A question of numbers (one for the maths fans)

It also works for any decimal expansion that has a repeating sequence. Only instead of 10x, its (whatever)*x, where the whatever is large enough to bring the repeating parts in line. For example,x = 0.123412341234...
10000x = 1234.123412341234...10000x-x = 9999x = 1234x=1234/9999This is why every decimal that consists of a repeating segment is a rational number, that is, can be written as an ordinary fraction of integers.

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"These monkeys are at once the ugliest and the most beautiful creatures on the planet./ And the monkeys don't want to be monkeys; they want to be something else./ But they're not."
-- Ernie Cline

quote:

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.9 is not equal to 1.
.99 is not equal to 1 but it is closer.
.999 is not equal to 1 but it is closer yet.
.9999 is not equal to 1 but it is still closer.
.999999999999999999 is not equal to 1 but it is very close indeed.

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.9... (the ... means an endless row of 9's. We need to be VERY careful with our notation) IS equal to 1 the difference is now zero.

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Heh. This is actually the correct way to prove this in mathematics, just without the technical language.In technical language, let a_n = 9*(1/10)^n. Then the correct way to write 0.99999... is sum_{i=0}^\infty a_nThe claim is that sum_{i=0}^\infty a_i = 1The proof is that sum_{i=0}^\infty a_n = \lim_{N\to \infty} \sum_{i=0}^N a_i (by definition of the infinite series)Now 1 - \lim_{N\to \infty} \sum_{i=0}^N a_i = \lim_{N\to \infty}( 1 - \sum_{i=0}^N a_i)= \lim{N\to \infty} (0.1)^N= 0So 1 - 0.99999... = 0 which means 1 = 0.9999....(Sorry for the LaTeX; I don't know any other way to write math symbols.)Of course, one should prove first that 0.99999... actually converges to a number, but this is obvious since the partial sums of the series are a Cauchy sequence.

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"These monkeys are at once the ugliest and the most beautiful creatures on the planet./ And the monkeys don't want to be monkeys; they want to be something else./ But they're not."
-- Ernie Cline