beginning trascendental

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5 The Beginning of Transcendental Numbers

We have dened a transcendental number (see Denition 3 of the Introduction), but so far we haveonly established that certain numbers are irrational. We now turn to the beginnings of transcen-dental numbers. Our rst theorem is

Theorem 11. Transcendental numbers exist.

Like many of our results so far, this will of course be a consequence of later results. The rstproof that there exist transcendental numbers was given by Liouville. Before we give his proof,we give a proof due to Cantor.

Proof 1. The essence of this proof is that the real algebraic numbers are countable whereas the setof all real numbers is uncountable, so there must exist real transcendental numbers. Dene

P (n) = f (x) =n

j =0

a j x j Z [x] : 1n

j =0

|a j | n .

Observe that P (n) is nite. Also, every non-zero polynomial in Z [x] belongs to some P (n). Byconsidering the real roots of polynomials in P (1), P (2), . . . (at the kth stage, consider the real rootsof polynomials in P (k) which have not occurred as a root of a polynomial in P ( j ) for j < k ), wecan order the algebraic numbers; hence, they are countable. Next, give the usual proof that the realnumbers are uncountable.

For Liouvilles proof, we dene

Denition 4. A real number is a Liouville number if for every positive integer n , there areintegers a and b with b > 1 such that

0 < ab

0 such that if a and b are integers with b > 0 ,then

ab >

Abn . (6)

Proof. Let M be the maximum value of |f (x)| on [ 1, +1] . Let 1, 2, . . . , m be the distinctroots of f (x) which are different from . Fix

A < min{1, 1/M, | 1| , | 2| , . . . , | m |} .

Assume (6) does not hold for some a and b integers with b > 0. Then

ab

Abn

A < min{1, | 1|, | 2|, . . . , | m |} .

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Hence,ab[ 1, + 1] and

ab { 1, . . . , m }.

By the Mean Value Theorem, there is an x0 between a/b and such that

f () f (a/b ) = ( a/b )f (x0)so that

ab

=f (a/b ) f ()

f (x0)=

|f (a/b )||f (x0)|

.

Since f (a/b ) = 0 , we deduce that

|f (a/b )| =n

j =0

a j a j bn j /b n 1/b n .

Thus, since |f (x0)| M, we obtain

ab

1

Mbn>

Abn

ab

,

giving a contradiction. Thus, the lemma follows.

Proof of Theorem 12. Let be a Liouville number. First, we show that must be irrational.Assume = c/d for some integers c and d with d > 0. Let n be a positive integer with 2n 1 > d .Then for any integers a and b with b > 1 and a/b = c/d , we have that

a

b=

c

d

a

b

1

bd>

1

2n

1b

1

bn .

It follows from the denition of a Liouville number that is not a Liouville number, giving acontradiction. Thus, is irrational.

Now, assume is an irrational algebraic number. By the lemma, there exist a real numberA > 0 and a positive integer n such that (6) holds for all integers a and b with b > 0. Let r be apositive integer for which 2r 1/A . Since is a Liouville number, there are integers a and b withb > 1 such that

ab

1 integers. Then

0 < ab

=

j = n +1

12 j !

0. Thenthere are at most nitely many pairs of integers (a, b) with b > 0 such that

ab 0. It sufces to show that the (Lebesgue) measure of the Liouville numbers in [0, 1]is < . Let n be a positive integer for which b=2 4/b

n 1 < (observe that this is possible since

b=2 4/bn 1 < (4/ 2n 3) b=2 1/b

2). If is a Liouville number in [0, 1], then there are integers aand b with b > 1 such that

ab < 1bn .

Since the right-hand side is 1/ 2 and [0, 1], we deduce that a/b ( 1/ 2, 3/ 2) so that b/2 < a < 3b/2 (a is in an open interval of length 2b). In particular, for a given integer b > 1,there are 2b possible values of a for which the above inequality can hold. For each b > 1, we getthat must be in one of 2b intervals of length 2/b n . Thus, the measure of the Liouville numbersmust be

b=2

4bbn

< ,

giving the desired result.

Corollary 1. There are transcendental numbers which are not Liouville numbers.

We now turn to a related discussion. Suppose that we have an sequence of positive integers{ak }k =1 with a1 < a 2 < . When will the argument given in the example above lead to a proof that

k =0 1/ 2a k is transcendental? It is not too difcult to see that what one essentially wants is

that ak increases fast enough where fast enough means that

lim inf k

ak +1ak

= .

On the other hand, it can be shown that sequences which increase much slower also lead to tran-scendental numbers, an observation apparently rst noticed by Erd os. The next theorem illustratesthe basic idea by showing a certain number of this form is transcendental; it can be shown also thatthis number is not a Liouville number.

Theorem 15. The number

k =0 1/ 22k is transcendental.

To prove Theorem 15, for positive integers k and m, we dene c(k, m) to be the number of m tuples ( j1, . . . , j m ) of non-negative integers for which

k = 2 j 1 + 2 j 2 + + 2 j m .

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Lemma 1. With the notation above, c(k, m) m2m .

Proof. We do induction on m . For m = 1 , c(k, m) {0, 1} for all k, so the result is clear. Supposem > 1. If k has more than m non-zero binary digits, then it is not too hard to see that c(k, m) = 0(note that the sum of two like powers of 2 is a power of 2). If k has exactly m non-zero binary

digits, then c(k, m) = m! mm

m2m

( j1 can correspond to any one of the m non-zero digits, j2 to any one of the remaining m 1 non-zero digits, and so on). If k has fewer than m non-zerobinary digits, then for some integers r and s with 1 r < s m , j r = j s . Since in this case,2 j r + 2 j s = 2 j r +1 , we deduce that

c(k, m) m2

c(k, m 1) m2(m 1)2m 2 m2m ,

establishing the lemma.

Lemma 2. Let t and m be positive integers. Then c(k, m) = 0 for every integer k (2t +1 + +2t + m , 2t + m +1 ).

Proof. This follows since each such k has > m non-zero binary digits (see the proof of Lemma1).

Proof of Theorem 15. Let = k =0 1/ 22k . Let m be a positive integer. Then the denition of

c(k, m) implies that

m =

k =0

122k

m

=

k =1

c(k, m)2k

.

Let t be a positive integer. Using Lemma 2 and then Lemma 1, we obtain

22t +1

+

+2t + m

m 22t +1

+

+2t + m

k =2 t +1 + +2 t + m +1

c(k, m)2k

22t +1 + +2 t + m

k =2 t + m +1

c(k, m)2k

22t +1 + +2 t + m m2m

k =2 t + m +1

12k

22t +1 + +2 t + m m2m 2 2

t + m +1 +1 = 2 1 2t +1

m2m .

If we view m as being xed and let t approach innity, we see that the binary expansion of m hasarbitrarily long strings of 0s. By the same reasoning, we see that more generally if b0, b1, . . . , bmare non-negative integers, then the binary expansion of b0 + b1 + + bm m has arbitrarily longstrings of 0s. In fact, if b0, b1, . . . , bm and c0, c1, . . . , cm are non-negative integers and if we havethe following binary expansions

{b0 + b1 + + bm m } = (0 .d1d2d3 . . . )2

and{c0 + c1 + + cm m } = (0 .d1d2d3 . . . )2,

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then for any positive integer N , there is a positive integer j for which

d j +1 = d j +2 = = d j + N = d j +1 = d j +2 = = d j + N = 0 .

Furthermore, we can take N and j so that

N = 2t

and j = 2t +1

+ + 2t + m

with t an integer as large (but perhaps not as small) as we wish.Now, assume is a root of f (x) = n j =0 a j x

jZ [x] with an > 0 (which would be possible

if were algebraic). Then we can write f (x) in the form

f (x) =n

j =0

b j x j n

j =0

c j x j ,

where the b j and c j are non-negative integers with b j c j = 0 for each j . In particular, bn = an > 0and cn = 0 . Take m = n 1, and let N = 2 t where t is a positive integer to be chosen momentarilyand j = j (t) is as above. Then

2 j +2t 2

n 1

i=0

bi i and 2 j +2t 2

n 1

i=0

ci i

both differ from an integer by 1/ 2N 2t 2

= 1 / 22t 1 +2 t 2 . If we write

2 j +2t 2

n 1

i=0

bi i = m1 + 1 and 2 j +2t 2

n 1

i=0

ci i = m2 + 2,

where m1 and m2 are the greatest integers in the above expressions, then we see that since f () =0,

2 j +2 t 2

bn n

= m3 + 3 with m3 Z and |3| = |1 2| 1

22t 1 +2 t 2 . (7)On the other hand, since j has the form given above, we get from the denition of c(k, m) thatc( j + 2 t 1, n ) 1 and c(k, n) = 0 for all k ( j + 2 t 2, j + 2 t 1). For t sufciently large, we havethat

2 j +2t 2

bn

k = j +2 t 2 +1

c(k, n)2k

= 2 j +2t 2

bn

k = j +2 t 1

c(k, n)2k

2 j +2t 2

bn n2n 2 j 2t 1 +1 = 2 1 2

t 2bn n2n

beginning trascendental

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