0.999...: Difference between revisions

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Revision as of 15:33, 8 May 2005 editBradBeattie (talk \| contribs)6,888 edits →See also ← Previous edit		Revision as of 15:34, 8 May 2005 edit undoBradBeattie (talk \| contribs)6,888 edits Moving alternative proofs out for brevity.Next edit →
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	:<math>\sum_{k=0}^\infty \left( \frac{1}{10} \right)^k = \frac{1}{1 - \frac{1}{10}}.</math>		:<math>\sum_{k=0}^\infty \left( \frac{1}{10} \right)^k = \frac{1}{1 - \frac{1}{10}}.</math>

	== Alternate proof ==

	{\|-
	\|<math>x\,</math>
	\|<math> = 0.999\ldots</math>
	\|-
	\|<math>10 \times x</math>
	\|<math> = 9.999\ldots</math>
	\|-
	\|<math>10 \times x - x</math>
	\|<math> = 9.999\ldots - 0.999\ldots</math>
	\|-
	\|<math>9 \times x</math>
	\|<math> = 9\,</math>
	\|-
	\|<math>x\,</math>
	\|<math>= 1\,</math>
	\|}

	Another: What is 1-0.99999... ? You get 0.000000... which is the same as zero.

	Try this: Divide one by three (one third) and you get .333333(an unending series of threes). Three one-thirds is one so three times .3333(an unending series of threes) is .99999999999(an unending series of nines).

	Finally: If you don't have a problem with 1.00000(an unending series of zeros), why should there be a problem with 0.9999(an unending series of nines) ?

	If you think there is a difference, in what way is that difference different from nil, nada, nothing, zilch, zero?

	== See also ==		== See also ==

Revision as of 15:34, 8 May 2005

In mathematics, one could easily fall in the trap of thinking that while 0.999... is certainly close to 1, nevertheless the two are not equal. Here's a proof that they actually are.

Proof

$0.999\ldots$	$={\frac {9}{10}}+{\frac {9}{100}}+{\frac {9}{1000}}+\cdots$
	$=-9+{\frac {9}{1}}+{\frac {9}{10}}+{\frac {9}{100}}+{\frac {9}{1000}}+\cdots$
	$=-9+9\times \sum _{k=0}^{\infty }\left({\frac {1}{10}}\right)^{k}$
	$=-9+9\times {\frac {1}{1-{\frac {1}{10}}}}$
	$=1.\,$