Page:A budget of paradoxes (IA cu31924103990507).pdf/510

496 Now, if possible, let $${a \over b} {{}\atop+}\ {c \over d} {{}\atop+}$$ &c. be $$\frac{A}{B}$$ at the limit; $$A$$ and $$B$$ being integers. Let

$$P$$, $$Q$$, $$R$$, &c. being integer or fractional, as may be. It is easily slown that all must be integer: for

&c., &c. Now, since $$a,B,b,A,$$ are integers, so also is $$P$$; and thence $$Q$$; and thence $$R$$, &c. But since $$\frac{A}{B}, \frac{P}{A}, \frac{Q}{P}, \frac{R}{Q},$$ &c. are all between $$-1$$ and $$+1$$, it follows that the unlimited succession of integers $$P,Q,R,$$ are each less in numerical value than the preceding. Now there can be no such unlimited succession of descending integers: consequently, it is impossible that $${a \over b} {{}\atop+}\ {c \over d} {{}\atop+}$$ &c. can have a commensurable limit.

It easily follows that the continued fraction is incommensurable if $$\frac{a}{b}, \frac{c}{d},$$ &c., being at first greater than unity become and continue less than unity after some one point. Say that $$\frac{i}{k}, \frac{l}{m}, \ldots$$

Let $$\phi z$$ represent

Let $$z$$ be positive: this series is convergent for all values of $$a$$, and