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

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

$P=A{\frac {c}{d+}}{\frac {e}{f+}}$ &c., $Q=P{\frac {e}{f+}}{\frac {g}{h+}}$ &c., $R=Q{\frac {e}{f+}}{\frac {g}{h+}}$ &c.,

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

${\frac {A}{B}}={\frac {a}{b+{\frac {P}{A}}}},$ or, $P=aB-bA$

${\frac {P}{A}}={\frac {c}{d+{\frac {Q}{P}}}},$ or, $Q=cA-dP$

${\frac {Q}{P}}={\frac {e}{f+{\frac {R}{Q}}}},$ or, $R=eP-fQ$

&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

$1+{\frac {a}{z}}+{\frac {a^{2}}{2z(z+1)}}+{\frac {a^{3}}{2\cdot 3\cdot z(z+1)(z+2)}}+\ldots$

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