A geometric proof that $e$ is irrational and a new measure of its
  irrationality
Microsoft Word - eIrrationalAddendum.doc
A Geometric Proof that e is Irrational 
and a New Measure of its Irrationality 
Jonathan Sondow 
1. INTRODUCTION. While there exist geometric proofs of irrationality for √2 [2], [27], 
no such proof for e, π , or ln 2  seems to be known. In section 2 we use a geometric 
construction to prove that e is irrational. (For other proofs, see [1, pp. 27-28], [3, p. 352], 
[6], [10, pp. 78-79], [15, p. 301], [16], [17, p. 11], [19], [20], and [21, p. 302].) The proof 
leads in section 3 to a new measure of irrationality for e, that is, a lower bound on the 
distance from e to a given rational number, as a function of its denominator. A 
connection with the greatest prime factor of a number is discussed in section 4. In section 
5 we compare the new irrationality measure for e with a known one, and state a number-
theoretic conjecture that implies the known measure is almost always stronger. The new 
measure is applied in section 6 to prove a special case of a result from [24], leading to 
another conjecture. Finally, in section 7 we recall a theorem of G. Cantor that can be 
proved by a similar construction. 
2. PROOF. The irrationality of e is a consequence of the following construction of a 
nested sequence of closed intervals In . Let I1 = [2, 3]. Proceeding inductively, divide the 
interval In−1 into n (≥ 2)  equal subintervals, and let the second one be In  (see Figure 1). 
For example, I2 =
2![ ] , I3 = 163!,173![ ] , and I4 = 654!,664![ ] . 
Figure 1. The intervals I1, I2, I3 , I4 . 
The intersection 
In = {e} (1) 
is then the geometric equivalent of the summation (see the Addendum) 
n!n=0
∑ = e . (2) 
When n > 1 the interval In+1 lies strictly between the endpoints of In , which are 
 and 
 for some integer a = a(n) . It follows that the point of intersection (1) is not a fraction 
with denominator n! for any n ≥ 1. Since a rational number p q with q > 0  can be 
written 
p ⋅(q −1)!
, (3) 
we conclude that e is irrational. •  
Question. The nested intervals In  intersect in a number—let's call it b. It is seen by the 
Taylor series (2) for e that b = e . Using only standard facts about the natural logarithm 
(including its definition as an integral), but not using any series representation for log, 
can one see directly from the given construction that log b = 1? 
3. A NEW IRRATIONALITY MEASURE FOR e. As a bonus, the proof leads to the 
following measure of irrationality for e. 
Theorem 1. For all integers p and q with q > 1 
(S(q) +1)!
, (4) 
where S(q)  is the smallest positive integer such that S(q)! is a multiple of q. 
For instance, S(q) = q  if 1 ≤ q ≤ 5 , while S(6) = 3 . In 19l8 A. J. Kempner [13] 
used the prime factorization of q to give the first algorithm for computing  
 S(q) = min{k > 0 : q k!}  (5) 
(the so-called Smarandache function [28]). We do not use the algorithm in this note. 
Proof of Theorem 1. For n > 1 the left endpoint of In  is the closest fraction to e with 
denominator not exceeding n!. Since e lies in the interior of the second subinterval of In , 
(n +1)!
 (6) 
for any integer m. Now given integers p and q with q > 1, let m = p ⋅S(q)! q  and 
n = S(q) . In view of (5), m and n are integers. Moreover, 
p ⋅S(q)! q
S(q)!
. (7) 
Therefore, (6) implies (4). •  
As an example, take q to be a prime. Clearly, S(q) = q . In this case, (4) is the 
(very weak) inequality 
(q +1)!
. (8) 
In fact, (4) implies that (8) holds for any integer q larger than 1, because S(q) ≤ q  always 
holds. But (4) is an improvement of (8), just as (7) is a refinement of (3). 
 Theorem 1 would be false if we replaced the denominator on the right side of (4) 
with a smaller factorial. To see this, let p q  be an endpoint of In , which has length 
. If 
we take q = n! , then since evidently 
 S(n!) = n  (9) 
and e lies in the interior of In , 
S(q)!
. (10) 
(If q < n! , then (10) still holds, since n > 2 , so p q  is not an endpoint of In−1, hence 
S(q) = n .) 
4. THE LARGEST PRIME FACTOR OF q. For q ≥ 2  let P(q)  denote the largest 
prime factor of q. Note that S(q) ≥ P(q) . Also, S(q) = P (q)  if and only if S(q)  is prime. 
(If S(q)  were prime but greater than P(q) , then since q divides S(q)!, it would also 
divide (S(q) −1)! , contradicting the minimality of S(q) .) 
 P. Er ʹ′ ʹ′ d os  and I. Kastanas [9] observed that 
 S(q) = P (q)      (almost all q).  (11) 
(Recall that a claim Cq  is true for almost all q if the counting function 
N(x) = #{q ≤ x :Cq is false}  satisfies the asymptotic condition N(x) x→ 0  as x→ ∞ .) 
It follows that Theorem 1 implies an irrationality measure for e involving the simpler 
function P(q) . 
Corollary 1. For almost all q, the following inequality holds with any integer p: 
(P(q) +1)!
. (12) 
When q is a factorial, the statement is more definite. 
Corollary 2. Fix q = n!>1. Then (12) holds for all p if and only if n is prime. 
Proof. If n is prime, then P(q) = n , so (4) and (9) imply (12) for all p. Conversely, if n is 
composite, then P(q) < n , and (10) shows that (12) fails for certain p. •  
 Thus when q > 1 is a factorial, (12) is true for all p if and only if S(q) = P (q) . To 
illustrate this, take p
 to be the left endpoint of I4 . Then P(q) = 3 < 4 = S(q) , and 
(12) does not hold, although of course (4) does: 
 0.00833 . . . =
= 0.00994 . . . <
= 0.04166 . . .  . 
5. A KNOWN IRRATIONALITY MEASURE FOR e. The following measure of 
irrationality for e is well known: given any ε > 0  there exists a positive constant q (ε)  
such that 
q2+ ε
 (13) 
for all p and q with q ≥ q(ε) . This follows easily from the continued fraction expansion 
of e. (See, for example, [23]. For sharper inequalities than (13), see [3, Corollary 11.1], 
[4], [7], [10, pp. 112-113], and especially the elegant [26].) 
Presumably, (13) is usually stronger than (4). We state this more precisely, and in 
a number-theoretic way that does not involve e. 
Conjecture 1. The inequality q2 < S(q)! holds for almost all q. Equivalently, q2 < P(q)!  
for almost all q. 
(The equivalence follows from (11).) This is no doubt true; the only thing lacking is a 
proof. (Compare [12], where A. Iv ʹ′ i c proves an asymptotic formula for the counting 
function N(x) = #{q ≤ x : P(q) < S(q)} and surveys earlier work, including [9].) 
Conjecture 1 implies that (13) is almost always a better measure of irrationality 
for e than those in Theorem 1 and Corollary 1. On the other hand, Theorem 1 applies to 
all q > 1. Moreover, (4) is stronger than (13) for certain q. For example, let q = n!  once 
more. Then (4) and (9) give (6), which is stronger than (13) if n > 2, since 
 (n +1)! < (n!)2      (n ≥ 3). (14) 
6. PARTIAL SUMS VS. CONVERGENTS. Theorem 1 yields other results on rational 
approximations to e [24]. One is that for almost all n, the n-th partial sum sn  of series (2) 
for e is not a convergent to the simple continued fraction for e. Here s 0 = 1 and sn  is the 
left endpoint of In  for n ≥ 1. (In 1840 J. Liouville [14] used the partial sums of the 
Taylor series for e2  and e−2  to prove that the equation ae2 + be−2 = c  is impossible if a, 
b, and c are integers with a ≠ 0 . In particular, e4  is irrational.) 
Let qn  be the denominator of sn  in lowest terms. When qn = n!  (see [22, 
sequence A102470]), the result is more definite, and the proof is easy. 
Corollary 3. If qn = n!  with n ≥ 3 , then sn  cannot be a convergent to e. 
Proof. Use (4), (9), (14), and the fact that every convergent satisfies the reverse of 
inequality (13) with ε = 0 [10, p. 24], [17, p. 61]. •  
When qn < n!  (for example, q19 =19! 4000—see [22, sequence A093101]), 
another argument is required, and we can only prove the assertion for almost all n. 
However, numerical evidence suggests that much more is true. 
Conjecture 2. Only two partial sums of series (2) for e are convergents to e, namely, 
s1 = 2  and s 3 = 8 3 . 
7. CANTOR'S THEOREM. A generalization of the construction in section 2 can be 
used to prove the following result of Cantor [5]. 
Theorem 2. Let a0, a1, . . .  and b1, b2, . . .  be integers satisfying the inequalities bn ≥ 2  
and 0 ≤ an ≤ bn −1  for all n ≥ 1. Assume that each prime divides infinitely many of the  
bn . Then the sum of the convergent series 
b1b2b3
+ ⋅ ⋅ ⋅  
is irrational if and only if both an > 0  and an < bn −1 hold infinitely often. 
For example, series (2) for e and all subseries (such as Σn≥0
(2n)!
= cosh1  and 
(2n+1)!
= sinh1) are irrational, but the sum Σn≥1
=1  is rational. 
An exposition of the "if" part of Cantor's theorem is given in [17, pp. 7-11]. For 
extensions of the theorem, see [8], [11], [18], and [25]. 
ADDENDUM. Here are some details on why the nested closed intervals In  constructed 
in section 2 have intersection e. Recall that I1 = [2, 3], and that for 
n ≥ 2 we get In  from 
In−1 by cutting it into n equal subintervals and taking the second one. The left-hand 
endpoints of   
I1, I2, I3,… are   
2, 2 + 12! , 2 +
3! ,…, which are also partial sums of the 
series (2) for e. Since the endpoints approach the intersection of the intervals, whose 
lengths tend to zero, the intersection is the single point e. 
ACKNOWLEDGMENTS. Stefan Krämer pointed out the lack of geometric proofs of 
irrationality. The referee suggested a version of the question in section 2. Yann Bugeaud 
and Wadim Zudilin supplied references on the known irrationality measures for e. 
Aleksandar Iv ʹ′ i c commented on Conjecture 1. Kyle Schalm did calculations [24] on 
Conjecture 2, and Yuri Nesterenko related it to Liouville's proof. I am grateful to them 
all. 
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jsondow@alumni.princeton.edu
