<p>the most interesting way to prove it is with logarithms, but I forgot how.</p>
<p>the point is, there are numerous ways to prove that .999 = 1, so skeptics just shut up lol. :D</p>
<p>
[quote]
x + 2 = x
-x -x
2 = 0
[/quote]
</p>
<p>i still don't know how that works. does x+2 ever equal x?</p>
<p>There was [thread=87881]this[/thread] thread in the CC Cafe and [thread=88044] this[/thread] thread in the MIT forum discussing this topic.</p>
<p>oh... i forgot to post a proof.</p>
<p>lim____ n
n->oo sigma 9/10u= .9 + .09 + .009 + .0009=.999....= .9/(1-.1)=1
______ u=1</p>
<p>.1=common ratio=r
.9=first term=a<em>1
using formula a</em>1/(1-r) [sum of infinite geometric series]</p>
<p>[oo=infinity]</p>
<p>lol</p>
<p>The way mathematicians use most often is simply:</p>
<p>.999... = sum( (9/10)^i) from 1 to infinity</p>
<p>we know that this series converges. We also know that to find the value of this series, we simply compute 1/(1-9/10) = 1. (infinite geometric series).</p>
<p>For the 1 = 2 business, one of the most devious ways is to expand log's into their taylor series, and then manipulate the taylor series in such a way that you end up with 1 = 2. Last I remember, my math dept head was publishing a paper showing how this is not valid (aggressive manipulation of the series and then back translation).</p>
<p>
[quote]
1+1=11
[/quote]
</p>
<p>1+1=10 </p>
<p>base 2 w00t</p>
<p>damn, surge beat me too it by seconds.</p>
<p>So then
lim .9^i = 1 is what you are saying?
i→∞</p>
<p>umm... .999=.999 :)</p>
<p>nooo...</p>
<p>summation of 9 x .1^i (where i is from 1 to infinity).</p>
<p>not .9^i</p>
<p>(and i is a bad variable to use, looks imaginary) :P</p>
<p>yes i =square root of -1....</p>
<p>oh i'm sorry... original post...
[quote]
oh... i forgot to post a proof.</p>
<p>lim____ n
n->oo sigma 9/10u= .9 + .09 + .009 + .0009=.999....= .9/(1-.1)=1
______ u=1</p>
<p>.1=common ratio=r
.9=first term=a<em>1
using formula a</em>1/(1-r) [sum of infinite geometric series]</p>
<p>[oo=infinity]
[/quote]
</p>
<p>I meant...
[quote]
oh... i forgot to post a proof.</p>
<p>lim____ n
n->oo sigma 9/(10*^u*)= .9 + .09 + .009 + .0009=.999....= .9/(1-.1)=1
______ u=1</p>
<p>.1=common ratio=r
.9=first term=a<em>1
using formula a</em>1/(1-r) [sum of infinite geometric series]</p>
<p>[oo=infinity]
[/quote]
</p>
<p>okay, that's enough kind nerds</p>
<p>
[quote]
umm... .999=.999
[/quote]
</p>
<p>Um, we're talking about .999... which does not equal .999</p>
<p>either way it will always be slightly less than 1...close but not 1</p>
<p>0.999 , infinite 9s, is 1. </p>
<p>0.99999...= 9/10+9/100+9/1000....and so on until infinity (n)
the ratio between this geometric series is= 1/10, since the ratios is smaller than 1 the series is convergent</p>
<p>so using the formula for the sum of a geometric series we can prove its 1</p>
<p>(A1(1-r^n))/(1-r) each time we increase the value of n r^n becomes smaller right? so in infinity r^n is zero</p>
<p>(A1(1))/(1-r) substitute the values we have:
(9/10)/(1-1/10)
(9/10)/(9/10)
1</p>
<p>ooopps sorry didn't see that surge had already posted this....sorry dude.</p>
<p>
[quote]
either way it will always be slightly less than 1...close but not 1
[/quote]
</p>
<p>So why are all these proofs already posted wrong? Becuase they have to be wrong if 0.999... doesn't eqaul 1, right?</p>
<p>Just wondering.</p>
<p>The intricacies of screwing with math formulas to produce a desired result are profound and do wondrous things.</p>
<p>Greendayfan, .999~=1. You have yet to prove anything that shows anything contrary to that. .999~ can be expressed as an infinte series that converges to 1. That has been shown.</p>