0.323132313231... ··· Ex.1
Assuming that the given number is a recurring decimal that repeats '3231' forever, we try to find an equivalent fraction.
Sol.1: 0.3231=3231x0.0001 Therefore,
0.323132313231...=3231x0.000100010001... While,
1/9=0.111..., 1/99=0.010101..., 1/999=0.001001001...,
1/9999=0.000100010001... Thus,
0.323132313231...=3231x(1/9999)=3231/9999=(3x3x359)/(3x3x11x101)=359/1111
The answer is: 0.323132313231...=359/1111
Sol.2: 10000·Ex.1=3231.323132313231... So that,
10000·Ex.1 - 1·Ex.1=(10000-1)·Ex.1=3231
That is: 9999·Ex.1=3231, so Ex.1=3231/9999=359/1111
The answer is: 0.323132313231...=359/1111