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Topic: Is .99999... = 1?  (Read 49603 times)

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Offline Dersan00

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Re: Is .99999... = 1?
« Reply #45 on: December 08, 2009, 09:53:43 PM »
The proof I've always seen associated with this involves fractions.

(1/3) = .3333...
3 x (1/3) = (3/3) = 1
3 x .3333... = .9999...

(3/3) = .9999....

Who the hell invented math anyways? I mean subtraction is the addition of negative numbers, division is multiplication of fractions, and all math can be done with just 1's and 0's.

CRAZY.

Offline juanrga

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Re: Is .99999... = 1?
« Reply #46 on: November 26, 2011, 03:35:28 PM »
let c = 0.99999...

10c = 9.9999...

10c - c = 9.9999... - 0.99999...

9c = 9

c = 1

0.99999... = 1

Fastest proof

A=1; B=0.99999...

A - B = 0.00000...
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Offline Benzene Martini

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Re: Is .99999... = 1?
« Reply #47 on: October 16, 2015, 10:54:20 PM »
In theory, no.

In reality, it depends.

It is all relative.

Offline SirReal

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Re: Is .99999... = 1?
« Reply #48 on: October 16, 2015, 10:56:12 PM »
You can't make that argument since the series of 9's after the decimal point is infinite.

Offline Benzene Martini

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Re: Is .99999... = 1?
« Reply #49 on: October 16, 2015, 11:07:58 PM »
You can't make that argument since the series of 9's after the decimal point is infinite.

Yes you can. You just have to know how to present the data.

Offline SirReal

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Re: Is .99999... = 1?
« Reply #50 on: October 18, 2015, 03:00:05 PM »
But 1/3 is = .33333, it is the infinite sum of:

Σ(n=1 :rarrow:∞) (3/10^n)  right?

So doesn't that mean that the limit of this series will indeed approach 1/3, and then the math is valid?  I think the problem is that the series is being treated as a number.  However, I am not a mathematician.  I will have to consult my calculus 2 professor from several terms ago, she is quite literally a genius who does research on algebraic topology.  Perhaps she will be able to shed light on this dilemma
« Last Edit: October 18, 2015, 03:28:21 PM by SirReal »

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