1. Introduction
A certain class of polynomials with integer coefficients displays divisibility properties which can be used to establish irrationality. The purpose of the present research is to exploit this property by evaluating such polynomials at zero and at the base of natural logarithms e, which, for the sake of contradiction, is assumed to be rational. Subsequent algebraic manipulations lead to a divisibility argument which forces a contradiction by producing an “integer” strictly between zero and one. This is a well-worn path in demonstrating irrationality, but the specifics for developing such contradictions are highly dependent on the number under consideration. Historically, irrationality proofs for the powers of e were developed for specific integer exponents. Euler [1] proved that both e and e2 were irrational in 1737, Liouville showed that e4 was irrational in 1840, and Hurwitz [2] proved that e3 was likewise irrational in 1891, however this was preceded by a comprehensive answer to all questions of this type when the transcendence of e
was established by Hermite [1] [3] in 1873. Observe that if
, then
, hence e solves
, contradicting transcendence. The
method below establishes the same result, namely the irrationality of all non-zero powers of e, more simply by avoiding the necessity of showing the transcendence of e.
2. Overview
Given some
which is to be shown irrational, construct the following polynomial function with the property that it can be translated from zero to
subject to a controllable error by a simple exponential transformation. Let
be the polynomial. Specifically, write
, where the error term can be dominated by a fixed multiple
of the degree of
. The trick in making this work is to define
so that it meshes with the pattern by which the Maclaurin expansion for
restores powers of
to
. The exact manner in which this can be done is the subject of a lemma below. The exponential transform is due to Hermite.
The form of
that emerges from the preceding considerations turns out to be the sum of the derivatives of a more basic polynomial, the Hurwitz polynomial of type
, defined as
. The primality of
is not required by the definition, but in the sequel it will be taken to be prime. So define
, where the sum runs from 0 to
, since all higher der-
ivatives vanish. Applying Leibnitz’ Rule for differentiating products to
, obtain
. Hence
, which is truly a cumbersome
expression to work with.
There is some computational relief provided by the fact that only
needs to be evaluated at the zeroes of
, namely 0 and
. Evaluation at these two points automatically eliminates a considerable number of terms.
The principal steps in the argument are:
1) Assume for the sake of contradiction that
, say
, where
and
2) Show
is an integer divisible by
3) Show
is an integer divisible by
but not by
(iv) Show
recall that the degree of
is
4) If
, say
, where
and
, then show
5) Show
6) Show
7) Establish contradiction by concluding
is a positive
integer strictly less than one for large
This contradiction yields the main theorem which asserts that
and
cannot both be positive rationals. This theorem is used by showing that if one is rational, the other is not. Interesting results flow immediately as corollaries. All positive rational powers of e must be irrational. Natural logarithms of positive rational numbers except 1 cannot be rational.
3. Main Theorem
Validation of the technical details of the preceding program follow. It is assumed
throughout that
and
are as defined above,
, with
and
, and
is a prime.
Lemma 1:
is an integer divisible by
.
Proof: Setting
in
would eliminate all terms except those where
has been differentiated
times. The surviving terms would be of the form
, corresponding to the situation where
has
also been differentiated
times. Hence
.
Simplifying, it is clear that
.
Rewriting
, evidently
,
which is an integer divisible by
, since it divides
for
and
is always an integer, establishing the result.
Lemma 2:
is an integer divisible by
but not by
for sufficiently large p.
Proof: Setting
in
would eliminate all terms except those where
has been differentiated
times. The surviving terms would be of the form
, corresponding to the situation where
has also been differentiated
times. Hence
. Simplifying further it follows
. Rewriting
as above, clearly
. Evidently
divides
, since it divides
for
and
is
always an integer. On the other hand, for the sake of contradiction, if
were to divide
, then it would divide
.
This is absurd, since it cannot divide the right hand side if
is chosen greater than
. Recall
, hence
, is fixed, but
can be an arbitrary prime. The contradiction establishes that
does not divide
and the result follows.
Lemma 3.1: Let
denote the infinite series
. Then
,
where
.
(Note that
is the series for
with the first
terms removed.)
Proof: Observe that
, which is equivalent to
.
Lemma 3.2: Suppose
. Let
be the corresponding sum-of-derivatives polynomial. Then
, with
as in Lemma 3.1.
Proof: By induction on
. If
, then
. There is only the zeroth derivative to sum, so
. Since
.,
evidently
, establishing the base case.
The induction hypothesis is that for any polynomial
of degree
, we have
. It is to be shown that for a polynomial of de- gree
, namely
,
it is true that. Now
, since the only derivatives of
missing from
are those attributable to the leading monomial
. Then
, as all the derivatives except the
of
evaluate to zero at zero. Forming
, we have
by the induction hypothesis. Observe that
. But
, which is precisely
. It follows that
where
as in Lemma 3.1. But the term
by Lemma 3.1, hence it can be absor- bed and evidently
. Recognizing that
, it is now clear that
, as required.
Lemma 3.3: With the notation of Lemma 3.2, for a given
of de-
gree
and fixed
, the function
is
.
Proof: It may be assumed that the fixed
. From Lemma 3.1
. Note that
since
is dominated by the absolutely convergent series
. Then
is bounded, and hence
. It follow that
, where
.
Lemma 3.4: With the notation above,
for some fixed
.
Proof: Lemma 3.2 shows
. Hence for
it must be that
by Lemma
3.3, since the degree of the Hurwitz polynomial
, which defines
via summation of derivatives, is
. The result is now immediate.
Lemma 4: If
, say
, where
and
, then
Moreover, if
as above, then
for sufficiently large p.
Proof: From Lemma 3.4,
, so
. Hence (since
)
Note that
is an integer that is not divisible by
for sufficiently large
by Lemma 2, but that
is an integer that is divisible by
for any large
by Lemma 1. Here
can be chosen large enough to not appear in any prime factorization of
. It follows that
would be a nonzero integer, establishing the claim.
Theorem 1: If
then
.
Proof: To the contrary, suppose
and
, then by the lemmas above,
would be an integer, and for sufficiently large
would satisfy
. In the expression on the right hand side, once
is fixed the terms
, and
are constant. Claim:
. The expression can be rewritten
and it is then apparent that the limit must be
, which is zero due to the factorial dominance over the exponential. So choose a prime
for which
. Now
. But since
divides both
and
but does not, by virtue of its selection, divide
, or
, it is clear that
is an integer. Moreover, it is apparently an integer strictly
between zero and one, which is absurd. The contradiction establishes the theorem.
4. Conclusions
Based on the preceding Theorem 1, the following conclusions are immediate.
Corollary 1: The nonzero rational powers of
are irrational
Proof: For the positive rational powers the statement is true immediately by
Theorem 1. If
, then
cannot be of the form
with
, otherwise
, contrary to Theorem 1.
Corollary 2: The natural logarithm of a positive rational number is irrational
Proof: By the contrapositive of Theorem 1, If
is rational (necessarily positive), then
is irrational.
Corollary 3: If
is rational, then either
is irrational or
Proof: If
is a nonzero rational, then
is irrational. If
, then
.