Page:Squaring the circle a history of the problem (IA squaringcirclehi00hobsuoft).djvu/59

Let the other roots be denoted by $x_{1},x_{2},\ldots x_{n-1}$; these may be real or complex. If ${\frac {p}{q}}$ be any rational fraction, we have

${\frac {p}{q}}-x={\frac {ap^{n}+bp^{n-1}q+cp^{n-2}q^{2}+\ldots }{q^{n}.a\left({\frac {p}{q}}-x_{1}\right)\left({\frac {p}{q}}-x_{2}\right)\ldots \left({\frac {p}{q}}-x_{n}\right)}}$.

If now we have a sequence of rational fractions converging to the value $x$ as limit, but none of them equal to $x$, and if ${\frac {p}{q}}$ be one of these fractions,

$\left({\frac {p}{q}}-x_{1}\right)\left({\frac {p}{q}}-x_{2}\right)\ldots \left({\frac {p}{q}}-x_{n}\right)$

approximates to the fixed number

$(x-x_{1})(x-x_{2})\ldots (x-x_{n})$.

We may therefore suppose that for all the fractions ${\frac {p}{q}}$,

$a\left({\frac {p}{q}}-x_{1}\right)\left({\frac {p}{q}}-x_{2}\right)\ldots \left({\frac {p}{q}}-x_{n}\right)$

is numerically less than some fixed positive number $A$. Also

$ap^{n}+bp^{n-1}q+\ldots$

is an integer numerically ≥ 1; therefore

$\left|{\frac {p}{q}}-x\right|>{\frac {1}{Aq^{n}}}$.

This must hold for all the fractions ${\frac {p}{q}}$ of such a sequence, from and after some fixed element of the sequence, for some fixed number $A$. If now a number $x$ can be so defined such that, however far we go in the sequence of fractions ${\frac {p}{q}}$, and however $A$ be chosen, there exist fractions belonging to the sequence for which $\left|{\frac {p}{q}}-x\right|<{\frac {1}{Aq^{n}}}$, it may be concluded that $x$ cannot be a root of an equation of degree $n$ with integral coefficients. Moreover, if we can shew that this is the case whatever value $n$ may have, we conclude that $x$ cannot be a root of any algebraic equation with rational coefficients.

Consider a number

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