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0.999... = 1

Elfenones

Elfenones

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x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1

----------------------------------------------------------------------------

1/3 + 1/3 + 1/3 = 3/3 = 1

1/3 = 0.333...

0.333... + 0.333... + 0.333... = 0.999 = 1

--------------------------------------------------------------------------

0.9999... = Sum 9/10^n
(n=1 -> Infinity)

= lim sum 9/10^n
(m -> Infinity) (n=1 -> m)

= lim .9(1-10^-(m+1))/(1-1/10)
(m -> Infinity)

= lim .9(1-10^-(m+1))/(9/10)
(m -> Infinity)

= .9/(9/10)

= 1

--------------------------------------------------------------------------

Discuss.
 
This is not my writing just something I found from another user's post that would answer the question.

Your second question gets into something that mathematicians call
"limits." You are very insightful to recognize that:

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

.9999... is equal to 1 because no matter how small a difference
between .9999... and 1 you ask for, I can write enough 9s to get
within that difference. So suppose you want it within .00001 of 1.
I can write .99999. This can go on and on.

Here is another way of looking at it. Let's say

n = .9999....

This means that 10 x n = 9.9999....

Now subtract 10n - n = 9.9999... - .9999...
So 9n = 9.0
n = 1

This works because that 9s keep repeating and .9999... - .9999... = 0

You can do this with any repeating decimal to find the fraction that
it is equivalent to. For example, what fraction is equivalent to
2.161616...?

Let n = 2.161616...
100n = 216.161616... (Notice that this time you multiply by 100.
Why?)
100n - n = 216.161616... - 2.161616... (Subtract)
99n = 214
n = 214/99

Plug 214/99 into your calculator and see what you get. Calculators
can go only one way, from fractions to decimals. You have to figure
out how to go from decimals to fractions.

-Written by, Doctor Gerald http://mathforum.org/library/drmath/view/57040.html
 
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It's not really that surprising; it works in all bases, and there are plenty of seemingly-odd results in math. For example, -1 > ø, -1 < ø, -1 = ø, and ø is blue.

Plus, it's not really a fitting SFA topic. SFA is for debate, not trivia.
 
Elfenones said:
x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1

----------------------------------------------------------------------------

1/3 + 1/3 + 1/3 = 3/3 = 1

1/3 = 0.333...

0.333... + 0.333... + 0.333... = 0.999 = 1

--------------------------------------------------------------------------

0.9999... = Sum 9/10^n
(n=1 -> Infinity)

= lim sum 9/10^n
(m -> Infinity) (n=1 -> m)

= lim .9(1-10^-(m+1))/(1-1/10)
(m -> Infinity)

= lim .9(1-10^-(m+1))/(9/10)
(m -> Infinity)

= .9/(9/10)

= 1

--------------------------------------------------------------------------

Discuss.
Click to expand...

10x - x = 9.999 - x. The rest of that problem is bullshit.

Second. Just, no.
 
I couldn't count the number of times I used to debate this on RS forums way back when. Thanks for posting this... good memories.
 
ask me im a professor i teach this at school :P it is simply a repeated fraction kids!
 
I remember my maths teacher use to go on about this all the time and I've read about it in the student room hundreds of times.
 
The Viper000 said:
Mainly because you're not explaining that .999... is an infinite number, which is imaginary.
Click to expand...
Also an incorrect explanation -.9999...

is not infinite, irrational, or imaginary. It has an exact equivalent of 9/9 or 1.

1/9 = .1111111...
2/9 = .2222222...
3/9 = .3333333...

The Viper000 said:
By that same logic, .555 = 1.8

x = 0.555
10x = 5.555
10x - x = 5.555 - 0.555
9x = 5
x = 1.8

This is a fool's proof, and a very old one.
Click to expand...
Unfortunately, you did the math wrong.

9x = 5
solves out to
x = 5/9, which does equal 5.555 (assuming you intended it as an infinite series).
 
I can use infinite series to "prove" that 0=1, or that 9999999999 = 0. Fool's proofs are very common and not so surprising.
 
KerokeroCola said:
I can use infinite series to "prove" that 0=1, or that 9999999999 = 0. Fool's proofs are very common and not so surprising.
Click to expand...

Do it! Would love to annoy my neighbours in next boring classes with it
 
This is ridiculous...
0.999... Approaches 1 If and Only If
0.999... = 0.9 + 0.09 + 0.009 + 0.0009...
= a/(1-r)
=0.9/(1-0.1)
=0.9/0.9
=1

You fail with your example. What you have basically done is assumed an infinite sum of a geometric series without denoting the relationship between a and r.

EDIT: Approaching symbols should have been used in that proof (-->) as opposed to (=) signs...
 
I really do despise morons like you that try to seem intellectual by posting plagiarized material from the internet, why would you do that?
 
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