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401
ALGEBRA
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401

divide n by p, let a_2, be the quotient and q the remainder; thus


divide p by q. let a_3 be the quotient and r the remainder; and soon. Thus


If m is less than n, the first quotient is zero, and we put


and proceed as before.

It will be observed that the above process is the same as that of finding the greatest common measure of m and n; hence if m and n are commensurable, we shall at length arrive at a stage where the division is exact and the process terminates. Thus every fraction whose numerator and denominator are positive integers can be converted into a terminating continued fraction.

Ex. Reduce 832 159 to a continued fraction.

Find the greatest common measure of 852 and 159 by the usual process, thus:

159)832(5 795 37)159(4 148 11)37(3 33 4)11(2 8 3)4(1 3 1)3(3 3 0

We have the successive quotients 5, 4, 3, 2, 1.3; hence

832 159 = 5 + 1 4+ 1 3+ 1 2+ 1 1+ 1 3.