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Extended Euclidian Algorithm Proofs

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Given positive integers a and b, the extended Euclidian algorithm constructs sequences qn, rn, sn and tn, which are defined recursively as follows:

q0=0, q1=0, qn= q└ rn-2/ rn-1 ┘ for n>=2;
r0=a, r1=b, rn= rn-2 - qnrn-1 for n>=2;
s0=1, s1=0, sn= sn-2 - qnsn-1 for n>=2;
t0=0, t1=1, tn= tn-2 - qntn-1 for n>=2;

the sequences terminating if rn=0 . Prove the following

i) The extended Euclidian algorithm always terminates, i.e., there exists n>=2 such that rn=0
(hint : first show that if k>0 and rk>0 then 0<= rk+1<rk.)
ii) If 0<=k<=n then ska+tkb = rk.
iii) If 1<=k<=n then gcd(rk-1,rk) = gcd(a,b).
iv) rn-1= sn-1a+tn-1b= gcd(a,b)

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a) For , we have , where means the greatest integer less than or equal to ...

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Extended Euclidian Algorithm Proofs are provided. The solution is detailed and well presented.