PlanetMath: $e^r$ is irrational for $r\in\mathbb{Q}\setminus\{0\}$

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is irrational for $r\in\mathbb{Q}\setminus\{0\}$

(Theorem)

We here present a proof of the following theorem:

Theorem 1 is irrational for all $r\in\mathbb{Q}\setminus\{0\}$

To begin with, note that it is sufficient to show that

is irrational for any positive integer ¹

(for if $e^r = e^{\frac{u}{v}}$ were rational, so would $( e^{\frac{u}{v}})^v = e^u$ ). Next, we look at some simple properties of polynomial $\displaystyle f_n(x) := \frac{x^n (1-x)^n}{n!}$ :

$\displaystyle f_n(x)=\frac{1}{n!}\sum_{i=n}^{2n} c_i x^i$ , with $c_i \in \mathbb{Z}$ for all .
$f_n^{(k)}(0)$ and $f_n^{(k)}(1)$ are integers for all $k\in\mathbb{N}_0$ : as is a root of order , $f_n^{(k)}(0)=0$ unless $n\leq k\leq 2n$ , in which case $f_n^{(k)}(0)=\frac{k!}{n!} c_k$ , an integer. Since $f_n^{(k)}(x)=(-1)^k f_n^{(k)}(1-x)$ , the same is true for $f_n^{(k)}(1)$ .
For all we have $0<f_n(x)<\frac{1}{n!}$ .

Now we can readily prove the theorem:

Proof. Assume that $e^u=\frac{a}{b}$ for some $(a,b)\in\mathbb{N}^2$ and let $$ F_n(x) := \sum_{k=0}^\infty (-1)^k u^{2n-k} f_n^{(k)}(x), $$ which is actually a finite sum since $f_n^{(k)}(x)=0$ for all

. Differentiating

yields $F_n'(x)= u^{2n+1} f_n(x) - u F_n(x)$ and thus: $$ \frac{d}{dx}\left[e^{ux} F_n(x)\right] = ue^{ux} F_n(x) + e^{ux} F'_n(x) = u^{2n+1} e^{ux} f_n(x). $$ Now consider the sequence $$ (w_n)_{n\in\bbN} := b\int_0^1 u^{2n+1}e^{ux}f_n(x)\,dx = b\left[ e^{ux} F_n(x) \right]^1_0 = a F_n(1) - b F_n(0). $$ Given the remarks on

should be an integer for all $n\in\mathbb{N}$ , yet it is clear that $w_n < b u^{2n+1}\frac{1}{n!} = \frac{a}{n!} u^{2n+1}$ and so $\lim\limits_{n\to\infty}w_n = 0$ , a contradiction. $\qedsymbol$

The result could also easily have been obtained by starting with and integrating by parts times. Note also that much stronger statements are known, such as `` is transcendental for all $a\in\mathbb{A}\setminus\{0\}$ ''². We conclude this entry with the following evident corollary:

Corollary 1 For all $r\in\mathbb{Q}^{+},\: \log r$ is irrational.

Bibliography

1: M. AIGNER & G. M. ZIEGLER: Proofs from THE BOOK, 3 $^\mathrm{rd}$ edition (2004), Springer-Verlag, 30-31.
2: G. H. HARDY & E. M. WRIGHT: An Introduction to the Theory of Numbers, 5 $^\mathrm{th}$ edition (1979), Oxford University Press, 46-47.

Footnotes

...http://planetmath.org/encyclopedia/NaturalNumber.html ¹: In this entry, $\mathbb{N}:= \{1,2,3,\dotsc\}$ and $\mathbb{N}_0 := \mathbb{N}\cup \{0\}$ .
...''²: $\mathbb{A}$ denotes the set of algebraic numbers.

is irrational for $r\in\mathbb{Q}\setminus\{0\}$ " is owned by Cosmin.

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Other names:

is irrational for non-zero rational r, irrationality of the exponential function on $\mathbb{Q}$

Keywords:

e, e irrational, e^r, rational, irrational, irrational exponential

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Cross-references: algebraic numbers, stronger, contradiction, clear, sequence, sum, finite, integers, polynomial, properties, rational, sufficient, irrational, theorem, proof
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This is version 9 of $e^r$

is irrational for $r\in\mathbb{Q}\setminus\{0\}$ , born on 2005-03-11, modified 2007-08-29.
Object id is 6872, canonical name is ErIsIrrationalForRinmathbbQsetminus0.
Accessed 3983 times total.

Classification:

AMS MSC:

11J72 (Number theory :: Diophantine approximation, transcendental number theory :: Irrationality; linear independence over a field)

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