http://www.scirp.org/journal/apm ISSN Online: 2160-0384

ISSN Print: 2160-0368

**On Irrational Points in the Plane **

**Thomas Beatty, Timothy Jones**

Florida Gulf Coast University, Fort Myers, Florida, USA

**Abstract **

This paper summarizes research intended to develop a pedagogically friendly argument that establishes the fact that

### ( )

*x e*,

*x*is never a rational point in the plane. A point

### ( )

2### ,

*x y*

### ∈

###

is rational if both x and y are rational. Applying a method based on Hurwitz polynomials, the research establishes simple irra-tionality proofs for nonzero rational powers of e and logarithms of positive rationals (excluding one).**Keywords **

Rational Point, Irrationality, Sum of Derivatives, Prime, Divisibility, Hurwitz Polynomial

**1. Introduction **

A certain class of polynomials with integer coefficients displays divisibility

prop-erties which can be used to establish irrationality. The purpose of the present

re-search 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

de-veloping such contradictions are highly dependent on the number under

considera-tion. Historically, irrationality proofs for the powers of *e* were developed for

spe-cific integer exponents. Euler [1] proved that both *e* and *e*2 were irrational in 1737,

Liouville showed that *e*4 was irrational in 1840, and Hurwitz [2] proved that *e*3

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

*m*
*n*

*e* ∈, then

*n*
*m*

*m*
*n*

*e* *e*

= ∈

, hence e solves 0

*m* *m*

*x* −*e* = , contradicting transcendence. The

How to cite this paper: Beatty, T. and Jones, T. (2017) On Irrational Points in the Plane. Advances in Pure Mathematics, 7, 299-305.

https://doi.org/10.4236/apm.2017.74016 Received: March 13, 2017

Accepted: April 17, 2017 Published: April 20, 2017 Copyright © 2017 by authors and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

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 α ∈ which is to be shown irrational, construct the following

po-lynomial function with the property that it can be translated from zero to α

sub-ject to a controllable error by a simple exponential transformation. Let

### ( )

### [ ]

*G x* ∈ *x* be the polynomial. Specifically, write e*x*

### ( )

0### ( ) (

error### )

*G* =*G x* + ,
where the error term can be dominated by a fixed multiple β of the degree of

### ( )

*G x* . The trick in making this work is to define *G x*

### ( )

so that it meshes with thepattern by which the Maclaurin expansion for e*x* restores powers of *x* to

### ( )

0*G* . The exact manner in which this can be done is the subject of a lemma

be-low. The exponential transform is due to Hermite.

The form of *G x*

### ( )

that emerges from the preceding considerations turns out tobe the sum of the derivatives of a more basic polynomial, the Hurwitz polynomial

of type *p*, defined as *g x*

### ( )

=*xp*−1

### (

*x*−

### α

### )

*p*. The primality of

*p*is not required by the definition, but in the sequel it will be taken to be prime. So define

### ( )

deg ( )_{( )}

0

*g* *n*
*n*

*G x* =

### ∑

_{=}

*g*

*x*, where the sum runs from 0 to 2

*p*−1, since all higher der-

ivatives vanish. Applying Leibnitz’ Rule for differentiating products to

### ( )

*p*1

### (

### )

*p*

*g x* =*x* − *x*−

### α

, obtain ( )### ( )

### ( )

1 ( )### (

### (

### )

### )

( ) 0*k*

*n k* _{p}

*n*

*n* *p*

*k*
*n*

*g* *x* *x* *x*

*k* α

− − =

= _{ } −

### ∑

. Hence### ( )

2 1### ( )

1 ( )### (

### (

### )

### )

( )0 0

*k*

*n k* *p*

*p* *n* *p*

*n* *k*

*n*

*G x* *x* *x*

*k* α

−

− −

= =

= _{} _{ } − _{}

### ∑

### ∑

, which is truly a cumbersomeexpression to work with.

There is some computational relief provided by the fact that only *G x*

### ( )

needsto be evaluated at the zeroes of *g x*

### ( )

, namely 0 and α . Evaluation at these twopoints automatically eliminates a considerable number of terms.

The principal steps in the argument are:

1) Assume for the sake of contradiction that α∈+, say α =*r s*, where
,

*r s*∈ and gcd

### ( )

*r s*, =1

2) Show *p*1

### ( )

*s* −*G* α is an integer divisible by *p*!
3) Show *p*

### ( )

0*s G* is an integer divisible by

### (

*p*−1 !

### )

but not by*p*!

(iv) Show eα*G*

### ( )

0 −*G*

### ( )

### α

≤### β

### (

2*p*−1

### )

recall that the degree of*g x*

### ( )

is 2*p*−1

4) If eα∈+, say e *u*

*v*

α _{=}

, where *u v*, ∈ and gcd

### ( )

*u v*, =1, then show

### ( )

0### ( )

### (

2 1### )

*uG* −*vG*

### α

≤### β

*p*−

*v*

5) Show 0<_{s uG}p

### ( )

0 −

_{vG}### ( )

### α

≤

_{s}p### β

### (

2*−1*

_{p}### )

_{v}6) Show

### (

### )

### (

### )

2 1

lim 0

1 !

*p*
*p*

*s* *p* *v*

*p*

β

→∞

− = −

7) Establish contradiction by concluding

### ( )

### ( )

### (

### )

0

1 !
*p*

*s uG* *vG*

*p*

### α

−

integer strictly less than one for large *p*

This contradiction yields the main theorem which asserts that α and eα

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

corolla-ries. 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 *g x*

### ( )

and*G x*

### ( )

are as defined above,*r*

*s*

### α

= , with*r s*, ∈

and gcd

### ( )

*r s*, =1, and

*p*is a prime.

**Lemma 1:** *p*1

### ( )

*s* −*G* α is an integer divisible by *p*!.

**Proof:** Setting *x*=α in

### ( )

2 1### ( )

1 ( )### (

### (

### )

### )

( )0 0

*k*

*n k* *p*

*p* *n* *p*

*n* *k*

*n*

*G x* *x* *x*

*k* α
−
− −
= =
= _{} _{ } − _{}

### ∑

### ∑

would eliminate all terms except those where

### (

*x*−

### α

### )

*p*has been differen-tiated

*p*times. The surviving terms would be of the form

### (

### )

### (

1 !### ) ( )

1 !1 !

*p k*

*p* *k* *p*

*p*

*p* *p* *k* α

− −

+ −

_{− −}

, corresponding to the situation where

1

*p*

*x* − has

also been differentiated *k* times. Hence

### ( )

_{(}

### (

### )

_{) ( )}

1
0
1 !
!
1 !
*p k*

*p*

*k*

*p* *k* *p*

*G* *p*

*k* *p* *k*

α α − −

=
+ −
= _{} _{}
− −

### ∑

.Simplifying, it is clear that

### ( )

1### (

### ) (

_{(}

### )

_{) ( )}

1 1### (

### )

### ( )

10 0

1

! 1 !

! !

! ! 1 !

*p k* *p k*

*p* *p*

*k* *k*

*p*

*p* *k* *p*

*G* *p* *p* *k*

*k*

*p k* *p* *k*

α − α − − − α − −

= =
−
+ −
= = + _{} _{}
− − _{} _{}

### ∑

### ∑

.Rewriting *r*

*s*

### α

= , evidently### ( )

### (

### )

1
1
1 1
0
1
!
*p k*
*p*
*p* *p*
*k*
*p* *r*

*s* *G* *p* *k* *s*

*k* *s*

α − − −

− −
=
−
= + _{} _{} _{ }

### ∑

,which is an integer divisible by *p*!, since it divides

### (

*p*+

*k*

### )

! for*k*≥0 and 1

*p*

*k*−

is always an integer, establishing the result.

**Lemma 2:** *p*

### ( )

0*s G* is an integer divisible by

### (

*p*−1 !

### )

but not by*p*! for sufficiently large p.

**Proof:** Setting *x*=0 in

### ( )

2 1### ( )

1 ( )### (

### (

### )

### )

( )0 0

*k*

*n k* *p*

*p* *n* *p*

*n* *k*

*n*

*G x* *x* *x*

*k* α
−
− −
= =
= _{} _{ } − _{}

### ∑

### ∑

would eliminate all terms except those where *p*1

*x* − has been differentiated

1

*p*− times. The surviving terms would be of the form

### (

### )

1_{(}

! _{) (}

### )

1 ! 0

!

*p k*

*p* *k* *p*

*p*

*k* *p* *k* α

− + −

− _{} _{} −

−

, corresponding to the situation where

### (

### )

*p*

*x*−

### α

has also been differentiated*k*times. Hence

### ( ) (

### )

_{0}1

_{(}

! _{) ( )}

0 1 !

!

*p k*
*p*

*k*

*p* *k* *p*

*G* *p*

*k* *p* *k* α

−
=
+ −
= − _{} _{} −
−

### ( ) (

### )

### ( ) (

_{(}

_{)}

### )

_{(}

_{) ( )}

### ( ) (

### )

### ( )

0

0

1 ! !

0 1 ! 1

1 tha ! ! ! 1 t 1 !

*p k* _{p k}

*p*
*k*

*p k* _{p k}

*p*
*k*

*p* *k* *p*

*G* *p*

*p* *k* *p* *k*

*p*
*p* *k*
*k*
α
α
− _{−}
=
− _{−}
=
+ −
= − −
− −
= − + − _{ }

### ∑

### ∑

. Rewriting *r*

*s*

### α

=as above, clearly

### ( )

### ( ) (

### )

0

0 1 1 !

*p k*
*p k*
*p*
*p* *p*
*k*
*p* *r*

*s G* *p* *k* *s*

*k* *s*
−
−
=
= − + − _{ } _{ }

### ∑

. Evidently### (

*p*−1 !

### )

divides*s Gp*

### ( )

0 , since it divides### (

*p*+ −

*k*1 !

### )

for*k*≥0 and

*p*

*k*

is
always an integer. On the other hand, for the sake of contradiction, if *p*! were to

divide *s Gp*

### ( )

0 , then it would divide### ( )

0 1### ( ) (

1 1 !### )

### ( ) (

1 1 !### )

_{0}

### ( )

*p k*

*p k* *p* _{p}

*p*

*p* *p*

*k*

*p* *r* *p*

*s G* *p* *k* *s* *p* *r*

*k* *s*
−
−
=
_{ } _{ } _{ }
− − + − = − −
_{} _{}

### ∑

.This is absurd, since it cannot divide the right hand side if *p* is chosen greater
than *r*. Recall α , hence *r*, is fixed, but *p* can be an arbitrary prime. The

contradiction establishes that *p*! does not divide *p*

### ( )

0*s G* and the result follows.

**Lemma 3.1: **Let ε_{k}

### ( )

*x*denote the infinite series

### (

### )(

### )

### (

### )

2

1

!

1 1 2 !

*j*

*j*

*x* *x* *k x*

*k* *k* *k* *k* *j*

∞ =

+ + =

+ + +

### ∑

+ . Then !### ( )

### ( )

*k*

*k* *k*

*x*

*x* *E* *x*

*k*

### ε

= ,where

### ( )

1!
*j*

*k* *j k*

*x*
*E* *x*
*j*
∞
= +
=

### ∑

.(Note that *E _{k}*

### ( )

*x*is the series for e

*x*with the first

*k*+1 terms removed.)

**Proof:** Observe that

### (

### )

### (

### )

1 1

!

! ! !

*k* *j* *j k*

*j* *j*

*x* *k x* *x*

*k* *k* *j* *k* *j*

+
∞ ∞
= =
=
_{+} _{+}

### ∑

### ∑

, which is equivalent to1
!
*j*
*j k*
*x*
*j*
∞
= +

### ∑

. **Lemma 3.2: **Suppose

### ( )

*n*

_{1}

_{0}

### [ ]

*n*

*g x* =*c x* ++*c x*+*c* ∈ *x* . Let *G x*

### ( )

be the corresponding sum-of-derivatives polynomial. Then### ( )

0### ( )

*n*

_{0}

### ( )

*x* *k*

*k* *k*
*k*

*e G* =*G x* +

### ∑

_{=}

*c x*

### ε

*x*, with ε

_{k}### ( )

*x*as in Lemma 3.1.

**Proof:** By induction on deg*g*. If deg*g*=0, then *g x*

### ( )

=*c*

_{0}. There is only the zeroth derivative to sum, so

*G*

### ( )

0 =*G x*

### ( )

=*c*

_{0}. Since

_{0}

### ( )

e*x*1

*x*

ε = − .,
evidently e*xG*

### ( )

0 = +### (

1### ε

_{0}

### ( )

*x c*

### )

_{0}= +

*c*

_{0}

### ∑

0

_{k}_{=}

_{0}

*c x*

_{0}0

### ε

_{0}

### ( )

*x*, establishing the base case.

The induction hypothesis is that for any polynomial

### ( )

11 1 0

*n*
*n*

*x* *c* *x* *c x* *c*

φ −

−

= ++ + of degree *n*−1, we have

### ( )

### ( )

1### ( )

0

e*x*Φ 0 = Φ *x* +

### ∑

*n*−

_{k}_{=}

*c x*

_{k}*k*

### ε

_{k}*x*. It is to be shown that for a polynomial of de- gree

*n*, namely

### ( )

### ( )

*n*

*n*

*g x* =φ *x* +*c x* , e*xG*

### ( )

0 =*G x*

### ( )

+### ∑

*n*

_{k}_{=}

_{0}

*c x*

_{k}*k*

### ε

_{k}### ( )

*x*it is true that. Now

### ( )

### ( )

### ( )

( )0

*j*

*n* *n*

*n*
*j*

*G x* = Φ *x* +

### ∑

_{=}

*c x*, since the only derivatives of

*G x*

### ( )

missing from Φ### ( )

*x*are those attributable to the leading monomial

*n*

*n*

*c x* . Then

### ( )

0### ( )

0 !*n*

*G* = Φ +*n c* , as all the derivatives except the *nth* of *n*
*n*

zero at zero. Forming e*xG*

### ( )

0 =e*x*

_{}Φ

### ( )

0 +*n c*!

**

_{n}_{}, we have

### ( )

### ( )

1### ( )

0

e*x*_{}Φ 0 +*n c*! _{n}_{}= Φ *x* +

### ∑

*n*−

_{k}_{=}

*c x*

_{k}*k*

### ε

_{k}*x*+e !

*xn c*by the induction hypothesis.

_{n}Observe that

### (

### )

2 1

e ! 1 !

1! 2! ! 1 !

*n* *n*

*x*

*n* *n*

*x* *x* *x* *x*

*n c* *n c*

*n* *n*

+

=_{} + + + + + _{+} + _{}

. But

### (

### )

2

1 2

1 ! 1 ! !

1! 2! !

*n*

*n* *n* *n*

*n* *n* *n* *n* *n* *n*

*x* *x* *x*

*n c* *c x* *nc x* *n n* *c x* *n c x* *n c*
*n*

− −

+ + + + = + + − + +

, which

is precisely

### ( )

( )0

*j*

*n* *n*

*n*
*j*= *c x*

### ∑

. It follows that### ( )

### ( )

### ( )

( ) 1### ( )

### ( )

0 0

e*xG* 0 *x* *nj* *c xn* *n* *j* *nk* *c xk* *k*ε*k* *x* *En* *x n c*! *n*

−

= =

= Φ_{} + _{}+ +

### ∑

### ∑

where

### ( )

1

!
*j*

*n* *j n*

*x*

*E* *x*

*j*

∞ = +

=

### ∑

as in Lemma 3.1. But the term### ( )

!### ( )

!### ( )

!*n*

*n*

*n* *n* *n* *n* *n* *n*

*x*

*E* *x n c* *x* *n c* *c x* *x*

*n* ε ε

=_{} _{} =

by Lemma 3.1, hence it can be absor-

bed and evidently 1

### ( )

### ( )

### ( )

0 ! 0

*n* _{k}*n* _{k}

*k* *k* *n* *n* *k* *k*

*k* *c x*

### ε

*x*

*E*

*x n c*

*k*

*c x*

### ε

*x*

−

= + = =

### ∑

### ∑

. Recognizingthat

### ( )

### ( )

( )### ( )

0

*j*

*n* *n*

*n*
*j*

*x* _{=} *c x* *G x*

Φ +

### ∑

= , it is now clear that### ( )

### ( )

_{0}

### ( )

e*xG* 0 =*G x* +

### ∑

*n*

_{k}_{=}

*c x*

_{k}*k*

### ε

_{k}*x*, as required.

**Lemma 3.3:** With the notation of Lemma 3.2, for a given *g x*

### ( )

∈### [ ]

*x*of de- gree

*n*and fixed

*x*, the function

### ε

### ( )

*n*=

### ∑

*n*

_{k}_{=}

_{0}

*c x*

_{k}_{0}

*k*

### ε

_{k}### ( )

*x*

_{0}is

*O n*

### ( )

.**Proof:** It may be assumed that the fixed *x*≠0. From Lemma 3.1

### ( )

*n*

_{0}

*k*

### ( )

*n*

_{0}

*k*

### ( )

*n*

_{0}

*k*

### ( )

*k* *k* *k* *k* *k* *k*

*k* *k* *k*

*n* *c x* *x* *c x* *x* *c* *x* *x*

ε =

### ∑

_{=}ε ≤

### ∑

_{=}ε =

### ∑

_{=}ε . Note that

### ( )

*k*

*E* *x* < ∞ since *E _{k}*

### ( )

*x*is dominated by the absolutely convergent series

*ex*. Then

_{k}### ( )

!

_{k}### ( )

*k*
*k*

*x* *E* *x*

*x*

ε = is bounded, and hence

### ( )

### ( )

0 max

*n* *k* *k*

*k* *k* *k n* *k* *k*

*k*= *c* *x* ε *x* ≤ ⋅*n* ≤ *c* *x* ε *x*

### ∑

. It follow that### ε

### ( )

*n*≤

*n*

### β

,where max *k*

### ( )

*k n* *ck* *x* *k* *x*

β = ≤ ε .

**Lemma 3.4:** With the notation above, eα*G*

### ( )

0 −*G*

### ( ) (

### α

≤ 2*p*−1

### )

### β

for some fixed β>0.**Proof:** Lemma 3.2 shows e*xG*

### ( )

0 −*G x*

### ( )

=### ∑

*n*

_{k}_{=}

_{0}

*c x*

_{k}*k*

### ε

_{k}### ( )

*x*. Hence for

*x*=α it

must be that

### ( )

### ( )

### ( )

### ( ) (

### )

0

eα*G* 0 −*G* α =

### ∑

*n*

_{k}_{=}

*c*α ε α

_{k}*k*

*= ε*

_{k}*p*≤ 2

*p*−1 β by Lemma 3.3, since the degree of the Hurwitz polynomial

*g x*

### ( )

, which defines*G x*

### ( )

viasummation of derivatives, is 2*p*−1. The result is now immediate.

**Lemma 4:** If eα∈+, say e *u*

*v*

α _{=}

, where *u v*, ∈ and gcd

### ( )

*u v*, =1, then

### ( )

0### ( )

### (

2 1### )

*uG* −*vG*

### α

≤### β

*p*−

*v*Moreover, if

*r*

*s*

### α

= as above, then### ( )

### ( )

### (

### )

0<* _{s uG}p* 0 −

_{vG}### α

≤

_{s}p### β

2*−1*

_{p}

_{v} for sufficiently large *p*.

**Proof:** From Lemma 3.4, e *G*

### ( )

0*G*

### ( )

*uG*

### ( )

0*G*

### ( ) (

2*p*1

### )

*v*

α _{−} _{α} _{=} _{−} _{α} _{≤} _{−} _{β}

, so

### ( )

0### ( )

### (

2 1### )

### ( )

0### ( )

### (

2 1### )

*p* *p*

*s uG* −*vG*

### α

≤*s*

### β

*p*−

*v*Note that

*p*

### ( )

0*s uG* is an integer that is
not divisible by *p*! for sufficiently large *p* by Lemma 2, but that

### ( )

*p*

*s vG* α is an integer that is divisible by *p*! for any large *p* by Lemma 1.
Here *p* can be chosen large enough to not appear in any prime factorization

of *s*. It follows that *s uGp*

### ( )

0 −*vG*

### ( )

### α

would be a nonzero integer, estab-lishing the claim. **Theorem 1:** If α∈+ then *e*α∉+.

**Proof:** To the contrary, suppose α∈+ and *e*α∈+, then by the lemmas
above, *s uGp*

### ( )

0 −*vG*

### ( )

### α

would be an integer, and for sufficiently large*p*

would satisfy 0<*s uGp*

### ( )

0 −*vG*

### ( )

### α

≤*sp*

### β

### (

2*p*−1

### )

*v*. In the expression on the right hand side, once α is fixed the terms

*s u v*

### , ,

, and β are constant.Claim:

### (

### )

### (

### )

2 1

lim 0

1 !

*p*
*p*

*s* *p* *v*

*p*

β

→∞

− =

− . The expression can be rewritten

### (

### )

2

2

2 1

2 ! 1

*p*

*s* *p*

*s* *v*

*p* *p* β

− _{−}

⋅ ⋅

− − and it is then apparent that the limit must be

### (

### )

2 2

2 lim

2 !

*p*
*p*

*s*
*s* *v*

*p*

β ⋅ →∞ _{−}− , which is zero due to the factorial dominance over the

exponential. So choose a prime *p*_{0} for which

### (

### )

### (

2 1 !### )

1 1*p*

*s* *p* *v*

*p*

β −

<

− . Now

### ( )

### ( )

### (

### )

0

0

0

0 1

1 !
*p*

*s* *uG* *vG*

*p*

### α

−

< <

− . But since

### (

*p*0−1 !

### )

divid es bo th*G*

### ( )

0 and### ( )

*G* α but does not, by virtue of its selection, divide

*s u*

### ,

, or*v*, it is clear that

### ( )

### ( )

### (

### )

0

0

0
1 !
*p*

*s* *uG* *vG*

*p*

### α

−

− 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 *e* are irrational

**Proof:** For the positive rational powers the statement is true immediately
by

**Theorem 1.** If γ >0, then e−γ cannot be of the form *r*

*s* with *r s*,

+

∈ ,

otherwise e *s*

*r*

γ _{=}

, contrary to Theorem 1.

**Corollary 2:** The natural logarithm of a positive rational number is irrational

**Proof:** By the contrapositive of Theorem 1, If ln

e γ

γ= is rational (necessarily positive), then lnγ is irrational.

**Corollary 3:** If lnγ +lnδ is rational, then either γδ is irrational or γδ =1

**Proof:** If lnγ +lnδ is a nonzero rational, then ln ln ln( )

e γ+ δ =e γδ =γδ is

**References **

[1] Herstein, I.N. (1975) Topics in Algebra. 2nd Edition, John Wiley & Sons, New York, 175-178.

[2] Hurwitz, A. (1893) Beweis der Transendenz der Zahl e. Mathematische Annalen, 43, 220-221.https://doi.org/10.1007/BF01443646

[3] Hermite, C. (1873) Sur la fonction exponientielle. Comptes Rendus de l'Academie des Sciences, 77, 18-24.

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