SECTION 5.1 Quick Survey of Limits 253

Show that LUB(A) = 1 and GLB(A) = 0.^4

9. (The irrationality of π.) This exercise will guide you through a
   proof of the irrationality ofπand follows roughly the proof of Ivan
   Niven^5. Assume, by way of contradiction, that we may express
   π=
   a
   b

,whereaandbare positive integers. For each integern≥ 1

define the function

fn(x) =

xn(a−bx)n

n!

.

Using the assumption thatπ=

a

b

, show that

(a) fn(0) = 0, fn(π) = 0;

(b) fn(x−π) =fn(x);

(c) fn(i)(0) is an integer for alli≥0;

(d) fn(i)(π) is an integer for alli≥0 (use (b)).

Next, define the new functions

Fn(x) =f(x)−f(2)(x) +f(4)(x)−···+ (−1)nf(2n)(x)

(^4) Showing that LUB(A) = 1 is pretty easy: defineP = ¶φ(pp)|pis prime©and show that
LUB(P) = 1. Showing that GLB(A) = 0 is a bit trickier. Try this: letp 1 = 2,p 2 = 3,p 3 = 5,···,
and sopkis thek-th prime. Note that
φ(p 1 p 2 ···pr)
p 1 p 2 ···pr =
Å
1 −p^11
ãÅ
1 −p^12
ã
···
Å
1 −p^1 r
ã
.
The trick is to show that φp(p 11 pp 22 ······pprr) →0 asr→ ∞. Next, one has the “harmonic series”
(see page 265)
∑∞
n=1
1
n, which is shown on page 265 to be infinite. However, from the Fundamental
Theorem of Arithmetic (see page 76), one has
∑∞
n=1
1
n =
Å
1 +^12 + 212 +···
ãÅ
1 +^13 + 312 ···
ãÅ
1 +^15 + 512 ···
ã
···

Å
1 −^12
ã− 1 Å
1 −^13
ã− 1 Å
1 −^15
ã− 1
···
From this one concludes that, as stated above,φ(pp 11 pp 22 ······pprr)→0 asr→∞.
(^5) Bull. Amer. Math. Soc.Volume 53, Number 6 (1947), 509.