SRDANOVA MATHEMATICS Revision of the current mathematics View attachment 4479 Marjanovic Srdan M.Biljanica 16201 Manojlovce Serbia ms.biljanica@gmail.com Introduction: I think that the current limited maths and sinful and should be reviewed with all new things that I discovered. I will explain the mathematical space with two starting points ( along the natural and real ). Natural Base: Natural along is what you see along the fig,1.Natural along has its beginning and its end , this property natural long we will contact points ( fig.2).Natural length along the ground ( natural meaning).Two more natural and longer merge points View attachment 4480 [S1]-along nature (fig. 1-a), [Sn]-mathematical facts [S2]-point ( natural meaning , Fig.2 -A(B)) The points will mark capital letters along the (length) small letters Definition of - teo points A,B , the length between points AB CM (current mathematics) - does not recognize the concept of nature along , the point is not defined so that all l everything _____________________________________________________________ Presupposition-Natural long - merge points in the direction 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 renaming of points, we get along ABC (AC) P2-ABC (AC) .. DE ..ABCD (AD) read it: 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) ... to read: natural along AB to point B, is connected to the natural long CD to point C, shall be renaming of points, we get along ABC (AC) P2-ABC (AC) .. DE ..ABCD (AD) read it: 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) ... View attachment 4481 [S3]-along (from natural long, two or more) Definition of the initial-and the last point, the length between the initial and final points. CM-I do not know the connection of natural longer, not along the natural base, but the real (the line , proof) ________________________________________________________________ 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 ),... The 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, ...} ... View attachment 4482 [S4]-numeric 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 of-numeric along a starting point, the last point at infinity -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-I know the term but long term numeric numeric rays (line) Natural numbers and zeros are given axiom
msbiljanica, your math seems far more limited than the mathematics of set theory, category theory or Euclidean geometry in that those subjects are capable of being communicated to other human beings. I think what you are saying is this: Definition 1: A segment has two end points. Definition 2: A segment has a length. Definition 3: A new segment may be constructed, given any two segments. Axiom 1: A segment of length 1 exists. Axiom 2: A new segment constructed from a given segment and a segment of length 1 has a length distinct from the given segment. Theorem: A unbounded sequence of segments with distinct lengths exists, constructed by adding a segment of length 1 to itself an arbitrary number of times. Definition 4: The length of a segment created by adding a segment of length 1 to itself N times is given as N+1. -- This seems less useful than the Peano axioms of arithmetic. 0 is a number if x is number, x = x if x and y are numbers, x = y means y = x if x, y and z are numbers, x = y and y = z means x = z if x is a number and x = y then you can be assured that y is a number if x is a number, S(x) is a number there is no number x such that S(x) = 0 if x and y are numbers, S(x) = S(y) means x = y if C is a collection of numbers and 0 is in C and if x is a number in C that means S(x) is in C then C contains every natural number http://en.wikipedia.org/wiki/Peano_axioms#The_axioms Given this, we can define P(,) so that P(x,0) = x P(x,S(y)) = S(P(x,y)) And we can define M(,) so that M(x,0) = 0 M(x,S(y)) = P(x,M(x,y)) Then we can prove: P(x,S(0)) = S(P(x,0)) = S(x) M(x,S(0)) = P(x,M(x,0)) = P(x,0) = x M(x,S(S(0))) = P(x,M(x,S(0))) = P(x,x) P(S(0),S(0)) = S(S(0)) P(S(S(0)),S(0)) = S(S(S(0))) M(x,S(S(S(0)))) = P(x, M(x,S(S(0))) = P(x, P(x, x)) Which justifies the definitions 1 = S(0) 2 = S(1) 3 = S(2) Reexpressing these theorems in terms of these definitions: P(x,1) = S(x) M(x,1) = x M(x,2) = P(x,x) P(1,1) = 2 P(2,1) = 3 M(x,3) = P(x, P(x, x))
rpenner-What's the score Z÷(10^n)=? Presupposition-point numbers have their The 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) ... View attachment 4488 [S6]-point number CM-I do not know the item number ________________________________________________________________ Presupposition-point numbers have opposite The process: P1-0 = (s.0) P2-1 = (s.0, 1) P3-2 = (s.0, 1.2) P4-3 = (s.0, 1,2,3) P5-4 = (s.0, 1,2,3,4) ... View attachment 4489 [S7]-opposite point of number SM-I do not know the opposite point of
Presupposition-numbers are comparable with each other The 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 View attachment 4492 P3-four numbers (a, b, c, d) are comparable with each other View attachment 4493 ... [S8]-comparability issues CM knows the comparability of two numbers, the comparability of three numbers (a number comparable with the numbers b and c) comparability of the other knows. __________________________________________________ _________ Presupposition-number ranges number along The process: P1-image P2-image P3-image ... View attachment 4494 [S9]-mobility of number CM does not know the number of mobility
Presupposition-number and mobile number of a contact The process: P1-3 + (.0) 2 = 3 P2-3 + (.1) 2 = 3 P3-3 + (.2) 2 = 4 P4-3 +2 = 5, same as the current sum View attachment 4495 [S10]-addition of CM knows only one type of addition is given as an axiom ____________________________________________________________ Presupposition-No I do not have a mobile contact number, except to point The process: P1-¤3(0)2¤ P2-¤3(1)2¤ P3-¤3(2)2¤ ... View attachment 4496 Next-gap number and mobile number have no contact, except to point ... [S11]-gap numbers Gn = {¤a(b)c¤,...,¤ (b )...( d) e ¤} [S12]-gap along the CM-gap does not know the numbers, gaps along
Presupposition-gap number is comparable with the gap number and The process: P1-¤a(b)c¤ , a+c=z P2-¤a(b)c(d)e¤ , a+c+e=z P3-¤a(b)c(d)e(f)g¤ , a+c+e+g=z ... Bringing the number of gaps in the number of z, as it compares the number of [S13]-comparability gap of CM-comparability gap does not know the number of ___________________________________________________________________________ Presupposition-two or more of the same numbers can be written in abbreviated form The process: P1-{a, a} = af2 P2-{a, a, a} = af3 P3-{a, a, a, a} = af4 ... [S14]-the same number of frequency CM does not know the frequency of the same number _____________________________________________________________________- Presupposition-this can be written in abbreviated form -growing (a, a + b, a + b + b,..., a + b + b +...+ b) -descending (a + b + b +...+ b, a + b + b, a + b, a) P1-abc, c = a + b, c = a + b +b, ..., c=a + b + b +...+ b-final P2-ab - infinity [S15]-srcko CM-does not know srcko
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
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¤ View attachment 4502 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 ___________________________________________________________ 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 ______________________________________________________ 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
msbiljanica: What use is your new number theory? What new results have you discovered using your theory? What are the advantages of your theory over conventional number theory? Where does your work fit into the general body of mathematical knowledge?
1.relationship and the existence of new types of numbers 2.existence of gap numbers 3.current numbers are set axiom, my numbers arise from something (along, the numbers are different name for the longer (except for the number 0, which is the point)) 4.when you compare my math (somewhere around May this year, when I show a lot of my math) and current mathematics get an answer to this question
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+(s.0))11 or d+(s.0)number (more) from 11 and 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<g1 a-(.b)c=d+(s.0)g , a+(.b)c<l a-(.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¤¤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 ... [S25]-left inequality subtraction CM-[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 11p and 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>g1l a-(.b)c=d-(s.0/s.0))g , a-(.b)c>l a-(.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¤¤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 ... [S26]-right inequality subtraction CM-[S26]-know , forms without any gaps numbers not known
from previous exposure to guess you realize what good gap numbers , for complex shapes that are separated by gap between them
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<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 ,... [S27]-function subtraction CM-[S27]-does no know _____________________________________ Presupposition-Parts number (a) and mobile number (b ) have a contact , contact remains , the rest is delete Process: View attachment 4504 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=1 General form a - (.0)b=c a - (.1)b=c ... a - (.d)b=c [S28]-opposite subtraction CM-[S28]-does no know -sign for. opposite subtraction (when you download the following PDF you will see how it looks)
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))1[sub]1[/sub] or d+(s.0)number (more) from 1[sub]1[/sub] and d+(.z)1[sub]1[/sub]¤¤e or d+(.z) number ( more ) from 1[sub]1[/sub]¤¤e , e={(f),(f)(f),(f)(f)(f),...} Process: P1 5 - (.0)5=5+(s.0)1[sub]1[/sub] , 5 - (.0)5<6[sub]1[/sub] ¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤+(.z)1[sub]1[/sub]¤¤(2) ¤2(2)2¤ - (.1)¤1(1)2¤<2¤¤(2)+(.z)1[sub]1[/sub]¤¤(2) , ¤2(2)2¤ - (.1)¤1/1)2¤<3[sub]1[/sub]¤¤(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)1[sub]1[/sub] ,a - (s.b)c<g[sub]1[/sub] a - (.b)c=d+(s.0)g , a - (.b)c<l a - (.b)c=d+(s.0)k[sub]p[/sub]g , a - (.b)c<h[sub]i[/sub]j ... a -(.b)c=d+(.z)1[sub]1[/sub]¤¤e , a - (.b)c<s¤¤e+(.z)1[sub]1[/sub]¤¤e , a - (s.b)c<g[sub]1[/sub]¤¤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)k[sub]p[/sub]g¤¤e , a - (.b)c<s¤¤e+(.z)k[sub]p[/sub]g¤¤e , a - (.b)c<h[sub]i[/sub]j¤¤e ... [S29]-left inequality opposite subtraction CM-[S29]-does no know _________________________________ Presupposition-In the expression a - (.b)c=d , d-(s.0/s.0))1[sub]1[/sub]p or d+(s.0/s.0)number (more) from 1[sub]1[/sub]p and d-(.z)1[sub]1[/sub]p¤¤e or d-(.z) number ( more ) from 1[sub]1[/sub]¤¤e , e={(f),(f)(f),(f)(f)(f),...} Process: P1 5 - (.0)5=5 -(s.0/s.0))1[sub]1[/sub]5 , 5 - (.0)5>0[sub]1[/sub]4 ¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤-(.z)2[sub]1[/sub]1¤¤(2) ¤2(2)2¤ - (.1)¤1(1)2¤>2¤¤(2)-(.z)2[sub]1[/sub]1¤¤(2) , ¤2(2)2¤ - (.1)¤1(1)2¤>0[sub]1[/sub]1¤¤(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)1[sub]1[/sub]p ,a - (s.b)c>g[sub]1[/sub]l a - (.b)c=d-(s.0/s.0))g , a - (.b)c>l a - (.b)c=d-(s.0/s.0))k[sub]s[/sub]g , a - (.b)c>h[sub]i[/sub]j ... a - (.b)c=d-(.z)1[sub]1[/sub]p¤¤e , a - (.b)c>s¤¤e-(.z)1[sub]1[/sub]p¤¤e , a - (s.b)c>g[sub]1[/sub]k¤¤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)k[sub]p[/sub]g¤¤e , a - (.b)c>s¤¤e-(.z)k[sub]p[/sub]g¤¤e , a - (.b)c>h[sub]i[/sub]j¤¤e ... [S30]-right inequality opposite subtraction CM-[S30]-does no know
Presupposition-Two ( more) opposite subtraction (left and right inequalities) can be short to write Process: P1 7 - (.0[sub]1[/sub]3)4=y , 7 - (.0[sub]1[/sub]3)4>y ,7 - (.0[sub]1[/sub]3)4<y P2 8 - (.2[sub]2[/sub]8)5=y , 8 - (.2[sub]2[/sub]8)5>y ,8 - (.2[sub]2[/sub]8)5<y ... General form a - (.b[sub]c[/sub]d)e=y ,a - (.b[sub]c[/sub]d)e>y , a - (.b[sub]c[/sub]d)e<y a - (.b[sub]c[/sub]d_e)f=y , a - (.b[sub]c[/sub]d_e)f>y , a - (.b[sub]c[/sub]d_e)f<y ,... [S31]-function opposite subtraction CM-[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)b P2 a+(s.c)a+(s.c)a=a×(s.c)3 , a×(s.c)b P3 a+(s.c)a+(s.c)a+(s.c)a=a×(s.c)4 , a×(s.c)b ... [S32]-multiplication CM-[S32]-know , axiom Presupposition-In terms a×(s.c)b , b can be number 0 (1) Process: P1 a×(s.c)0 P2 a×(s.c)1 [S32a]-multiplication-amendment