1549901369-Elements_of_Real_Analysis__Denlinger_

2.5 Monotone Sequences 93

Then, 0 :::; 100 [x - K.d 1 ] < 10. We repeat the above process: :3 unique

integer d2 E {1, 2, · · · , 9} such that

d2 :::; lOO(x - K.d1) < d2 + 1

0 :::; lOOx - lOOK.d1 - d2 < 1

0 < 100 [x -(K +di + ~)] < 1

• 10 100
  0:::; 100 [x - K.d1d2] < l.

Continuing in this way, we obtain a sequence { dn} of "digits"

dn E {1, 2, · · · , 9} 3 Vn E N,

0 :::; ion [x - K.d1d2 · · · dn] < l;

1

i.e., 0 :S X - K.d1d2 · · · dn < lOn

1

K.d1d2 ... d n < - x < K.d1d2 ... d n + -. ion

(5)

(6)

In words, inequality (6) tells us that corresponding to every given n E N,
there is a unique n-place decimal K.d 1 d2 · · · dn :::; x closest to x.
Inequ ality (5), a long with the first squeeze principle, tells us that

lim K.d1d2 · · · dn = x.
n->oo
In that sense, we say that every nonnegative real number is "represented"
by an infinite decimal:

Note: the decimal expansion of a real number is not necessarily unique.
For example,

1 = { 1.0000 ... 0 ...

0.9999 ... 9 ...

and -^1 = {0.5000···^0 ···

2 0.4999 ... 9 ...

• Example 2.5.6 Prove that 0.999 · · ·^9 · · · = 1.000 · · · 0 · · ·.

Proof. By definition, 0.9999 · · · 9 · · · = n->oo lim O.d1d2 · · · dn, where di= 9, for

all i. Now,Vn EN,

O.d1d2 · .. dn = .9999 · · · 9 (n nines)

= 1-0.00···01 (n-lzeros)

1

= 1--. 10n