Page:Squaring the circle a history of the problem (IA squaringcirclehi00hobsuoft).djvu/71

 or than

or than

where $$\bar{\beta}$$ denotes the greatest of the numbers $$|\beta_1|, |\beta_2|, \ldots |\beta_n|$$. It thus appears that the modulus of

is less than a number of the form $$P Q^P/(p-1)!$$, where $$P$$ and $$Q$$ are independent of $$p$$ and of $$m$$.

We have now

(A + T A) - KPA + T #<*> (ft,)

T

m=l m=l

where $$K_p A$$ is not a multiple of $$p$$, the second term is an integer divisible by $$p$$, and $$L$$ is less than $$nP Q^P/(p-1)!$$. The prime $$p$$ may be chosen so large that $$nP Q^P/(p-1)!$$ is numerically less than unity. Since $$\textstyle K_p (A + \sum\limits^{m=n}_{m=1} e^{\beta_m})$$ is expressed as the sum of an integer which does not vanish and of a number numerically less than unity, it is impossible that it can vanish. Having now shewn that no such equation as

can subsist, we see that $$\pi i$$ cannot be a root of an algebraic equation with integral coefficients, and thus that $$\pi$$ is transcendental.

It has thus been proved that $$\pi$$ is a transcendental number, and hence, taking into account the theorem proved on page 50, the im- possibility of "squaring the circle" has been effectively established.