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Stufe 0: Definieren von 0,9p und zeigen per Induktion, das summe(i=1;n;9/(10^i)) = 1 - 1/10^i)
∑
i
=
1
n
9
10
i
{\displaystyle \sum _{i=1}^{n}{\frac {9}{10^{i}}}}
a
1
=
0.9
,
a
2
=
0.99
,
a
3
=
0.999
,
a
4
=
0.9999
,
…
{\displaystyle a_{1}=0.9,\ a_{2}=0.99,\ a_{3}=0.999,\ a_{4}=0.9999,\ \dots }
b
1
=
1
−
0.9
=
0.1
,
b
2
=
1
−
0.99
=
0.01
,
b
3
=
1
−
0.999
=
0.001
,
b
4
=
1
−
0.9999
=
0.0001
{\displaystyle b_{1}=1-0.9=0.1,\ b_{2}=1-0.99=0.01,\ b_{3}=1-0.999=0.001,\ b_{4}=1-0.9999=0.0001}
∑
i
=
1
n
9
10
i
=
1
−
1
10
n
{\displaystyle \sum _{i=1}^{n}{\frac {9}{10^{i}}}=1-{\frac {1}{10^{n}}}}
∑
i
=
1
n
9
10
i
+
9
10
n
+
1
=
∑
i
=
1
n
+
1
9
10
i
{\displaystyle \sum _{i=1}^{n}{\frac {9}{10^{i}}}+{\frac {9}{10^{n+1}}}=\sum _{i=1}^{n+1}{\frac {9}{10^{i}}}}
∑
i
=
1
n
9
10
i
+
9
10
n
+
1
=
1
−
1
10
n
+
9
10
n
+
1
{\displaystyle \sum _{i=1}^{n}{\frac {9}{10^{i}}}+{\frac {9}{10^{n+1}}}=1-{\frac {1}{10^{n}}}+{\frac {9}{10^{n+1}}}}
∑
n
+
1
i
=
1
9
10
i
=
1
−
1
10
n
+
1
{\displaystyle \sum ^{n+1}{i=1}{\frac {9}{10^{i}}}=1-{\frac {1}{10^{n+1}}}}
1
−
1
10
n
+
9
10
n
+
1
=
1
−
1
10
n
+
1
{\displaystyle 1-{\frac {1}{10^{n}}}+{\frac {9}{10^{n+1}}}=1-{\frac {1}{10^{n+1}}}}
1
−
1
10
n
+
9
10
n
+
1
=
1
−
1
10
n
+
1
{\displaystyle 1-{\frac {1}{10^{n}}}+{\frac {9}{10^{n+1}}}=1-{\frac {1}{10^{n+1}}}}
−
1
10
n
+
9
10
n
+
1
=
−
1
10
n
+
1
{\displaystyle -{\frac {1}{10^{n}}}+{\frac {9}{10^{n+1}}}=-{\frac {1}{10^{n+1}}}}
9
10
n
+
1
−
1
10
n
=
−
1
10
n
+
1
{\displaystyle {\frac {9}{10^{n+1}}}-{\frac {1}{10^{n}}}=-{\frac {1}{10^{n+1}}}}
9
10
n
+
1
−
10
10
n
+
1
=
−
1
10
n
+
1
{\displaystyle {\frac {9}{10^{n+1}}}-{\frac {10}{10^{n+1}}}=-{\frac {1}{10^{n+1}}}}
9
−
10
10
n
+
1
=
−
1
10
n
+
1
{\displaystyle {\frac {9-10}{10^{n+1}}}=-{\frac {1}{10^{n+1}}}}
−
1
10
n
+
1
=
−
1
10
n
+
1
{\displaystyle {\frac {-1}{10^{n+1}}}=-{\frac {1}{10^{n+1}}}}
−
1
10
n
+
1
=
−
1
10
n
+
1
{\displaystyle -{\frac {1}{10^{n+1}}}=-{\frac {1}{10^{n+1}}}}
∑
i
=
1
n
9
10
i
=
1
−
1
10
n
{\displaystyle \sum _{i=1}^{n}{\frac {9}{10^{i}}}=1-{\frac {1}{10^{n}}}}
0
,
9
p
=
lim
n
→
∞
∑
i
=
1
n
9
10
i
=
lim
n
→
∞
1
−
1
10
n
{\displaystyle 0,9p=\lim _{n\rightarrow \infty }\sum _{i=1}^{n}{\frac {9}{10^{i}}}=\lim _{n\rightarrow \infty }1-{\frac {1}{10^{n}}}}
Stufe 1: 0,9p = 1 auf die schnelle Art
0.9p = n | multiplizieren mit 10 10 * 0.9p = 10 * n 9.9p = 10n | 0.9p = n subtrahieren 9.9p - 0.9p = 10n - n 9 = 9n | division durch 9 1 = n | n = 0.9p 1 = 0.9p
Stufe 2: für Skeptiker durch Induktion zeigen, daß 0,9p * 10 = 9,9p
10
∗
0
,
9
p
=
10
∗
l
i
m
n
→
∞
1
−
1
10
n
{\displaystyle 10*0,9p=10*lim_{n\rightarrow \infty }1-{\frac {1}{10^{n}}}}
9
,
9
p
=
lim
n
→
∞
10
−
1
10
n
{\displaystyle 9,9p=\lim _{n\rightarrow \infty }10-{\frac {1}{10^{n}}}}
Stufe 3: für Skeptiker durch Induktion zeigen, daß 9,9p - 0,9p = 9p
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