# Extended Euclidian Algorithm Proofs

(See attached file for full problem description)

---

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)

---

https://brainmass.com/math/discrete-math/extended-euclidian-algorithm-proofs-65120

#### Solution Preview

Please see the attached file for the complete solution.

Thanks for using BrainMass.

Proof:

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

#### Solution Summary

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