Page:Squaring the circle a history of the problem (IA squaringcirclehi00hobsuoft).djvu/60

where the integers $k_{1},k_{2},\ldots k_{m},\ldots$ are all less than the integer r, and do not all vanish from and after a fixed value of m.

Let

${\frac {p}{q}}={\frac {k_{1}}{r^{1!}}}+{\frac {k_{2}}{r^{2!}}}+\ldots +{\frac {k_{m}}{r^{m!}}}$,

then ${\frac {p}{q}}$ continually approaches # as m is increased. We have

${\begin{aligned}x-{\frac {p}{q}}&={\frac {k_{m+1}}{r^{(m+1)!}}}+{\frac {k_{m+2}}{r^{(m+2)!}}}+\ldots \\&<r\left({\frac {1}{r^{(m+1)!}}}+{\frac {1}{r^{(m+2)!}}}+\ldots \right)\\&<{\frac {2r}{q^{m+1}}},{\text{ since }}q=r^{m!}.\end{aligned}}$

It is clear that, whatever values $A$ and $n$ may have, if $m$, and therefore $q$, is large enough, we have ${\frac {2r}{q^{m+1}}}<{\frac {1}{Aq^{n}}}$; and thus the relation $\left|{\frac {p}{q}}-x\right|>{\frac {1}{Aq^{n}}}$ is not satisfied for all the fractions ${\frac {p}{q}}$. The numbers $x$ so defined are therefore transcendental. If we take $r=10$, we see how to define transcendental numbers that are expressed as decimals.

This important result provided a complete justification of the division of numbers into two classes, algebraical numbers, and transcendental numbers; the latter being characterized by the property that such a number cannot be a root of an algebraical equation of any degree whatever, of which the coefficients are rational numbers.

A proof of this fundamentally important distinction, depending on entirely different principles, was given by G. Cantor (Crelle's Journal, vol. 77, 1874) who shewed that the algebraical numbers form an enumerable aggregate, that is to say that they are capable of being counted by means of the integer sequence 1, 2, 3,…, whereas the aggregate of all real numbers is not enumerable. He shewed how numbers can be defined which certainly do not belong to the sequence of algebraic numbers, and are therefore transcendental.

This distinction between algebraic and transcendental numbers being recognized, the question now arose, as regards any particular number defined in an analytical manner, to which of the two classes it belongs; in particular whether $\pi$ and $e$ are algebraic or transcendental. The difficulty of answering such a question arises from the fact that the recognition of the distinction between the two classes of numbers