<p>Does it im just wondering</p>
<p>no, it equals .999~...</p>
<p>Depends on your definition of "equals." .999~ is actually an infinite series (a geometric series, to be precise) that converges to 1. Generally, for a mathematician, "converges to" is as good as "equals," although there are some technicalities that you'd want to avoid. I doubt anyone's that interested, so I won't discuss them.</p>
<p>Hmmm...so there are details that complicates the value of a convergent infinite series? I'd like to hear about them...</p>
<p>On the other hand, the original post reeks of pointless trolling. Where's Onyxia when you need her?</p>
<p>The answer is yes. Anybody that has taken a course in precalc should know this.</p>
<p>Well, say you have a function f(x) that equals 1 at x=1 but 0 everywhere else. Now say that you have a sequence of numbers y<em>1=0.9, y</em>2=0.99, y<em>3=0.999... y</em>i=1-10^(-i). The sequence of all y<em>i converges to 1 as i approaches infinity, but f(y</em>i) does not converge to 1 - it's always 0.</p>
<p>This is why you can't be sloppy and just say something like "y_infinity=1".</p>
<p>gxing, when you hit calculus, you'll realize that the answer is no</p>
<p>looking at limits, a certain function may not exist at "1" --an assymptote, but will exist at .999~</p>
<p>I see it sort of like the english language...synonyms for example</p>
<p>irritate and bother mean just about the same thing, but they aren't EXACTLY the same thing...if they were, then one of those two words wouldn't exist.</p>
<p>Anonymous when you hit calculus you will realize the answer is yes.</p>
<p>.999~ = 1. </p>
<p>I'd take the time to prove it, which I can, but I can't really use the symbols I would need here. If I had latex that would be a different story.</p>
<p>umm Ok...</p>
<p>X = .99999999999
10x = 9.999999999
10x - x = 9
9x = 9
x = 1 </p>
<p>or </p>
<p>1/3 = .33333333333333
2/3 = .66666666666666
3/3 = .99999999999999 or 3/3 = 1 </p>
<p>N.O.M.O.H.O.</p>
<p>Gospy, what's the flaw in my logic?</p>
<p>I realize that you can't explain your logic w/out the proper symbols, so lets stick to what we have... (sorry if that sounded mean...i didn't mean for it to!)</p>
<p>Is my logic flawed somewhere? I think of scientific notation when I see this...maybe it could be 1 when there's only one precise digit or something, but I still don't see what's wrong with my logic.</p>
<p>^ Don't do that. That's insulting our intelligence.
I'm referring to two posts up.</p>
<p>Consider this: lim b -> inf sum n=1 to b 9/(10^n).
(9/10 + 9/100 + 9/1000 + ...)</p>
<p>We can all agree that is an accurate representation of .999~.</p>
<p>This is a geometric series in which a= 9/10 r=1/10</p>
<p>We all know the sum of a geometric infinite series is a/(1-r) which in this case is equal to (9/10)/(1-1/10) or 1.</p>
<p>If you want to argue that .999~ never reaches that infinite 9, then fine, it't not equal to 1, but then it is not equal to .999~ either.</p>
<p>.999 repeating = 1</p>
<p>Final Answer.</p>
<p>.9999= 1</p>
<p>yes</p>
<p>
[quote]
If you want to argue that .999~ never reaches that infinite 9, then fine, it't not equal to 1, but then it is not equal to .999~ either.
[/quote]
</p>
<p>that was insightful! I never actually thought about that...</p>
<p>
[quote]
Final Answer.
[/quote]
</p>
<p>not to sound mean, but I hope that you meant that that was YOUR final answer...you didn't provide any help to the thread at all...
But, if you meant that that was the "final answer" to the thread, then I'm just going to ignore you seeing that I don't believe you have the authority to "close" or end a thread just through your wise words</p>
<p>.999~ =! .999~? I DONT GET IT!</p>
<p>Explain?</p>
<p>It equals 1 because it'd just impractical. And everything works out.</p>
<p>There's a more "mathematical" reasoning, but you know what? That's good enough.</p>
<p>My proof is good enough.</p>
<p>i think mine was better haha I took that from a friend short sweet and to the point and apparently insulting! That's called multitalented.</p>
<p>By multiplying by 10 you are adding an extra number that you are not accounting for.</p>
<p>10(.999~) = 9.999~ but it stops one position short of the the last infite term. That's the only way I can think to explain that.</p>
<p>That only works by coincidence.</p>
<p>I'm not be the best math student out there, but I didn't understand Tony's first proof with a flip...did you just randomly add stuff to random sides of your equation? I've always heard that what you do to one side, you must do to the other side; in addition, 3/3 = 1 always, it is never .9999999</p>
<p>In his position 10x-x = 9.0000000~1, not 9</p>