After reading all the kind answers for this previous question question of mine,
I wonder... How do we get a fraction whose decimal expansion is the simple 0.¯9?
I don't mean to look like kidding or joking (of course, one can teach math with fun so it becomes more interesting), but this series has really raised a flag here, because 99 won't solve this case, although it solves for all other digits (e.g. 0.¯8=89 and so on).
Thanks!
Beco.
Answer
The number 0.9999⋯ is in fact equal to 1, which is why you get 99. See this previous question.
To see it is equal to 1, you can use any number of ideas:
The hand-wavy but convincing one: Let x=0.999⋯. Then 10x=9.999⋯=9+x. So 9x=9, hence x=1.
The formal one. The decimal expansion describes an infinite series. Here we have that
x=∞∑n=1910n.
This is a geometric series with common ration 110 and initial term 910, so
x=∞∑n=1910n=9101−110=910910=1.
In general, a number whose decimal expansion terminates (has a "tail of 0s") always has two decimal expansions, one with a tail of 9s. So:
1.0000⋯=0.9999⋯2.480000⋯=2.4799999⋯1938.01936180000⋯=1938.019361799999⋯
etc.
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