Re: .999~ = 1
 

Hi guys,
never posted before because i'm afraid i'm not as technically proficient with FIRST robotics as a lot of you, but, if there's one thing i know, it's calculus.

.999~ (.999 recurring or repeating as its commonly known) does in fact equal one. The "Multiply by Ten" Proof is true, but difficult to grasp for the common mind, so suffice to say that this proves it once and for all: (granted someone already did mention this)

A infinite geometric series is a series of numbers where one begins with a set point (the First element) and finds the next point by multiplying the First element by some number X. Then, to find the next number in the sequence, you keep on taking the number you have and multiplying by X (X is called the common ratio, in fact). Odd as it may seem, such infinite series of numbers do in fact add up to definite finite real values. The sum of an infinite geometric series is A / (1 - R) where A is the First element of the series and R is the common ratio. In this case, The series is .9 + .09 + .009 + . 0009 ad infinitum. Thus, the First element "A" is clearly .9 and the Common Ratio "R" is clearly .1 , or One-Tenth. Now, consider: in this case, A / (1 - R) = .9 / (1 - .1) which equals .9/.9 which of course equals 1. Ta da! For those of you who don't believe this geometric series sum rule i am using, there are myriad other ways to show the fact, such as the calculus' Ratio test, etc.

Meanwhile, this is a fascinating discussion, and I enjoyed reading everyone's thoughts. Someone mentioned the idea of 0/0. There's a name for it, and I seem to forget exactly what the name is (the name does make sense once you hear it). Anyway it's not 0; 0/0 is equal to INF/INF or 1^INF or 0^0 or any other myriad devilish expressions : it simply does not exist. I think it might be called an infinite discontinuity but that's just a guess. anyhow, there's math for ya!

captain dan
team 1168
www.friarbot.com rocks my world too hard

Re: .999~ = 1
 

note - only for R < 1.

One of the best known applications of this is L'Hopital's rule, which is used for finding the integrals of certain functions that can be transformed into something that resembles one of those patterns - 0/0, inf/inf, inf-inf, 1^inf, 0^0, et cetera.

Re: .999~ = 1
 

L'Hopital's Rule is actually usually used for computing limits. For example, to compute

lim x/sin(x) as x-->0, we can apply L'Hopitals rule to find that

lim x/sin(x) = lim 1/cos x = 1.

Re: .999~ = 1
 

I believe you are looking for "indeterminate form." Check out mathworld

Re: .999~ = 1
 

also thus used in calculus for finding the limits of infinite integrals (happened to be doing that a couple of days ago in calculus, so I mentioned it in that context). :)

Re: .999~ = 1
 

a couple of things
Firstly:
i can prove that 1=0


ok, assume its an identity (cant do that on a keyboard but assume the = is an identity sign)
if you say 1=0
if you take 1 from each side
0=-1
multiply by 2
0=-2
add one to both sides
1=-1
Square
1=1
voila :D
or do it backwards to get it if you like.

also, i can prove mathamatically the first part
1/9 = 0.1111... (recurring)
10/9 = 1.1111... (recurring)
10/9 - 1/9 = 9/9 = 0.9999999999, or 1 as 9/9 is 1 but 9 x 0.11111111111 = 0.9 revcurring. therefore 0.9999.... doesn't exist


its marvalous that e^(pi i )=-1

Re: .999~ = 1
 

squaring is not a 1 to 1 operation, thus you cannot do the proof backwards, and thus it is not valid.

x^2 = y^2 --/--> x = y
(x squared = y squared) does not imply (x = y)

Re: .999~ = 1
 

You can't get the square root of 1 equal to -1 so it fails.

Re: .999~ = 1
 

At least when it is not defined. arccos x is defined from 0 to 1 and arcsin x from -pi/2 to pi/2. This makes it one to one and gives it an inverse function.

Re: .999~ = 1
 

can't divide by zero....3rd to last step

Re: .999~ = 1
 

at least someone found the correct reason for it being incorrect :)


What is the sqrt of 1?

Consider that if you square -1 you get 1 and so the sqrt of 1 is -1 or 1

Re: .999~ = 1
 

Anyone want to help me with this one?

Find the fallacy in the following argument that 0 = 1.

dv = dx --> v = Integral (dx) = x
u = 1/x --> du = -1/(x squared) dx

0 + Integral (dx/x) = (1/x)(x) - Integral ((-1/(x squared)(x)) dx = 1 + Integral (dx/x)

So, 0 = 1

It is integration by parts ( integral (u dv) = uv - integral (v du)). I should be figuring it out myself but when I saw the problem I thought of all of you.

Re: .999~ = 1
 

Given that you're working with reals, it's true that you can square a negative number, but the square root function only outputs positive numbers.

sqrt(1) /= 1 or -1
+/-1^2 = 1

Re: .999~ = 1
 

constant of integration.

Re: .999~ = 1
 

Everyone's proving stuff...
Well... 2 = 1
Why ??

Check this out... nothing illegal and you can go backwards :)

a = b
ab = b^2
ab - a^2 = b^2 - a^2
a(b-a) = (b-a)*(b+a)
a = b+a

So if a=1, b=1 too (since the first thing I assumed was a = b )

1 = 1+1 ???

Have fun...
Don't flame... I am putting this to have some fun :p