Euclidean algorithm

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January 17th 2010, 11:54 PM
kingwinner

Euclidean algorithm

http://sites.google.com/site/asdfasdf23135/nt2.JPG

All I think can of is the following:
Since the reaminder become strictly smaller each step, the worst case is when the remainders go down by 1 each step, in which case the process will take "b" steps in total to complete. Therefore, the process will terminate in at most "b" steps.

But how can we prove that all reaminders satisfy the inequality r(i+2) < r(i)/2? By induction? or...? This seems to provide a much better upper bound on the number of steps to complete the process.

Any help is much appreciated!

[also under discussion in s.o.s. math board]
January 18th 2010, 04:43 PM
Bruno J.

Well you have $r_i = r_{i+1}q_{i+2}+r_{i+2}$ . So $r_i-2r_{i+2} = r_{i+1}q_{i+2}-r_{i+2}$ . Since $q_{i+2}\geq 1$ and $0<r_{i+1}<r_{i+2}$ you have $r_i-2r_{i+2} = r_{i+1}q_{i+2}-r_{i+2} \geq r_{i+1}-r_{i+2} > 0$ .
January 19th 2010, 09:58 AM
kingwinner

Quote:

Originally Posted by Bruno J.

Well you have $r_i = r_{i+1}q_{i+2}+r_{i+2}$ . So $r_i-2r_{i+2} = r_{i+1}q_{i+2}-r_{i+2}$ . Since $q_{i+2}\geq 1$ and $0<r_{i+1}<r_{i+2}$ you have $r_i-2r_{i+2} = r_{i+1}q_{i+2}-r_{i+2} \geq r_{i+1}-r_{i+2} > 0$ .

But I think q(i+2) can be 0?

Also, I think we should have 0<r(i+2)<r(i+1)?

thanks.
January 19th 2010, 11:13 AM
Jhevon

Quote:

Originally Posted by kingwinner

But I think q(i+2) can be 0?

Remember, $r_i \ge r_{i + 1}$ (if equal, the algorithm only has one step). If the algorithm starts with $r_{i + 1} > r_i$ , then they are switched in the very next line. So, if our algorithm is to have more than one line, it is just assumed that $r_i > r_{i + 1}$ right off the bat. In this case, $q_{i + 2} \ge 1$ is forced to happen.

Quote:

Also, I think we should have $0 < r_{i + 2} < r_{i + 1}$ ?

yes, BrunoJ had a typo. But if you look at what he did, you can see that he assumed just as you wrote here.