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[Tylui]

Knowledge=Power=Energy=Matter=Mass.

I'm not fat, I'm knowledgeable!

[Christes]

Ok, first of all, 1.000...1 is not a number.

Think about it this way - you're saying there is an infinite (note: infinite means unending) amount of zeros with a 1 at the end of it.  It
doesn't end - so how can there be a 1 "at the end"?

The decimal system assigns a value to every digit.  There simply can't be a value "at infinity" as that would make no sense.

To answer the original question (which could very well have been done to death by now - I didn't read the last couple posts to closely), there is no closest real number to 1.  Take any two real numbers X and Y.  (X+Y)/2 is between them and is also a real number.

So even if I had a real number closest to 1, I could just average it with 1 to get something closer.  That's what we would call a contradiction.

[Virex]

As far as I know, using limits is valid in this case. We were looking for the minimum value of a given function. The minimum value of a function is the lowest value for which it exists. If we, for example, would like to know the minimum value of Y=C*e^(K+D/X), we would solve:
Y(min) = lim(X -> X(min)) Y                     => We know the function is continuous over R, so X(min) = -infinity
Y(min) = lim(X-> -infinity) C*e^(K+D/X)    => D/X -> 0 for X -> -infinity, so
Y(min) = C*e^K                                    => Note that having X go to 0 would yield 2 awnsers depending on from where we approach
(Yes I know I'm throwing infinity arround like it's a number, but since I'm using limits it's a bit safer then normal)

So Y aproaches C*e^K if X goes to negative infinity. As far as I know this is actualy a proper use for limits (the very definition of a derivative uses a limit similar to this).
Now if I would also have imposed the conditions that Y = 0 at X = -10 and C =!= 0 (For example, because that follows from the theory behind the equation), then the set of equations and conditions is contradictory. The same happens in the example I gave. Now if you think it's still an improper use of limits, then I would like you to explain to me how to approach such a problem without using limits.

Now comming back to my earlier calculations, to calculate the lowest possible value for the given variable, we would need to take the limit of one variable aproaching the other. But the conditions explicitly prohibit us not from taking the limit, but from using the obtained value as an awnser, since said awnser contradicts with the conditions.
So unless one would have another definition for the minimal value of a function then
f_x(min) = lim(X -> x(min)) f_x 
with X(min) the lowest value X can assume where f_x is either defined, or approaches a singularity (You can't take a limit over an area where the function doesn't exist, but you can approach a singularity or the edge of such an area)

[bjlong]

As to the .999...9=1, use .999...99998, as the "proof" fails in that case.

what is this I don't even

[eerr]

If we use the equation Y = (Z+X)/2 to define the value of Y, which is by definition halfway between Z and X, Then I think we can take the limit.



Then Z>(Z+X)/2>X
The limit of Z as Z approaches X From the right. =X
Therefore, Z>X when Y is defined as between Z and X, with Z as its ceiling and X as it's floor.

I'm sensing some order of operations shenanigins here.

[Virex]

The problem is that the < and > operators arn't compatible with limits.

[eerr]

The problem is that the < and > operators arn't compatible with limits.

Well I had a forumula that meant close to the same thing, but everybody was obessed with infinity.

[Christes]

Then Z>(Z+X)/2>X
The limit of Z as Z approaches X From the right. =X
Therefore, Z>X when Y is defined as between Z and X, with Z as its ceiling and X as it's floor.

Would you mind being a little more precise?  I'm not sure exactly what's being limited here.

The limit (as Z goes to X) of Z is X.  The limit (as Z goes to X) of (Z+X)/2 is also X.  In the limit as Z goes to X, all three of the expressions you wrote go to X. 

[eerr]

Then Z>(Z+X)/2>X
The limit of Z as Z approaches X From the right. =X
Therefore, Z>X when Y is defined as between Z and X, with Z as its ceiling and X as it's floor.

Would you mind being a little more precise?  I'm not sure exactly what's being limited here.

The limit (as Z goes to X) of Z is X.  The limit (as Z goes to X) of (Z+X)/2 is also X.  In the limit as Z goes to X, all three of the expressions you wrote go to X. 

If you don't assume some sort of order, then you get the weird result that Z can be anything, Y and X just need to be less than Z. Order of operations? idk.

I redefined Y. Because We assume the value of X is set first. Then Y. Z is only supposed to be defined last.


But if we define that Y is between Z and X... that works a little better.