epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

32 The Nuts and Bolts of Proof, Third Edition
with ^1 > 0 and 0 < ri <fo.
Then divide b by ri, obtaining:
b = riq2-hr2
with q2>0 and 0 < r2<r\.
Continuing, we can divide ri by r2 to obtain:
with ^3 > 0 and 0 < rs <r2.
Continue this process as long as ri / 0.
Then, the greatest common divisor of a and b, denoted as {a,b) or
GCD{a,b) is the last nonzero remainder. (See front of the book for the
definition of the greatest common divisor of two numbers.)
Proof: If we use the process described in the statement, we obtain:
a = bqi-\- ri
r„_3 = rn-2qn-i + rn-i
rn-2 = rn-1 qn + rn
r„_i =r„(j„+i +0.
The process will take at most b steps because /7>ri>r2>....> 0.
The last of the equahties written above implies that r„ = GCD(r„_i,r„).
(Explain why.)
Because
rn-2 = rn-1 Qn + ^n
= rn Qn+i <?n + rn = r„ ti
with ti > 0, it follows that r„ divides r„_2 and r„_i. So, r„ is a common
divisor of r„_2 and r„_i.
If d is another positive integer divisor of r„_2 and r„_i, then J will
divide r„. (Check this claim.) Therefore,
r„ = GCD(r„_i,r„_2).