0.999... = 1

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.

 

http://mathforum.org/dr.math/faq/faq.0.9999.html

 

Look's like a copy paste. You could have copied the text or added your own. Looks like you don't even know what your posting tbh, because it lacks any sort of explanation from your view.


EDIT:

Nice find, I guess it is a copy paste.

 

bump.

 

why?



are you trying to teach us how to round?

 

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

 

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.

 

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.

 

??

if x - 0.999

then 10x = 9.99... NOT 9.999

therefore.. x does = 0.999

 

Do you really know how stupid you are?

C'mon man this is basic re arranging.

 

ask me im a professor i teach this at school it is simply a repeated fraction kids!

 

bump.

 

Thread doesn't seem to be worth bumping. Old news is old. This is not a new or particularly surprising mathematical phenomena.

If you are inclined to investigate more interesting phenomena, I suggest http://en.wikipedia.org/wiki/Millennium_Prize_Problems

 

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.

 

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...

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.

 

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?