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Topic: Is .99999... = 1?  (Read 49182 times)

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Offline Donaldson Tan

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Is .99999... = 1?
« on: October 27, 2006, 09:42:45 PM »
let c = 0.99999...

10c = 9.9999...

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

9c = 9

c = 1

0.99999... = 1
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Offline Mitch

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Re: Is .99999... = 1?
« Reply #1 on: October 27, 2006, 11:25:12 PM »
You did the math wrong, this line: 9c = 9

should of been 9c = 8.9999....
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Offline Yggdrasil

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Re: Is .99999... = 1?
« Reply #2 on: October 27, 2006, 11:50:30 PM »
0.999... = 0.9 + 0.09 + 0.009 + ...
               = 9(0.1 + 0.01 + 0.001 + ...)
               = 9(-1 + 1 + 0.1 + 0.01 + 0.001 + ...)
               = 9(-1 + ??n=0(1/10)n)
Since ??n=0rn = 1 / (1 - r) for |r| < 1,
0.999... = 9(-1 + 1/(1 - 1/10))
               = 9(-1 + 10/9)
               = 9(1/9)
               = 1
qed

Offline Borek

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Re: Is .99999... = 1?
« Reply #3 on: October 28, 2006, 03:43:47 AM »
http://en.wikipedia.org/wiki/Infinite_geometric_series

sum = a/(1-r)

0.9 + 0.09 + 0.009 + ...

a = 0.9, r = 0.1

0.1/(1-0.1) = 1

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Offline pantone159

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Re: Is .99999... = 1?
« Reply #4 on: October 28, 2006, 02:35:24 PM »
0.99999... means 'the limit approached as you add more and more nines to the end' which is exactly 1.0.

Offline Donaldson Tan

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Re: Is .99999... = 1?
« Reply #5 on: October 28, 2006, 03:55:56 PM »
Mark, No.

"0.99999 = 1" is not an approximation.

That's the whole point of bringing this up.
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Offline Donaldson Tan

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Re: Is .99999... = 1?
« Reply #6 on: October 29, 2006, 04:34:19 PM »
You did the math wrong, this line: 9c = 9

should of been 9c = 8.9999....

The workings first assume 0.9999... != 1

However, the subsequent steps reveal that 0.999... = 1
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Offline FeLiXe

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Re: Is .99999... = 1?
« Reply #7 on: October 29, 2006, 06:18:02 PM »
it's a correct proof. My math professor said that a couple of times that .999.... is exactly the same as 1
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Offline Borek

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Re: Is .99999... = 1?
« Reply #8 on: October 29, 2006, 06:20:47 PM »
1/9 = 0.1111111...

9*1/9 = 0.9999999...

9*1/9 = 1
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Offline Donaldson Tan

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Re: Is .99999... = 1?
« Reply #9 on: October 29, 2006, 06:32:23 PM »
This is one of the odd but interesting results in Math..
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Offline constant thinker

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Re: Is .99999... = 1?
« Reply #10 on: October 29, 2006, 08:25:04 PM »
The concept of .9999... being equal to 1 is difficult to grasp. Seeing the proof though make me believe.

I fully agree with geodome in how it's an odd but interesting result.

When I first saw that someone prooved it equaled one, I was like what the...
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Re: Is .99999... = 1?
« Reply #11 on: October 30, 2006, 10:02:08 PM »
I don't understand what all the fuss is about...

And what's wrong with what Mark said??

Offline Borek

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Re: Is .99999... = 1?
« Reply #12 on: October 31, 2006, 02:36:56 AM »
And what's wrong with what Mark said??

Nothing :)
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Offline xiankai

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Re: Is .99999... = 1?
« Reply #13 on: December 03, 2006, 07:03:36 AM »
"0.99999 = 1" is not an approximation.

if it is not an approximation,

then if c = 0.999...

10 c = 9.99...0

because the last number cannot be left out.

therefore 10 c - c = 8.999...1

9 c = 8.999...1

there are similar problems involving infinity, like how

1 + 2 + 3 ... = ?

2 + 4 + 6 ... = ?

even though both have the same number of terms, the second series is obviously twice that of the first series. however, two times infinity still equals infinity.

This is one of the odd but interesting results in Math..

so it looks like that's the way we structure the number 'infinity'.
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Offline FeLiXe

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Re: Is .99999... = 1?
« Reply #14 on: December 03, 2006, 01:35:47 PM »

therefore 10 c - c = 8.999...1

9 c = 8.999...1

That's the wrong concept of infinity. All digits are 9 and there is no last digit.

there are similar problems involving infinity, like how

1 + 2 + 3 ... = ?

2 + 4 + 6 ... = ?

even though both have the same number of terms, the second series is obviously twice that of the first series. however, two times infinity still equals infinity.

so it looks like that's the way we structure the number 'infinity'.

actually there are different kinds of infinity. for example there are more real numbers then there are whole numbers.


in fact you can easily prove that there are infinitely many types infinity.

but on the other hand from a mathematically useful point of view the amount of rational numbers is the same as whole numbers. the rational numbers are a "countable set", you can put them in order: (1, -1/2, 1/2, -1/3, 1/3, -2/3, 2/3, -1/4, 1/4, -3/4, 3/4, ...). you can define a distinct order where every rational number has its index. this would not work with the real numbers (can be proven easily too)

another question is if there is a cardinality step in between the whole and real numbers. no one has found one. but no one has proven that it does not exist either. the continuum hypothesis states there isn't one.
« Last Edit: December 03, 2006, 01:47:56 PM by FeLiXe »
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