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

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#### 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.

#### 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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#### 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.

#### 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.

#### 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.

#### 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 ∞) (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 »