Die Eulersche Zahl
e
{\displaystyle e}
a
e
2
+
b
e
+
c
=
0
{\displaystyle ae^{2}+be+c=0\,}
für
a
,
b
,
c
∈
Z
{\displaystyle a,b,c\in \mathbb {Z} }
a
,
c
≠
0
{\displaystyle a,c\neq 0}
Wir führen einen Widerspruchsbeweis.
Angenommen es gäbe
a
,
b
,
c
∈
Z
{\displaystyle a,b,c\in \mathbb {Z} }
a
,
c
≠
0
{\displaystyle a,c\neq 0\,}
a
e
2
+
b
e
+
c
=
0
{\displaystyle ae^{2}+be+c=0\,}
Für ein beliebiges
n
>
|
a
|
+
|
c
|
+
1
{\displaystyle n>|a|+|c|+1}
0
=
n
!
(
a
e
+
b
+
c
e
)
=
a
∑
k
=
0
∞
n
!
k
!
+
n
!
b
+
c
∑
k
=
0
∞
(
−
1
)
k
n
!
k
!
=
n
!
b
⏟
∈
Z
+
∑
k
=
0
n
(
a
+
c
(
−
1
)
k
)
n
!
k
!
⏟
∈
Z
+
∑
k
=
n
+
1
∞
(
a
+
c
(
−
1
)
k
)
n
!
k
!
⏟
=:
R
n
.
{\displaystyle {\begin{aligned}0&=n!\,\left(ae+b+{\frac {c}{e}}\right)\\\\&=a\sum _{k=0}^{\infty }{\frac {n!}{k!}}+n!\,b+c\sum _{k=0}^{\infty }(-1)^{k}{\frac {n!}{k!}}\\\\&=\underbrace {n!\,b} _{\in \mathbb {Z} }+\underbrace {\sum _{k=0}^{n}{\Big (}a+c\,(-1)^{k}{\Big )}{\frac {n!}{k!}}} _{\in \mathbb {Z} }+\underbrace {\sum _{k=n+1}^{\infty }{\Big (}a+c\,(-1)^{k}{\Big )}{\frac {n!}{k!}}} _{=:R_{n}}.\end{aligned}}}
Hier wurde von der ersten auf die zweite Zeile die Reihenentwicklung
e
−
1
=
∑
k
=
0
∞
(
−
1
)
k
k
!
{\displaystyle e^{-1}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k!}}}
verwendet.
Also muss
R
n
{\displaystyle R_{n}\,}
|
R
n
|
=
|
∑
k
=
1
∞
(
a
+
c
(
−
1
)
n
+
k
)
n
!
(
n
+
k
)
!
|
≤
(
|
a
|
+
|
c
|
)
∑
k
=
1
∞
n
!
n
k
(
n
+
k
)
!
1
n
k
<
(
|
a
|
+
|
c
|
)
∑
k
=
1
∞
1
n
k
=
|
a
|
+
|
c
|
n
−
1
<
1
{\displaystyle {\begin{aligned}|R_{n}|&=\left|\sum _{k=1}^{\infty }{\Big (}a+c\,(-1)^{n+k}{\Big )}{\frac {n!}{(n+k)!}}\right|\\&\leq (|a|+|c|)\,\sum _{k=1}^{\infty }{\frac {n!\,n^{k}}{(n+k)!}}\,{\frac {1}{n^{k}}}\\&<(|a|+|c|)\,\sum _{k=1}^{\infty }{\frac {1}{n^{k}}}={\frac {|a|+|c|}{n-1}}\\&<1\end{aligned}}}
Also ist
R
n
=
0
{\displaystyle R_{n}=0\,}
n
{\displaystyle n\,}
Damit wäre
0
=
n
R
n
−
1
−
R
n
=
a
+
c
(
−
1
)
n
⇒
a
=
±
c
⇒
a
=
c
=
0
{\displaystyle 0=nR_{n-1}-R_{n}=a+c(-1)^{n}\,\Rightarrow \,a=\pm c\,\Rightarrow \,a=c=0}
◻
{\displaystyle \Box }
Insbesondere kann
e
{\displaystyle e}
quadratischen Zahlkörpers aufgefasst werden.
Rationale Potenzen von e sind irrational
Bearbeiten
Für jede rationale Zahl
a
∈
Q
∖
{
0
}
{\displaystyle a\in \mathbb {Q} \setminus \{0\}}
e
a
{\displaystyle e^{a}}
Wir nehmen zunächst an
a
{\displaystyle a}
Angenommen
e
a
{\displaystyle e^{a}}
p
∈
Z
,
q
∈
N
+
{\displaystyle p\in \mathbb {Z} ,q\in \mathbb {N} _{+}}
e
a
=
p
q
{\displaystyle e^{a}={\frac {p}{q}}}
Sei nun
f
(
x
)
:=
x
n
(
1
−
x
)
n
n
!
{\displaystyle \textstyle f(x):={\frac {x^{n}(1-x)^{n}}{n!}}}
n
{\displaystyle n\,}
Niven-Polynom und
F
(
x
)
:=
∑
k
=
0
∞
(
−
1
)
k
a
2
n
−
k
f
(
k
)
(
x
)
{\displaystyle F(x):=\sum _{k=0}^{\infty }(-1)^{k}\,a^{2n-k}\,f^{(k)}(x)}
Die Ableitung von
F
{\displaystyle F}
F
′
(
x
)
=
∑
k
=
0
∞
(
−
1
)
k
a
2
n
−
k
f
(
k
+
1
)
(
x
)
=
∑
k
=
1
∞
(
−
1
)
k
−
1
a
2
n
+
1
−
k
f
(
k
)
(
x
)
=
a
2
n
+
1
f
(
x
)
−
a
F
(
x
)
{\displaystyle F'(x)=\sum _{k=0}^{\infty }(-1)^{k}\,a^{2n-k}\,f^{(k+1)}(x)=\sum _{k=1}^{\infty }(-1)^{k-1}\,a^{2n+1-k}\,f^{(k)}(x)=a^{2n+1}f(x)-a\,F(x)}
Demzufolge ist
F
′
(
x
)
+
a
F
(
x
)
=
a
2
n
+
1
f
(
x
)
{\displaystyle F'(x)+a\,F(x)=a^{2n+1}f(x)\,}
∫
0
1
F
′
(
x
)
e
a
x
d
x
=
[
F
(
x
)
e
a
x
]
0
1
−
a
∫
0
1
F
(
x
)
e
a
x
d
x
{\displaystyle \int _{0}^{1}F'(x)\,e^{ax}\,dx=\left[F(x)\,e^{ax}\right]_{0}^{1}-a\int _{0}^{1}F(x)\,e^{ax}\,dx}
⇒
∫
0
1
(
F
′
(
x
)
+
a
F
(
x
)
)
e
a
x
d
x
=
F
(
1
)
p
q
−
F
(
0
)
{\displaystyle \Rightarrow \int _{0}^{1}\left(F'(x)+a\,F(x)\right)e^{ax}\,dx=F(1)\,{\frac {p}{q}}-F(0)}
q
⋅
a
2
n
+
1
∫
0
1
f
(
x
)
e
a
x
d
x
=
F
(
1
)
⋅
p
−
F
(
0
)
⋅
q
{\displaystyle \textstyle q\cdot a^{2n+1}\int _{0}^{1}f(x)\,e^{ax}\,dx=F(1)\cdot p-F(0)\cdot q}
∫
0
1
f
(
x
)
e
a
x
d
x
<
1
n
!
e
a
{\displaystyle \textstyle \int _{0}^{1}f(x)\,e^{ax}\,dx<{\frac {1}{n!}}\,e^{a}}
n
→
∞
{\displaystyle n\to \infty \,}
Also ist
e
a
{\displaystyle e^{a}}
e
−
a
{\displaystyle e^{-a}}
Wäre
e
a
b
{\displaystyle e^{\frac {a}{b}}}
b
∈
N
+
{\displaystyle b\in \mathbb {N} _{+}}
e
a
{\displaystyle e^{a}\,}
◻
{\displaystyle \Box }
Die Eulersche Zahl
e
{\displaystyle e}
transzendent über
Q
{\displaystyle \mathbb {Q} }
e
{\displaystyle e}
a
0
+
a
1
e
+
.
.
.
+
a
n
e
n
=
0
{\displaystyle \,a_{0}+a_{1}e+...+a_{n}e^{n}=0}
mit
n
∈
N
+
{\displaystyle n\in \mathbb {N} _{+}}
a
k
∈
Z
{\displaystyle a_{k}\in \mathbb {Z} }
a
0
,
a
n
≠
0
{\displaystyle a_{0},a_{n}\neq 0\,}
Unter der Annahme
e
{\displaystyle e}
algebraisch gibt es eine Gleichung
a
0
+
a
1
x
+
.
.
.
+
a
n
x
n
=
0
{\displaystyle \,a_{0}+a_{1}x+...+a_{n}x^{n}=0}
a
k
∈
Z
{\displaystyle a_{k}\in \mathbb {Z} }
a
0
,
a
n
≠
0
{\displaystyle a_{0},a_{n}\neq 0\,}
die
x
=
e
{\displaystyle \,x=e}
α
:=
2
(
n
+
1
)
max
|
a
k
|
{\displaystyle \alpha :=2\,(n+1)\,\max |a_{k}|}
Q
(
x
)
:=
∏
k
=
1
n
(
x
−
k
)
=
x
n
±
.
.
.
±
n
!
[
⟹
Q
p
(
x
)
=
x
n
p
±
.
.
.
±
n
!
p
]
.
{\displaystyle Q(x):=\prod _{k=1}^{n}(x-k)=x^{n}\pm ...\pm n!\quad {\Big [}\Longrightarrow Q^{p}(x)=x^{np}\pm ...\pm n!^{p}{\Big ]}.}
Für variables
p
∈
P
{\displaystyle p\in \mathbb {P} }
F
(
x
)
:=
1
(
p
−
1
)
!
x
p
−
1
Q
p
(
x
)
e
−
x
{\displaystyle F(x):={\frac {1}{(p-1)!}}x^{p-1}Q^{p}(x)e^{-x}}
q
∈
{
0
,
1
,
.
.
.
,
n
}
{\displaystyle \,q\in \{0,1,...,n\}}
ε
p
q
:=
e
q
∫
0
q
F
d
x
{\displaystyle \varepsilon _{pq}:=e^{q}\int _{0}^{q}F\,dx}
M
p
q
:=
e
q
∫
q
∞
F
d
x
⟹
ε
p
k
+
M
p
k
=
e
k
∫
0
∞
F
d
x
=
e
k
M
p
0
{\displaystyle M_{pq}:=e^{q}\int _{q}^{\infty }F\,dx\;\Longrightarrow \varepsilon _{pk}+M_{pk}=e^{k}\int _{0}^{\infty }F\,dx=e^{k}M_{p0}}
∀
x
∈
[
0
,
n
]
{\displaystyle \forall x\in [0,n]}
F
(
x
)
→
0
{\displaystyle F(x)\to 0\,}
p
→
∞
{\displaystyle p\to \infty \,}
|
ε
p
q
|
→
0
{\displaystyle |\varepsilon _{pq}|\to 0}
Man wähle die Primzahl
p
{\displaystyle \,p}
p
>
α
{\displaystyle \,p>\alpha }
|
ε
p
q
|
<
1
α
∀
q
{\displaystyle |\varepsilon _{pq}|<{\frac {1}{\alpha }}\;\forall q}
Es ist:
M
p
0
=
∫
0
∞
F
d
x
=
1
(
p
−
1
)
!
∫
0
∞
x
p
−
1
(
x
n
p
±
.
.
.
±
n
!
p
)
e
−
x
d
x
=
1
(
p
−
1
)
!
(
(
p
−
1
+
n
p
)
!
±
.
.
.
±
n
!
p
(
p
−
1
)
!
)
≡
n
!
p
≡
n
!
≢
0
mod
p
{\displaystyle {\begin{aligned}M_{p0}&=\int _{0}^{\infty }F\,dx={\frac {1}{(p-1)!}}\int _{0}^{\infty }x^{p-1}(x^{np}\pm ...\pm n!^{p})e^{-x}dx\\&={\frac {1}{(p-1)!}}{\Big (}(p-1+np)!\pm ...\pm n!^{p}(p-1)!{\Big )}\equiv n!^{p}\equiv n!\not \equiv 0\operatorname {mod} \,p\end{aligned}}}
Für
q
=
1
,
.
.
.
,
n
{\displaystyle \,q=1,...,n}
Q
(
x
+
q
)
=
∏
k
=
1
n
(
x
+
q
−
k
)
{\displaystyle Q(x+q)=\prod _{k=1}^{n}(x+q-k)}
x
{\displaystyle \,x}
M
p
q
=
e
q
∫
q
∞
F
d
x
=
e
q
∫
0
∞
F
(
x
+
q
)
d
x
=
1
(
p
−
1
)
!
∫
0
∞
(
x
+
q
)
p
−
1
Q
p
(
x
+
q
)
e
−
x
d
x
=
1
(
p
−
1
)
!
∫
0
∞
x
p
R
(
x
)
e
−
x
d
x
=
1
(
p
−
1
)
!
(
r
n
p
−
1
(
n
p
−
1
)
!
±
.
.
.
±
r
0
p
!
)
≡
0
mod
p
.
{\displaystyle {\begin{aligned}M_{pq}&=e^{q}\int _{q}^{\infty }F\,dx=e^{q}\int _{0}^{\infty }F(x+q)\,dx\\&={\frac {1}{(p-1)!}}\int _{0}^{\infty }(x+q)^{p-1}Q^{p}(x+q)e^{-x}\,dx\\&={\frac {1}{(p-1)!}}\int _{0}^{\infty }x^{p}R(x)e^{-x}dx={\frac {1}{(p-1)!}}{\Big (}r_{np-1}(np-1)!\pm ...\pm r_{0}p!{\Big )}\equiv 0\operatorname {mod} \,p\,.\end{aligned}}}
0
=
M
p
0
∑
k
=
0
n
a
k
e
k
=
∑
k
=
0
n
a
k
(
ε
p
k
+
M
p
k
)
=
∑
k
=
0
n
a
k
ε
p
k
+
a
0
M
p
0
+
∑
k
=
1
n
a
k
M
p
k
{\displaystyle 0=M_{p0}\sum _{k=0}^{n}a_{k}e^{k}=\sum _{k=0}^{n}a_{k}{\big (}\varepsilon _{pk}+M_{pk}{\big )}=\sum _{k=0}^{n}a_{k}\varepsilon _{pk}+a_{0}M_{p0}+\sum _{k=1}^{n}a_{k}M_{pk}}
⟹
−
∑
k
=
0
n
a
k
ε
p
k
≡
a
0
M
p
0
≢
0
mod
p
{\displaystyle \Longrightarrow -\sum _{k=0}^{n}a_{k}\varepsilon _{pk}\equiv a_{0}M_{p0}\not \equiv 0\operatorname {mod} p}
Aber
|
∑
k
=
0
n
a
k
ε
p
k
|
≤
∑
k
=
0
n
|
a
k
|
1
α
≤
(
n
+
1
)
max
|
a
k
|
α
=
1
2
{\displaystyle \left|\sum _{k=0}^{n}a_{k}\varepsilon _{pk}\right|\leq \sum _{k=0}^{n}|a_{k}|{\frac {1}{\alpha }}\leq {\frac {(n+1)\max |a_{k}|}{\alpha }}={\frac {1}{2}}}
◻
{\displaystyle \Box }
![{\displaystyle \int _{0}^{1}F'(x)\,e^{ax}\,dx=\left[F(x)\,e^{ax}\right]_{0}^{1}-a\int _{0}^{1}F(x)\,e^{ax}\,dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d845163ae0aaf3abb18bb34c1316ef47007898a0)
![{\displaystyle Q(x):=\prod _{k=1}^{n}(x-k)=x^{n}\pm ...\pm n!\quad {\Big [}\Longrightarrow Q^{p}(x)=x^{np}\pm ...\pm n!^{p}{\Big ]}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3019f0f0f6b141444dc2fab14d437109eac1b6a)
![{\displaystyle \forall x\in [0,n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7700237f9aaeba4f62b88045b8e1d90a31fccf4)
