Ein System
K
{\displaystyle \mathbb {K} }
Körper genau dann, wenn es zu je zwei Elementen
a
,
b
∈
K
{\displaystyle a,b\in \mathbb {K} }
a
+
b
∈
K
{\displaystyle a+b\in \mathbb {K} }
a
b
∈
K
{\displaystyle ab\in \mathbb {K} }
(
K
1
)
,
(
K
2
)
,
(
K
3
)
{\displaystyle (K_{1}),(K_{2}),(K_{3})}
1. Axiome der Addition
(
K
1
)
{\displaystyle (K_{1})}
a) Assoziativgesetz: Für alle
a
,
b
,
c
∈
K
{\displaystyle a,b,c\in \mathbb {K} }
(
a
+
b
)
+
c
=
a
+
(
b
+
c
)
{\displaystyle (a+b)+c=a+(b+c)}
b) Kommutativgesetz: Für alle
a
,
b
∈
K
{\displaystyle a,b\in \mathbb {K} }
a
+
b
=
b
+
a
{\displaystyle a+b=b+a}
c) Existenz des additiv neutralen (Null-)Elements: Es existiert ein neutrales Element
0
∈
K
{\displaystyle 0\in \mathbb {K} }
a
∈
K
{\displaystyle a\in \mathbb {K} }
a
+
0
=
a
{\displaystyle a+0=a}
d) Existenz des additiv inversen (negativen) Elements: Zu jedem
x
∈
K
{\displaystyle x\in \mathbb {K} }
y
∈
K
{\displaystyle y\in \mathbb {K} }
x
+
y
=
0
{\displaystyle x+y=0}
y
=:
−
x
{\displaystyle y=:-x}
2. Axiome der Multiplikation
(
K
2
)
{\displaystyle (K_{2})}
a) Assoziativgesetz: Für alle
a
,
b
,
c
∈
K
{\displaystyle a,b,c\in \mathbb {K} }
(
a
b
)
c
=
a
(
b
c
)
{\displaystyle (ab)c=a(bc)}
b) Kommutativgesetz: Für alle
a
,
b
∈
K
{\displaystyle a,b\in \mathbb {K} }
a
b
=
b
a
{\displaystyle ab=ba}
c) Existenz des multiplikativ neutralen (Eins-)elements: Es existiert ein neutrales Element
1
∈
K
{\displaystyle 1\in \mathbb {K} }
a
∈
K
{\displaystyle a\in \mathbb {K} }
a
⋅
1
=
a
{\displaystyle a\cdot 1=a}
d) Existenz des multiplikativ inversen (reziproken) Elements: Zu jedem
x
∈
K
∖
{
0
}
{\displaystyle x\in \mathbb {K} \setminus \{0\}}
y
∈
K
{\displaystyle y\in \mathbb {K} }
x
⋅
y
=
1
{\displaystyle x\cdot y=1}
y
=:
x
−
1
{\displaystyle y=:x^{-1}}
3. Distributivgesetz
(
K
3
)
{\displaystyle (K_{3})}
Für alle
a
,
b
,
c
∈
K
{\displaystyle a,b,c\in \mathbb {K} }
(
a
+
b
)
c
=
a
c
+
b
c
{\displaystyle (a+b)c=ac+bc}
Aus den Körperaxiomen lassen sich weitere Eigenschaften der Elemente von
K
{\displaystyle \mathbb {K} }
(1) Für beliebige
a
,
b
∈
K
{\displaystyle a,b\in \mathbb {K} }
a
+
x
=
b
{\displaystyle a+x=b}
(2) Für beliebige
a
∈
K
∖
{
0
}
{\displaystyle a\in \mathbb {K} \setminus \{0\}}
b
∈
K
{\displaystyle b\in \mathbb {K} }
a
⋅
y
=
b
{\displaystyle a\cdot y=b}
(3) Für alle
x
∈
K
{\displaystyle x\in \mathbb {K} }
x
⋅
0
=
0
{\displaystyle x\cdot 0=0}
(
−
1
)
⋅
x
=
−
x
{\displaystyle (-1)\cdot x=-x}
(4) Für alle
x
∈
K
{\displaystyle x\in \mathbb {K} }
−
(
−
x
)
=
x
{\displaystyle -(-x)=x}
(5) Für alle
x
,
y
∈
K
∖
{
0
}
{\displaystyle x,y\in \mathbb {K} \setminus \{0\}}
x
y
≠
0
{\displaystyle xy\neq 0}
Nach
(
K
1
)
{\displaystyle (K_{1})}
a
∈
K
{\displaystyle a\in \mathbb {K} }
−
a
∈
K
{\displaystyle -a\in \mathbb {K} }
a
+
x
=
b
{\displaystyle a+x=b}
(
−
a
)
{\displaystyle (-a)}
(
−
a
)
+
a
+
x
=
(
−
a
)
+
b
{\displaystyle (-a)+a+x=(-a)+b}
(
K
1
)
{\displaystyle (K_{1})}
0
+
x
=
x
=
b
+
(
−
a
)
=:
b
−
a
{\displaystyle 0+x=x=b+(-a)=:b-a}
x
=
b
−
a
{\displaystyle x=b-a}
a
+
x
=
b
{\displaystyle a+x=b}
(
K
1
)
{\displaystyle (K_{1})}
a
+
x
=
a
+
[
b
+
(
−
a
)
]
=
(
a
+
b
)
+
(
−
a
)
=
(
b
+
a
)
+
(
−
a
)
=
b
+
[
a
+
(
−
a
)
]
=
b
+
0
=
b
{\displaystyle a+x=a+[b+(-a)]=(a+b)+(-a)=(b+a)+(-a)=b+[a+(-a)]=b+0=b}
.
Damit hat man die Existenz einer Lösung nachgewiesen.
q.e.d.
Satz 4 (Ungleichung von Cauchy-Schwarz)
Bearbeiten
Wenn
a
k
,
b
k
∈
R
{\displaystyle a_{k},b_{k}\in \mathbb {R} }
k
=
1
,
2
,
.
.
.
,
n
{\displaystyle k=1,2,...,n}
(
∑
k
=
1
n
a
k
b
k
)
2
≤
(
∑
k
=
1
n
a
k
2
)
⋅
(
∑
k
=
1
n
b
k
2
)
{\displaystyle \left(\sum _{k=1}^{n}a_{k}b_{k}\right)^{2}\leq \left(\sum _{k=1}^{n}a_{k}^{2}\right)\cdot \left(\sum _{k=1}^{n}b_{k}^{2}\right)}
.
Wir betrachten die Funktion
t
↦
f
(
t
)
=
∑
k
=
1
n
(
a
k
t
+
b
k
)
2
.
{\displaystyle t\mapsto f(t)=\sum _{k=1}^{n}(a_{k}t+b_{k})^{2}.}
Nach Definition ist klar, dass
f
(
t
)
≥
0
{\displaystyle f(t)\geq 0}
t
{\displaystyle t}
∑
k
=
1
n
(
a
k
t
+
b
k
)
2
=
t
2
⋅
∑
k
=
1
n
a
k
2
+
2
t
⋅
∑
k
=
1
n
a
k
b
k
+
∑
k
=
1
n
b
k
2
=
t
2
⋅
U
+
2
t
⋅
V
+
W
{\displaystyle \sum _{k=1}^{n}(a_{k}t+b_{k})^{2}=t^{2}\cdot \sum _{k=1}^{n}a_{k}^{2}+2t\cdot \sum _{k=1}^{n}a_{k}b_{k}+\sum _{k=1}^{n}b_{k}^{2}=t^{2}\cdot U+2t\cdot V+W}
zeigt, dass
f
{\displaystyle f}
(
2
V
)
2
−
4
U
W
=
4
(
V
2
−
U
W
)
{\displaystyle (2V)^{2}-4UW=4(V^{2}-UW)}
nichtpositiv ist, wenn also
V
2
≤
U
W
{\displaystyle V^{2}\leq UW}
q.e.d.
Für
n
∈
N
0
{\displaystyle n\in \mathbb {N} _{0}}
n
{\displaystyle n}
0
!
:=
1
{\displaystyle 0!:=1}
1
!
:=
1
{\displaystyle 1!:=1}
2
!
:=
2
⋅
1
=
2
{\displaystyle 2!:=2\cdot 1=2}
.
.
.
{\displaystyle ...}
n
!
:=
∏
k
=
1
n
k
{\displaystyle n!:=\prod _{k=1}^{n}k}
Weiter erklären wir für
k
,
n
∈
N
0
{\displaystyle k,n\in \mathbb {N} _{0}}
Binomialkoeffizienten
(23)
(
n
k
)
:=
n
!
k
!
(
n
−
k
)
!
=
n
⋅
(
n
−
1
)
⋅
.
.
.
⋅
(
n
−
k
+
1
)
k
!
{\displaystyle {\begin{pmatrix}n\\k\end{pmatrix}}:={\frac {n!}{k!(n-k)!}}={\frac {n\cdot (n-1)\cdot ...\cdot (n-k+1)}{k!}}}
.
Wegen
(
n
k
)
+
(
n
k
−
1
)
=
(
23
)
n
!
k
!
(
n
−
k
)
!
+
n
!
(
k
−
1
)
!
(
n
−
k
+
1
)
!
{\displaystyle {\begin{pmatrix}n\\k\end{pmatrix}}+{\begin{pmatrix}n\\k-1\end{pmatrix}}{\stackrel {(23)}{=}}{\frac {n!}{k!(n-k)!}}+{\frac {n!}{(k-1)!(n-k+1)!}}}
=
n
!
k
!
(
n
−
k
+
1
)
!
[
(
n
−
k
+
1
)
+
k
]
=
(
n
+
1
)
!
k
!
[
(
n
+
1
)
−
k
]
!
{\displaystyle ={\frac {n!}{k!(n-k+1)!}}[(n-k+1)+k]={\frac {(n+1)!}{k![(n+1)-k]!}}}
=
(
n
+
1
k
)
{\displaystyle ={\begin{pmatrix}n+1\\k\end{pmatrix}}}
gilt für alle
k
,
n
∈
N
{\displaystyle k,n\in \mathbb {N} }
Additionstheorem für die Binomialkoeffizienten :
(24)
1
≤
k
≤
n
⇒
(
n
k
)
=
(
n
−
1
k
−
1
)
+
(
n
−
1
k
)
{\displaystyle 1\leq k\leq n\Rightarrow {\begin{pmatrix}n\\k\end{pmatrix}}={\begin{pmatrix}n-1\\k-1\end{pmatrix}}+{\begin{pmatrix}n-1\\k\end{pmatrix}}}
.
Satz 5 (Binomischer Lehrsatz)
Bearbeiten
Für alle
n
∈
N
{\displaystyle n\in \mathbb {N} }
a
,
b
∈
K
{\displaystyle a,b\in \mathbb {K} }
(25)
(
a
+
b
)
n
=
∑
k
=
0
n
(
n
k
)
a
k
b
n
−
k
{\displaystyle (a+b)^{n}=\sum _{k=0}^{n}{\begin{pmatrix}n\\k\end{pmatrix}}a^{k}b^{n-k}}
.
Sei
b
=
0
{\displaystyle b=0}
(
a
+
0
)
n
=
∑
k
=
0
n
(
n
k
)
a
k
0
n
−
k
=
(
n
n
)
a
n
0
0
=
a
n
{\displaystyle (a+0)^{n}=\sum _{k=0}^{n}{\begin{pmatrix}n\\k\end{pmatrix}}a^{k}0^{n-k}={\begin{pmatrix}n\\n\end{pmatrix}}a^{n}0^{0}=a^{n}}
offenbar erfüllt. Sei also
b
≠
0
{\displaystyle b\neq 0}
b
−
n
{\displaystyle b^{-n}}
(
a
+
b
)
n
⋅
b
−
n
=
[
b
(
a
b
+
1
)
]
n
⋅
b
−
n
=
(
a
b
+
1
)
n
{\displaystyle (a+b)^{n}\cdot b^{-n}=\left[b\left({\frac {a}{b}}+1\right)\right]^{n}\cdot b^{-n}=\left({\frac {a}{b}}+1\right)^{n}}
=
b
−
n
⋅
∑
k
=
0
n
(
n
k
)
a
k
b
n
−
k
=
∑
k
=
0
n
(
n
k
)
a
k
b
−
k
=
∑
k
=
0
n
(
n
k
)
(
a
b
)
k
{\displaystyle =b^{-n}\cdot \sum _{k=0}^{n}{\begin{pmatrix}n\\k\end{pmatrix}}a^{k}b^{n-k}=\sum _{k=0}^{n}{\begin{pmatrix}n\\k\end{pmatrix}}a^{k}b^{-k}=\sum _{k=0}^{n}{\begin{pmatrix}n\\k\end{pmatrix}}\left({\frac {a}{b}}\right)^{k}}
.
Mit Hilfe der Substitution
z
:=
a
b
∈
K
{\displaystyle z:={\frac {a}{b}}\in \mathbb {K} }
(26)
(
z
+
1
)
n
=
∑
k
=
0
n
(
n
k
)
z
k
{\displaystyle (z+1)^{n}=\sum _{k=0}^{n}{\begin{pmatrix}n\\k\end{pmatrix}}z^{k}}
zu zeigen.
Wir beweisen nun durch vollständige Induktion, dass (26) für alle
n
∈
N
{\displaystyle n\in \mathbb {N} }
H
(
n
)
{\displaystyle H(n)}
(IA) Für
n
0
=
1
{\displaystyle n_{0}=1}
z
∈
K
{\displaystyle z\in \mathbb {K} }
(
1
+
z
)
1
=
∑
k
=
0
1
(
1
k
)
z
k
=
(
1
0
)
z
0
+
(
1
1
)
z
1
=
1
+
z
{\displaystyle (1+z)^{1}=\sum _{k=0}^{1}{\begin{pmatrix}1\\k\end{pmatrix}}z^{k}={\begin{pmatrix}1\\0\end{pmatrix}}z^{0}+{\begin{pmatrix}1\\1\end{pmatrix}}z^{1}=1+z}
.
(IS) Nach Induktionsvoraussetzung gilt (26) für ein beliebiges
n
∈
N
{\displaystyle n\in \mathbb {N} }
(
z
+
1
)
n
+
1
=
(
z
+
1
)
⋅
(
z
+
1
)
n
=
(
I
V
)
(
z
+
1
)
⋅
[
∑
k
=
0
n
(
n
k
)
z
k
]
{\displaystyle (z+1)^{n+1}=(z+1)\cdot (z+1)^{n}{\stackrel {(IV)}{=}}(z+1)\cdot \left[\sum _{k=0}^{n}{\begin{pmatrix}n\\k\end{pmatrix}}z^{k}\right]}
=
(
K
3
)
∑
k
=
0
n
(
n
k
)
z
k
+
∑
k
=
0
n
(
n
k
)
z
k
+
1
{\displaystyle {\stackrel {(K_{3})}{=}}\sum _{k=0}^{n}{\begin{pmatrix}n\\k\end{pmatrix}}z^{k}+\sum _{k=0}^{n}{\begin{pmatrix}n\\k\end{pmatrix}}z^{k+1}}
=
l
:=
k
+
1
∑
k
=
0
n
(
n
k
)
z
k
+
∑
l
=
1
n
(
n
l
−
1
)
z
l
+
(
n
n
)
z
n
+
1
{\displaystyle {\stackrel {l:=k+1}{=}}\sum _{k=0}^{n}{\begin{pmatrix}n\\k\end{pmatrix}}z^{k}+\sum _{l=1}^{n}{\begin{pmatrix}n\\l-1\end{pmatrix}}z^{l}+{\begin{pmatrix}n\\n\end{pmatrix}}z^{n+1}}
=
(
n
0
)
z
0
+
∑
k
=
1
n
[
(
n
k
)
+
(
n
k
−
1
)
]
z
k
+
(
n
n
)
z
n
+
1
{\displaystyle ={\begin{pmatrix}n\\0\end{pmatrix}}z^{0}+\sum _{k=1}^{n}\left[{\begin{pmatrix}n\\k\end{pmatrix}}+{\begin{pmatrix}n\\k-1\end{pmatrix}}\right]z^{k}+{\begin{pmatrix}n\\n\end{pmatrix}}z^{n+1}}
=
(
24
)
1
+
∑
k
=
1
n
(
n
k
)
z
k
+
(
n
+
1
n
+
1
)
z
n
+
1
=
∑
k
=
1
n
+
1
(
n
+
1
k
)
z
k
{\displaystyle {\stackrel {(24)}{=}}1+\sum _{k=1}^{n}{\begin{pmatrix}n\\k\end{pmatrix}}z^{k}+{\begin{pmatrix}n+1\\n+1\end{pmatrix}}z^{n+1}=\sum _{k=1}^{n+1}{\begin{pmatrix}n+1\\k\end{pmatrix}}z^{k}}
.
Damit ist Satz 5 nach dem Prinzip der vollständigen Induktion bewiesen.
q.e.d.
Beispiel 6 (Teleskopsummen)
Bearbeiten
Seien
m
,
n
∈
N
{\displaystyle m,n\in \mathbb {N} }
m
<
n
{\displaystyle m<n}
{
b
i
}
{\displaystyle \{b_{i}\}}
i
=
m
,
m
+
1
,
.
.
.
,
n
+
1
{\displaystyle i=m,m+1,...,n+1}
{
a
i
}
{\displaystyle \{a_{i}\}}
a
i
:=
b
i
+
1
−
b
i
{\displaystyle a_{i}:=b_{i+1}-b_{i}}
m
≤
i
≤
n
{\displaystyle m\leq i\leq n}
(27)
∑
i
=
m
n
a
i
=
∑
i
=
m
n
(
b
i
+
1
−
b
i
)
=
∑
i
=
m
n
b
i
+
1
−
∑
i
=
m
n
b
i
{\displaystyle \sum _{i=m}^{n}a_{i}=\sum _{i=m}^{n}(b_{i+1}-b_{i})=\sum _{i=m}^{n}b_{i+1}-\sum _{i=m}^{n}b_{i}}
=
(
b
m
+
1
+
b
m
+
2
+
.
.
.
+
b
n
+
1
)
−
(
b
m
+
b
m
+
1
+
b
m
+
2
+
.
.
.
+
b
n
)
=
b
n
+
1
−
b
m
{\displaystyle =(b_{m+1}+b_{m+2}+...+b_{n+1})-(b_{m}+b_{m+1}+b_{m+2}+...+b_{n})=b_{n+1}-b_{m}}
Für
b
i
:=
i
2
{\displaystyle b_{i}:=i^{2}}
m
:=
1
{\displaystyle m:=1}
a
i
=
(
i
+
1
)
2
−
i
2
=
2
i
+
1
{\displaystyle a_{i}=(i+1)^{2}-i^{2}=2i+1}
und
∑
i
=
1
n
a
i
=
∑
i
=
1
n
(
2
i
+
1
)
=
n
+
2
⋅
(
∑
i
=
1
n
i
)
{\displaystyle \sum _{i=1}^{n}a_{i}=\sum _{i=1}^{n}(2i+1)=n+2\cdot \left(\sum _{i=1}^{n}i\right)}
.
Andererseits ist nach (27)
∑
i
=
1
n
(
2
i
+
1
)
=
(
n
+
1
)
2
−
1
=
n
2
+
2
n
{\displaystyle \sum _{i=1}^{n}(2i+1)=(n+1)^{2}-1=n^{2}+2n}
,
woraus sich unmittelbar die Gauß-Formel
∑
k
=
1
n
k
=
n
2
(
n
+
1
)
{\displaystyle \sum _{k=1}^{n}k={\frac {n}{2}}(n+1)}
ergibt.
![{\displaystyle a+x=a+[b+(-a)]=(a+b)+(-a)=(b+a)+(-a)=b+[a+(-a)]=b+0=b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c42fe83de4aad028a42c475d1f700ec5a420283a)
![{\displaystyle ={\frac {n!}{k!(n-k+1)!}}[(n-k+1)+k]={\frac {(n+1)!}{k![(n+1)-k]!}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb0e878cdaae90cd81f42b9adfba2061cea0ce7d)
![{\displaystyle (a+b)^{n}\cdot b^{-n}=\left[b\left({\frac {a}{b}}+1\right)\right]^{n}\cdot b^{-n}=\left({\frac {a}{b}}+1\right)^{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0fc3f57664af4ffa0f12cd80693252a1813e5915)
![{\displaystyle (z+1)^{n+1}=(z+1)\cdot (z+1)^{n}{\stackrel {(IV)}{=}}(z+1)\cdot \left[\sum _{k=0}^{n}{\begin{pmatrix}n\\k\end{pmatrix}}z^{k}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c2f7b2049485661fad15a756d2649c0e44dbeb7)
![{\displaystyle ={\begin{pmatrix}n\\0\end{pmatrix}}z^{0}+\sum _{k=1}^{n}\left[{\begin{pmatrix}n\\k\end{pmatrix}}+{\begin{pmatrix}n\\k-1\end{pmatrix}}\right]z^{k}+{\begin{pmatrix}n\\n\end{pmatrix}}z^{n+1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/58104c890fd4b6870c032c7d066128b29ccd4c71)
