1549901369-Elements_of_Real_Analysis__Denlinger_

416 Chapter 7 • The Riemann Integral

Thus, by the fundamental theorem of calculus,

1

1

ePxp^2 n+l'l/Jn(x)dx = [ ePx Fn(x) t = eP Fn(l) - Fn(O), so

b 1

1

ePxp^2 n+l'l/Jn(x)dx = beP Fn(l) - bFn(O) = aFn(l) - bFn(O), (31)

which is an integer since a, b, Fn(l), and Fn(O) are integers.

By property (b) of 'l/Jn(x) listed in Theorem 7.6.20 above, for all x E [O, l],

ePp2n+l

0 :S: ePxp2n+l'l/Jn(x) < I

n.

1

1 11 ePp2n+l

So, 0 < b ePxp^2 n+l'l/Jn(x)dx < b dx

o o n!

bePp2n+l

n!

= bePp (p2r = ap (p2r.

n! n!

By Corollary 2.3.11, lim (p

2

)

1

n = 0. Hence we can find n EN such that

n~oo n.

0 <b1

1

ePxp^2 n+l'l/Jn( x )dx < 1.

But we noted in (31) above that b J 01 ePxp^2 n+^1 'lfJn(x)dx is an integer. Thus,

we have an integer between 0 and 1. Since this is impossible, we must reject
our assumption that ex is rational. Therefore, ex is irrational. •

Theorem 7.6.22 7r is irrational.

*Proof. If 7r is rational, then n^2 is rational, so it suffices to prove that n^2 is

a

irrational. For contradiction, suppose n^2 is rational. Then 3 a, b E N 3 n^2 = b.

As in the proof of Theorem 7.6.21, we shall make use of the functions 'I/Jn defined

in 7.6.19. This time, we define the functions Fn by

n

= bn 2)-l)kn2n-2k'l/J~2k)(x).

k=O