msbiljanica Posted January 22, 2012 Report Share Posted January 22, 2012 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 [email protected]natural axiomWhat 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 Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted January 22, 2012 Author Report Share Posted January 22, 2012 NATURAL MATHEMATICS Presupposition-natural long merge points in the direction of the first natural along ABProcess: 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 Quote Link to comment Share on other sites More sharing options...
sigurdV Posted January 23, 2012 Report Share Posted January 23, 2012 I dont understand, please make a step by step repeat. Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted January 23, 2012 Author Report Share Posted January 23, 2012 I dont understand, please make a step by step repeat.This presentation is a step by step, each new piece of evidence derives from "natural along" or previous evidence, it can be easier than thisPresupposition-Numbers have their pointsProcess: 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___________________________________________________________Presupposition-numbers have opposite pointsProcess: 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 Quote Link to comment Share on other sites More sharing options...
sigurdV Posted January 23, 2012 Report Share Posted January 23, 2012 This presentation is a step by step, each new piece of evidence derives from "natural along" or previous evidence, it can be easier than thisPresupposition-Numbers have their pointsProcess: 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___________________________________________________________Presupposition-numbers have opposite pointsProcess: 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 I still dont understand it all, but now i begin to think there might be something to understand. Since you have an interest in maths, perhaps you also care for logic? If so, I invite you to the thread "The Final Solution of the Liar".(#1) Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted January 24, 2012 Author Report Share Posted January 24, 2012 Presupposition-numbers are comparable with each otherProcess: 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 otherP3-four numbers (a, b, c, d) are comparable with each other... [s8]-comparability numbersCM-[s8]known two of comparability, comparability of three numbers(a number comparable with the numbers b and c), comparability of the other knows.___________________________________________________________________________________________Presupposition-number ranges for number alongProcess:P1-imageP2-imageP3-image[s9]-mobility of numberCM-[s9]-does not know Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted January 25, 2012 Author Report Share Posted January 25, 2012 Presupposition-Number (a) and mobile number (b ) of a contat ,mergeProces:P1 3+(.0/.0)2=3P2 3+(.1/.0)2=3P3 3+(.2/.0)2=4 - imageP4 3+(.3/.0)2=5 designation means the more we write (.0/.0) , unless specifically indicated to beP1 3+(.0)2=3P2 3+(.1)2=3P3 3+(.2)2=4P4 3+(.3)2=5 General forma+(.0)b=ca+(.1)b=c...a+(.d)b=c[s10]-additionCM-only form a+(s.0/.0)b=c , others do not know , axiom______________________________________Presupposition-Number (a) and mobile number (b ) but have no contact with the itemProcess:P1 ¤3(0)2¤P2 ¤3(1)2¤P3 ¤3(2)2¤...Next - gap number and mobile number have no contact , except to point...[s11]-gap number GN={¤a(b)c¤,...,¤g(f)...(d)e¤}[s12]-gap alongDefinition[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______________________________________________Third Now we can describe ( c ) the number of gaps ,¤10m(5m)5m¤PDF - http://www.mediafire.com/file/2c7vvarvvqaqp63/srdanova math.pdf Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted January 26, 2012 Author Report Share Posted January 26, 2012 Presupposition-Gap number is comparable with the gap number and numberProcess:P1 ¤a(b)c¤ , a+(s.0)c=zP2 ¤a(b)c(d)e¤ , a+(s.0)c+(s.0)e=zP3 ¤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 numbersCM-[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)bb ) a+(s.0)b+...+(s.0)b,...,a+(s.0)b+(s.0)b, a+(s.0)b , a+(s.0)0Process: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 , 743General form -abc , ab[s14]-srckoCM-[s14]-does no know Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted January 27, 2012 Author Report Share Posted January 27, 2012 Presupposition-Srcko can join a number not that can not be in the structure srckoProcess:P1 101070 and 5 , 5_101070P2 5520 and 22 ,5520_22P3 75 and 25 , 75_25P4 68 and 2 ,2_68...General form -abc_d , d_abc , ab_d ,d_ab...[s15]-pendant srckoCM-[s15]-does no knowNote-only one number can be pendand , number two goes into a complex srcko__________________________________________________________Presupposition-Two ( more ) srcko (pendand srcko) are combined into one unitProcess:P1 106 and 118 , 106118P2 10565 and 703 ,10565_703P3 30360 and 45277_78 ,30360_45277_78...General form -abcd , abc_de ,abc_def_g ,...[s16]-two ( more) srckoCM-[s16]-does no know Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted January 28, 2012 Author Report Share Posted January 28, 2012 Presupposition-Two ( more ) srcko have the first ( last) common numberProcess: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 srckoCM-[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<91P2 5+(.0)5=5+(s.0)224 , 5+(.0)5<729...General form - a+(.b)c=d+(s.0)11 ,a+(s.b)c<e1a+(.b)c=d+(s.0)e , a+(.b)c<fa+(.b)c=d+(s.0)efg , a+(.b)c<hij ...[s18]-left inequalityCM-[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 srcko5550={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 srcko55383(_50_)1090792remains part of the function, when we come to it Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted January 29, 2012 Author Report Share Posted January 29, 2012 Presupposition-Parts number (a) and mobile number (b ) have a contact , the contact is deleteProcess:P1 4-(.0)2=2P2 4-(.1)2=¤1(2)1¤ imageP3 4-(.2)2=2P4 4-(.3)2=¤3(1)1¤P5 4-(.4)2=¤4(0)2¤General forma-(.0)b=ca-(.1)b=c...a-(.d)b=c[s19]-subtractionCM-only form a-(s.0/s.0)b=c , others do not know , axiom_______________________________________________________________________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>017P2 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>01ea+(.b)c=d-(s.0/s.0)e , a+(.b)c>fa+(.b)c=d-(s.0/s.0)efg , a+(.b)c>hij ...[s20]-right inequality additionCM-[s20]-know Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted January 30, 2012 Author Report Share Posted January 30, 2012 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<yP2 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<ya+(.bcd_e)f=y , a+(.bcd_e)f>y , a+(.bcd_e)f<y ,...[s21]-function additionCM-[s21]-does no know______________________________________________________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 numberCM-[s22]-does no know___________________________PDF - http://www.mediafire.com/?jzheej0qu93jjkm Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted January 31, 2012 Author Report Share Posted January 31, 2012 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 additionCM-[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 subtractionCM-[s24]-does no know Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted February 1, 2012 Author Report Share Posted February 1, 2012 Presupposition-In the expression a-(.b)c=d , d+(s.0))11 or d+(s.0)number (more) from 11and d+(.z)11¤¤e or d+(.z) number ( more ) from 11¤¤e , e={(f),(f)(f),(f)(f)(f),...}Process: P1 4-(.2)2=2+(s.0)11 , 4-(.2)2<31 4-(.3)2=¤1(2)1¤+(.z)11¤¤(2) , 4-(.3)2<2¤¤(2)+(.z)11¤¤(2) , 4-(.3)2<31¤¤(2)P2 4-(.2)2=2+(s.0)4 , 4-(.2)2<6 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-(s.0)11 ,a+(s.b)c<g1a-(.b)c=d+(s.0)g , a+(.b)c<la-(.b)c=d+(s.0)kpg , a+(.b)c<hij ...a-(.b)c=d+(.z)11¤¤e , a-(.b)c<s¤¤e+(.z)11¤¤e , a+(s.b)c<g1¤¤ea-(.b)c=d+(.z)g¤¤e , a-(.b)c<s¤¤¤e+(.z)g¤¤e , a+(.b)c<l¤¤ea-(.b)c=d+(.z)kpg¤¤e , a-(.b)c<s¤¤e+(.z)kpg¤¤e , a+(.b)c<hij¤¤e ...[s25]-left inequality subtractionCM-[s25]-know , forms without any gaps numbers not known____________________________________________________________________Presupposition-In the expression a-(.b)c=d , d-(s.0/s.0))11p or d+(s.0/s.0)number (more) from 11pand d-(.z)11p¤¤e or d-(.z) number ( more ) from 11¤¤e , e={(f),(f)(f),(f)(f)(f),...}Process: P1 4-(.2)2=2-(s.0/s.0))112 , 4-(.2)2>011 4-(.3)2=¤1(2)1¤-(.z)211¤¤(2) , 4-(.3)2>2¤¤(2)-(.z)211¤¤(2) , 4-(.3)2>011¤¤(2)P2 4-(.2)2=2-(s.0/s.0))1 , 4-(.2)2>1 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-(s.0/s.0)11p ,a-(s.b)c>g1la-(.b)c=d-(s.0/s.0))g , a-(.b)c>la-(.b)c=d-(s.0/s.0))ksg , a-(.b)c>hij ...a-(.b)c=d-(.z)11p¤¤e , a-(.b)c>s¤¤e-(.z)11p¤¤e , a-(s.b)c>g1k¤¤ea-(.b)c=d-(.z)g¤¤e , a-(.b)c>s¤¤e-(.z)g¤¤e , a-(.b)c>l¤¤ea-(.b)c=d-(.z)kpg¤¤e , a-(.b)c>s¤¤e-(.z)kpg¤¤e , a-(.b)c>hij¤¤e ...[s26]-right inequality subtractionCM-[s26]-know , forms without any gaps numbers not known Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted February 2, 2012 Author Report Share Posted February 2, 2012 Presupposition-Two ( more) subtraction (left and right inequalities) can be short to write Process:P1 7-(.013)4=y , 7-(.013)4>y ,7-(.013)4<yP2 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<ya-(.bcd_e)f=y , a-(.bcd_e)f>y , a-(.bcd_e)f<y ,...[s27]-function subtractionCM-[s27]-does no know_____________________________________Presupposition-Parts number (a) and mobile number (b ) have a contact , contact remains , therest is deleteProcess:P1 4 - (.0)2=2P2 4 - (.1)2=2 imageP3 4 - (.2)2=2P4 4 - (.3)2=1P5 4 - (.4)2=1General forma - (.0)b=ca - (.1)b=c...a - (.d)b=c [s28]-opposite subtractionCM-[s28]-does no know-sign for. opposite subtraction (when you download the following PDF you will see how it looks) Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted February 3, 2012 Author Report Share Posted February 3, 2012 3.how to solve this current knowledge of mathematics: along a (20m) ,deleted between 10 m and 15 m (b=5m) , wet get c 20m-(.5m)=¤10m(5m)5m¤___________________________________________________Presupposition-In the expression a - (.b)c=d , d+(s.0))11 or d+(s.0)number (more) from 11and d+(.z)11¤¤e or d+(.z) number ( more ) from 11¤¤e , e={(f),(f)(f),(f)(f)(f),...}Process: P1 5 - (.0)5=5+(s.0)11 , 5 - (.0)5<61 ¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤+(.z)11¤¤(2) ¤2(2)2¤ - (.1)¤1(1)2¤<2¤¤(2)+(.z)11¤¤(2) , ¤2(2)2¤ - (.1)¤1/1)2¤<31¤¤(2)P2 5 - (.0)5=5+(s.0)4 , 5 - (.0)5<9 ¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤+(.z)7¤¤(2) ¤2(2)2¤ - (.1)¤1(1)2)¤<2¤¤(2)+(.z)7¤¤(2), ¤2(2)2¤ - (.1)¤1(1)2¤<9¤¤(2)...General form a - (.b)c=d-(s.0)11 ,a - (s.b)c<g1a - (.b)c=d+(s.0)g , a - (.b)c<la - (.b)c=d+(s.0)kpg , a - (.b)c<hij ...a -(.b)c=d+(.z)11¤¤e , a - (.b)c<s¤¤e+(.z)11¤¤e , a - (s.b)c<g1¤¤ea - (.b)c=d+(.z)g¤¤e , a - (.b)c<s¤¤¤e+(.z)g¤¤e , a - (.b)c<l¤¤ea - (.b)c=d+(.z)kpg¤¤e , a - (.b)c<s¤¤e+(.z)kpg¤¤e , a - (.b)c<hij¤¤e ...[s29]-left inequality opposite subtractionCM-[s29]-does no know_________________________________Presupposition-In the expression a - (.b)c=d , d-(s.0/s.0))11p or d+(s.0/s.0)number (more) from 11p and d-(.z)11p¤¤e or d-(.z) number ( more ) from 11¤¤e , e={(f),(f)(f),(f)(f)(f),...}Process: P1 5 - (.0)5=5 -(s.0/s.0))115 , 5 - (.0)5>014 ¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤-(.z)211¤¤(2) ¤2(2)2¤ - (.1)¤1(1)2¤>2¤¤(2)-(.z)211¤¤(2) , 4¤2(2)¤ - (.1)¤1(1)2¤>011¤¤(2)P2 5 - (.0)5=5-(s.0/s.0))1 , 5 - (.0)5>1 ¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤-(.z)1¤¤(2) ¤2(2)2¤ - (.1)¤1(1)2¤>2¤¤(2)-(.z)1¤¤(2) , ¤2(2)2¤ - (.1)¤1(1)2¤>1¤¤(2)...General form a - (.b)c=d-(s.0/s.0)11p ,a - (s.b)c>g1la - (.b)c=d-(s.0/s.0))g , a - (.b)c>la - (.b)c=d-(s.0/s.0))ksg , a - (.b)c>hij ...a - (.b)c=d-(.z)11p¤¤e , a - (.b)c>s¤¤e-(.z)11p¤¤e , a - (s.b)c>g1k¤¤ea - (.b)c=d-(.z)g¤¤e , a - (.b)c>s¤¤e-(.z)g¤¤e , a - (.b)c>l¤¤ea - (.b)c=d-(.z)kpg¤¤e , a - (.b)c>s¤¤e-(.z)kpg¤¤e , a - (.b)c>hij¤¤e ...[s30]-right inequality opposite subtractionCM-[s30]-does no know_______________________________-PDF - http://www.mediafire.com/?qle11bvr2yuey9o Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted February 4, 2012 Author Report Share Posted February 4, 2012 Presupposition-Two ( more) opposite subtraction (left and right inequalities) can be short to write Process:P1 7 - (.013)4=y , 7 - (.013)4>y ,7 - (.013)4<yP2 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<ya - (.bcd_e)f=y , a - (.bcd_e)f>y , a - (.bcd_e)f<y ,...[s31]-function opposite subtractionCM-[s31]-does no know__________________________________Presupposition-Two ( more) addition the same number can be written in shorted form Process:P1 a+(s.c)a=a×(s.c)2 , a×(s.c)bP2 a+(s.c)a+(s.c)a=a×(s.c)3 , a×(s.c)bP3 a+(s.c)a+(s.c)a+(s.c)a=a×(s.c)4 , a×(s.c)b...[s32]-multiplicationCM-[s32]-know , axiomPresupposition-In terms a×(s.c)b , b can be number 0 (1) Process:P1 a×(s.c)0P2 a×(s.c)1[s32a]-multiplication-amendment Quote Link to comment Share on other sites More sharing options...
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