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

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,

approximates to the fixed number

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

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

is an integer numerically ≥ 1; therefore

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