questions 1.Z÷(10^n)=?,Z-integers 2.write in abbreviated form (if the function can be final and natural) 2+5=7 , 2+10=12 , 2+15=17, 2+20=22 , 2+25=27 , 2+30=32 , 2+35=37 , 2+38=40, 2+40=42, 2+41=43 , 2+44=46 , 2+45=47, 2+47=49 , 2+50=52 ,2+57=59 , 2+60=62 , 2+64=66, 2+70=72, 2+71=73 , 2+78=80 , 2+80=82 , 2+85=87 , 2+90=92 ,2+92=94 3.how to solve this current knowledge of mathematics: along a (20m) ,deleted between 10 m and 15 m (b=5m) , wet get c (image) Can mathematics explain the only two axiom that the rest are just evidence (experiments), if you think so join me show you Srdanova math, see you ________________________________________________________________ I figure this way, from education school has 12 years, I was always the subject of mathematics and physics had the best grades, math deal amateur, studying mathematics I came to know that mathematics can be simplified and be connected (to be explained only with two axiom) and extend the mathematics that can solve math problems that present no solution. Marjanovic Srdan M.Biljanica 16201 Manojlovce Serbia <SNIP> natural axiom What is " nature along "? -nature along in figure 1 What is "point"? -start (end) natural long in figure 2 What is the " basic rule "? -basic rule is determined that the two ( more) longer only have to connect the points [Sn]-mathematical facts [S1]-nature along [S2]-point (natural meaning of) Definition[natural along]-two points , distance between two points CM (current mathematics)-[S1]-does not know , [S2]-point is not defined , so anything and everything
Hello msbiljanica and welcome to All About Circuits. I am not quite sure what points you are making here. Could you separate them and post them one at a time?
Thank you for the welcome, this is like a like Euclid's Elements, in the simplest form, as geometric objects (the first natural geometrical object along the notion of point, the basic rule), all other evidence (slightly different than the current mathematics), which arise as a ratio of two (more) geometrical object formed from natural or along previous geometrical object. the point is that I think all of their math geometry, and that the numbers are a calculation function is actually another name for the related geometrical object ___________________ NATURAL MATHEMATICS Presupposition-natural long merge points in the direction of the first natural along AB Process: P1-AB..CD..ABC(AC) to read- natural along AB to point B, is connected to the natural long CD to point C, shall be P2-ABC(AC)..DE..ABCD(AD) read- along the ABC(AC) to point C , connecting with the natural long DE to point D is done renaming of points , we get along ABCD(AD) P3-ABCD(AD)..EF..ABCDE(AE) ... [S3]-along (natural basis) Definition[along]-the first and last point and the distance between points CM-[S3]-does not know __________________________________________________ Presupposition - All points of a longer (the infinite form) can be replaced with labels: (0), (0,1 ),..., (0,1,2,3,4,5,6,7,8,9 ),... Process: P1-N (0) = {0,00,000,0000,...} P2-N (0,1) = {0,1,10,11,100,...} ... P10-N (0,1,2,3,4,5,6,7,8,9) = {0,1,2,3,4,5,6,7,8,9,10,11, ...} ... [S4]-number along [S5]-set of natural numbers N We will use N (0,1,2,3,4,5,6,7,8,9) = {0,1,2,3,4,5,6,7,8,9,10,11,12,...} Definition[number along]- a starting point (0), the last point at infinity [number N]-The number 0 is the point 0 -Other numbers are longer, the first item is 0, the last point is the point of the name (number) CM-[S4].does not know , [S5]-axiom ___________________________________ Presupposition-Numbers have their points Process: P1 0=(.0) P2 1={(.0),(.1)} P3 2={(.0),(.1),(.2)} P4 3={(.0),(.1),(.2),(.3)} P5 4={(.0),(.1),(.2),(.3),(.4)} ... [S6]-number points CM-[S6]does not know
In addition to the axioms of Euclid's Geometry, which were stated without proof there were a number of other supporting statements, also without proof. In all there were 23 Definitions 5 Postulates 5 Common Notions or Axioms There were also axioms added after Euclid, eg "two straight lines do not enclose or contain a space." Any new system will need to provide equivalents to all of these, which I suspect will add up to rather more than two statements made without proof. You seem to be putting points on a line into one to one correspondence with the integers. It is easy to show that there are point on a continuous line for which this cannot be done.
Presupposition-numbers have opposite points Process: P1 0=(s.0) P2 1={(s.0),(s.1)} P3 2={(s.0),(s.1),(s.2)} P4 3={(s.0),(s.1),(s.2),(s.3)} P5 4={(s.0),(s.1),(s.2),(s.3),(s.4)} ... ... [S7]-number opposite points CM-[S7]does not know _______________________________ Presupposition-numbers are comparable with each other Process: P1-two numbers (a, b ) are comparable with each other - a> b, a =b, a <b, ).(=(>,=,<) P2-three numbers (a, b, c) are comparable with each other P3-four numbers (a, b, c, d) are comparable with each other ... [S8]-comparability numbers CM-[S8]known two of comparability, comparability of three numbers(a number comparable with the numbers b and c), comparability of the other knows.
Please clear up something for me. By "does not know" are you referring to what we call an open or closed set or interval? An interval is part of a line or the set of all numbers in that part of a line. An open interval does not contain its end points A closed interval does So for instance the open interval 0, 1 contains all the numbers between 0 and 1 but not 0 or 1 themselves. So a closed interval contains all the numbers between 0 and 1 and 0 and 1 themselves. We write the open inteval 0,1 as ] 0 , 1 [ and the closed interval 0,1 as [ 0 , 1 ] to distinguish them.
calculation, or the name of a geometry object, and the rest, which does not exist in the current mathematics
Thanks for your last comments. Reading your other posts again is helping a little but I am still not sure why you are trying to identify natural numbers with lines or is it points on a line? As I pointed out in post#6 although all the numbers in N have equal significance, there are two types of points on a line and these types do not have equal significance.
numbers are another name for the point (0) and longer (1,2,3,4, ....) of my math ____________________________ Presupposition-number ranges for number along Process: P1-image P2-image P3-image [S9]-mobility of number CM-[S9]-does not know ______________ Presupposition-Number (a) and mobile number (b ) but have no contact with the item Process: P1 ¤3(0)2¤ P2 ¤3(1)2¤ P3 ¤3(2)2¤ ... Next - gap number and mobile number have no contact , except to point ... [S10]-gap number GN={¤a(b)c¤,...,¤g(f)...(d)e¤} [S11]-gap along Definition[gap along] a>0-¤0(0)0¤-point -¤0(a)0¤-two point ,separated by a gap -¤a(0)0¤,¤0(0)a¤,¤a(0)a¤-along , two points -¤a(a)a¤-two along , 4 points ... CM-[S11],[S12] -does no know ________________ Presupposition-Number (a) and mobile number (b ) of a contat ,merge Proces: P1 3+(.0/.0)2=3 P2 3+(.1/.0)2=3 P3 3+(.2/.0)2=4 - image P4 3+(.3/.0)2=5 form (.a/0) and (s.a/0) well continue this write (.a) and (s.a) P1 3+(.0)2=3 P2 3+(.1)2=3 P3 3+(.2)2=4 P4 3+(.3)2=5 General form a+(.0)b=c a+(.1)b=c ... a+(.d)b=c [S12]-addition CM-only form a+(s.0/.0)b=c , others do not know , axiom
What I understand you are saying is Take a stick of length 4 units. Place another stick of length 2 units on one end or altenatively cut the first stick into two pieces and place the second stick between the pieces of the first. Either way the result will be a length of 6 units? Please correct this if I have missed something. As regards the motivation for this, You have used (assumed) several underlying properties from set theory about the natural numbers. 1) If a and b are members of the set N then a+b = c is also a member of the set. Furthermore c is unique. There is no other member, say d, for which a + b = d. 2) N contains a member 0 such that a + 0 = a for every member of N There is only one member 0 is unique. 3) N contains a member (-a) such that a + (-a) = 0 for every member of N Again (-a) is unique
Presupposition-Gap number is comparable with the gap number and number Process: P1 ¤a(b)c¤ , a+(s.0)c=z P2 ¤a(b)c(d)e¤ , a+(s.0)c+(s.0)e=z P3 ¤a(b)c(d)e(f)g¤ ,a+(s.0)c+(s.0)e+(s.0)g=z ... number z as it compares as a number of [S13]-comparability gap numbers CM-[S13]-does no know __________________________________________________ Presupposition-Adding the result can be written in short form: a) a+(s.0)0 , a+(s.0)b , a+(s.0)b+(s.0)b , ...,a+(s.0)b+...+(s.0)b b ) a+(s.0)b+...+(s.0)b,...,a+(s.0)b+(s.0)b, a+(s.0)b , a+(s.0)0 Process: P1 - 3+(s.0)0=3 , 3+(s.0)4=7 , 347 3+(s.0)0=3 , 3+(s.0)4=7 , 3+(s.0)4+(s.0)4=11 , 3411 3+(s.0)0=3 , 3+(s.0)4=7 , 3+(s.0)4+(s.0)4=11 , 3+(s.0)4+(s.0)4+(s.0)4=15 , 3415 ... 3+(s.0)0=3 , ... , 3+(s.0)4+...+(s.0)4=d , 34 P2 - 3+(s.0)4+(s.0)4+(s.0)4=15 , 3+(s.0)4+(s.0)4=11 , 3+(s.0)4=7 , 3+(s.0)0=3 , 1543 ... 3+(s.0)4=7 , 3+(s.0)0=3 , 743 General form -abc , ab [S14]-srcko CM-[S14]-does no know _______________________________________ Presupposition-Srcko can join a number not that can not be in the structure srcko Process: P1 101070 and 5 , 5_101070 P2 5520 and 22 ,5520_22 P3 75 and 25 , 75_25 P4 68 and 2 ,2_68 ... General form -abc_d , d_abc , ab_d ,d_ab... [S15]-pendant srcko CM-[S15]-does no know Note-only one number can be pendand , number two goes into a complex srcko
Presupposition-Two ( more ) srcko (pendand srcko) are combined into one unit Process: P1 106 and 118 , 106118 P2 10565 and 703 ,10565_703 P3 30360 and 45277_78 ,30360_45277_78 ... General form -abcd , abc_de ,abc_def_g ,... [S16]-two ( more) srcko CM-[S16]-does no know _______________________________ Presupposition-Two ( more ) srcko have the first ( last) common number Process: P1 10530 and 3330 , 10533(_30) P2 4444 and 441094 and 44256 , 44(_44_)1094256 ... General form -abcd(_e) , ab(_c_)defg , ... [S17]-two ( more) first-last srcko CM-[S17]-does no know ______________________________________________________ Presupposition-In the expression a+(.b)c=d , d+(s.0)11 or d+(s.0)number (more) from 11 Process: P1 3+(.s.0)5=8+(s.0)11 , 3+(s.0)5<91 P2 5+(.0)5=5+(s.0)224 , 5+(.0)5<729 ... General form - a+(.b)c=d+(s.0)11 ,a+(s.b)c<e1 a+(.b)c=d+(s.0)e , a+(.b)c<f a+(.b)c=d+(s.0)efg , a+(.b)c<hij ... [S18]-left inequality CM-[S18]-know _______________________________________ 2+5=7 , 2+10=12 , 2+15=17, 2+20=22 , 2+25=27 , 2+30=32 , 2+35=37 , 2+38=40, 2+40=42, 2+41=43 , 2+44=46 , 2+45=47, 2+47=49 , 2+50=52 ,2+57=59 , 2+60=62 , 2+64=66, 2+70=72, 2+71=73 , 2+78=80 , 2+80=82 , 2+85=87 , 2+90=92 ,2+92=94 srcko 5550={5,10,15,20,25,30,35,40,45,50} 38350={38,41,44,47,50} 501090={50,60,70,80,90} 50792={50,57,64,71,78,85,92} two(more) first-last srcko 55383(_50_)1090792 remains part of the function, when we come to it
I'm sorry but you have lost me somewhere along the way. It would be very helpful if you could answer one my basic questions rather than ploughing on with more of your maths. Otherwise I fear you are doing a lot of work for nothing. Please state what these two axioms are and how far they extend. Please also state what problems conmventional maths cannot solve, but your system can, so that a comparison can be made. That is pose a conventional maths problem in a conventional way.
You know the first axiom is called the natural along ,term point, the basic rule for linking the two (more) natural along. second axiom when we come to it all I ever show starts along the natural, or what became of natural along 2.write in abbreviated form (if the function can be final and natural) 2+5=7 , 2+10=12 , 2+15=17, 2+20=22 , 2+25=27 , 2+30=32 , 2+35=37 , 2+38=40, 2+40=42, 2+41=43 , 2+44=46 , 2+45=47, 2+47=49 , 2+50=52 ,2+57=59 , 2+60=62 , 2+64=66, 2+70=72, 2+71=73 , 2+78=80 , 2+80=82 , 2+85=87 , 2+90=92 ,2+92=94 5550={5,10,15,20,25,30,35,40,45,50} 38350={38,41,44,47,50} 501090={50,60,70,80,90} 50792={50,57,64,71,78,85,92} two(more) first-last srcko 55383(_50_)1090792 remains part of the function, when we come to it a question for you if a function can be final and natural ?advantage of my mathematics 1.Z÷(10^n)=?,Z-integers a={0,1,2,3,4,5,6,7,8,9} , b={1,2,3,4,5,6,7,8,9} n=1 , Z÷10={...,(-2÷10),(-1÷10),(0÷10),(1÷10),(2÷10),...}={...,-0.2,-0.1,0,0.1,0.2,...}={Z,Z.b} n=2 , Z÷100={Z,Z.b,Z.ab} n=3 , Z÷1000={Z,Z.b,Zab,Zaab} n=4 , Z÷10000={Z,Z.b,Z.ab,Zaab,Zaaab} ... Z÷(10^n)=R This evidence shows that the same rational and real numbers, that there are irrational numbers
Presupposition-Parts number (a) and mobile number (b ) have a contact , the contact is delete Process: P1 4-(.0)2=2 P2 4-(.1)2=¤1(2)1¤ image P3 4-(.2)2=2 P4 4-(.3)2=¤3(1)1¤ P5 4-(.4)2=¤4(0)2¤ General form a-(.0)b=c a-(.1)b=c ... a-(.d)b=c [S19]-subtraction CM-only form a-(s.0/s.0)b=c , others do not know , axiom _______________________________ 3.how to solve this current knowledge of mathematics: along a (20m) ,deleted between 10 m and 15 m (b=5m) , wet get c (image) 20m-(.10m)5m=¤10m(5m)5m¤ How would you solve with current knowledge of mathematics _________________-- Presupposition-In the expression a+(.b)c=d , d-(s.0/s.0))11f or d-(s.0/s.0))number (more) from 11f Process: P1 3+(.s.0)5=8-(s.0/s.0))118 , 3+(s.0)5>017 P2 5+(.0)5=5-(s.0/s.0))224 , 5+(.0)5>321 ... General form - a+(.b)c=d-(s.0/s.0))11f ,a+(s.b)c>01e a+(.b)c=d-(s.0/s.0)e , a+(.b)c>f a+(.b)c=d-(s.0/s.0)efg , a+(.b)c>hij ... [S20]-right inequality addition CM-[S20]-know _________________________ Presupposition-Two ( more) addition (left and right inequalities) can be short to write Process: P1 3+(.013)4=y , 3+(.013)4>y ,3+(.013)4<y P2 8+(.228)5=y , 8+(.228)5>y ,8+(.228)5<y ... General form - a+(.bcd)e=y ,a+(.bcd)e>y , a+(.bcd)e<y a+(.bcd_e)f=y , a+(.bcd_e)f>y , a+(.bcd_e)f<y ,... [S21]-function addition CM-[S21]-does no know
Do I understand correctly by saying that I think you mean Find an expression for Z such that Z is an integer or Find all integers, Z, such that What do you mean by a function, a final function and a natural function?
I'm sorry I don't understand your statement. If you are asserting that the set of all integers, whether divided by some power of 10 or not, is isomorphic to the set of all real numbers This is not the case. There are more numbers in R than there are in Z.
Presupposition-Gap number ( value z) is a variable (same value) with the gaps that is constant Process: P1 ¤5(2)0¤,¤4(2)1¤,3(2)2¤,¤2(2)3¤,¤1(2)4¤,¤0(2)5¤ --5¤¤(2) P2 ¤3(2)0¤,¤2(2)1¤,¤1(2)2¤,¤0(2)3¤ ¤2(2)0¤,¤1(2)1¤.¤0(2)2¤ ¤1(2)0¤,¤0(2)1¤ ¤0(2)0¤ --013¤¤(2) ¤2(2)0¤,¤1(2)1¤,¤0(2)2¤ ¤3(2)0¤,¤2(2)1¤,¤1(2)2¤,¤0(2)3¤ ... 21¤¤(2) [S22]-variability of z number CM-[S22]-does no know _______________ Presupposition-Translation of gap number in the variability z ( with constant gap ) can addition (s.0) Process: P1 ¤3(2)3¤+(.z)¤4(2)4=6¤¤(2)+(.z)8¤¤(2)=14¤¤(2) P2 ¤1(6)1(9)1¤+(.z)¤3(6)2(9)1¤=3¤¤(6)(9)+(.z)6¤¤(6)(9)=9¤¤(6)(9) ... General form ¤a(b)c¤+(.z)¤d(b)e¤=f¤¤(b )+(.z)g¤¤(b )=h¤¤(b ) ... [S23]-z addition CM-[S23]-does no know __________________________________________ Presupposition-Translation of gap number in the variability z ( with constant gap ) can subtraction (s.0/s.0) Process: P1 ¤3(2)3¤-(.z)¤1(2)1=6¤¤(2)-(.z)2¤¤(2)=4¤¤(2) P2 ¤3(6)2(9)1¤-(.z)¤1(6)1(9)1¤=6¤¤(6)(9)-(.z)3¤¤(6)(9)=3¤¤(6)(9) ... General form ¤a(b)c¤-(.z)¤d(b)e¤=f¤¤(b )-(.z)g¤¤(b )=h¤¤(b ) ... [S24]-z subtraction CM-[S24]-does no know
Presupposition-In the expression a-(.b)c=d , d+(.z)11¤¤e or d+(.z) number ( more ) from 11¤¤e , e={(f),(f)(f),(f)(f)(f),...} Process: P1 4-(.3)2=¤1(2)1¤+(.z)11¤¤(2) , 4-(.3)2<2¤¤(2)+(.z)11¤¤(2) , 4-(.3)2<31¤¤(2) P2 4-(.3)2=¤1(2)1¤+(.z)7¤¤(2) , 4-(.3)2<2¤¤(2)+(.z)7¤¤(2), 4-(.3)<9¤¤(2) ... General form a#(.b)c=d+(.z)11¤¤e , a#(.b)c<s¤¤e+(.z)11¤¤e , a#(s.b)c<g1¤¤e a#(.b)c=d+(.z)g¤¤e , a#(.b)c<s¤¤¤e+(.z)g¤¤e , a#(.b)c<l¤¤e a#(.b)c=d+(.z)kpg¤¤e , a#(.b)c<s¤¤e+(.z)kpg¤¤e , a#(.b)c<hij¤¤e ..., #-calculation operations (+,-,×,..) [S25]-left inequality gap number CM-[S25]-does no know _____________________________________ Presupposition-In the expression a-(.b)c=d , d-(.z)11p¤¤e or d-(.z) number ( more ) from 11p¤¤e ,e={(f),(f)(f),(f)(f)(f),...} Process: P1 4-(.3)2=¤1(2)1¤-(.z)211¤¤(2) , 4-(.3)2>2¤¤(2)-(.z)211¤¤(2) , 4-(.3)2>011¤¤(2) P2 4-(.3)2=¤1(2)1¤-(.z)1¤¤(2) , 4-(.3)2>2¤¤(2)-(.z)1¤¤(2) , 4-(.3)>1¤¤(2) ... General form a-(.b)c=d-(.z)11p¤¤e , a-(.b)c>s¤¤e-(.z)11p¤¤e , a-(s.b)c>g1k¤¤e a-(.b)c=d-(.z)g¤¤e , a-(.b)c>s¤¤e-(.z)g¤¤e , a-(.b)c>l¤¤e a-(.b)c=d-(.z)kpg¤¤e , a-(.b)c>s¤¤e-(.z)kpg¤¤e , a-(.b)c>hij¤¤e .. ,#-calculation operations (+,-,×,...) [S26]-right inequality gap number CM-[S26]-does no know _______________________________ Presupposition-The numbers are added , contact remains the rest is deleted Process: P1 4 - (.0)2=2 P2 4 - (.1)2=2 image P3 4 - (.2)2=2 P4 4 - (.3)2=1 P5 4 - (.4)2=0 P1 ¤1(1)2¤ - (.0)2=1 P2 ¤1(1)2¤ - (.1)2=1 image P3 ¤1(1)2¤ - (.2)2=2 P4 ¤1(1)2¤ - (.3)2=1 P5 ¤1(1)2¤ - (.4)2=0 General form a - (.0)b=c a - (.1)b=c ... a - (.d)b=c [S27]-opposite subtraction CM-[S27]-does no know