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0.999... = 1

smiles · 2005-11-26 22:57:41

I like this one. To prove 0.999.... = 1 simply rewrite 0.999... as (9/10) + (9/100) + (9/1000) + .... Which is a simple series. Rewrite: (9/10) [1 + (1/10) + (1/10)^2 + (1/10)^3 + ...] Apply sum of series formula, with ratio of (1/10). (9/10) [ 1 / (1-(1/10))] = (9/10) * [10/9] = 1. Thus 0.999.... = 1.

26-11-2005 10:57pm

#1

smiles

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I like this one.

To prove 0.999.... = 1 simply rewrite 0.999... as (9/10) + (9/100) + (9/1000) + ....
Which is a simple series. Rewrite:
(9/10) [1 + (1/10) + (1/10)^2 + (1/10)^3 + ...]
Apply sum of series formula, with ratio of (1/10).
(9/10) [ 1 / (1-(1/10))]
= (9/10) * [10/9] = 1.

Thus 0.999.... = 1.

0

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#2 26-11-2005 11:00pm

whiskeyman

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do you work for AIB foreign exchange?

:v:

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#3 26-11-2005 11:04pm

johnnyrotten

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whiskeyman wrote:

do you work for AIB foreign exchange?

:v:

ROFL!

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#4 26-11-2005 11:11pm

smiles

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whiskeyman wrote:

do you work for AIB foreign exchange?

:v:

Not the foreign exchange.....

0
#5 27-11-2005 9:10am

Pherekydes

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smiles wrote:

....

What's that? It's an AIB ellipsis: 33% markup.

0
#6 27-11-2005 3:12pm

Tar.Aldarion

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Back on topic, It's a sweet proof!

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