"Bezout" --- did you mean "Euclid's algorithm"? https://en.wikipedia.org/wiki/Euclidean_algorithm But since x:y apparently denotes division x/y , it's not clear to me how this procedure improves on the usual one! WFL On 2/20/20, françois mendzina essomba2 via math-fun <math-fun@mailman.xmission.com> wrote:
Solve without using Bezout let's take this example
879*U[0]-13723*V[0]=1
Step1
13723 : 879 = 15,6120
879(15)-13723(1)=-532
879(15,6*10)-13723(1*10)=-106
879(15,62*100)-13723(1*100)=698
Step 2
698 :106=6,5
We take as integers 6 and 7
a)
879(156*6)-13723(10*6)=-636
879(1562)-13723(100)=698
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879(2498)-13723(160)=62
b)
879(156*7)-13723(10*7)=-742
879(1562)-13723(100)=698
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879(2654)-13723(170)=-44
a+b)
879(2498)-13723(160)=62
879(2654)-13723(170)=-44
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879(5152)-13723(330)=18
Step 3
44 :18=2,4
We use integers 2 and 3
c)
879(5152*2)-13723(330*2)=36
879(2654)-13723(170)=-44
-------------------------------------
879(12958)-13723(830)=-8
d)
879(5152*3)-13723(330*3)=54
879(2654)-13723(170)=-44
-------------------------------------
879(18110)-13723(1160)=10
c+d)
879(12958)-13723(830)=-8
879(18110)-13723(1160)=10
--------------------------------------
879(31068)-13723(1990)=2
After simplification by 2
879(15534)-13723(995)=1
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