numbers (there is this form, but not numbers {0,a}{c,c},{0,0}{a,b},{0,b}{c,d},{0,0}{a,b}{c,c},...\{0,a\}\cup\{c,c\},\{0,0\}\cup\{a,b\},\{0,b\}\cup\{c,d\},\{0,0\}\cup\{a,b\}\cup\{c,c\},... )

Smaug123 - to current mathematics defined dup ?

I don't know what you mean by "dup", I'm afraid - do you mean the ordered pair (a,a)(a, a), for some aa? If so, it can be implemented in set notation as {a,{a}}\{a, \{a\} \}.
Reply 10
Original post by Smaug123
I don't know what you mean by "dup", I'm afraid - do you mean the ordered pair (a,a)(a, a), for some aa? If so, it can be implemented in set notation as {a,{a}}\{a, \{a\} \}.

dup=straights lines + gaps
Original post by point.ms
dup=straights lines + gaps

I don't see how a dup differs from the tuple (point1, point2, line between point1 and point2)?
Reply 12
Original post by Smaug123
I don't see how a dup differs from the tuple (point1, point2, line between point1 and point2)?



Theorem - Two numbers have contact, their condition is described counts of first number


PROOF - number 3 and number 2 have a contact at point 0
3023^{\underline{0}} 2



- number 3 and number 2 have a contact at point 1
3123^{\underline{1}}2



- number 3 and number 2 have a contact at point 2
3223^{\underline{2}}2



- number 3 and number 2 have a contact at point 3
3323^{\underline{3}}2



(CM.) - Knows no contact numbers
Reply 13
Theorem - The contact number is sorted horizontally only be a natural straight line that gives a natural straight line


proof -11 1\rightarrow 1

4+10{+_1^{\underline0}}2=2


4+11{+_1^{\underline1}}2=(1,1)


4+12{+_1^{\underline2}}2=2


4+13{+_1^{\underline3}}2=(3,1)


4+14{+_1^{\underline4}}2=6 or 4+2=6


+1+_1 - addition rule 1


(CM.) - There are no "addition rule 1" only when the contact point number, Axiom


we were the first form of addition, the advantage of my mathematics
-----------


question for trained mathematicians, which is a procedure to be applied in order to get (a sign?)
4 ? 2=2
4 ? 2=(1,1)
4 ? 2 =2
4 ? 2=(3,1)
4+2=6 This is known
more complex
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = (1,1,1,2) my solution and notation 42322+103254\underline23\underline22{+_{1}^{\underline0}}3 \underline25=(1,1,1,2)
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = (1,4) my solution and notation 42322+113254\underline23\underline22{+_{1}^{\underline1}}3 \underline25=(1,4)
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = (2,1,1,2,1) my solution and notation 42322+123254\underline23\underline22{+_{1}^{\underline2}}3 \underline25=(2,1,1,2,1)
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = (3,4,2) my solution and notation 42322+133254\underline23\underline22{+_{1}^{\underline3}}3 \underline25 =(3,4,2)
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = (6,4,1) my solution and notation 42322+143254\underline23\underline22{+_{1}^{\underline4}}3 \underline25=(6,4,1)
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = (4,1,1,1,2) my solution and notation 42322+153254\underline23\underline22{+_{1}^{\underline5}}3 \underline25=(4,1,1,1,2)
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = (4,3) my solution and notation 42322+163254\underline23\underline22{+_{1}^{\underline6}}3 \underline25=(4,3)
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = (4,1,1,4) my solution and notation 42322+173254\underline23\underline22{+_{1}^{\underline7}}3 \underline25=(4,1,1,4)
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = (4,2,2,5) my solution and notation 42322+183254\underline23\underline22{+_{1}^{\underline8}}3 \underline25=(4,2,2,5)
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = (4,5,5) my solution and notation 42322+193254\underline23\underline22{+_{1}^{\underline9}}3 \underline25=(4,5,5)
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = (4,3,1,5) my solution and notation 42322+1103254\underline23\underline22{+_{1}^{\underline{10}}}3 \underline25=(4,3,1,5)
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = (4,3,1,5) my solution and notation 42322+1113254\underline23\underline22{+_{1}^{\underline{11}}}3 \underline25=(4,3,1,5)
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = (4,3,2,5) my solution and notation 42322+1123254\underline23\underline22{+_{1}^{\underline{12}}}3 \underline25=(4,3,2,5)
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = (4,3,3,5) my solution and notation 42322+1133254\underline23\underline22{+_{1}^{\underline{13}}}3 \underline25=(4,3,3,5)
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = (4,3,3,5) my solution and notation 42322+1143254\underline23\underline22{+_{1}^{\underline{14}}}3 \underline25=(4,3,3,5)
Original post by point.ms
Theorem - The contact number is sorted horizontally only be a natural straight line that gives a natural straight line


proof -11 1\rightarrow 1

4+10{+_1^{\underline0}}2=2


4+11{+_1^{\underline1}}2=(1,1)


4+12{+_1^{\underline2}}2=2


4+13{+_1^{\underline3}}2=(3,1)


4+14{+_1^{\underline4}}2=6 or 4+2=6


+1+_1 - addition rule 1

This operation, I think, is a restricted version of the binary xor. In your notation, a+1bca +_1^{\underline b}c is "2a xor (2c1)×2b2^a \text{ xor } (2^c-1) \times 2^b".

Mathematica program to calculate a+1bca +_1^{\underline b}c:
Block[{a = 4, b = 1, c = 2}, Length /@
Select[Split[
IntegerDigits[BitXor[2^a - 1, (2^c - 1) 2^b], 2] // Reverse],
First[#] == 1 &]]
Original post by point.ms
.

You're basically doing the usual arithmetic, but in binary and using a very unusual notation.
Ah, I'm starting to understand your notation now. You're performing run-length encoding on binary numbers.


Options[plus1] = {"Output" -> "Standard"};


plus1[a_, b_, c_, OptionsPattern[]] :=
If[OptionValue["Output"] == "Standard", Identity, toEncoded]@
BitXor[a, c 2^b]


fromEncoded[a_] :=
FromDigits[
Reverse[Flatten[
ConstantArray @@@
Transpose[{PadRight[{}, Length[a], {1, 0}], a}]]], 2]


toEncoded[n_] :=
Length /@ Select[Split[Reverse@IntegerDigits[n, 2]], First[#] == 1 &]


It can be invoked as follows: the toEncoded and fromEncoded functions move from standard notation to your notation and back, respectively:


toEncoded[5] (*outputs {1,1}*)
fromEncoded[{1,1}] (*outputs 5*)


The plus1 function:


plus1[fromEncoded[{4, 2, 3, 2, 2}], 5, fromEncoded[{3, 2, 5}],
"Output" -> "Nonstandard"]

outputs {4, 1, 1, 1, 2}, consistent with your assertion that 42322+15325=(4,1,1,1,2)4\underline23\underline22{+_{1}^{\underline5}}3 \underline25=(4,1,1,1,2).
Reply 17
Theorem - The contact numbers is sorted horizontally to be the only one natural straight line that gives a natural straight line , when there are two (more) results between them becomes a gap.


PROOF -11(1)1\rightarrow1(\underline{1})

4+202=24{+_2^{\underline0}}2=2
4+212=1214{+_2^{\underline1}}2=1 \underline{2}1
4+222=24{+_2^{\underline2}}2=2
4+232=3114{+_2^{\underline3}}2=3 \underline{1}1
4+242=64{+_2^{\underline4}}2=6


+2 - addition rule 2


(CM.) - No "addition rule 2"
more complex
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = {0,1}\cup{2,3}\cup{6,7}\cup{8,10} my solution and notation 42322+203254\underline23\underline22{+_{2}^{\underline0}}3 \underline25=111311​2
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = {0,1}\cup{9,13} my solution and notation 42322+213254\underline23\underline22{+_{2}^{\underline1}}3 \underline25=18​4
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} ={0,2}\cup{4,5}\cup{6,7}\cup{9,11}\cup{12,13} my solution and notation 42322+223254\underline23\underline22{+_{2}^{\underline2}}3 \underline25=22111221​1
...

{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} ={0,4}\cup{6,9}\cup{11,16} \cup{18,23} my solution and notation 42322+2133254\underline23\underline22{+_{2}^{\underline{13}}}3 \underline25=423252​5
Original post by point.ms
Theorem - The contact numbers is sorted horizontally to be the only one natural straight line that gives a natural straight line , when there are two (more) results between them becomes a gap.


PROOF -11(1)1\rightarrow1(\underline{1})

4+202=24{+_2^{\underline0}}2=2
4+212=1214{+_2^{\underline1}}2=1 \underline{2}1
4+222=24{+_2^{\underline2}}2=2
4+232=3114{+_2^{\underline3}}2=3 \underline{1}1
4+242=64{+_2^{\underline4}}2=6


+2 - addition rule 2


(CM.) - No "addition rule 2"
more complex
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = {0,1}\cup{2,3}\cup{6,7}\cup{8,10} my solution and notation 42322+203254\underline23\underline22{+_{2}^{\underline0}}3 \underline25=111311​2
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} = {0,1}\cup{9,13} my solution and notation 42322+213254\underline23\underline22{+_{2}^{\underline1}}3 \underline25=18​4
{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} ={0,2}\cup{4,5}\cup{6,7}\cup{9,11}\cup{12,13} my solution and notation 42322+223254\underline23\underline22{+_{2}^{\underline2}}3 \underline25=22111221​1
...

{0,4}\cup{6,9}\cup{11,13} ? {0,3}\cup{5,10} ={0,4}\cup{6,9}\cup{11,16} \cup{18,23} my solution and notation 42322+2133254\underline23\underline22{+_{2}^{\underline{13}}}3 \underline25=423252​5



Options[plus] = {"Output" -> "Standard", "Operation" -> 1};


plus[a_, b_, c_, OptionsPattern[]] :=

With[{ans = BitXor[a, c 2^b]}, Switch[OptionValue["Operation"],
1, If[OptionValue["Output"] == "Standard", Identity, toEncoded]@ans,
2, If[Mod[ans, 2] == 0, Rest, Identity]@
runlength[IntegerDigits[ans, 2] // Reverse]
]
]


runlength[bitlist_] := Length /@ Split[bitlist]


toEncoded[n_] := runlength[Reverse@IntegerDigits[n, 2]][[1 ;; -1 ;; 2]]


fromEncoded[a_] :=
FromDigits[
Reverse[Flatten[
ConstantArray @@@
Transpose[{PadRight[{}, Length[a], {1, 0}], a}]]], 2]


toEncoded[n_] :=
Length /@ Select[Split[Reverse@IntegerDigits[n, 2]], First[#] == 1 &]

Example, matching your last example:

In[120]:= plus[fromEncoded[{4, 2, 3, 2, 2}], 13,
fromEncoded[{3, 2, 5}], "Operation" -> 2]


Out[120]= {4, 2, 3, 2, 5, 2, 5}


You are still computing the same quantity; just giving your answer in a slightly different way.
I'm glad you two are having fun.

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