c is 1/infinity

balki, in order for it to be the number, there has to be no difference, not just an infinitesmally small difference.

y2k, sorry, I realized that i did not directly resond to your "proof" from last page. here is something that i said earlier that should help to disprove it.

"the x=.999 10x-x thing is also flawed as whenever you multiply anything by 10 it will move the number to the left, and there will be 1 less number after the decimal point. For example, 3.3(10)-3.3 does not equal 30, it equals 29.7 because there is one less number after it. Before you start saying, "infinity minus 1 is still infinity" try setting it up in an equation (ex, inf-1=inf, subtract inf) and you will see that, unless the laws of math are not true, then it would prove that it would be true in that situation. Frankly speaking there are no logical/ mathematical/ tested reasons of why your assertion would be true so it would be a way better policy to go with an assertion that is literally universal."

tvrocks:

A = 0.9999999....
B = 1
A + D^(1/2) = B

Solve for D.

(1/inf)^2

Do you know what infinity means?

endless/ unlimited.

3.3*10 - 3.3 = 29.7 and this / by 9 is 3.3 so you are by your start, like 0.99999....is 1.
D=(B-A)^2 so D=0, the iphone with 15 decimal places calculate 1e-30, take a better machine and you get 1e-300 or 1e-30000 but with infinity decimal places you will get 0 exact.

And I am done. Enjoy!

@tvrocks, yeah, but it's pretty close.

gerry, it seems that you don't get why i was mentioning that. I used that example in order to prove that when you multiply by 10 and then subtract by the original number it will not be a whole number in situations like this. This would mean that 10x-x would be less than 9 and thus the equation should not work.

balki, i agree, close does not mean it is the number.

You guys are missing the point. We solved with a long time ago.
@tvrocks. what is 1/2+1/4+1/8+1/16+...?
what is 9/10+9/100+9/1000+9/10000+...

.999999... is not a number approaching 1, it literally is the end result, which is WHY you use infinite series to solve it because it literally IS an infinite geometric series.

@tvrocks
is .9999... an infinite series?
is it arithmetic or geometric?
if it is a geometric infinite series, why would you not use the formula for infinite geometric series?

@balki,
It is not close. It is exactly 1.

tvrocks: I dont think 10x-x is less than 9x, I also dont think that 9.999.... - 0.999.....is less than 9. Inf+1 is inf , so inf -1 is also inf , so 9.9999....-0.99999....is 9 exact or?
We dont lose a decimal place by multiplying with 10, I think.

...and 3.3 is 3.29999......so btw :-)

@tvrocks
When you multiply by 10 and then subtract the original number, it will be a whole number. This is what you are failing to grasp. The reason you are failing to grasp is it that you are failing to grasp the concept of infinity. I don't want to ask for your education in the field of mathematics, but I am assuming it is not going much higher than high-school. So, listen carefully and learn. Be humble in a field you clearly don't have enough knowledge in.

Infinity + 1 = Infinity
Infinity + any number = Infinity
10 * Infinity = Infinity
Any number * Infinity = Infinity

If you agree with what I have written above, continue reading. If not, read an algebra book.

Now, in 0.999... you have an infinite number of 9's after 0. Adding or removing one or ten or gazillion 9's will not change the fact that there are infinite number of 9's after 0. It is going to be EXACTLY the same number. Not approximately. Not sort of. EXACTLY.

Now, if you are able to put a number between two numbers, they are not equal. If you are not able to put a number between two numbers, they are equal. You can put a number between 1 and 2 (1.3, for example). You can put a number between 0.9 and 1 (0.95, for example). You can even put a number between 0.99999 and 1 (0.999999, for example). You CANNOT put a number between 0.999... and 1. It will either be SMALLER than 0.999... or LARGER than 1. Thus, 0.999... = 1.

Now, stop being troll and go learn some basic algebra. Seriously.

Lol you guys are way too passionate about this. For all we know, tvrocks could be just fooling around for amusement.

tvrocks, this method works. take .4545454545454... for instance.
.454545...=x
45.454545...=100x
45=99x
.454545...=x
surely you agree that .454545...=.454545

Same with .3333...

3.3333...=10x
3=9x
.333333=x

why would it be any different for .99999

So now I have 2 withstanding arguments:
Why would the formula for infinite geometeric series not work for an infinite geometry series?
Why would a method of multiplying and dividing an infinitely long decimal work for all infinite decimals except .9999...?

a=a

a = b
a*a = b*a
a*a + a*a = b*a + a*a
2*a*a = b*a + a*a
2*a*a - 2*a*b = b*a + a*a - 2*a*b
2*a*a - 2*a*b = a*a - a*b
2*(a*a - a*b) = 1*(a*a - a*b)
2 = 1

error: divide by zero
@y2k
you just divided by 0, that's why is doesn't work. if a=b, them (a*a-a*b)=0
therefore you't simplify that last step like that. You just get 2(0)=1(0) which is 0=0

But a and b are different letters so that can't be zero!

Right tvrocks??

if different letters can't add up to zero, then they'd not be able to be equal either, so your argment would be wrong in the first line, a = b

principian, you're not adding the numbers, you're subtracting them. two numbers that are equivalent subtracted would be zero.

a = a
a / inf = a / inf
(a / inf) * inf = (a / inf) * inf
(a * inf) / inf = 0 * inf
a * (inf / inf) = 0
a * 1 = 0
a = 0!

TVROCKS IS RIGHT!

Jeff, I really hope you're joking

Of course it's a joke. tvrocks doesn't understand that some functions approach the same limit at different rates.
1 / x
1 / x^2
1 / (2^x)
1 / (e^x)
1 / (e^x^2)
etc.

yeah, I should have said "can't substract", but don't understand your point, I was answering to y2kjbk:
if different letters cann't equivalent enough to give zero when you "substract" them, then they should not be able to be equivalent enough to put them around a "="

He was being sarcastic to tvrocks.