Friday, 6 December 2019

decimal expansion - I'm puzzled with 0.99999











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:




  1. The hand-wavy but convincing one: Let x=0.999. Then 10x=9.999=9+x. So 9x=9, hence x=1.


  2. 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=9101110=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=0.99992.480000=2.47999991938.01936180000=1938.019361799999


etc.


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