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'$e^r$ is irrational for $r\in\mathbb{Q}\setminus\{0\}$'
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| Title of object: |
$e^r$ is irrational for $r\in\mathbb{Q}\setminus\{0\}$ |
| Canonical Name: |
ErIsIrrationalForRinmathbbQsetminus0 |
| Type: |
Theorem |
| Created on: |
2005-03-11 04:47:51 |
| Modified on: |
2005-03-11 15:15:52 |
| Classification: |
msc:11J72 |
| Keywords: |
e, e irrational, e^r, rational, irrational, irrational exponential |
| Synonyms: |
$e^r$ is irrational for $r\in\mathbb{Q}\setminus\{0\}$=$e^r$ is irrational for non-zero rational r
$e^r$ is irrational for $r\in\mathbb{Q}\setminus\{0\}$=irrationality of the exponential function on $\mathbb{Q}$ |
Revision comment (for changes between this and next version):
| Changes for correction #6345 ('Contains own proof'). |
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Content:
\PMlinkescapeword{observation}
\PMlinkescapeword{order}
\PMlinkescapeword{times}
\PMlinkescapeword{simple}
We here present a proof of the following theorem:
\begin{thm} \( e^r \) is irrational for all \( r\in\bbQ\setminus\{0\} \) \end{thm}
To begin with, note that it is sufficient to show that \( e^u \) is irrational for any \PMlinkname{positive integer}{NaturalNumber}\footnote{In this entry, \( \bbN := \{1,2,3,\dotsc\} \) and \(\bbN_0 := \bbN \cup \{0\}\).} \( u \) (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 \( \displaystyle f_n(x) := \frac{x^n (1-x)^n}{n!} \):
\begin{itemize}
\item \( \displaystyle f_n(x)=\frac{1}{n!}\sum_{i=n}^{2n} c_i x^i \), with \( c_i \in \bbZ \) for all \( i \).
\item \(f_n^{(k)}(0)\) and \(f_n^{(k)}(1)\) are integers for all \( k\in\bbN_0 \): as \( 0 \) is a \PMlinkname{root}{Root} of order \( n \), \( f_n^{(k)}(0)=0 \) unless \( n\leq k\leq 2n \), in which case \( f_n^{(k)}(0)=\frac{k!}{n!} c_k \), an integer. Since \( f_n^{(k)}(x)=(-1)^m f_n^{(k)}(1-x) \), the same is true for \( f_n^{(k)}(1) \).
\item For all \( 0<x<1 \) we have \( 0<f_n(x)<\frac{1}{n!} \).
\end{itemize}
Now we can readily prove the theorem:
\begin{proof}
Assume that \( e^u=\frac{a}{b} \) for some \( (a,b)\in\bbN^2 \) 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 \( k>2n \). Differentiating \( F_n(x) \) 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 \( f_n(x) \), \( w_n \) should be an integer for all \( n\in\bbN \), 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.
\end{proof}
The result could also easily have been obtained by starting with \( w_n \) and integrating by parts \( 2n \) times. Note also that much stronger statements are known, such as ``\( e^a \) is transcendental for all \( a\in\bbA\setminus\{0\} \)''\footnote{\(\bbA\) denotes the set of algebraic numbers.}. We conclude this entry with the following evident corollary:
\begin{cor} For all \( r\in\bbQ^{+},\: \log r\) is irrational. \end{cor}
\begin{thebibliography}{2}
\bibitem{AZ} M. Aigner and G. M. Ziegler: \emph{Proofs from THE BOOK}, 3\(^\mathrm{rd}\) edition (2004), Springer-Verlag, 30--31.
\bibitem{HW} G. H. Hardy and E. M. Wright: \emph{An Introduction to the Theory of Numbers}, 5\(^\mathrm{th}\) edition (1979), Oxford University Press, 46--47.
\end{thebibliography} |
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