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Dezimalentwicklung/Periodisch/Summe/Aufgabe/Lösung
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Dezimalentwicklung/Periodisch/Summe/Aufgabe
Es ist
0
,
2
13
¯
=
2
10
+
0
,
0
13
¯
=
2
10
+
1
10
⋅
0
,
13
¯
=
2
10
+
1
10
⋅
0
,
131313
¯
=
2
10
+
1
10
⋅
131313
999999
{\displaystyle {}{\begin{aligned}0{,}2{\overline {13}}&={\frac {2}{10}}+0{,}0{\overline {13}}\\&={\frac {2}{10}}+{\frac {1}{10}}\cdot 0{,}{\overline {13}}\\&={\frac {2}{10}}+{\frac {1}{10}}\cdot 0{,}{\overline {131313}}\\&={\frac {2}{10}}+{\frac {1}{10}}\cdot {\frac {131313}{999999}}\end{aligned}}}
und
0
,
127
¯
=
0
,
127127
¯
=
127127
999999
.
{\displaystyle {}{\begin{aligned}0{,}{\overline {127}}&=0{,}{\overline {127127}}\\&={\frac {127127}{999999}}.\end{aligned}}}
Somit ist
0
,
2
13
¯
+
0
,
127
¯
=
2
10
+
1
10
⋅
131313
999999
+
127127
999999
=
2
10
+
1
10
⋅
(
131313
999999
+
1271270
999999
)
=
2
10
+
1
10
⋅
1402583
999999
=
2
10
+
1
10
+
1
10
⋅
402584
999999
=
3
10
+
1
10
⋅
0
,
402584
¯
=
0
,
3
402584
¯
.
{\displaystyle {}{\begin{aligned}0{,}2{\overline {13}}+0{,}{\overline {127}}&={\frac {2}{10}}+{\frac {1}{10}}\cdot {\frac {131313}{999999}}+{\frac {127127}{999999}}\\&={\frac {2}{10}}+{\frac {1}{10}}\cdot {\left({\frac {131313}{999999}}+{\frac {1271270}{999999}}\right)}\\&={\frac {2}{10}}+{\frac {1}{10}}\cdot {\frac {1402583}{999999}}\\&={\frac {2}{10}}+{\frac {1}{10}}+{\frac {1}{10}}\cdot {\frac {402584}{999999}}\\&={\frac {3}{10}}+{\frac {1}{10}}\cdot 0{,}{\overline {402584}}\\&=0{,}3{\overline {402584}}.\end{aligned}}}
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